Post by: Twinkle Charms on September 18, 2010, 01:29:30 pm
just been two days i started chem A level (both AS & A2) and any doubts i find ill be postin here...
so my first doubt,
can someone please clearly explain me what orbitals are ???
accordin to the definition, it says tht they are space around the nucleus where there is max. probablity of findin the electron within its orbit.
for example when n=3; l=0,1,2
so by 0,1,2 does it mean 0 from the nucleus (like coverin all the orbits in between as well> n=1 and n=2) ???
Post by: elemis on September 18, 2010, 01:50:44 pm
I wish Nid was here to explain this stuff. She's the Chem Masta. :(
Post by: Twinkle Charms on September 18, 2010, 02:08:30 pm
I too have just started my AS Levels.ohh sad =[
I wish Nid was here to explain this stuff. She's the Chem Masta. :(
yeah i posted on her fb to get back soon, hope the lady checks it out =|
Post by: Aadeez || Zafar on September 18, 2010, 05:30:16 pm
for eg.
shell one contains oribtal - 1s(2e-) (grand Total 2e-)
shell two contains orbital - 2s (2e-), 2p(2*3 e-(px,py,pz) (grand total 8e=(2e-+6e-)
shell three contains ornital - 3s(2e-), 3p(2*3e-), 4s(2e-), 3d(2*5e-) (element in which electrons are filled in 3d would be transition metal) [grand total 20e-)
soory!
my explain skill is very poor
Post by: nid404 on September 18, 2010, 05:41:18 pm
And check www.knockhardy.co.uk for ppts.
If you still don't understand(rare case) then I shall do my bit :)
Post by: Twinkle Charms on September 18, 2010, 05:53:14 pm
@Harsh
we havnt reached tht bit yet, we'll be doin it tom hopefully
buh im really thankful to you for atleast sparing a min here :)
Post by: Aadeez || Zafar on September 18, 2010, 06:00:18 pm
Post by: Twinkle Charms on September 18, 2010, 07:17:30 pm
At which session did u gave ur IGCSE?june 2009
Post by: Arthur Bon Zavi on September 19, 2010, 05:36:03 am
june 2009
Too Early and it's too late to start AS/A Levels, don't u think so?
Post by: Twinkle Charms on September 19, 2010, 07:09:25 pm
Too Early and it's too late to start AS/A Levels, don't u think so?ohh yeaa too late NOT actually :P
hahaa
i gave my physics & maths A level (AS & A2) in june 2010
n now in june 2011 ill be giving chem Alevel and ict AS ;)
Post by: Arthur Bon Zavi on September 20, 2010, 01:29:16 pm
ohh yeaa too late NOT actually :P
hahaa
i gave my physics & maths A level (AS & A2) in june 2010
n now in june 2011 ill be giving chem Alevel and ict AS ;)
Ohh!! Best Of luck with your subjects and keep posting the doubts right here!! :D
Post by: Twinkle Charms on September 24, 2010, 08:02:48 pm
Ohh!! Best Of luck with your subjects and keep posting the doubts right here!! :Dthanks theonlyone =]
So can anyone suggest why the relative atomic mass of Martian oxygen is different from that of oxygen obtained from earth?
Post by: Deadly_king on September 25, 2010, 05:47:48 am
thanks theonlyone =]
So can anyone suggest why the relative atomic mass of Martian oxygen is different from that of oxygen obtained from earth?
Quote
The analysis of a sample of Martian air collected by the Phoenix Lander shows that the ratio of oxygen and carbon dioxide isotopes present in the atmosphere of Mars is very similar to that of Earth. Due to Mars’ low gravity and lack of magnetic field, lighter isotopes are lost to space and so an older atmosphere will consist of mainly heavier isotopes.
In other words the oxygen on Mars would consist mainly of O18 while on Earth our oxygen consist mainly of O16.
Hope it helps clear your mind about that :)
Post by: Twinkle Charms on September 25, 2010, 03:06:35 pm
In other words the oxygen on Mars would consist mainly of O18 while on Earth our oxygen consist mainly of O16.oohhh thanks ;D
Hope it helps clear your mind about that :)
The first 4 ionisation energies of an element are 590, 1150, 4940 and 6480 KJ mol-1. Suggest the group in periodic table to which this element belongs. [I have the answer from the back of the text, buh dont understand it =\ so i want explanation please =] )
Post by: Deadly_king on September 25, 2010, 03:55:58 pm
oohhh thanks ;D
The first 4 ionisation energies of an element are 590, 1150, 4940 and 6480 KJ mol-1. Suggest the group in periodic table to which this element belongs. [I have the answer from the back of the text, buh dont understand it =\ so i want explanation please =] )
From the given values, we can note a very significant rise (more than twice) from the second ionisation energies to the third ionisation energies. This suggest that there has been a change in orbitals(more precisely shell number) which proves that the element has two electrons in its outermost shell.
Therefore the element is in group 2.
Did you get me??
If not......let me know and i'll try to elaborate more :)
Post by: Twinkle Charms on September 25, 2010, 04:02:18 pm
From the given values, we can note a very significant rise (more than twice) from the second ionisation energies to the third ionisation energies. This suggest that there has been a change in orbitals(more precisely shell number) which proves that the element has two electrons in its outermost shell.you mean to say because there is a large change in ionisation energies between 2nd and 3rd, which shows that the 3rd electron is in other shell and the first two in the valence shell??
Therefore the element is in group 2.
Did you get me??
If not......let me know and i'll try to elaborate more :)
Post by: Deadly_king on September 25, 2010, 04:08:35 pm
you mean to say because there is a large change in ionisation energies between 2nd and 3rd, which shows that the 3rd electron is in other shell and the first two in the valence shell??
Yeah you got it right.
Normally the ionisation energies always rises from the first to the last since the force of attraction between the electrons and the nucleus will be greater with less electrons present.
But if the change in ionisation energies is much larger that the previous one, it implies a change in shell number.
Post by: Twinkle Charms on September 25, 2010, 04:10:27 pm
Yeah you got it right.yo im gettin smarter haha ;)
Normally the ionisation energies always rises from the first to the last since the force of attraction between the electrons and the nucleus will be greater with less electrons present.
But if the change in ionisation energies is much larger that the previous one, it implies a change in shell number.
thank yooouuu :)
Post by: Deadly_king on September 25, 2010, 04:19:19 pm
yo im gettin smarter haha ;)
thank yooouuu :)
Hahaha.....very good :)
keep up the good work ;)
You are most welcome :)
Post by: Twinkle Charms on September 25, 2010, 05:50:27 pm
ok so i got the mass of Sn used by subtracting 0.1230 from 0.4650 =0.342g, now for iodine how do i go about? subtract the mass of Sn used from the total mass of the product that is 1.8020? or i find it thru the mole ratio thing? if tht is so then how do i know the number of moles of each and how can i write a balanced equation out of it when we have SnIx ??
Post by: Deadly_king on September 25, 2010, 06:13:35 pm
Q) Tin reacts with iodine in an organic solvent to form a covalent compound SnIx when 0.4650g of Sn (in excess) was used; all the iodine reacted. A mass of 0.1230g of unreacted Sn and 1.8020g of SnIx were obtained. Find the value of x.
ok so i got the mass of Sn used by subtracting 0.1230 from 0.4650 =0.342g, now for iodine how do i go about? subtract the mass of Sn used from the total mass of the product that is 1.8020? or i find it thru the mole ratio thing? if tht is so then how do i know the number of moles of each and how can i write a balanced equation out of it when we have SnIx ??
Equation is as follows :
Sn + (x/2)I2 ----> SnIx
Forget about the number of moles of iodine to be used.
From equation whatever be the value of x,
1 mole of Sn will form 1 mole of SnIx
Number of moles of Sn used = Mass/Ar = 0.342/119
Therefore same number of moles of SnIx must be formed.
Number of moles of SnIx formed = Mass/Mr = 1.8020/(119+127x)
In other words :
1.8020/(119+127x) = 0.342/119
Solve this equation and obtain x = 4
Hope you understand :)
Feel free to ask if you have any doubt :)
Post by: Twinkle Charms on September 25, 2010, 07:03:15 pm
Equation is as follows :wow thanks =)
Sn + (x/2)I2 ----> SnIx
Forget about the number of moles of iodine to be used.
From equation whatever be the value of x,
1 mole of Sn will form 1 mole of SnIx
Number of moles of Sn used = Mass/Ar = 0.342/119
Therefore same number of moles of SnIx must be formed.
Number of moles of SnIx formed = Mass/Mr = 1.8020/(119+127x)
In other words :
1.8020/(119+127x) = 0.342/119
Solve this equation and obtain x = 4
Hope you understand :)
Feel free to ask if you have any doubt :)
Post by: Deadly_king on September 26, 2010, 07:31:26 am
wow thanks =)Mention not :)
Post by: $!$RatJumper$!$ on September 26, 2010, 11:41:16 am
Just a question in chem. For AS inorganic 9.1(e), in the syllabus it says:
"describe the reactions, if any, of the elements with oxygen and chlorine (to give Na2O etc)"
Does describe mean we just learn the equations or also the conditions under which these oxides and chlorides are made?
Post by: Deadly_king on September 26, 2010, 11:45:42 am
Hey guys
Just a question in chem. For AS inorganic 9.1(e), in the syllabus it says:
"describe the reactions, if any, of the elements with oxygen and chlorine (to give Na2O etc)"
Does describe mean we just learn the equations or also the conditions under which these oxides and chlorides are made?
You need to know their conditions as well. Sometimes you might be asked the colour of their flames when the metals are burnt.
NOTE : The conditions are not difficult to remember. Most of them are just heating.
But you need to know the reactions that the oxides or chlorides undergo. I mean you may be asked to describe their trends.
Here's a link which may be helpful :
http://www.chemguide.co.uk/inorganic/group1/reacto2.html#top
Post by: $!$RatJumper$!$ on September 26, 2010, 11:50:30 am
One more thing, it says "describe the reactions of the oxides with water". Again, is it fine if we just state if the oxides will be soluble or insoluble in water... or would we have to elaborate more?
Post by: Arthur Bon Zavi on September 26, 2010, 12:20:50 pm
Thank you for the info buddy :)
One more thing, it says "describe the reactions of the oxides with water". Again, is it fine if we just state if the oxides will be soluble or insoluble in water... or would we have to elaborate more?
describe means, u have to elaborate! :D
Post by: moon on September 26, 2010, 08:45:50 pm
Post by: Deadly_king on September 27, 2010, 05:30:42 am
Thank you for the info buddy :)
One more thing, it says "describe the reactions of the oxides with water". Again, is it fine if we just state if the oxides will be soluble or insoluble in water... or would we have to elaborate more?
Mention not :)
Hmm........sometimes it may be just enough depending on the number of marks and the trend the specific question is leading to. But truthfully, I don't think you can risk losing marks here. So to be on the safe side, mention all that you know about these reactions, relative to the question though.
Like I said earlier, the conditions are not that difficult. It won't take you much time to remember all of them. Trust me! :)
Post by: Deadly_king on September 27, 2010, 07:27:14 am
can anyone pls explain Q2 (b), Q4 (b)(iii), (iv) & Q7 (b)Nov.2008 ppr4? thanx in advance. ???
Nov 08 P4
2.(b) Given rate= k[H2O2]a[I-]b[H+]c
a represents the order of reaction with respect to [H2O2.
b represents the order of reaction with respect to [I-].
c represents the order of reaction with respect to [H+].
NOTE : 1. The slowest step is the rate determining step.
2. The number of moles of a reactant in the rate determining step gives the order with respect to that of the reactant.
When step 1 is slowest overall : a=1, b=1 and c=0
This is because in step 1, the number of moles of H2O2 is 1 and so is I-. However H+ is not present in this equation which means it has order zero.(Reaction is independent of H+)
When step 2 is slowest overall: a=1, b=1 and c=1
IO- is formed from H2O2 and I-. Replace the first equation in the second and you'll get :
H2O2 + I- + H+ ---> HOI + H2O
From this equation note the number of moles of the respective reactants.
When step 3 is slowest overall : a=1, b=2 and c=2.
Replace the new equation found above in this third step and you'll get :
H2O2 + 2I- + 2H+ ---> I2 + 2H2O
From this equation note the number of moles of the respective reactants.
3.(b)(iii). Malachite contains copper. Copper burns to form a black solid CuO.(Copper(II) Oxide)
Form an equation for the combustion of malachite which will form carbon dioxide, steam and CuO in absence of oxygen.
Cu2O5CH2 ---> CO2 + H2O + 2CuO
Mr of Cu2O5CH2 = 221
Mr of CuO : 79.5
From equation :
1 mole of malachite produces 2 moles of CuO
221g of malachite produces 2(79.5)g of CuO.
Hence 10g of malachite will produce (2*79.5)/221 * 10 = 7.19g of CuO.
(iv)
E is copper since Iron will displace copper from copper sulfate to form iron sulfate.
CuSO4 + Fe ----> FeSO4 + Cu
Data booklet is used to see if reaction will occur by finding the electrode potential value of the cell.
Electrode potential value : +0.44 + 0.34 = +0.78V
Since the value is positive, the reaction will take place.
I sincerely hope my explanations are clear enough for you to understand. :)
Post by: Deadly_king on September 27, 2010, 07:37:55 am
Condensation reaction occurs between the NH2 group of one amino acid and the CO2H group of another amino acid to form a peptide linkage as described in the first picture.
However the questions ask about the formation about a tripeptide. Hence it will be a reaction involving 3 amino acids. The displayed formula of a tripeptide is shown in the second picture.
NOTE : The R groups of atoms may be replaced by almost any atom. However it will be much easier and simpler to replace it by H atom. Your answer should not consist of R.
Post by: moon on September 27, 2010, 09:20:46 pm
Post by: Deadly_king on September 28, 2010, 05:25:14 am
Thank u very much for ur reply as well as ur explanation. It is really helpful, but can u explain Q4(b)(iii), of the same year Nov. 2008? You have explained Q3 not Q4 and what is the formula of the tripeptide as shown in the markscheme?....I really do appreciate ur co-operation.
Oops ....... sorry for the confusion dude.
4(b)(iii)
That will depend on the equation you wrote for part (b)(i).
Let's take it to be : C8H18 -----> 2C2H4 + C4H10
Now you need to find the enthalpy change for this reaction.
Number of bonds to be broken in the reactant : 18C-H bonds and 7C-C bonds.
Total number of bonds to be formed in the products : 2C=C bonds + 8C-H bonds and 3C-C bonds + 10C-H bonds.
Use the values from the Data Booklet to find :
(i) Energy required for bond-breaking : 18(410) + 7(350) = 9830 KJ/mol
(ii) Energy given out for bond formation : 2(610) + 8(410) + 3(350) + 10(410) = 9650 KJ/mol
Standard enthalpy change of reaction : 9830 - 9650 = +180 KJ/mol
NOTE : Bond breaking is endothermic(+ve) while bond forming is exothermic(-ve).
Same method is applied whatever the equation you wrote in (b)(i).
(iv)
The conditions include a temperature of about 600oC. Hence this suggests that the reaction is endothermic since heat must be supplied for the reaction to occur and this is confirmed by the entahlpy change obtained in (b)(iii).
7(b)
The formula of the tripeptide can vary. This is why the marking scheme only mentioned 'correct tripeptide'. The product in the second picture I uploaded earlier is the right answer. Except that it is asking for displayed formula, so you need to elaborate on the bonds and replace all the R groups by H for the simplest tripeptide. :)
Post by: moon on September 28, 2010, 12:27:33 pm
Post by: Deadly_king on September 29, 2010, 04:42:36 am
Thank you very much for ur help. I got it... :)
You are welcome pal :)
Post by: HUSH1994 on October 01, 2010, 01:50:32 pm
State the factors affecting the ionisation energy and state the action of each one of them
Post by: Twinkle Charms on October 01, 2010, 02:15:51 pm
Anyone can answer this question urgently:1) Size of the positive nucleur charge -- as the atomic number increases, there is an increase in the nucleur charge which causes an increase in the ionisation energy.
State the factors affecting the ionisation energy and state the action of each one of them
2)Distance of the electron from the nucleus -- when the distance of electron increases from the nucleus, the attractive force decreases therefore ionisation energy decreases.
3)Shielding effect of the inner electrons -- all the electrons are negatively charged and repel each other. Electrons in filled inner shells repel electrons in the outer shell and reduce the effect of positive nucleur charge. This is called the sheilding effect. The greater the shielding effect on an electron, lower the ionisation energy.
Post by: Arthur Bon Zavi on October 01, 2010, 02:57:17 pm
1) Size of the positive nucleur charge -- as the atomic number increases, there is an increase in the nucleur charge which causes an increase in the ionisation energy.
2)Distance of the electron from the nucleus -- when the distance of electron increases from the nucleus, the attractive force decreases therefore ionisation energy decreases.
3)Shielding effect of the inner electrons -- all the electrons are negatively charged and repel each other. Electrons in filled inner shells repel electrons in the outer shell and reduce the effect of positive nucleur charge. This is called the sheilding effect. The greater the shielding effect on an electron, lower the ionisation energy.
8) 8) This!!
Post by: Twinkle Charms on October 01, 2010, 04:03:31 pm
8) 8) This!!You have to but in everything yeah? :P :D
Post by: Deadly_king on October 01, 2010, 05:37:26 pm
1) Size of the positive nucleur charge -- as the atomic number increases, there is an increase in the nucleur charge which causes an increase in the ionisation energy.
2)Distance of the electron from the nucleus -- when the distance of electron increases from the nucleus, the attractive force decreases therefore ionisation energy decreases.
3)Shielding effect of the inner electrons -- all the electrons are negatively charged and repel each other. Electrons in filled inner shells repel electrons in the outer shell and reduce the effect of positive nucleur charge. This is called the sheilding effect. The greater the shielding effect on an electron, lower the ionisation energy.
rightly said Twinkle :)
+rep
Post by: HUSH1994 on October 02, 2010, 04:58:28 am
rightly said Twinkle :)Thank You twinkle,+rep as well ;)
+rep
Post by: ruby92 on October 03, 2010, 04:45:14 pm
Post by: Deadly_king on October 04, 2010, 05:23:30 am
given that the bond energy needed to break an O2 bond is 150 what is thae standard enthalpy of atomisation?According to your statement, when breaking the O2 molecules, we get 2 oxygen atoms. For this 150KJ/mol energy is required. (Bond energy)
The definition of enthalpy change of atomisation :
It is the heat energy change when one mole of separate gaseous atom of the element is formed from the element under standard state conditions.
By breaking the O2 bond we obtain 2 moles of oxygen atom. But enthalpy change of atomisation requires only one mole of gaseous atom to be formed.
Hence the answer will be :150/2 = 75 KJ/mol
Post by: Adzel on October 05, 2010, 03:43:05 pm
Post by: nid404 on October 05, 2010, 03:51:16 pm
Post by: Adzel on October 05, 2010, 03:54:19 pm
and take your time.
Post by: Deadly_king on October 05, 2010, 04:09:51 pm
Can someone plz answer these questions for me :)...
It would surely have been quicker if you specified your problems. I mean, you would not gain anything if we do everything for you.
Anyway I'll be starting by the last number. I guess we'll be meeting midway Garfield ;)
Post by: Adzel on October 05, 2010, 04:16:39 pm
Post by: nid404 on October 05, 2010, 05:17:57 pm
a) The first ionization energy is the energy required to remove one electron from each atom in one mole of gaseous atoms of an element.
b)
kJ/mol log10I.E
1st I.En 801 2.9
2nd I.E 2427 3.4
3rd I.E 3660 3.6
4th I.E 25026 4.4
5th I.E 32808 4.5
Boron has electronic configuration 1s22s22p1
c)As you can see (graph attached), there is a general increasing trend of I.Es. It is easier to remove the further most electron and hence a low First I.E. Once the p electron is removed, shielding effect decreases and electrons are more closely held by the nucleus and hence an increase in the 2nd and 3rd I.E. Following,there are only two electrons in the first shell and they are closest to the nucleus. Hence a huge change between the 3rd and 4th I.E.
(since they haven't provided the marks, I have explained what I think you would ought to know)
d) i) Sulphur
(http://www.webelements.com/_media/elements/ionisation_energies/S.gif)
ii)Potassium
(http://210.34.15.15/webelements/webelements/K-5.gif)
Post by: nid404 on October 05, 2010, 05:20:20 pm
Post by: Adzel on October 05, 2010, 05:29:48 pm
Post by: Deadly_king on October 05, 2010, 05:42:19 pm
I guess Ari must have already helped with part (a)
But anyway here's a link which might help:
http://www.s-cool.co.uk/alevel/chemistry/transition-metals/electronic-configuration.html
(b)
(i) The 4s orbital is at a lower energy level than 3d. Hence we fill the 4s orbital before the 3d one.
(ii) Chromium is a transition element and has only 4 electrons int its 3d sub-shell while 2 electrons are present in the 4s sub-shell. The latter suffer repulsion such that one of them moves into one of the empty sub-shells of the 3d orbital. This results in greater stability of the element.
(iii) Copper experiences greater stability with a filled 3d-orbital rather than a filled 4s orbital. Hence it contains only one electron in 4s.
Post by: nid404 on October 05, 2010, 06:43:15 pm
Since they don't tell you how much you need to answer I will just say whatever is required
a)The Atomic Spectra of an element are the lines which are present at characteristic wavelengths in the light emitted from an electric arc between electrodes of that element. Each line in the spectrum corresponds to the energy difference between two orbitals in the atom. According to the quantum theory, the electrons in an atom can only reside in specific orbitals, each of which has a fixed energy level, and one photon is emitted for each electronic transition between orbitals.
When an electron drops from an orbital of high energy to one of lower energy, a photon (i.e. a packet of light) is emitted corresponding to the energy difference between the orbitals.
b) If electrons possessed energy, it would be a continuous spectrum
The graph I'm not sure if it's plotted in terms of frequency or wavelength, so I cannot be of much help.
http://www.chemguide.co.uk/atoms/properties/hspectrum.html
Both possibilities are mentioned here....should help :)
Post by: moon on October 05, 2010, 10:46:50 pm
Post by: Freaked12 on October 05, 2010, 11:42:55 pm
SiO2+2NaOH->Na2SiO3+H2O
Post by: Deadly_king on October 06, 2010, 05:05:41 am
Rightly said Requiem :)
Post by: Dania on October 07, 2010, 02:41:53 pm
E.g. OHCH(CN)CH(CN)OH + H2SO4 = ?
Post by: nid404 on October 07, 2010, 02:44:16 pm
Post by: Dania on October 07, 2010, 05:39:09 pm
Post by: Deadly_king on October 07, 2010, 05:44:54 pm
What happens when you add dilute sulfuric acid to a cyanohidrin?
E.g. OHCH(CN)CH(CN)OH + H2SO4 = ?
Hmm.......the CN groups will get converted to CO2H
In other words the nitrile is oxidised to a carbolxylic group
Therefore resulting compound will be OHC(CO2H)HC(CO2H)HOH
Hope it helps :)
Post by: Dania on October 07, 2010, 05:47:44 pm
The density of ice is 1.00 g cm–3.
What is the volume of steam produced when 1.00 cm3 of ice is heated to 323 °C (596 K) at a
pressure of one atmosphere (101 kPa)?
[1 mol of a gas occupies 24.0 dm3 at 25 °C (298 K) and one atmosphere.]
A 0.267 dm3
B 1.33 dm3
C 2.67 dm3
D 48.0 dm3
The answer is C. How would I obtain it? I'm not sure what equation to use.
Post by: Deadly_king on October 07, 2010, 06:04:49 pm
Thanks :) I have a MCQ also:
The density of ice is 1.00 g cm–3.
What is the volume of steam produced when 1.00 cm3 of ice is heated to 323 °C (596 K) at a
pressure of one atmosphere (101 kPa)?
[1 mol of a gas occupies 24.0 dm3 at 25 °C (298 K) and one atmosphere.]
A 0.267 dm3
B 1.33 dm3
C 2.67 dm3
D 48.0 dm3
The answer is C. How would I obtain it? I'm not sure what equation to use.
First you need to find the number of moles present in 1cm3 of ice.
Using density = 1gcm-3, we take the mass present in 1 cm3 of ice to be 1g.
Number of moles = Mass/Mr ----> Number of moles = 1/18 = 0.0556
Now we use the formula for ideal gas ---> PV = nRT
V = (0.0556 x 8.31 x 596)/(101 x 103) = 0.0027m3
This is equal to 2.7dm3
Am not sure though.........since I did not obtain 2.67 dm3 :-[
Post by: nid404 on October 07, 2010, 06:05:34 pm
Post by: Deadly_king on October 08, 2010, 07:50:11 am
You've become faster :P
Hehe......yeah I guess so ;)
Anyway it would be nice if you could confirm the method since I didn't obtain the exact answer ???
Post by: Dania on October 08, 2010, 10:04:34 am
Why?
Post by: Dania on October 08, 2010, 11:17:00 am
Post by: Deadly_king on October 08, 2010, 12:08:56 pm
The answer is D.Since it is unreactive to mild oxidising agents, the compound should have been a tertiary alcohol before dehydration. Only D offers this possibility.
Why?
NOTE : Tertiary alcohols cannot be oxidised.
Post by: ashish on October 08, 2010, 02:02:59 pm
The answer is B. Why?
CaCO3(limestone) -----> CaO + CO2
from equation 1 mole CaCO3 produced 1 mole CO2
Mr CaCO3 =100.1
Mr CaO=56.1
Mr CO2=44
1000mt CaCO3 will give 560.1mt Cao and 440mt CO2
200mt CaCO3 will give 112.02mt Cao and 88mt CO2
mt=million tonnes
total CO2 = 440+88
=528 --- near to 527
IS that CIE i never worked out such question before
i am not sure whether the method is good or not
Post by: Dania on October 08, 2010, 06:33:51 pm
Post by: Deadly_king on October 08, 2010, 06:38:34 pm
CaCO3(limestone) -----> CaO + CO2
from equation 1 mole CaCO3 produced 1 mole CO2
Mr CaCO3 =100.1
Mr CaO=56.1
Mr CO2=44
1000mt CaCO3 will give 560.1mt Cao and 440mt CO2
200mt CaCO3 will give 112.02mt Cao and 88mt CO2
mt=million tonnes
total CO2 = 440+88
=528 --- near to 527
IS that CIE i never worked out such question before
i am not sure whether the method is good or not
Good job pal........though it was a bit long ;)
+ rep
Post by: moon on October 08, 2010, 09:24:05 pm
also Nov.2009 ppr42 Q8(c) (ii), d(ii),(iv)
thanx in advanc.
Post by: moon on October 09, 2010, 01:01:15 am
complete combustion of a hydrocarbon gave 0.352gm of CO2 & 0.072gm of water. What is the formula of hydrocarbon? ??? ???
Post by: elemis on October 09, 2010, 04:57:58 am
can u plz another question?
complete combustion of a hydrocarbon gave 0.352gm of CO2 & 0.072gm of water. What is the formula of hydrocarbon? ??? ???
Your question is incomplete I need the mass of the original hydrocarbon.
But the methodology is as follows :
If there are 2 grams of Hydrogen in 18 grams of water THEN
there are X grams of Hydrogen in 0.072 grams of water
Hence, X = 8*10-3
If there are 12 grams of Carbon in 44 grams of CO2 THEN
there are Y grams of Carbon in 0.352 grams of CO2
Hence, Y = 0.096
At this point you would add X and Y and subtract that value from the mass of the Hydrocarbon to find the mass of oxygen.
Then you would have the mass of C H and O
You could then calculate the Empirical formula.
Post by: Deadly_king on October 09, 2010, 05:13:44 am
can u plz another question?
complete combustion of a hydrocarbon gave 0.352gm of CO2 & 0.072gm of water. What is the formula of hydrocarbon? ??? ???
Nan Ari......the data provided is sufficient to find the answer.
CxHy + (x + y/4)O2 ----> xCO2 + (y/2) H2O
Number of moles CO2 = Mass / Mr = 0.352/44 = 0.008
Number of moles of H2O = 0.072/18 = 0.004
1 mole of the hydrocarbon gives 0.008 moles of CO2 and 0.004 moles of H2O
Therefore from equation above we can note that x = 0.008 while y/2 = 0.004
Empirical formula will hence be C8H8
If you don't understand, let me know :)
Post by: elemis on October 09, 2010, 05:17:13 am
+rep.
Post by: Deadly_king on October 09, 2010, 05:23:17 am
Ah, damn. I didnt think of it in that way. :-X
+rep.
It's alright dude ;)
Post by: Deadly_king on October 09, 2010, 06:37:06 am
plz help in Nov.2009 ppr22 Q2(v) &
also Nov.2009 ppr42 Q8(c) (ii), d(ii),(iv)
thanx in advanc.
Nov 09 P2 Q2(v)
In most cases, Hydrogen has an oxidation number of -1.
Therefore in the first 3 cases. take oxidation number of H to be -1. Since they are all neutral molecules, resultant charge should become to zero. This principle is applied to ionic compounds only.
MgH2 :
Let oxidation number of Mg be x.
x + 2(-1) = 0 ---> x = +2
Apply same principle for AlH3 and you'll get +3.
As for the other two, they are covalently bonded such that oxidation state of hydrogen is +1
PH3
Let oxidation no of P be x.
x + 3(+1) = 0
x = -3
same method for H2S. ;)
Post by: nid404 on October 09, 2010, 06:54:45 am
Hehe......yeah I guess so ;)
Anyway it would be nice if you could confirm the method since I didn't obtain the exact answer ???
Your method is not wrong really...
what I would have done instead
at 298K 1 mole occupies 24dm3
so At 596 K 1 mole occupies 48 dm3
1g of steam= 1/18 moles
volume of steam= 1/18 X 48= 2.67 dm3
Post by: Deadly_king on October 09, 2010, 07:04:13 am
Your method is not wrong really...
what I would have done instead
at 298K 1 mole occupies 24dm3
so At 596 K 1 mole occupies 48 dm3
1g of steam= 1/18 moles
volume of steam= 1/18 X 48= 2.67 dm3
Oooh.........yeah.
I guess I chose the long way :P
It's far easier to understand like you described ;)
Thanks :)
Post by: nid404 on October 09, 2010, 07:08:37 am
Oooh.........yeah.
I guess I chose the long way :P
It's far easier to understand like you described ;)
Thanks :)
Considering it's an MCQ , you wouldn't want to do it the long way ;)
My pleasure :)
Post by: Deadly_king on October 09, 2010, 07:27:26 am
c)(ii) If two chlorine atoms are present, the three possibilities are : Cl35-Cl35, Cl35-Cl37 and Cl37-Cl37
Cl35 and Cl37 exist in the ratio 3 : 1
Probability of obtaining Cl35-Cl35 = 3/4 x 3/4 = 9/16 ---> M peak
Probability of obtaining Cl35-Cl37 = 2(3/4 x 1/4) = 6/16 ---> M+2 peak
Probability of obtaining Cl37-Cl37 = 1/4 x 1/4 = 1/16 ----> M+4 peak
Therefore the ratio will be 9 : 6 : 1
NOTE : The probability for M+2 peak is multiplied by two since it can exist both as Cl35-Cl37 and Cl37-Cl35
d)(ii) PCBs.
This is because when you compare the graphs upstream and downstream, you will note that PCBs is already in very high concentration unlike the others which are present in minimal amount.
Therefore after passing the mill, the concentration of the other three chemicals increase in concentration showing that the mill releases these chemicals.
(iv) Four chlorine atoms will result in 5 possible peakss.
1. Cl35-Cl35-Cl35-Cl35
2. Cl37-Cl37-Cl37-Cl37
3. Cl35-Cl37-Cl37-Cl37
4. Cl35-Cl37-Cl35-Cl37
5. Cl37-Cl37-Cl37-Cl35
NOTE : The arrangement of the atoms in each case may differ but they will form the same peak.
Hope it helps :)
Post by: Deadly_king on October 09, 2010, 07:29:15 am
Considering it's an MCQ , you wouldn't want to do it the long way ;)Absolutely right madam ;)
My pleasure :)
Post by: ashish on October 09, 2010, 07:33:40 am
Good job pal........though it was a bit long ;)
+ rep
^_^ thank you !
Post by: $!$RatJumper$!$ on October 09, 2010, 11:29:17 am
Post by: elemis on October 09, 2010, 03:09:16 pm
For chemistry practical.. what is the best way to study for it other than actually doing it in school? We're on study leave so dont have access to lab equipment. How is the best way to do it at home?
Watch videos on youtube.
Post by: $!$RatJumper$!$ on October 09, 2010, 05:19:42 pm
Watch videos on youtube.
Can you please provide me a link to get started?
Post by: Freaked12 on October 09, 2010, 05:20:13 pm
What the heck
moles CO2 = 0.352 g / 44.009 g/mol=0.00800
moles H = 2 x 0.072 /18.02 g/mol=0.00800
Post by: Freaked12 on October 09, 2010, 05:20:54 pm
AlH3 is Al3+ + 3 H- ions so oxidation number of Al = +3 (Group 3 - loses 3 electrons to form Al3+ ion 3 - 3 empties outer shell)
PH3 Is P3- and3 H+ so oxidation number of P = -3 (Group 5 - gains 3 electrons to form P3- ion : 5 + 3 = 8 fills up outer shell)
H2S is 2 H+ and S2- ions so oxidation number of S is -2 (Group 6 - gains 2 electrons to form S2- ion ; 6 + 2 = 8 fills up outer shell)
Sorry
Just wanted to get involved
Post by: moon on October 10, 2010, 08:34:52 am
Nan Ari......the data provided is sufficient to find the answer.
CxHy + (x + y/4)O2 ----> xCO2 + (y/2) H2O
Number of moles CO2 = Mass / Mr = 0.352/44 = 0.008
Number of moles of H2O = 0.072/18 = 0.004
1 mole of the hydrocarbon gives 0.008 moles of CO2 and 0.004 moles of H2O
Therefore from equation above we can note that x = 0.008 while y/2 = 0.004
Empirical formula will hence be C8H8
why didn't we find the mole ratio CO2: H2O
so that the formula can be C2H2...
Post by: moon on October 10, 2010, 08:42:05 am
The oxidation state of hydrogen in the metal hydrides is always -1 so why in PH3 the oxidation state is +1 and hence x+3(-1)=0 so x=+3 really need help..Thanks n advance
Post by: Deadly_king on October 10, 2010, 08:45:29 am
for nov 2009 ppr 2
The oxidation state of hydrogen in the metal hydrides is always -1 so why in PH3 the oxidation state is +1 and hence x+3(-1)=0 so x=+3 really need help..Thanks n advance
Phosphorus is not a metal.
Being a non-metal, it cannot give electrons to turn onto +3 oxidation states. It has to accept electrons which is why it has a negative oxidation number :)
Post by: moon on October 10, 2010, 08:54:49 am
why didn't we find the mole ratio CO2: H2O
so that the formula can be C2H2...
please i really need help in this question and thank a lot 4ur ppr4 explanation....
Post by: Deadly_king on October 10, 2010, 09:26:24 am
why didn't we find the mole ratio CO2: H2O
so that the formula can be C2H2...
Hmmm........it's not possible to do it using the ratio CO2 : H2O
But you must realise that :
1 mole of hydrocarbon will form x moles CO2 and y/2 moles of H2O
Ratio CO2 : H2O will hence be x : y/2
These are two unknown values which are not linked in any way. So we cannot solve it.
First you need the ratio hydrocarbon : CO2 ---> 1 : x
Then use ratio of hydrocarbon to H2O ---> 1 : y
Only then will you be able to find the values of x and y :)
Post by: moon on October 10, 2010, 09:43:59 am
Post by: Deadly_king on October 10, 2010, 10:14:33 am
thank u very much 4ur explanation... :)
You're welcome buddy.
I just hope my explanations were clear enough :)
Post by: moon on October 13, 2010, 01:00:10 am
I would appreciate ur help, thanx in advance.
Post by: Deadly_king on October 13, 2010, 12:06:33 pm
can somebody plz explain Q5 d,Q8 b(ii),c & Q9a(i) in June 2009 ppr4
I would appreciate ur help, thanx in advance.
Jun 09 p4 No 5
Hmmm......part (d) will depend on the alcohols you drew in part (a).
If it is a primary alcohol, the latter will get oxidised to a carboxylic acid.
If it is a secondary alcohol, it will get oxidised to a ketone.
If it is a tertiary alcohol, it will not get oxidised. (stays the same)
However you won't be having any tertiary alcohols, since the question asks for unbranched compounds.
Post by: moon on October 13, 2010, 12:21:54 pm
Jun 09 p4 No 5
If it is a tertiary alcohol, it will not get oxidised. (stays the same)
Do u mean that teertiary alcohol on oxidation will be a tertiary alcohol and not ketone????????????? because the markscheme's answer was CH3CH2COCH2CH3
Post by: Deadly_king on October 13, 2010, 12:27:50 pm
Do u mean that teertiary alcohol on oxidation will be a tertiary alcohol and not ketone????????????? because the markscheme's answer was CH3CH2COCH2CH3
Yeah.........but the compound you drew is not a tertiary alcohol. It is a secondary one. That's why it gets converted into a ketone.
NOTE : A tertiary alcohol is one whose OH group is bonded to a carbon atom which is already bonded to three other carbon atoms. In other words the carbon atom is not attached to any hydrogen atom ;)
P.S : I edited my previous post. Do have a look, it might clear your doubt.
Post by: moon on October 13, 2010, 12:49:44 pm
Post by: Deadly_king on October 13, 2010, 12:55:46 pm
b)(ii) KPC = Concentartion of PCBs in fats / Concetration of PCBs in water
The above mentioned KPC should have a high value since from the paragraph written above, we can note that PCBs are highly soluble in fats (the main components of the Inuit diet). Therefore we can conclude that PCBs will be more soluble in fat than in water. Since PCBs is more soluble in fats, it will have a greater concentration in fats rather than water.
c)(i) Four spots were observed after the first solvent has been used. Spots which are just vertically above another spot s counted as one spot only.
(ii) Spot at the bottom.
(iii) Spot at the very top.
If you need more explanation, let me know :)
Post by: moon on October 13, 2010, 01:10:28 pm
Post by: Deadly_king on October 13, 2010, 02:23:44 pm
a)(i) You just need to draw an exact copy of the kevlar structure just below it. Then you need to show how they are bonded by hyfrogen bonds which occur between on O atom from C=O and one H atom from the opposite N-H.
In other words the bond will be formed from an O atom from the already drawn kevlar structure to a H atom from the structure you'll be drawing.
NOTE : Hydrogen bonds take place only between a H atom and an electronegative atom like oxygen, nitrogen and chlorine.
Post by: cs on October 15, 2010, 05:43:26 am
Post by: Deadly_king on October 15, 2010, 06:00:27 am
I have a doubt, what is the difference between elimination and reduction in organic chemistry..?
That may depend upon the different reactions. Do you have any example ???
Reduction -----> Redox reaction where a substance is reduced.
Elimination -----> A molecule is formed as by-product from the reaction. Usually it's water.
Examples :
i) When an aldehyde is reacted with a reducing agent like lithium aluminium hydride dissolved in ether, a primary alcohol is formed.
This is reduction as the aldehyde is reduced to a primary alcohol. :)
ii) When an alcohol is reacted with concentrated sulfuric acid at a temperature of 180oC, an alkene is formed with water as by-product.
We say water is being eliminated from the alcohol and the reaction is known as elimination ;)
Post by: cs on October 15, 2010, 12:40:37 pm
Post by: Deadly_king on October 15, 2010, 02:53:58 pm
Yes Deadly King, the examples are exactly what i asked, so water is "eliminated", and the other one is reduced. Thank you so much. +REP as usual. Great help
Yeah......most of the time, water is eliminated but in some case HCl or HBr may also be eliminated, depending on the structure of the reacting compound :)
Anytime dear ;)
Post by: $!$RatJumper$!$ on October 16, 2010, 07:54:03 pm
If this isn't clear, i am basically referring to S10 P33 Chemistry (9701) question 1(c)(i).
Thankx
Post by: ashish on October 17, 2010, 02:25:35 am
Guys, for AS chemistry in practicals, when they tell us to calculate answers in which we have to use q=mc(delta)t, i know that c=4.2, (delta)t is easy to calculate from our results, but m, i have no idea where to get it from :/ All i have are the volumes and not masses. Can i use volumes for the value of m in the equation?
If this isn't clear, i am basically referring to S10 P33 Chemistry (9701) question 1(c)(i).
Thankx
yes you must take the total volumes as M
Post by: Deadly_king on October 17, 2010, 08:18:12 am
yes you must take the total volumes as M
Yupz.......m is taken as the total volume of solutions which you used when measuring the temperature change. ;)
Post by: $!$RatJumper$!$ on October 17, 2010, 11:19:44 am
Post by: moon on October 17, 2010, 02:37:21 pm
thanx in advance.
Post by: $!$RatJumper$!$ on October 17, 2010, 03:08:28 pm
plz can somebody help me in calculation ppr2 Nov.07 Q1 (e)
thanx in advance.
The general equation for enthalpy change of formation is: (delta)H(formation) = Products - Reactants
Now, the information they have given us for the reaction, is all in (delta)H(combustion). So what you have to do, is simply reverse the signs on each of the values to change them to the formation values. In this case, since they are all negative, we make them positive.
Once done that, now identify your products and your reactants.
Products: C2H4
Reactants: 2C, 2H2
Then plug in their values for formation into the equation above like this:
(delta)H(formation) = (–1411.0) - [(2 * –285.9) + (2 * –393.7)]
Therefore, (delta)H(formation) = 51.8 kJ mol–1
P.S We multiplied them by 2 because the equation said 2C and 2H2. The 2 represents the moles so we take them into account.
Hope this helped
Post by: Deadly_king on October 17, 2010, 03:17:32 pm
The general equation for enthalpy change of formation is: (delta)H(formation) = Products - Reactants
Now, the information they have given us for the reaction, is all in (delta)H(combustion). So what you have to do, is simply reverse the signs on each of the values to change them to the formation values. In this case, since they are all negative, we make them positive.
Once done that, now identify your products and your reactants.
Products: C2H4
Reactants: 2C, 2H2
Then plug in their values for formation into the equation above like this:
(delta)H(formation) = (–1411.0) - [(2 * –285.9) + (2 * –393.7)]
Therefore, (delta)H(formation) = 51.8 kJ mol–1
P.S We multiplied them by 2 because the equation said 2C and 2H2. The 2 represents the moles so we take them into account.
Hope this helped
Good work buddy ;)
+rep
Post by: moon on October 17, 2010, 03:25:58 pm
The general equation for enthalpy change of formation is: (delta)H(formation) = Products - Reactants
Now, the information they have given us for the reaction, is all in (delta)H(combustion). So what you have to do, is simply reverse the signs on each of the values to change them to the formation values. In this case, since they are all negative, we make them positive.
Once done that, now identify your products and your reactants.
Products: C2H4
Reactants: 2C, 2H2
Then plug in their values for formation into the equation above like this:
(delta)H(formation) = (–1411.0) - [(2 * –285.9) + (2 * –393.7)]
Therefore, (delta)H(formation) = 51.8 kJ mol–1
P.S We multiplied them by 2 because the equation said 2C and 2H2. The 2 represents the moles so we take them into account.
Hope this helped
I got the same answer -51.8 KJmol-1 but the markscheme's answer is +51.8KJmol-1 How come???
Post by: Deadly_king on October 17, 2010, 03:32:00 pm
I got the same answer -51.8 KJmol-1 but the markscheme's answer is +51.8KJmol-1 How come???
The answer is indeed +51.8 KJmol-1
Delta H = 2(-393.7) + 2(-285.9) - (-1411) = +51.8
2C + 2H2 ----> C2H4
Since the enthalpy changes of combustion has been given to you, you also need to write the equation relating each substance and its respective combustion.
1. C + O2 ----> CO2 (Enthalpy change of combustion of carbon or graphite)
2. H2 + 0.5O2 ----> H2O (Enthalpy change of combustion of hydrogen)
3. C2H4 + 3O2 ----> 2CO2 + 2H2O (Enthalpy change of combustion of butane)
As you may observe, the combustion of both hydrogen and carbon leads to carbon dioxide and hydrogen as well as the combustion of butane.
Hence, according to Hess's law, we may opt for another route in the conversion of carbon and hydrogen to butane.
Let's say we'll convert 2 moles of carbon and 2 moles of Hydrogen gas to carbon dioxide and water. Then these two products are converted to ethane.
In practise these reactions are not feasible. But in theory we may opt for that method to find standard enthalpy changes of a particular reaction.
Let ^C be the enthalpy change of combustion of carbon.
So Enthalpy change of formation of butane = 2(^C) + 2(^H) - (^C2H4)
NOTE : We need to subtract the enthalpy change of combustion since we are converting carbon dioxide and water back to butane whereas the enthalpy change of combustion of butane actually is the energy evolved when 1 mole of butane is completely burnt to carbon dioxide and water. In other words we are doing just the contrary. that's why we need to subtract it.
If you don't understand, let me know :)
Post by: $!$RatJumper$!$ on October 17, 2010, 03:41:29 pm
I got the same answer -51.8 KJmol-1 but the markscheme's answer is +51.8KJmol-1 How come???
You do get +51.8 when you do the calculation i mentioned
Post by: Deadly_king on October 17, 2010, 03:45:17 pm
You do get +51.8 when you do the calculation i mentioned
Yeah.......you've done it correctly br0 :)
Post by: $!$RatJumper$!$ on October 17, 2010, 03:50:13 pm
The answer is indeed +51.8 KJmol-1
Delta H = 2(-393.7) + 2(-285.9) - (-1411) = +51.8
2C + 2H2 ----> C2H4
Since the enthalpy changes of combustion has been given to you, you also need to write the equation relating each substance and its respective combustion.
1. C + O2 ----> CO2 (Enthalpy change of combustion of carbon or graphite)
2. H2 + 0.5O2 ----> H2O (Enthalpy change of combustion of hydrogen)
3. C2H4 + 13/2 O2 ----> 2CO2 + 2H2O (Enthalpy change of combustion of butane)
As you may observe, the combustion of both hydrogen and carbon leads to carbon dioxide and hydrogen as well as the combustion of butane.
Hence, according to Hess's law, we may opt for another route in the conversion of carbon and hydrogen to butane.
Let's say we'll convert 2 moles of carbon and 2 moles of Hydrogen gas to carbon dioxide and water. Then these two products are converted to ethane.
In practise these reactions are not feasible. But in theory we may opt for that method to find standard enthalpy changes of a particular reaction.
Let ^C be the enthalpy change of combustion of carbon.
So Enthalpy change of formation of butane = 2(^C) + 2(^H) - (^C2H4)
NOTE : We need to subtract the enthalpy change of combustion since we are converting carbon dioxide and water back to butane whereas the enthalpy change of combustion of butane actually is the energy evolved when 1 mole of butane is completely burnt to carbon dioxide and water. In other words we are doing just the contrary. that's why we need to subtract it.
If you don't understand, let me know :)
good one :)
Post by: moon on October 17, 2010, 04:09:43 pm
Post by: Deadly_king on October 17, 2010, 04:24:19 pm
I understood it now thanks Deadly_king and thanks 4 u too $!$RatJumper$!$
Anytime dude :)
Post by: $!$RatJumper$!$ on October 17, 2010, 04:28:05 pm
I dont understand how to do the the second step.
Post by: Ghost Of Highbury on October 18, 2010, 11:32:50 am
2) A typical fertilizer contains the elements N,P and K in the ratio 15g : 30g : 15g per 100g of fertilizer.
The recommended usage of fertilizer is 14g of fertilizer per 5dm^3 of water.
What is the concentration of nitrogen atoms in this solution?
Post by: SpongeBob on October 18, 2010, 11:48:06 am
1) How many lone pairs and bond pairs of electrons surround XE in XeF4?
2) A typical fertilizer contains the elements N,P and K in the ratio 15g : 30g : 15g per 100g of fertilizer.
The recommended usage of fertilizer is 14g of fertilizer per 5dm^3 of water.
What is the concentration of nitrogen atoms in this solution?
(http://www.grandinetti.org/Teaching/Chem121/Lectures/Hybridization/assets/XeF4.gif)
4 co-ordinate bonds.
So,
4 bond pairs. 2 lone pairs.
15 g in 100
x g in 14
cross multiply
x= 2.1 g
moles= 2.1/14
conc= mol/ dm3
conc= (2.1/14)/ 5
=0.03 mol/dm3
Post by: Ghost Of Highbury on October 18, 2010, 01:09:47 pm
(http://www.grandinetti.org/Teaching/Chem121/Lectures/Hybridization/assets/XeF4.gif)
4 co-ordinate bonds.
So,
4 bond pairs. 2 lone pairs.
15 g in 100
x g in 14
cross multiply
x= 2.1 g
moles= 2.1/14
conc= mol/ dm3
conc= (2.1/14)/ 5
=0.03 mol/dm3
thanks a lot
i have a doubt though
the 2nd q says the elements N,P and K are in the "RATIO" 15:30:15 in 100g of the fertilizer
this means that N in 100g of fertilizers is 15/60 * 100 = 25g , rite??
Post by: Arthur Bon Zavi on October 18, 2010, 01:13:10 pm
thanks a lot
i have a doubt though
the 2nd q says the elements N,P and K are in the "RATIO" 15:30:15 in 100g of the fertilizer
this means that N in 100g of fertilizers is 15/60 * 100 = 25g , rite??
Right!! :D Go ahead, any more doubts?
Post by: nid404 on October 18, 2010, 01:53:32 pm
thanks a lot
i have a doubt though
the 2nd q says the elements N,P and K are in the "RATIO" 15:30:15 in 100g of the fertilizer
this means that N in 100g of fertilizers is 15/60 * 100 = 25g , rite??
It says ratio of 15g: 30g : 15g They've mentioned the units
it means 15g of N, 30g of P and 15g of K in 100 g of the fertilizer. The fertilizer is not only made up of N,P and K...there are other compounds/elements/mixtures which make up the rest of the 40g.
Post by: Arthur Bon Zavi on October 18, 2010, 01:58:57 pm
It says ratio of 15g: 30g : 15g They've mentioned the units
it means 15g of N, 30g of P and 15g of K in 100 g of the fertilizer. The fertilizer is not only made up of N,P and K...there are other compounds/elements/mixtures which make up the rest of the 40g.
Dumb Girl, refer the question carefully. :P
Tell me overall what is it and what's wrong?
Post by: nid404 on October 18, 2010, 02:01:56 pm
Dumb Girl, refer the question carefully. :P
Tell me overall what is it and what's wrong?
I'm not a girl you moron. ::)
Heh? Nothing is wrong. The person answered it correctly. I'm just clarifying Ghost Of Highbury's doubt. ::)
Post by: Arthur Bon Zavi on October 18, 2010, 02:06:19 pm
I'm not a girl you moron. ::)
Heh? Nothing is wrong. The person answered it correctly. I'm just clarifying Ghost Of Highbury's doubt. ::)
Can I call u bhaibandh, then. :P
Yes, thank you for your explanation. That answer was apparently right. :D
Post by: Ghost Of Highbury on October 18, 2010, 04:04:23 pm
It says ratio of 15g: 30g : 15g They've mentioned the units
it means 15g of N, 30g of P and 15g of K in 100 g of the fertilizer. The fertilizer is not only made up of N,P and K...there are other compounds/elements/mixtures which make up the rest of the 40g.
Ahh the unit, darn. Thanks both of u...
Post by: $!$RatJumper$!$ on October 18, 2010, 05:13:50 pm
I realise its something to do with ratios, but i dont get what volumes to use. Please any help would be appreciated asap as my chem prac exam is tomorrow :) thankx
Post by: SpongeBob on October 19, 2010, 10:58:29 am
Post by: $!$RatJumper$!$ on October 19, 2010, 11:28:07 am
Post by: SpongeBob on October 19, 2010, 01:29:49 pm
Post by: Dania on October 29, 2010, 04:37:02 pm
Refer to the attachment.
Thanks in advance.
Post by: Deadly_king on October 29, 2010, 05:55:13 pm
Can someone please help and explain this question for me?
Refer to the attachment.
Thanks in advance.
The diagram represents a mass spectrum for HCl.
NOTE : Only positive ions produce peaks in a mass spectrum since the substance is initially bombarded with electrons such that it loses one electron from a shared pair or a lone pair due to collision.
Hence for part 1 => 35Cl+
Second answer is 1H37Cl+
As for the third part, we must use the relative abundance of the two isotopes and multiply them by the Ar of the respective isotope. Then the sum of these two values is divided by the total amount of substance, in terms of abundance.
Relative atomic mass of Cl = (75(35) + 25(37))/(75 + 25) = 35.5
Just showing the above equation should be enough to score all the marks ;)
Post by: Dania on October 30, 2010, 05:57:57 pm
Please explain any substitutions and/or additions.
Thanks in advance.
Post by: Deadly_king on October 30, 2010, 06:13:05 pm
What would I get if I had to react these 2 organic compounds?
Please explain any substitutions and/or additions.
Thanks in advance.
That may depend on the conditions.
But the most appropriate reaction might be between the hydroxyl group of Y to the carboxylic group of Z to form an ester.
An alcohol reacts with a carboxylic acid to give an ester and water. Therefore it will be a condensation reaction whereby the H from alcohol and the OH from carboxylic acid combine to form water while the remaining compound joins together to give an ester. ;)
Post by: $!$RatJumper$!$ on October 30, 2010, 06:23:27 pm
What would I get if I had to react these 2 organic compounds?
Please explain any substitutions and/or additions.
Thanks in advance.
Thats right, it would form an ester and water for sure
Post by: Dania on October 30, 2010, 08:54:40 pm
Post by: missbeautiful789 on October 30, 2010, 09:58:09 pm
It makes sence to work backwards from U.
Add an -OH and an -H to the alkene.....then your done
I can't explain the mechanism..if they say they react, then i believe them...i just know how to find the product
Aldehydes react with each other apparently.
Post by: Deadly_king on October 31, 2010, 08:17:27 am
Crap. Sorry about that. It's question 5 from May/June 2009 Paper 2, Variant 2. Please check it out and get back to me.
Yeah..........you have only one choice here. You need to work back from U.
It has been said that T contains one OH group and form U, we may easily identify the presence of an aldeyde group. However it cannot be the conversion of a primary alcohol to an aldehyde since we can note that T has 2 oxygen atoms.
So it is bound to contain an OH group on one of the carbon carrying the C=C.
T is converted into U by using concentrated sulfuric acid which is a dehydrating agent and will cause the elimination of water from T top form U
Type of reaction : Elimination reaction or dehydration.
b) U contains two functional groups. You can name any one of them and identify them using the appropriate substance.
Example :
Alkene -----> It decolourises bromine.
Ketone -----> It gives an orange precipitate with 2,4 DNPH
Hope it helps :)
Post by: Dania on October 31, 2010, 01:23:44 pm
Post by: Aadeez || Zafar on November 01, 2010, 05:22:18 pm
Post by: Deadly_king on November 01, 2010, 05:33:12 pm
No more doubts frm u all
Why are you saying this?
Post by: Aadeez || Zafar on November 01, 2010, 05:59:47 pm
Why are you saying this?i think that i know much of the chemistry so can solve any doubts
Post by: Deadly_king on November 02, 2010, 04:53:18 am
i think that i know much of the chemistry so can solve any doubts
Ooh...........glad to have members like you who are so interested to help. :)
But you'll have to be patient, wait for other members to post their doubts ;)
Post by: moon on November 02, 2010, 01:00:13 pm
element Atomic radius/pm Ionic radius/pm
lithium 152 ??????
sodium 186 95
potassium 227 138
Post by: Arthur Bon Zavi on November 02, 2010, 01:17:43 pm
plz help me to solve this question. thanx in advance.
element Atomic radius/pm Ionic radius/pm
lithium 152 ??????
sodium 186 95
potassium 227 138
43 - 7 = 36
Therefore 95 - 36 = 59 pm. IDK :D
Post by: moon on November 02, 2010, 01:38:20 pm
43 - 7 = 36
Therefore 95 - 36 = 59 pm. IDK :D
sorry Ancestor I didn't get it.
Post by: Arthur Bon Zavi on November 02, 2010, 01:45:34 pm
sorry Ancestor I didn't get it.
Do you know the answer?
Post by: moon on November 02, 2010, 01:47:43 pm
Do you know the answer?yes, the answer is 78 but how did we find out this value ??? ???
Post by: Arthur Bon Zavi on November 02, 2010, 01:49:22 pm
yes, the answer is 78 but how did we find out this value ??? ???
Sorry, that was wrong method. >:(
Post by: Dania on November 02, 2010, 04:16:24 pm
Post by: SpongeBob on November 03, 2010, 02:07:32 pm
Post by: Deadly_king on November 03, 2010, 03:48:34 pm
I guess I'm late =(
I don't think so. ;)
Do you mind to help Dania for her chem number, please?
Am pretty busy for now koz i've got exams tomorrow :(
Post by: ashish on November 04, 2010, 02:33:25 am
Can someone please help me on October November 2006 Paper 2 no. 5 (e)?
her is your answer check out the attachment
Post by: Uchia on November 04, 2010, 08:01:34 am
PCl5(g) backward and forward reaction PCL3(g) + CL2 (g)
could you please help me to solve this question and thanks
Post by: Uchia on November 04, 2010, 08:02:43 am
Post by: Master_Key on November 04, 2010, 08:05:42 am
Post by: Arthur Bon Zavi on November 04, 2010, 08:49:39 am
when some phosphorus penta chloride was heated to a temperature T in a container 78% OF IT decomposed and the pressure in the container was 2.2
PCl5(g) backward and forward reaction PCL3(g) + CL2 (g)
could you please help me to solve this question and thanks
PCl5 <--------------> PCl3(g) + Cl2(g)
1 mole of phosphorous-penta-chloride was heated, i.e 100 %, 78 % decomposed ! i.e 78 % of 1 mole..
Therefore : 78 / 100 = 0.78 moles
[PCl3] X [Cl2]
Kp = ----------------------------------------
[PCl5]
Total : 0.78 + 0.78 + 0.78 = 2.34
Fraction of PCl5 = 0.78 1
------- = --- (Same is for PCl3 and Cl2).
2.34 3
2.2 MPa X 1000 = 2200 KPa
(1/3) (2200) = 733.3 ----------------------> Partial pressure for the three.
Therefore Kp = (733.3)2 / 733.3 = 733.3 KPa
Is this the answer ?
Post by: Uchia on November 04, 2010, 10:00:03 am
anyways thnxs for your help I found the answer
Initial PCl5 PCl3 Cl2
1 0 0
Change
1-0.78=0.22 0.78 0.78
Total mole 0.22 + 0.78 + 0.78 = 1.78
(0.22/1.78) * 2.2
(0.78/1.78) * 2.2
(0.78/1.78) * 2.2
then continue
see the problem was that for PCl5 you take the amount left the one that was unreacted this is done by subtraction because at equilibrium you should use the amount of both the reactants and the products in the experimen and using the amount that decomposed could not be used because it already changed to products
Post by: Arthur Bon Zavi on November 04, 2010, 10:07:00 am
mm no
anyways thnxs for your help I found the answer
Initial PCl5 PCl3 Cl2
1 0 0
Change
1-0.78=0.22 0.78 0.78
Total mole 0.22 + 0.78 + 0.78 = 1.78
(0.22/1.78) * 2.2
(0.78/1.78) * 2.2
(0.78/1.78) * 2.2
then continue
see the problem was that for PCl5 you take the amount left the one that was unreacted this is done by subtraction because at equilibrium you should use the amount of both the reactants and the products in the experimen and using the amount that decomposed could not be used because it already changed to products
Oh! Yeah. Sorry. Just near. :(
But was that 2.2 MPa Or KPa ?
Post by: Uchia on November 04, 2010, 12:28:17 pm
Post by: Greed444 on November 04, 2010, 03:49:35 pm
Question 1(b). need help and explanation to get the answer please
Post by: Deadly_king on November 05, 2010, 06:05:24 am
i have a doubt on May/June 2007 paper 4
Question 1(b). need help and explanation to get the answer please
Conditions Product at anode Product at cathode
ZnCl2(l) Cholrine Zinc
ZnCl2(concentrated aqueous) Chlorine Hydrogen
ZnCl2(diltue aqueous) Oxygen Hydrogen
NOTE : Positive ions get discharged at the cathode while negative ones are discharged at the anode.
i) Zinc Chloride in liquid form is composed of Zn2+ and Cl- ions which are discharged at the electrodes forming the respective compounds.
ii) Concentrated aqueous Zinc Chloride contains the following ions : Zn2+, Cl-, H+ and OH-.
For cations, you should compare them from the reactivity series. The lower the cation is in the reactivity series the earlier it will be discharged in preference to the most reactive cation.
This is why hydrogen gets preferentially discharged at the expense of Zn2+
For anions, it's about the same thing but you should note that the second part is concentrated aqueous, which means that Cl- is abundant while OH- is present in very small amount.
Hence oxygen will be formed in both cases but it in so minimal amount that you won't even notice and Cl- will immediately be discharged after that.
Post by: moon on November 05, 2010, 08:50:13 pm
Compounds containing manganese oxo-anions often disproportionate in aqueous solution.
(i) Use the following half-equation, and other data from the Data Booklet, to construct an
overall equation for the disproportionation of MnO42- in strongly acidic solution.
Calculate the E o cell for the process.
MnO42- + 8H+ + 4e <=> ? Mn2+ + 4H2O Eo = +1.74 V
(ii) When potassium manganate(VII) is reduced with aqueous sodium sulphite, the bright
blue salt K3MnO4 is produced. The salt readily disproportionates in acidic solution, giving
a brown precipitate of MnO2(s) and a purple solution.
Calculate the oxidation number of manganese in the blue salt, and construct a balanced
ionic equation for its disproportionation.
Post by: moon on November 05, 2010, 11:45:16 pm
June 08 Q2 a (ii) explain and draw the diagram. thank u very much for ur help.
Post by: Deadly_king on November 06, 2010, 05:36:15 am
plz help me in solving this question. thanx in advance.
Compounds containing manganese oxo-anions often disproportionate in aqueous solution.
(i) Use the following half-equation, and other data from the Data Booklet, to construct an
overall equation for the disproportionation of MnO42- in strongly acidic solution.
Calculate the E o cell for the process.
MnO42- + 8H+ + 4e <=> ? Mn2+ + 4H2O Eo = +1.74 V
(ii) When potassium manganate(VII) is reduced with aqueous sodium sulphite, the bright
blue salt K3MnO4 is produced. The salt readily disproportionates in acidic solution, giving
a brown precipitate of MnO2(s) and a purple solution.
Calculate the oxidation number of manganese in the blue salt, and construct a balanced
ionic equation for its disproportionation.
i)Acidic conditions:
3MnO42- + 4H+ –> MnO2 + 2MnO4- + 2H2O
This disproportionation can be though of as two half equations:
1. MnO42- + 4H+ + 2e –> MnO2 + 2H2O ……….. Eº = +2.26V ( the Manganate ion is reduced)
2. MnO42- –> MnO4- + 1e ……………………….Eº = +0.56V ( the Manganate ion is oxidised)
Calculating Eº = E(reduced state) – E(oxidised state) = +2.26V – +0.56V = + 1.70V
This is a large positive value (remember that any Eº value greater than 0.3v means that the reaction is spontaneous as shown) so the forward reaction proceeds.
Here (http://www.ibchem.com/faq/?tag=mno42) is the reference.
ii) The blue salt is K3MnO4
Let the oxidation number of Mn in the salt be
Oxidation number of O is -2 while that of K is +1. Overall charge is zero since it is an uncharged salt.
Hence 3(+1) +
2MnO43- + 4H+ ---> MnO2 + 2H2O + MnO42-
Post by: Deadly_king on November 06, 2010, 06:28:44 am
can u help me in another Question plz.
June 08 Q2 a (ii) explain and draw the diagram. thank u very much for ur help.
::O=:Cl.=O::
Valence electrons = 19
Chlorine attaches to the oxygens through double bonds, and and each oxygen has two lone pairs to complete its octet. The chlorine also one lone pair and one electron. Chlorine in this structure obviously does not obey octet rule but in chlorine there is 3d subshell in which electrons are promoted and hence chlorine can also have 3,5 and 7 bonds.
Since chlorine has one lone pair, the ClO2 molecule will be a bent(v-shaped) one but not the same angle as H2O since it has only one lone pair.
For more details click here (http://en.wikipedia.org/wiki/Chlorine_dioxide).
Post by: moon on November 06, 2010, 11:53:08 am
::O=:Cl.=O::
Valence electrons = 19
Chlorine in this structure obviously does not obey octet rule but in chlorine there is 3d subshell in which electrons are promoted and hence chlorine can also have 3,5 and 7 bonds.
How can we know the number of valence electrons?????and how can we know that chlorine is not obeying octet rule???
how do we determine the number of bonds formed by the chlorine in this case????
For the question above
how to construct a balanced ionic equation for its disproportionation.
Thanx in advance
Post by: $H00t!N& $t@r on November 06, 2010, 12:02:24 pm
Q) A flask containing a mixture of 0.200 mol of ethanoic acid and 0.11 mol of ethanol was maintained at 25 degrees Celsius until the following equilibrium had been established.
CH3C00H + C2H5OH <-> CH3COOC2H5 + H2O
The ethanoic present at equilibrium required 72.5 cm3 of a 1.50 mol dm-3 solution of sodium hydroxide for a complete reaction.
- Calculate the value of the equilibrium constant Kc for this reaction at 25 degrees Celsius
- The enthalpy change for this reaction is quite small. By reference to the number and type of bonds broken and made, explain how this might have been predicted.
Thanks in advance :)
and sorry i dont have the answer to this question because i found it on a website :-\
Post by: Deadly_king on November 06, 2010, 12:33:54 pm
How can we know the number of valence electrons?????and how can we know that chlorine is not obeying octet rule???
how do we determine the number of bonds formed by the chlorine in this case????
For the question above
how to construct a balanced ionic equation for its disproportionation.
Thanx in advance
Valence electrons is the total number of electrons from the outermost shells of the atoms forming the molecule.
Number of electrons in the outermost shell of O : 6
Number of electrons in the outermost shell of Cl : 7
Hence valence electrons : 2(6) + 7 = 19
Hmm............in this case ClO2 is a neutral molecule which means that oxygen has a stable electronic configuration. Otherwise there would have been a negative charge.
For oxygen to have a stable electronic configuration, it should form two covalent bonds or gain two electrons. Cl is a non-metal which does not give electrons. The only plausible solution is that it formed two bonds with each O atom.
Hence structure of ClO2 => ::O=:Cl.=O::
Moreover in the question itself, it is stated that the bonds between Cl and O is double bond. ;)
I need to go now. I'll complete the answer asap :)
Post by: moon on November 06, 2010, 12:54:14 pm
Post by: Hypernova on November 06, 2010, 01:30:47 pm
number of moles=concentration*volume
=1.5 * (72.5/1000)
=0.10875 moles of CH3C00H at equilibrium
When 1 mole of CH3C00H reacts with 1 mole of C2H5OH 1 mole of CH3COOC2H5 and H2O is produced.
So if 0.09125 moles of CH3C00H reacts, then 0.09125 of CH3COOC2H5 and H2O is formed.
CH3C00H + C2H5OH <-> CH3COOC2H5 + H2O
Before equilibrium 0.2 0.11 0 0
At equilibrium 0.10875 0.01875 0.09125 0.09125
KC= 0.09125 * 0.09125
0.10875 * 0.01875
=4.08
Note: the concentration should be used to find Kc, but since they all in the same container, the volumes is the same and will cancel out in this case, so i didn't need to put in volumes.
For the second part, you need to find enthalpy change of reaction. enthalpy change = bonds broken - bonds made
but thats long.
Post by: $H00t!N& $t@r on November 06, 2010, 01:44:52 pm
+rep :)
Post by: Greed444 on November 06, 2010, 02:39:32 pm
Thanks Bro! i Completely understand now :)
Conditions Product at anode Product at cathode
ZnCl2(l) Cholrine Zinc
ZnCl2(concentrated aqueous) Chlorine Hydrogen
ZnCl2(diltue aqueous) Oxygen Hydrogen
NOTE : Positive ions get discharged at the cathode while negative ones are discharged at the anode.
i) Zinc Chloride in liquid form is composed of Zn2+ and Cl- ions which are discharged at the electrodes forming the respective compounds.
ii) Concentrated aqueous Zinc Chloride contains the following ions : Zn2+, Cl-, H+ and OH-.
For cations, you should compare them from the reactivity series. The lower the cation is in the reactivity series the earlier it will be discharged in preference to the most reactive cation.
This is why hydrogen gets preferentially discharged at the expense of Zn2+
For anions, it's about the same thing but you should note that the second part is concentrated aqueous, which means that Cl- is abundant while OH- is present in very small amount.
Hence oxygen will be formed in both cases but it in so minimal amount that you won't even notice and Cl- will immediately be discharged after that.
Post by: Greed444 on November 06, 2010, 02:46:00 pm
1) E(std)cell = E(+ve terminal) - E(-ve terminal)
2) E(std)cell = E(right hand terminal) - E(left hand terminal)
in my class, we used the first one. but when i attempt June 2008 paper4 Q1, they used the 2nd one to get the answer. im confused :S
Post by: moon on November 06, 2010, 03:06:49 pm
does anyone know why some books say
1) E(std)cell = E(+ve terminal) - E(-ve terminal)
2) E(std)cell = E(right hand terminal) - E(left hand terminal)
in my class, we used the first one. but when i attempt June 2008 paper4 Q1, they used the 2nd one to get the answer. im confused :S
Yes I am confused too . I would really appreciate any help Thanks in advance
Post by: Deadly_king on November 07, 2010, 04:35:59 pm
does anyone know why some books say
1) E(std)cell = E(+ve terminal) - E(-ve terminal)
2) E(std)cell = E(right hand terminal) - E(left hand terminal)
in my class, we used the first one. but when i attempt June 2008 paper4 Q1, they used the 2nd one to get the answer. im confused :S
Usually both equations should result in the same answer, since the conventional way of drawing an electrode potential cell is to draw the positive terminal to the right.
The question however, seems to be telling us otherwise. Personally I think the marking scheme is wrong. :-[
This is the only CIE question that arose some concerns. The rest are all pretty easy and abide by the laws of electrode potential. ;)
For current to flow in the wires, the reactions in the half cells should occur. And for the reactions to occur, the standard electrode potential value MUST be positive. In this case the value is negative but still it is asking the direction for the flow of electrons. So it appears to be contradicting the rule of electrode potential. :-\
Hence the answer for this question should be +0.59V.
Nevertheless this is my theory. As a student, I may make mistakes. :-[
This is why i'll advice you to consult someone more experienced in chemistry, like your teacher. :)
Post by: moon on November 07, 2010, 05:21:01 pm
Post by: Deadly_king on November 08, 2010, 03:48:30 am
Thanks a lot Deadly_King for ur effort ,explanation and advice. :) :)
My pleasure buddy ;)
Post by: Greed444 on November 08, 2010, 07:28:15 am
Usually both equations should result in the same answer, since the conventional way of drawing an electrode potential cell is to draw the positive terminal to the right.
The question however, seems to be telling us otherwise. Personally I think the marking scheme is wrong. :-[
This is the only CIE question that arose some concerns. The rest are all pretty easy and abide by the laws of electrode potential. ;)
For current to flow in the wires, the reactions in the half cells should occur. And for the reactions to occur, the standard electrode potential value MUST be positive. In this case the value is negative but still it is asking the direction for the flow of electrons. So it appears to be contradicting the rule of electrode potential. :-\
Hence the answer for this question should be +0.59V.
Nevertheless this is my theory. As a student, I may make mistakes. :-[
This is why i'll advice you to consult someone more experienced in chemistry, like your teacher. :)
Thankyou Dk. eventhough u say that you are a student, but still your explanations and advices are like coming from a true expert :D
Post by: Deadly_king on November 08, 2010, 01:40:33 pm
Thankyou Dk. eventhough u say that you are a student, but still your explanations and advices are like coming from a true expert :D
Hahaha.............sorry to disappoint you but I am a student. ;)
Anyway thanks for the kind words :)
Post by: moon on November 10, 2010, 10:56:00 pm
Post by: Deadly_king on November 11, 2010, 06:52:19 am
plz can u help me in ppr 1 Nov.03 Q2,6,13,31,40. thanx in advance :)
2. First you need to find the % of phosphorus in P2O5
Mr of P2O5 = 2(31) + 5(16) = 142
Hence % of phosphorus in P2O5 = 2(31)/142 x 100 = 43.6%
Since P2O5 forms part of only 30% of the fertiliser, % of phosphorus in P2O5 in the fertiliser will be (30/100 x 43.6) = 13.1%
Answer is B
6. That's kind of obvious. ;)
Six bonding electrons imply a molecule which is composed of only 3 bonds since a bond carries 2 electrons.
A C2H4 ------> 6 bonds (12 bonding electrons)
B C2F6 ------> 7 bonds (14 e)
C H2O -------> 2 bonds (4 e)
D NF3 -------> 3 bonds (6 e)
NOTE : A double bond carries 2 pairs of electrons.
Hence Answer is D
13. You should be aware of the trends of ionisation energies for period 3. Take a look here (http://www.chemguide.co.uk/inorganic/period3/elementsphys.html#top).
There is a fall at Aluminium since there is a change in sub-shells namely from 3s to 3p. That's why aluminium has a lower 1st I.E. ;)
Answer is B
31. So let's examine the answers one by one.
1. An acid- base reaction : That's true since SO3 and water combines to form sulfuric acid. Sulfuric acid then reacts with basic NH3 to form the corresponding salt. ;)
2. Ionic bond formation : That's true as well. The salt exists in the form of NH4+ and SO42- which are ions involved in ionic bonds.
3. Oxidation and reduction : You need to calculate the oxidation numbers of the different elements in order to determine this.
If we just look at the equation, we might have the impression that SO3 is being oxidised but this is not the case.
Let oxidation number of sulfur in SO3 be
Now for SO42-.
Since oxidation number remains the same, there has been neither oxidation nor reduction. I'll let you calculate the oxidation numbers of the other elements yourself for confirmation. ;)
Hence answer is B since only 1 and 2 are good.
40. Heating under reflux with NaOH is an alkaline hydrolysis, i.e there is the breaking of bonds usually COO.
Therefore the ester bond will be broken and two compounds, namely an alcohol and a carboxylic acid will be formed.
Since NaOH is used the carboxylate will be formed instead of the carboxylic acid, but upon distillation the carboxylate will get back to the acid.
Product 3 cannot be formed since it involves the breaking of a C=O which is not possible.
Hence answer is D.
Hope it helps :)
Post by: moon on November 11, 2010, 03:59:33 pm
Post by: Deadly_king on November 11, 2010, 05:10:49 pm
Ur explanation really helped me a lot. thanx so much 4 ur co-operation :) ;)
Hehe.........Anytime buddy ;)
Post by: moon on November 12, 2010, 01:37:06 am
June 01 Q6,11,37
June 2002 Q 38
I would really apreciate any help. thanx a lot. :)
Post by: Hypernova on November 12, 2010, 01:45:24 pm
plzzzzzzzzz help me in these questions. :-[ :-[June 07 Q 1, 9
June 01 Q6,11,37
June 2002 Q 38
I would really apreciate any help. thanx a lot. :)
June 07 Q 1, 9
2.
Relative atomic mass is the average mass of an element compared to that of carbon 12
You need to find the average mass of the element using its reative abundance, so
Ar=10 x 1/5 + 11 x 4/5
=10.8
9.
Kc of a reaction is equal to the product of consentration of products over the product of concentration of reactants all raised to their stochiometric coefficients
so Kc for reaction 1 is 2
Kc=[X2Y]2
[X2]2x[Y2]
Know you have an equation for kc of reaction 1
The queation is find Kc for reaction 2
? = [X2]x[Y2]1/2
[X2Y]
Take your equation for kc of reaction 1 and make it look like the Kc equation for reaction 2
2=[X2Y]2
[X2]2x[Y2]
the reciprocal looks like
1 = [X2]2x[Y2]
2 [X2Y]2
now square root it
1 = [X2]x[Y2]1/2
sqrt(2) [X2Y]
so Ans is A
Post by: Hypernova on November 12, 2010, 02:10:53 pm
plzzzzzzzzz help me in these questions. :-[ :-[June 07 Q 1, 9
June 01 Q6,11,37
June 2002 Q 38
I would really apreciate any help. thanx a lot. :)
I dont have June 01 paper 1, could you please upload it
Q38
For this reaction you need to know what is happening
Br2 will be attracted to the double bonds and it will become polar. Then 1 Br atom binds across the double bond, and the other has a negative charge.
This is the bromination mechanism used in the test for alkenes, I assume you know it, I will elaborate if you ask.
So then the transition state (attached) for alkene is formed.
There are charged particles in solution:
the Cl- from NaCl
The other Br- atom from Br2
these atoms are negatively charged
Notice the positively charged C+ atom in the ethene transition state (attached). It will bind to one of these charged particles.
So
1 CH2ClCH2Cl impossible
2 CH2BrCH2Cl possible
3 CH2BrCH2Br possilbe
Ans is C
Post by: moon on November 12, 2010, 05:12:50 pm
Post by: Deadly_king on November 12, 2010, 05:18:28 pm
Thank u so much hypernova By the way for june 2001 i have hard copy so i can't upload it sorry :-\.
Post the whole question then ;)
We'll try to help as mush as we can. :)
Post by: moon on November 13, 2010, 11:49:49 am
Post by: Deadly_king on November 13, 2010, 12:00:02 pm
I have attached June 2001 as well as November 2001.It seems that these exams are difficult to find so i thought these might help you practice more questions. Best wishes ;) :)
Thank you very much for thinking about us dude. :D
Really appreciated. :)
+rep
Post by: moon on November 13, 2010, 12:04:33 pm
Post by: moon on November 13, 2010, 04:12:14 pm
Post by: moon on November 13, 2010, 04:25:56 pm
June 01 Q6,11,37
thanks for ur help. :) :)
Post by: Hypernova on November 13, 2010, 10:16:13 pm
plzzzzzzzzz help me in these questions. :-[ :-[June 07 Q 1, 9
June 01 Q6,11,37
June 2002 Q 38
I would really apreciate any help. thanx a lot. :)
Was waiting for the marking scheme..i don't post untill im 100% sure
June 01 Q6
See the attached file
Using Hess's law, the enthalpy change of a reaction is the same, irrespective of the route taken
So we choose the route I drew.
If we move in the direction of the reaction we add the enthalpy change, if we move in the opposite direction to the reaction, we subract the enthalpy change, so
the equation looks like this
-109 - -111 + -395
=-393 KJ/mol
Ans is C
11)
Its a basice math question that tests your ability to read of the graph.
They want the fraction of molecules higher than the activation energy
That is: the total number of molecules above Ea
total number of molecules
so the total number of molecules above Ea is reprisented by the areas shaded R and Q
and
total number of molecules is represented by the areas shaded P and Q
which is: Q + R
P + Q
37)
Can't do it, its got something to do with oxinumbers
thanks for the papers
+Reputation
Post by: Deadly_king on November 14, 2010, 08:04:25 am
pls I need help in Nov.06 Q9, 11 12 31, 32.
thanks for ur help. :) :)
Nov 06 p1
9. Reaction : CO2(g) + H2(g) -----> CO(g) + H2O(g)
So we are converting CO2 to CO while H2 is being oxidised to H2O. However both H2 and H2O are in gaseous state. ;)
Conversion of CO to CO2 is given as -283 KJmol-1. Since we're converting CO2 to CO, it will be +283 KJmol-1.
Converting H2 to water in liquid is given -286 KJmol-1. But we have to convert it back to gaseous state, hence net energy will be (-286 + 44) = -242 KJmol-1
Hence Change in enthalpy of reaction : (+283 - 242) = +41 KJmol-1
Answer turns out to be C
11. For the reaction Kc = [CH3CO2C2H5][H2O] / [C2H5OH][CH3CO2H]
Since from the reaction mole ratio is 1:1 for every reactant and products, it becomes easier.
[C2H5OH] [CH3CO2H] [CH3CO2C2H5] [H2O]
Initially 1.0 1.0 0.0 0.0
At equilibrium 1-
Therefore Kc = (
It also implies [
Now you simplify and you'll get
Hence number of moles of ethyl ethanoate formed is 2/3
Answer is B
12. At low temperature, less molecules will have have reached activation energy to undergo the reaction. Hence the graph with lower peak is to be used.
Energy of such a graph is found on the x-axis. ;)
Hence answer is D
31. The chlorine oxide radical is ClO..
It is made up of a covalent bond between the Cl and O atom. Since the oxygen atom carries only one bond, it has one more free electron available for bonding.
Hence the oxygen atom contains an odd number of electrons. However no dative bond is present in this molecule.
Hence answer is B
32. Strong acid imply that it dissociates rapidly to form the corresponding ions while a weak acid does not dissociate much.
Hence sulfuric acid is a strong one which implies it will form much H+ and HSO4-. The latter being a weak acid will not dissociate much such that SO4- will only be present in minimal amount. ;)
Answer is D.
Post by: haris94 on November 14, 2010, 02:14:50 pm
pls explain
thanx
Post by: Deadly_king on November 14, 2010, 03:01:40 pm
state 2 reasons why noble gases behave less ideally as you go down the group 8?
pls explain
thanx
1. As you go down the group, the atoms increase in size such that their volume become large enough. Hence their volume are no more negligible. ;)
2. As you go down the group the atoms of the noble gases become larger due to increase in number of shells. Hence the outermost electron is further away from the nucleus. Therefore the force of attraction decreases such that it is easier to remove an electron from it. It becomes less stable and it is easier to form attraction.
Quote from: http://www.glencove.k12.ny.us/highschool/ChemRev/ChemRev.html
Ideal gases [How gases should behave but don’t]-
1. No attraction between molecules/atoms
2. Molecules have a negligible volume
3. Collisions are elastic
4. Particle movement is random
Real gases VERY RARELY BEHAVE LIKE IDEAL GASES since
1. There IS an attraction between particle (van der Waals)
2. The volume of particles are NOT negligible, esp. at low temps & high-pressure since atoms/molecules are close together
***HYDROGEN and HELIUM are the most IDEAL gases.*** Also, Diatomic molecules and nonsymmetrical molecules & noble gases act the most ideal. THE SMALLER THEY ARE THE MORE IDEAL THEY BEHAVE.
Hope it help :)
Post by: birchy33 on November 15, 2010, 09:48:55 am
according to the following equation.
N2O4 2NO2
What is the value of the equilibrium constant, Kp, for this reaction at 60oC?
A 1/3 atm
B 2/3 atm
C 4/3 atm
D 2 atm
Please explain. Its from MJ/06 P1 thank-you. Oh and please can you held with 18 and 20 from the same exam.
Post by: moon on November 15, 2010, 01:16:31 pm
Post by: Master_Key on November 15, 2010, 01:28:35 pm
Q3 - Ans:- D
One electron in S orbital means that it is in S-block of periodic table but the first group.
Post by: elemis on November 15, 2010, 01:32:27 pm
Question 5
A and D are eliminated as they are symmetrical and hence have no overall dipole moments.
In C Chlorine is very electronegative and hence tugs hard at the electrons. Oxygen which is even more electronegative tugs even harder so the net dipole moment is very small.
In B we see that hydrogen can offer little resistance to the effect of the highly electronegative oxygen causing a large dipole moment in the direction of oxygen
Ans = b
Post by: Deadly_king on November 15, 2010, 02:21:38 pm
pls can u help in Nov.08 Q3,30,39
Nov 08 p1
30. First you need to use the mass of the respective molecules given to find the number of moles of each.
No of moles = Mass / Mr
No of moles of ethanol : 30/46 = 0.652
No of moles of ethanoic acid : 30/60 = 0.50
Ratio of moles of ethanol to ethanoic acid was supposed to be 1:1. Hence we can deduce that ethanol is in excess. Only 0.5 mole of each reactant will react.
No of moles of ethyl ethanoate : 22/88 = 0.25.
Hence it can be noted that only 0.25 mole of ester is formed when 0.5 mole of reactants react together.
% Yield is therefore : 0.25/0.5 x 100 = 50%
Answer is C
39. From the information given above, you can note that for methane which contains only one carbon atom, ethane with two carbon atoms can be obtained as product. In other words the number of carbon atoms double.
So for propane which contains 3 carbon atoms, the resulting product may contain maximum 6 carbon atoms.
Hence we can reject the last suggestion leaving us with only 1 and 2 as potential answers.
Answer is B
Post by: Deadly_king on November 15, 2010, 04:04:43 pm
10. At a total pressure of 1.0 atm, dinitrogen tetraoxide is 50% dissociated at a temperature of 60 C,
according to the following equation.
N2O4 2NO2
What is the value of the equilibrium constant, Kp, for this reaction at 60oC?
A 1/3 atm
B 2/3 atm
C 4/3 atm
D 2 atm
Please explain. Its from MJ/06 P1 thank-you. Oh and please can you held with 18 and 20 from the same exam.
Jun 06 p1
10. Kp = [
At equilibrium, pressure of N2O4 is
Therefore
Kp = (2/3)2 / (1/3) = 4/3
Answer is C.
18. C is the most appropriate solution since it is cheaper and the reaction is not exothermic.
20. First reaction is removal of hydrogen atom, i.e oxidation of an alcohol to an aldehyde.
Second reaction is a nucleophilic reaction as an aldehyde is made to react with an alcohol to form a ketone.
Third reaction is again oxidation.
Answer is C.
Post by: birchy33 on November 15, 2010, 11:30:56 pm
Jun 06 p1
10. Kp = [(NO2)]2 /
At equilibrium, pressure of N2O4 iswhile that of NO2 is 2
Therefore
Kp = (2/3)2 / (1/3) = 4/3
Answer is C.
18. C is the most appropriate solution since it is cheaper and the reaction is not exothermic.
20. First reaction is removal of hydrogen atom, i.e oxidation of an alcohol to an aldehyde.
Second reaction is a nucleophilic reaction as an aldehyde is made to react with an alcohol to form a ketone.
Third reaction is again oxidation.
Answer is C.
Cheers mate, I'm just a bit unsure about the 2x + x = 1 bit. Secondly There's another equilibrium question: Question 9 from MJ/07. Your help would be much appreciated. Cheers + rep.
Post by: cs on November 16, 2010, 09:13:19 am
Edit: found my answer for the second one in previous page, so left the 1st and 3rd one.. Thanks
Post by: $!$RatJumper$!$ on November 16, 2010, 09:23:00 am
pls I need help in Nov.06 Q9, 11 12 31, 32.
June 01 Q6,11,37
thanks for ur help. :) :)
Nov 06 p1
12. At low temperature, less molecules will have have reached activation energy to undergo the reaction. Hence the graph with lower peak is to be used.
Energy of such a graph is found on the x-axis. ;)
Hence answer is D
Energy at lower temperatures mean we take the graph with the higher peak and not the lower. Thus the answer is C and not D
Post by: elemis on November 16, 2010, 09:32:38 am
For the 1st question :
500/24000 = 0.020833333333 moles of oxygen.
Using avogadro's constant : 0.020833333333 divided by 1/(6.02*1023)
gives an answer of 1.25 *10^22
Post by: elemis on November 16, 2010, 09:45:05 am
For the Second question :
Imagine there is 1 mole of N2O4 hence there must be 2 moles of NO2
Therefore the partial pressures of each gas are 1/3 (N2O4) and 2/3 for NO2
inputting these into the initial equation at the top gives 4/3
Post by: $!$RatJumper$!$ on November 16, 2010, 11:13:38 am
pls can u help in Nov.08 Q3,30,39 and Nov. 09 Varient 1 Q5,21,22,28,29,31. Thanx in advance.
Nov 08 p1
30. First you need to use the mass of the respective molecules given to find the number of moles of each.
No of moles = Mass / Mr
No of moles of ethanol : 30/46 = 0.652
No of moles of ethanoic acid : 30/60 = 0.50
Ratio of moles of ethanol to ethanoic acid was supposed to be 1:1. Hence we can deduce that ethanol is in excess. Only 0.5 mole of each reactant will react.
No of moles of ethyl ethanoate : 22/88 = 0.25.
Hence it can be noted that only 0.25 mole of ester is formed when 0.5 mole of reactants react together.
% Yield is therefore : 0.25/0.5 x 100 = 50%
Answer is C
39. From the information given above, you can note that for methane which contains only one carbon atom, ethane with two carbon atoms can be obtained as product. In other words the number of carbon atoms double.
So for propane which contains 3 carbon atoms, the resulting product may contain maximum 6 carbon atoms.
Hence we can reject the last suggestion leaving us with only 1 and 2 as potential answers.
Answer is B
For Q39 why do we reject the last one? Doesnt it also have 6 carbon atoms?
Post by: cs on November 16, 2010, 11:43:45 am
@CS
For the Second question :![\frac{[NO_2]^2}{[N_2O_4]}](https://studentforums.biz/cgi-bin/mimetex.cgi?\frac{[NO_2]^2}{[N_2O_4]})
Imagine there is 1 mole of N2O4 hence there must be 2 moles of NO2
Therefore the partial pressures of each gas are 1/3 (N2O4) and 2/3 for NO2
inputting these into the initial equation at the top gives 4/3
What about the 50% thing?
Post by: cs on November 16, 2010, 11:46:22 am
@CS
For the 1st question :
500/24000 = 0.020833333333 moles of oxygen.
Using avogadro's constant : 0.020833333333 divided by 1/(6.02*1023)
gives an answer of 1.25 *10^22
ohh so we don't need to use Pv= nRT?
Post by: Deadly_king on November 16, 2010, 11:50:48 am
For Q39 why do we reject the last one? Doesnt it also have 6 carbon atoms?
Oops........sorry I forgot to precise.
The third one indeed contains 6 carbon atoms. But if you break it to form two radicals, you'll find that you'll be getting two compounds one with 4 carbon atoms and the other with only two carbon atoms.
This cannot be since propane contains 3 carbon atoms. So both radicals should contain 3 carbon atoms as well. ;)
Post by: Deadly_king on November 16, 2010, 11:53:39 am
What about the 50% thing?
The 50% indicates that the equilibrium is at dynamic equilibrium. ;)
Since 50% of reactant becomes products while the other 50% remains as reactants. ;D
Post by: $!$RatJumper$!$ on November 16, 2010, 12:01:38 pm
Oops........sorry I forgot to precise.
The third one indeed contains 6 carbon atoms. But if you break it to form two radicals, you'll find that you'll be getting two compounds one with 4 carbon atoms and the other with only two carbon atoms.
This cannot be since propane contains 3 carbon atoms. So both radicals should contain 3 carbon atoms as well. ;)
Thankx makes sense now :)
Gosh how do you people remember such small details while doing questions... wish i was as smart :/
Post by: $!$RatJumper$!$ on November 16, 2010, 12:03:06 pm
Post by: Deadly_king on November 16, 2010, 12:04:40 pm
Cheers mate, I'm just a bit unsure about the 2x + x = 1 bit. Secondly There's another equilibrium question: Question 9 from MJ/07. Your help would be much appreciated. Cheers + rep.
JUn 07 p1 No 9
Kc for reaction 1 is given to be equal to 2.
Kc = [X2Y]2 / [X2]2[Y2] = 2
For reaction 2 ---> Kc = [X2][Y2]1/2 / [X2Y]
If you square the Kc equation for reaction 2, you'll find that it becomes the reciprocal of the Kc equation for reaction 1. ;)
Hence for reaction 2 ---> Kc2 = 1/2.
So Kc will come to be (1/2)1/2 ----> Kc = 1/(21/2)
Answer is A.
Hope it's clear now :D
Post by: cs on November 16, 2010, 12:06:18 pm
The 50% indicates that the equilibrium is at dynamic equilibrium. ;)
Since 50% of reactant becomes products while the other 50% remains as reactants. ;D
ok, got it finally, my third question please, anyone.. Thank you!
I have two days to do chem and i'm done. ;D
Post by: Deadly_king on November 16, 2010, 12:11:51 pm
ok, got it finally, my third question please, anyone.. Thank you!
I have two days to do chem and i'm done. ;D
No problem ;)
Same for me. Last exam on Friday though i'll be having Bio on Thursday. ;)
Jun 06 No 25
Radicals result from chlorofluorocarbon by the loss of a chlorine atom. ;)
This is because it is the most unstable and most weak bond within any CFCS. C-F is a very strong and stable bond which is not affected by U.V and so is C-H bond.
So you just need to look for an answer which lacks a chlorine atom.
Only C shows this. ;D
Post by: cs on November 16, 2010, 12:18:00 pm
No problem ;)
Same for me. Last exam on Friday though i'll be having Bio on Thursday. ;)
Jun 06 No 25
Radicals result from chlorofluorocarbon by the loss of a chlorine atom. ;)
This is because it is the most unstable and most weak bond within any CFCS. C-F is a very strong and stable bond which is not affected by U.V and so is C-H bond.
So you just need to look for an answer which lacks a chlorine atom.
Only C shows this. ;D
understood, i thought of that too, but not so sure.Thanks! My friends are sitting for Bio on Thursday too.. I'm not :P Good Luck!
Post by: Deadly_king on November 16, 2010, 12:19:53 pm
Energy at lower temperatures mean we take the graph with the higher peak and not the lower. Thus the answer is C and not D
Yupz........you are right.
I guess I was not very attentive :-[
Anyway the higher the temperature the more particles will be having activation energy.
The number of particles is represented as the area after activation energy. In this case area is greater for the lower peak.
Hence at lower temperatures, the lower peak should be used. ;)
Post by: Deadly_king on November 16, 2010, 12:22:21 pm
understood, i thought of that too, but not so sure.Thanks! My friends are sitting for Bio on Thursday too.. I'm not :P Good Luck!
Anytime ;)
So you have only one more paper and it's an MCQ. ;D
Thanks :D
Post by: $!$RatJumper$!$ on November 16, 2010, 01:56:29 pm
12, (I get C. Why D)
15, (I get D)
16,
18,
24, (Why A and not C)
25, (I know it either A or D but cant choose which one)
30, (I dont see how this would be hydrolised)
31,
34,
36, (what does disproportionation mean)
39, (How are hydrocarbons oxidised in a convertor)
40 (What does optically active mean)
Thank you :)
Post by: Deadly_king on November 17, 2010, 06:19:13 am
W05
12, (I get C. Why D)
15, (I get D)
16,
18,
24, (Why A and not C)
25, (I know it either A or D but cant choose which one)
30, (I dont see how this would be hydrolised)
31,
34,
36, (what does disproportionation mean)
39, (How are hydrocarbons oxidised in a convertor)
40 (What does optically active mean)
Thank you :)
W0w.............all these doubts in only one paper. :o
Man, you should review your notes again.
Anyway i'll try to help but I might be taking some more time since am pretty busy today. I'll try as hard as possible. ;)
Post by: $!$RatJumper$!$ on November 17, 2010, 06:56:28 am
W0w.............all these doubts in only one paper. :o
Man, you should review your notes again.
Anyway i'll try to help but I might be taking some more time since am pretty busy today. I'll try as hard as possible. ;)
Yeah some of us arent as clever as you guys hey :)
But thankx.. no worries if you're busy...
i got how to do Q16, 25, 30, 31 and 34 .. so no need for those :)
Post by: Deadly_king on November 17, 2010, 08:07:33 am
Yeah some of us arent as clever as you guys hey :)
But thankx.. no worries if you're busy...
i got how to do Q16, 25, 30, 31 and 34 .. so no need for those :)
It's not about being clever dude. You're going to have the same exams in 2 days, so you should have been ready. :)
Nov 05 p1
12. You just need to write the balanced equations for the complete combustion of one mole of each element.
Mg + 1/2O2 ----> Mg0
Al + 3/4O2 ----> 1/2Al2O3
S + 3/2O2 ----> SO3
Hence we can note that the number of moles of oxygen required increases but not constantly from Magnesium to Sulfur. ;)
Answer is D
15. You should first write the balanced equation representing the decomposition of the nitrate. That will be:
Ca(NO3)2 -----> CaO + 2NO2 + 1/2O2
From equation:
1 mole of nitrate gives 1/2 mole of oxygen gas.
164g of nitrate produces (24/2) dm3 of oxygen gas.
Hence 8.2g of nitrate will emit (12/164 x 8.2) = 0.6dm3 = 600cm3
Answer is C
19. Chiral centres are shown by a carbon atom carrying 4 different group of atoms. No carbon involved in a C=C will be able to show chirality.
Hence this leaves us with only two chiral centres. Starting from the right, they are the second and the third carbon atom
Answer is B
24. I'll proceed by elimination.
Propan-2-ol and 2-methyl-propan-2-ol are first to be rejected since they can form only one type of alkene due to free rotation. ;)
2-methylbutan-2-ol can form only 2 types of alkenes due to free rotation. It is connected to three other carbon atoms and should have been able to form 3 alkenes but two of them are identical.
Butan-2-ol should have formed only two types of alkenes but one of them show cis-trans isomerism. Hence it forms 3 types of alkenes.
Answer is A.
36. A reaction in which the same element is both oxidized and reduced is called a disproportionation reaction.
Is that enough or should I go in more details?
39.
Unburnt hydrocarbons are made to burn completely to form carbon dioxide and water. Hence this is a form of oxidation. Oxygen comes from the oxides of nitrogen which in turn gets reduced. ;)
Post by: $!$RatJumper$!$ on November 17, 2010, 09:30:41 am
I understood all that you mentioned.. but i think you did no 19 instead of no 18 :)
If you have time please do it, otherwise its fine. Thankx again +rep
Post by: Deadly_king on November 17, 2010, 09:56:10 am
Thankx a ton for the help buddy! :)
I understood all that you mentioned.. but i think you did no 19 instead of no 18 :)
If you have time please do it, otherwise its fine. Thankx again +rep
Anytime ;)
Ooh yeah.......I'm sorry for the confusion.
18. CaO is a basic oxide while ammonium sulfate is some kind of acidic as H+ can be liberated from NH4+. So they'll react together to give off ammonia, which being a gas will rise into the atmosphere. Hence the soil will lose its nitrogen which is required by plants.
Answer is C.
Post by: $!$RatJumper$!$ on November 17, 2010, 10:01:33 am
Post by: Hypernova on November 17, 2010, 10:13:13 am
Post by: $!$RatJumper$!$ on November 17, 2010, 09:37:05 pm
Thank you kindly :)
Post by: birchy33 on November 18, 2010, 01:04:56 am
O/N 09 QP1 variant 2
Q20
Q23
Q27
and to my much embarrassment ... Q1: I know its either A or D but I've completely forgot how to pick which one.
Cheers brudas.
Post by: Master_Key on November 18, 2010, 06:32:18 am
You guys have been terrific and I hate to bundle you with more questions ... but ;D
O/N 09 QP1 variant 2
Q20
Q23
Q27
and to my much embarrassment ... Q1: I know its either A or D but I've completely forgot how to pick which one.
Cheers brudas.
O/N 09 QP1 Variant 2.
Q1. Answer is D.
It is because 0.2 Moles of HC burn to give 35.4 g of CO2.
So, RFM of CO2=(12+(16*2)=44 g.
35.4/44 = 0.8
0.8 = 0.2
? = 1
0.8/0.2=4
This is the simplest method. We get 4 Carbon atoms in one molecule of HC.
Do the Same with Water and we will get 8 Hydrogen atoms.
So the formula is C4H8
Post by: Master_Key on November 18, 2010, 06:38:19 am
You guys have been terrific and I hate to bundle you with more questions ... but ;D
O/N 09 QP1 variant 2
Q20
Q23
Q27
and to my much embarrassment ... Q1: I know its either A or D but I've completely forgot how to pick which one.
Cheers brudas.
O/N 09 QP1 Variant 2.
Q23. Answer is A.
The diagram shows 2 double bonds between C=C atoms. These are the bonds which are broken and Hydrogen or Bromine added. The C=O will not break. Both the double bonds break and form a single bond C-C. 2 Hydrogen atoms are bonded with two Carbon atoms. There are four carbon atoms so 4 Hydrogen atoms.
This gives to 2 moles of both substances as both exist as diatomic molecules.
Post by: elemis on November 18, 2010, 06:48:29 am
S06 Q 10, 12, 17, 21, 25
Thank you kindly :)
Question 17
Chlorine displaces Iodine from the salt. So the Iodide anions lose electrons to chlorine i.e. they are oxidised.
Post by: elemis on November 18, 2010, 06:52:44 am
For question 10
@CS
For the Second question :
Imagine there is 1 mole of N2O4 hence there must be 2 moles of NO2
Therefore the partial pressures of each gas are 1/3 (N2O4) and 2/3 for NO2
inputting these into the initial equation at the top gives 4/3
Post by: komailnaqvi on November 18, 2010, 07:42:06 am
Post by: elemis on November 18, 2010, 08:27:19 am
Plz anyone explain N08 Q2 P1. Im having probs in these volume related ques and nothing else! Thanks.
CS2(g) + 3O2(g) -----> CO2(g) + 2SO2(g)
10 cm3 of CS2 was used along with 50cm3 of O2. From the above equation we can see that only 30 cm3 of O2 is needed.
Thus, 20 cm3 of O2 is in excess and will remain unreacted at the end.
Since only 10cm3 of CS2 was used it is the LIMITING AGENT. Hence, only 10cm3 of CO2 is produced along with 20cm3 of SO2.
Thus, 20 + 10 + 20 = 50cm3
Hence, answer must be C or D
However, when the acidic gases (SO2 and CO2) are passed through NaOH they will be absorbed leaving only the excess O2 behind.
Hence, after addition of NaOH only 20cm3 of gases are left.
ANSWER = C
Post by: komailnaqvi on November 18, 2010, 09:28:19 am
N09 Q20
Post by: yasser37 on November 18, 2010, 09:30:59 am
in June 10 paper 11
can someone explain questions 13 , 22 ,28 , and 29
Post by: komailnaqvi on November 18, 2010, 09:36:10 am
Both Alcohols are primary alcohols and so both will give same results with KmnO4, Na, and PCl5 while if you notice with Conc.H2so4 it is meant dehydration and hence two different reactions will take place one alcohol will go through dehydration while the other will not. hence
making B the Ans.
Post by: komailnaqvi on November 18, 2010, 10:52:01 am
and also in Q24 why cant it be 1-bromobutane???
Post by: Deadly_king on November 18, 2010, 10:55:19 am
Q.22 J10 p11
Both Alcohols are primary alcohols and so both will give same results with KmnO4, Na, and PCl5 while if you notice with Conc.H2so4 it is meant dehydration and hence two different reactions will take place one alcohol will go through dehydration while the other will not. hence
making B the Ans.
Good explanation buddy :)
+rep
Post by: Deadly_king on November 18, 2010, 11:08:38 am
P1 Nov02 Q1 Plz. Thanx.
and also in Q24 why cant it be 1-bromobutane???
Nov 02 p1
1. Write the balanced equations for the reactions.
CH4 + 2O2 -----> CO2 + 2H2O
C2H6 + 7/2O2 -----> 2CO2 + 3H2O
From 1st equation, 10 cm3 of methane will produce 10 cm3 of CO2 which is the only product absorbed by the alkali.
From 2nd equation, 10 cm3 of ethane will produce 20 cm3 of CO2.
So, in all 30 cm3 pf CO2 is produced.
Answer is C
24. It cannot be 1-bromobutane since upon oxidation the primary alcohol resulting from the first reaction will be oxidised into a carboxylic acid which does not have the same formula. ;)
Post by: moon on November 18, 2010, 12:39:18 pm
Post by: Master_Key on November 18, 2010, 12:40:42 pm
Can someone please explain me nov.2008 q3,33,39 thanx in advance
Which paper and variant?
Post by: Hypernova on November 18, 2010, 01:17:59 pm
Could someone do it. no explanation, Just the stochiometry calculations needed. Thanks
Post by: yasser37 on November 18, 2010, 01:24:06 pm
21
37
39
Post by: Deadly_king on November 18, 2010, 01:57:33 pm
November 06
21
37
39
Nov 06 p1
21. This one, you need to do it step by step. Find the number of chloroethanes when n=1, 2, 3 and 4 in the formula C2H6-nCln
When n=1 ----> C2H5Cl : You can have only one type.
When n=2 ----> C2H4Cl2 : You can have 2 types, one when both chlorine atoms are found on the same carbon and two, when each carbon atoms in the compound carries one chlorine.
When n=3 ----> C2H3Cl3 : You can have 2 types, one when all three chlorine atoms are found on the same carbon and two when one carbon carries 2 chlorine while the other carbon carries one chlorine.
When n=4 -----> C2H2Cl4 : Again 2 types. One when one carbon carries 3 Cl while the other carries one Cl. and two when both carbon atoms carries 2 Cl.
So in all that makes 7 types.
Answer is C
37. To have two organic products upon oxidation, you need to have two double bonds so that when both breaks two different compounds are obtained.
So number two is rejected since only one product will be formed. ;)
Now, to have a ketone, you need to have a secondary alcohol formed upon oxidation of the alkenes.
The answer is D
This is because the double bonds are arranged in such a way that upon oxidation you'll be getting two different compounds each carrying a ketone.
The third option however will produce a compound carrying two ketone groups while the other compound will form a carboxylic acid. So only compound 1 is appropriate. :)
39. Decolourise KMnO4 ----> he's talking about alkenes.
So it's going to be a dehydration process. Therefore number 3 being a carboxylic acid will not be dehydrated.
Both number 1 and 2 are alcohols and will be dehydrated by sulfuric acid to form alkenes.
Hence answer is B
NOTE : I've not had time to confirm my answers with the marking scheme. Do check it out and let me know if there are mistakes. ;)
Post by: $!$RatJumper$!$ on November 18, 2010, 02:04:53 pm
:)
Post by: Deadly_king on November 18, 2010, 02:23:18 pm
W04 Q3 pls
:)
It's just like Ari explained the previous number.
10CH2SH + 30O2 -----> 10CO2 + 10SO2 + 20H2O
When cooled, the water will condense. Hence it will no more be in gaseous state.
NOTE: 60cm3 of oxygen was available and only 30cm3 were used . So there are 30cm3 in excess.
In all it makes 50cm3
Therefore answer is C.
Post by: komailnaqvi on November 18, 2010, 02:24:57 pm
Right so , X is supposed to be 2.920g
and the Xcl2 is 5.287
subtract the two to get the mass of Chloride used which will come out to be = 2.367g
Mass/Mr = 2.367/71 will get you moles of Chloride which are = 0.333333
Mole Ratio is 1:1
if 0.3333 mole of X is 2.920
then how much will be in one mole = 2.920/0.3333 = 87.9g
Thus Ans is C.
Strontium
Post by: cs on November 18, 2010, 02:40:07 pm
June 2010 P11, Q1 please...
Post by: Hypernova on November 18, 2010, 02:41:52 pm
then how much will be in one mole = 2.920/0.3333 = 87.9g
Thus Ans is C.
Strontium
Nah, thats what i did. You get the MR=87.9, but the Mr of strontium is 38
Post by: komailnaqvi on November 18, 2010, 02:45:03 pm
Post by: Hypernova on November 18, 2010, 02:49:03 pm
buddy Mr is 88 , atomic number is 38.
holywillywonker.
your right, +rep
Sorry, using a diff periodic table,and not used to it. I read it wrong.
Post by: komailnaqvi on November 18, 2010, 02:50:01 pm
An urgent one:
June 2010 P11, Q1 please...
Ans is C.
Due to the fact that it will have the config 1s2 2s2 2p6 3s2 3p3 in each atom the p orbital has 3 unpaired!
Post by: komailnaqvi on November 18, 2010, 02:56:15 pm
Post by: komailnaqvi on November 18, 2010, 03:11:11 pm
Post by: Deadly_king on November 18, 2010, 03:14:12 pm
Plz anyone J04 Q28 P1
Compound X is unreactive to mild oxidising agents ----> It should be a tertiary alcohol which will produce an alkene upon dehydration.
Only D offers this possiblity. ;)
Post by: moon on November 18, 2010, 04:16:17 pm
Post by: $!$RatJumper$!$ on November 18, 2010, 04:27:38 pm
W06 Q 4
W04 Q 9, 10, 23
Any help would be humbly appreciated :) Thank you very much :)
Post by: Dania on November 18, 2010, 04:34:07 pm
Post by: Deadly_king on November 18, 2010, 05:00:09 pm
Can someone please explain me nov.2008 q3,39 ??? ::) i need help in these questions urgently :'( I can't get them at all I would really appreciate any help.....thanx
3. Only one electron in s-orbital -----> should be a group I metal.
So everyone apart D is rejected.
Li is a Group I metal. So it has its outermost electron in an s-orbital. Cr is a transition metal. One of its s-orbital goes into it's partially filled d-orbital for stability. ;)
39. From the information given above, you can note that for methane which contains only one carbon atom, ethane with two carbon atoms can be obtained as product. In other words the number of carbon atoms double.
So for propane which contains 3 carbon atoms, the resulting product may contain maximum 6 carbon atoms with as reactants two radicals carrying 3 carbon atoms.
The third one indeed contains 6 carbon atoms. But if you break it to form two radicals, you'll find that you'll be getting two compounds one with 4 carbon atoms and the other with only two carbon atoms.
This cannot be since propane contains 3 carbon atoms. So both radicals should contain 3 carbon atoms as well. ;)
Answer is B.
Post by: Deadly_king on November 18, 2010, 05:11:51 pm
The answer is D
Ammonium nitrate exists as ions : NH4+ and NO3-
So now you can calculate its oxidation number.
Let oxidation number of nitrogen be x.
NH4+ => x + 4(+1) = +1 -----> x = -3
NO3- => x + 3(-2) = -1 -----> x = +5
N2O = > 2x + (-2) = 0 ----> x = +1
So change in oxidation number is from -3 to +1 ---> +4 and from +5 to +1 ---> -4
SO answer is D.
Post by: moon on November 18, 2010, 05:14:01 pm
I really appreciate ur help and effort..thanx a lot. :-[ :-[ :-[
Post by: Deadly_king on November 18, 2010, 05:28:52 pm
W07 Q 9, 10
W06 Q 4
W04 Q 9, 10, 23
Any help would be humbly appreciated :) Thank you very much :)
Nov 07 p1
9. From the information above we can note that ratio of metal to SO32- is 2 : 1.
So now we calculate the oxidation number of S in SO32- and SO42-. You'll see that it is +4 and +6 respectively.
So we can observe that there is a change of +2 in oxidation number. Or you could have observed that by the 2 electrons emitted. So the metal will on the contrary decrease in oxidation number by 2. But since the ratio of metal was twice compared to the sulfite, change in oxidation number for each metal cation will be -2/2 = -1.
SO it will decrease from +3 to +2. (+3 - 1)
Answer is B
10.
Initial number of moles of NO2 : 4 while initial number of the products are nil.
At equilibrium number of moles of oxygen is given to be 0.8. So that of the other product NO will be 1.6
Now we can determine the number of moles of NO2 at equilibrium : (4 - 2(0.8)) = 2.4
Now Kc will be 1.62 x 0.8 / 2.42
So answer is D
Nov 06 No 4
10CH4 + 20O2 -----> 10CO2 + 20H2O
So residual gas at r.t.p will exclude volume of H2O but include the 50cm3 excess of oxygen. ;)
Hence for CH4, total volume of residual gas will be 60cm3. This eliminates A, B and C.
Answer is D
Nov 04 p1
9. The alkane will release most energy since the others already carry an oxygen atom, meaning that they are already a bit oxidised, while the alkane needs to be completely burned. ;)
So answer is B
23. A mixture of bromine is allowed to react with ethane. ---> Free radical substitution reaction.
I believe the bromine will be in excess, so it can be bromoethane, dibromoethane(2 types), tribromoethane(2 types), tetrabromoethane(2 types), pentabromoethane and heptabromoethane.
So answer is D.
NOTE : AM sorry but i've not had time to check marking schemes, so forgive me if some of the answers are wrong. Do tell me and i'll try my best to correct my mistakes. :-[
Post by: Dania on November 18, 2010, 05:32:17 pm
thank u soooooooooo much really don't know to thank u and if don't mind can u explain me june 2010 varient1 Q40 please???
I really appreciate ur help and effort..thanx a lot. :-[ :-[ :-[
I also need an explanation to the same question! Anyone?
Post by: $!$RatJumper$!$ on November 18, 2010, 05:44:03 pm
Nov 07 p1
9. From the information above we can note that ratio of metal to SO32- is 2 : 1.
So now we calculate the oxidation number of S in SO32- and SO42-. You'll see that it is +4 and +6 respectively.
So we can observe that there is a change of +2 in oxidation number. Or you could have observed that by the 2 electrons emitted. So the metal will on the contrary decrease in oxidation number by 2. But since the ratio of metal was twice compared to the sulfite, change in oxidation number for each metal cation will be -2/2 = -1.
SO it will decrease from +3 to +2. (+3 - 1)
Answer is B
10.
Initial number of moles of NO2 : 4 while initial number of the products are nil.
At equilibrium number of moles of oxygen is given to be 0.8. So that of the other product NO will be 1.6
Now we can determine the number of moles of NO2 at equilibrium : (4 - 2(0.8)) = 2.4
Now Kc will be 1.62 x 0.8 / 2.42
So answer is D
Many thankx.. may god bless you DK :)
I jus dont understand, why did you divide -2/2? as in where did the -2 come from?
Post by: Dania on November 18, 2010, 06:52:20 pm
Post by: Deadly_king on November 19, 2010, 03:28:22 am
Many thankx.. may god bless you DK :)
I jus dont understand, why did you divide -2/2? as in where did the -2 come from?
No problem buddy :)
So first you should realise that this is a redox reaction such that one substance must get oxidised and another one reduced.
We have found that the change in oxidation number of S is +2, i.e it has been oxidised. So there should be another substance which has undergone same oxidation state but in the other direction, i.e which has been reduced. In this case it's the metal M.
So the change in oxidation number of the metal should be -2. Since there are two atoms of it, we divide by 2. ;)
Hope it helps :)
Post by: Deadly_king on November 19, 2010, 03:46:18 am
The answer is A.
Both reactions, i.e with bromine and hydrogen under respective conditions will occur at the C=C. Both of them will break the 2 C=C bonds, each with one molecule.
One molecule of H2 is required to break one C=C bond to form the corresponding alkane. Since there are two C=C, two moles of H2 will be required.
The same applies for bromine. ;)
Hope it helps :)
Post by: komailnaqvi on November 19, 2010, 04:06:25 am
Post by: Deadly_king on November 19, 2010, 04:12:26 am
2H2O(g) <==> 2H2(g) + O2(g)
Initially there is only pressure of 1 atm due to steam.
Initially 1 0 0
At equilibrium 1 - 2
Since we have been told that 20% of the steam has been converted, this implies that there still remains 80% at equilibrium.
So 1 - 2
So the total pressure at equilibrium will be [(1 - 2
Now we can find each partial pressure at equilibrium by dividing by total pressure.
I'll let you find the other by yourself. ;)
Answer is D.
Jun 10 p11 No 40
Bromoethane has a lower boiling point(39oC) than the solution from which it is being made, so as it is formed, the heat supplied will cause it to evaporate and we'll be able to collect it using the apparatus shown.
Ethanal also has a lower boiling point(21oC) that the reactants producing it. So it can be successfully separated using the apparatus.
1, dibromoethane however has a much higher boiling points(132oC) than the other substances in the reacting mixture. Hence we'll not be getting it through this apparatus.
So answer is B.
Hope it helps :)
Post by: komailnaqvi on November 19, 2010, 04:19:48 am
J05 Q2 plz
+rep
Post by: Deadly_king on November 19, 2010, 04:34:55 am
thanx man.
J05 Q2 plz
+rep
Anytime dude :)
Hmm.................you have been given an equation where 2 moles of Sodium azide forms 2 moles of sodium and 3 moles of nitrogen.
Divide everything by 2 so as to get the equation resulting from one mole of sodium azide. You'll get 1.5 moles of nitrogen and 1 moles of sodium.
But from second equation provided, you'll note that 10 moles of sodium form one mole of N2. So the one mole of sodium formed above will form only 0.1 mole of N2.
Total number of moles of nitrogen produced will be 1.5 + 0.1 = 1.6
Answer is B
Post by: komailnaqvi on November 19, 2010, 04:36:55 am
Post by: $!$RatJumper$!$ on November 19, 2010, 05:15:42 am
Nov 04 No 10
2H2O(g) <==> 2H2(g) + O2(g)
Initially there is only pressure of 1 atm due to steam.
Initially 1 0 0
At equilibrium 1 - 2
Since we have been told that 20% of the steam has been converted, this implies that there still remains 80% at equilibrium.
So 1 - 2
So the total pressure at equilibrium will be [(1 - 2
Now we can find each partial pressure at equilibrium by dividing by total pressure.
I'll let you find the other by yourself. ;)
Answer is D.
Jun 10 p11 No 40
Bromoethane has a lower boiling point(39oC) than the solution from which it is being made, so as it is formed, the heat supplied will cause it to evaporate and we'll be able to collect it using the apparatus shown.
Ethanal also has a lower boiling point(21oC) that the reactants producing it. So it can be successfully separated using the apparatus.
1, dibromoethane however has a much higher boiling points(132oC) than the other substances in the reacting mixture. Hence we'll not be getting it through this apparatus.
So answer is B.
Hope it helps :)
Excellent! :) Thank you very much! +rep
Post by: Deadly_king on November 19, 2010, 05:21:33 am
Just ace the paper today ;D
Good luck to you all ;)
Post by: cs on November 19, 2010, 06:29:46 am
Post by: Twinkle Charms on November 19, 2010, 07:36:16 pm
(ii) Suggest why X needs to be heated strongly?
(iii)The material X becomes black in color. Suggest what substance is responsible for this black color and how it arises in the reaction?
Post by: Hissa on November 20, 2010, 02:41:58 pm
1) Increasing the temperature of the propanoic acid solution causes the pH to decrease.
What does this tell you about the enthalpy of dissociation?
2) Why is an indicator usually added in iodine/thiosulphate titrations but not in titrations involving potassium managanate (VII)?
Post by: Alpha on November 20, 2010, 02:54:36 pm
Post by: Arthur Bon Zavi on November 20, 2010, 03:05:40 pm
(i) The material X can be pumice or ceramic used such as brick or broken crockery. Write the chemical name of the substance that might be present in one of these materials.
(ii) Suggest why X needs to be heated strongly?
(iii)The material X becomes black in color. Suggest what substance is responsible for this black color and how it arises in the reaction?
I am trying :
(i) Silicon(IV) carbide (SiC) / Silicon(IV) nitride (Si3N4).
(ii) To overcome it's melting point (their melting points are considerably high)/ To have it in the liquid state.
(iii) Oxygen (or oxygen compound) from atmosphere, by heating.
Post by: Ivo on November 21, 2010, 11:58:07 am
Many thanks in advance!
Post by: Deadly_king on November 22, 2010, 08:20:36 am
I am trying :
(i) Silicon(IV) carbide (SiC) / Silicon(IV) nitride (Si3N4).
(ii) To overcome it's melting point (their melting points are considerably high)/ To have it in the liquid state.
(iii) Oxygen (or oxygen compound) from atmosphere, by heating.
Nice try but I don't think your answer to the second question is correct. It's not necessary that it turns into liquid for combustion to take place.
(ii) For it to reach activation energy.
The rest is good :D
Post by: Deadly_king on November 22, 2010, 08:32:15 am
I have some few questions :(
1) Increasing the temperature of the propanoic acid solution causes the pH to decrease.
What does this tell you about the enthalpy of dissociation?
2) Why is an indicator usually added in iodine/thiosulphate titrations but not in titrations involving potassium managanate (VII)?
1) According to the information given an increase in temperature causes the pH to decrease. Decrease in pH implies the solution becomes more acidic, i.e there is more H+ present in the solution.
Hence we can deduce that an increase in temperature has caused propanoic acid to dissociate more. In other words the enthalpy change is endothermic. this is because according to Hess's law, an increase in temperature will cause the equilibrium to shift so as to oppose the change which is in this case decrease temperature.
Since the forward reaction for dissociation was favoured, dissociation is an endothermic process. ;)
2) Iodine and thiosulfate are colourless compounds which will react together to form colourless solutions. Hence we will never be able to identify the end-point unless an indicator is used.
Potassium Manganate however is purple in colour and will turn colourless as it is being titrated. Therefore the point at which the solution changes from purple to colourless will be the end-point. This can be observed with our own bare eyes. ;D
Hope it helps :)
Post by: Twinkle Charms on November 22, 2010, 05:07:12 pm
Post by: ~ Miss Relina ~ on November 29, 2010, 10:38:48 pm
alsowhat is the electron density maps ???
??? ???
need help asap
Post by: Deadly_king on November 30, 2010, 04:56:14 am
guys can anyone explain what is the ground state of an atom (related to the elctronic configuration)
alsowhat is the electron density maps ???
??? ???
need help asap
Atoms can exist in 2 states: Ground state and Excited state
The ground state of an atom is when its electronic configurations are at their lowest energy level. In other words all the electrons orbiting around the nucleus will be in the lowest sub-shells.
Excited state is just the opposite as the electrons may be found in higher energy sub-shells. An atom is more likely to undergo reactions in the excited state rather than the ground state since it has more energy.
Take Sodium for example, its electron configuration is: 1s2 2s2 2p6 3s1
Electrons that are excited have moved to another shell. So if a Sodium atom was excited it would still have the same amount of electrons but a different configuration. Example: 1s2 2s2 2p6 3p1
Electron density maps are used by scientists during an experiment called the X-ray crystallography so as to estimate the location of electrons around the nucleus of a molecule.
If you need more information about it, click here (http://www.proteopedia.org/wiki/index.php/Electron_density_maps). ;)
Post by: ruby92 on November 30, 2010, 11:12:55 am
Q2 b and c
Post by: Master_Key on November 30, 2010, 01:30:51 pm
For Sr(OH)2 = 2467 - (1414 + (2*550)) = -47 kJ mol-1
(ii) Sr(OH)2 is more soluble in water as it releases energy when it dissolves in water (Exothermic).
(iii) Sr(OH)2 is less soluble in hot water as it releases energy when it dissolves in water (Exothermic). The same principle as in equilibrium.
Post by: elemis on December 02, 2010, 10:54:42 am
Question 6, 9 and 16, please.
Post by: Deadly_king on December 02, 2010, 11:45:28 am
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_s02_qp_1.pdf (http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_s02_qp_1.pdf)
Question 6, 9 and 16, please.
Jun 02 p1
6. Make use of the formula PV = nRT
It is said in the question that temperature is constant and so is the number of moles. In other words P is inversely proportional to V.
Therefore P1V1 = P2V2
P1V1 is given by the sum of the initial PxVx + PyVy while the final volume V2 will be Vx + Vy.
So (2 x 1) + ( 1 x 2) = P2(1 + 2) ----> P2 = 4/3
Answer is A.
9. Enthalpy change of atomisation is the energy required to convert one mole of hydrazine to its respective atoms.
In other words we just have to break all the bonds involved within its structure.
Energy required to break 1 N-N bond + 4 N-H bond. Use values from the data booklet to find the required answer. ;)
Delta H = 160 + 4(390) = 1720 KJ
Answer is B.
16. You need to know this equation by heart. ;)
NaCl + 2H2O -----> Cl2 + H2 + 2NaOH
Answer is B.
Post by: elemis on December 06, 2010, 02:16:50 pm
I dont understand how the answer is D for Question 11.
Please help me.
Post by: tmisterr on December 06, 2010, 02:42:46 pm
Now Calculate the partial pressures gas, in total we have b-x+0.5x+0.5x which is b moles of gases. partial pressure HI is therefore b-x/b for H2 is 0.5x/b and for I2 it is again 0.5x/b.
Replace these values into the Kp equation = p[H2]*p[I2]/p[HI]2
we get (0.5xp/b)2/((b-x)p/b)2
from here it is just simple simplification, the p2 and b2 will cancel out top and bottom and in the end you get 0.25x2/(b-x)2 which is = x2/4(b-x)2
Post by: elemis on December 06, 2010, 03:01:59 pm
Thanks ;)
Post by: elemis on December 07, 2010, 09:39:08 am
Question 17.
For Question 36 Ammonia is a reducing agent because it loses Hydrogen, correct ?
Post by: Deadly_king on December 07, 2010, 12:52:24 pm
http://www.freeexampapers.com/download.php?l=A%20Level/Chemistry/CIE/2002%20Nov/9701_w02_qp_1.pdf&t1=2wypmop&t2=31e3abol (http://www.freeexampapers.com/download.php?l=A%20Level/Chemistry/CIE/2002%20Nov/9701_w02_qp_1.pdf&t1=2wypmop&t2=31e3abol)
Question 17.
For Question 36 Ammonia is a reducing agent because it loses Hydrogen, correct ?
Nov 02 p1
17. In this question, all the four answers sounds good from the first view. But there's once best and we need to find out which one.
The question is asking about the enthalpy change of formation, that is the energy required when HCl is formed from its reactants at rtp or it could also be the reverse, that is the conversion of HCl to its reactants.
For this we just need to break the C-Cl bond and H-I bond.
Electronegativity does not affect enthalpy changes. Activation energy will not be the reason since it's only the energy required for the reaction to start. Since we're doing the reverse bond energy of the halogens is not required.
I'll take only HCl for explanation. Same principle is applied for HI.
Option C refers to the energy required to break the bond between Hydrogen and the chlorine atom, that H-Cl . Upon breaking this bonds we obtain the atoms of hydrogen and chlorine which will combine together to form hydrogen gas and chlorine molecule.
In other words we have converted H-Cl to H2 and Cl2 and the energy required to do so is the same as the enthalpy change of formation of the molecule except for the sign which will be opposite. This is we're converting H-Cl to H2 and Cl2 while enthalpy change of formation is the conversion of Cl2 and H2 to HCl.
Bond energy for H-Cl is greater since Cl is a smaller atom, hence force of attraction will be greater. ;)
So answer is C.
36. You're right. ;)
Post by: elemis on December 07, 2010, 01:04:00 pm
Anyways, how is it C ? What does option C even mean ?
Post by: Deadly_king on December 07, 2010, 01:24:55 pm
Dude, the answer is 17 C.
Anyways, how is it C ? What does option C even mean ?
Ooh yeah.......i'll modify my post. :-[
Post by: elemis on December 07, 2010, 01:50:35 pm
Bond energy for H-Cl is greater since Cl is a smaller atom, hence force of attraction will be greater. ;)
H-Cl has a shorter bond length than H-I therefore it is stronger and hence requires greater energy to break the bond, right ?
Post by: Deadly_king on December 07, 2010, 02:29:20 pm
H-Cl has a shorter bond length than H-I therefore it is stronger and hence requires greater energy to break the bond, right ?
You're absolutely correct br0 :D
Post by: elemis on December 08, 2010, 11:03:07 am
Q33
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w03_qp_1.pdf (http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w03_qp_1.pdf)
Q2
Post by: Deadly_king on December 08, 2010, 12:25:09 pm
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_s03_qp_1.pdf (http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_s03_qp_1.pdf)
Q33
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w03_qp_1.pdf (http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w03_qp_1.pdf)
Q2
Jun 03 p1 No 33
It is given that the conversion of graphite to diamond is endothermic.
1. The enthalpy change of atomisation of diamond is smaller than that of graphite -----> It is correct since converting graphite to diamond is endothermic meaning graphite requires more energy.
2. The bond energy of the C-C bonds in graphite is greater than that in diamond -----> It is correct as well since in diamond, each carbon carries 4 bonds while in graphite they carry only 3 bonds.
3. The enthalpy change of combustion of diamond is greater than that of graphite -----> It is correct too dor the same reason as the first one.
So answer is A.
Nov 03 p1 No 2
First you need to find the % of phosphorus in P2O5 ----> 2(31)/142 x 100 = 43.7 %
It is given that P2O5 is 30% of the fertiliser. Hence phosphorus will be 43.7% of this 30%.
% by mass of phosphorus = (43.7/100 x 30/100)/1 x 100 = 13.1%
Therefore answer is B.
Post by: elemis on December 09, 2010, 04:51:45 am
Jun 03 p1 No 33
It is given that the conversion of graphite to diamond is endothermic.
1. The enthalpy change of atomisation of diamond is smaller than that of graphite -----> It is correct since converting graphite to diamond is endothermic meaning graphite requires more energy.
2. The bond energy of the C-C bonds in graphite is greater than that in diamond -----> It is correct as well since in diamond, each carbon carries 4 bonds while in graphite they carry only 3 bonds.
3. The enthalpy change of combustion of diamond is greater than that of graphite -----> It is correct too dor the same reason as the first one.
So answer is A.
You are contradicting yourself. If Carbon has 4 bonds and its atoms are arranged tetrahedrally then it should have a LARGER BOND ENERGY than graphite.
Therefore, wouldnt this eliminate option 2 ?
Can you explain how option 1 is correct in more detail ?
Post by: Deadly_king on December 09, 2010, 05:19:34 am
You are contradicting yourself. If Carbon has 4 bonds and its atoms are arranged tetrahedrally then it should have a LARGER BOND ENERGY than graphite.
Therefore, wouldnt this eliminate option 2 ?
Can you explain how option 1 is correct in more detail ?
Nope........am not contradicting myself dude. Let me get it clear for you.
Bond energy is not the 4 C-C bonds present but only one of them. A carbon atom in carbon carries 4 bonds, so the force of attraction needs to be divided exactly in 4. While in graphite the force of attraction is divided in only 3.
So the bond energy in graphite will be larger. ;)
Option 1.
Enthalpy change of Atomisation can also be considered as the energy required for breaking of bonds to form C(g). So since bond energy for graphite is larger, so will be atomisation energy.
Post by: elemis on December 09, 2010, 05:26:37 am
Quote
It is correct since converting graphite to diamond is endothermic meaning graphite requires more energy.
So what if its endothermic ? How did you realise graphite requires more energy ?
Post by: Deadly_king on December 09, 2010, 05:39:45 am
Okay, I see.
So what if its endothermic ? How did you realise graphite requires more energy ?
Hmm.........I had the same explanation I gave you, in mind.
Post by: elemis on December 09, 2010, 07:08:09 am
Question 13. I understand the answer is D and that you need a strong acid and base, but why ? What is the chemistry behind all of this ?
Why wouldnt a strong acid and weak base give a SHARP colour change ?
What chapter does this appear in the chem book by Brian Ratcliff ?
Post by: Deadly_king on December 09, 2010, 08:09:48 am
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w04_qp_1.pdf (http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w04_qp_1.pdf)
Question 13. I understand the answer is D and that you need a strong acid and base, but why ? What is the chemistry behind all of this ?
Why wouldnt a strong acid and weak base give a SHARP colour change ?
What chapter does this appear in the chem book by Brian Ratcliff ?
A strong acid and a strong alkali gives the sharpest end-points. Weak bases or acids take more time to dissociate and hence the end-point is not as sharp as that using strong solutions.
It appears in the chapter Equilibria ---> The part in A2 where you deal with indicators.
Post by: elemis on December 10, 2010, 01:00:41 pm
Question 3
I know the answer is B... I actually got this question correct. However, I want to check if my reasoning is correct and that I didnt get this by some fluke.
Once the p electron in option B is removed a fully filled s orbital is left. This orbital has a stable arrangement of electrons because it is filled and hence requires a larger IE to remove these electrons.
Is this correct ?
Post by: The Golden Girl =D on December 10, 2010, 01:11:29 pm
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w05_qp_1.pdf (http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w05_qp_1.pdf)
Question 3
I know the answer is B... I actually got this question correct. However, I want to check if my reasoning is correct and that I didnt get this by some fluke.
Once the p electron in option B is removed a fully filled s orbital is left. This orbital has a stable arrangement of electrons because it is filled and hence requires a larger IE to remove these electrons.
Is this correct ?
Yes , I chose B too because the Element is in Group 1 , ur explanation ,hence the Highest I.E. among the other =]
Post by: Master_Key on December 10, 2010, 01:29:39 pm
Yes , I chose B too because the Element is in Group 1 , ur explanation ,hence the Highest I.E. among the other =]
To remove the first electron is easy as it is the only electron, but next shell is full, sheilding effect is higher and thus requires higher energy.
Post by: elemis on December 21, 2010, 08:07:15 am
can be represented by the following equation.
2NCl3(l) + 6NaOH(aq) ---> N2(g) + 3NaCl (aq) + 3NaOCl (aq) + 3H2O(l)
What will be the result of this reaction?
1 The nitrogen is oxidised.
2 A bleaching solution remains after the reaction.
3 The final solution gives a precipitate with acidified silver nitrate.
The MS says 2 and 3 are correct only.
Why isnt 1 also correct ? It is obvious that Nitrogen goes from an oxidation state of -3 in NCl3 to 0 in N2. Hasnt the Nitrogen been oxidised?
Post by: Deadly_king on December 21, 2010, 11:15:16 am
When the yellow liquid NCl3 is stirred into aqueous sodium hydroxide, the reaction that occurs
can be represented by the following equation.
2NCl3(l) + 6NaOH(aq) ---> N2(g) + 3NaCl (aq) + 3NaOCl (aq) + 3H2O(l)
What will be the result of this reaction?
1 The nitrogen is oxidised.
2 A bleaching solution remains after the reaction.
3 The final solution gives a precipitate with acidified silver nitrate.
The MS says 2 and 3 are correct only.
Why isnt 1 also correct ? It is obvious that Nitrogen goes from an oxidation state of -3 in NCl3 to 0 in N2. Hasnt the Nitrogen been oxidised?
Dude the oxidation number of nitrogen in NCl3 is +3 and not -3. Therefore it gets converted to 0 which means it is reduced. ;)
Post by: elemis on December 21, 2010, 11:15:45 am
I must be really daft.... but I cant figure out why the answer is as stated in the MS.
Question 1
Post by: elemis on December 21, 2010, 11:18:24 am
Dude the oxidation number of nitrogen in NCl3 is +3 and not -3. Therefore it gets converted to 0 which means it is reduced. ;)
Damn, right. Can you answer my other question ?
Post by: Deadly_king on December 21, 2010, 11:55:00 am
http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_s10_qp_11.pdf (http://www.xtremepapers.net/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_s10_qp_11.pdf)
I must be really daft.... but I cant figure out why the answer is as stated in the MS.
Question 1
1.What could be the proton number of an element that has three unpaired electrons in each of its
atoms?
A 5
B 13
C 15
D 21
You just have to write the electronic configuration of the elements.
5 -----> 1s22s22p1 : Only one unpaired electron.
13 -----> 1s22s22p63s23p1 : Only one unpaired electron at all.
15 ------> 1s22s22p63s23p3 : 3 unpaired electrons only.
21 ------> 1s22s22p63s23p64s23d1 : Only one unpaired electrons.
So answer is C. ;)
Is it right?
Post by: elemis on December 21, 2010, 12:03:02 pm
Post by: Deadly_king on December 21, 2010, 12:05:58 pm
I got answer C too.... I think I was looking at the wrong MS at the time -_-
Hehe.........no worries br0. :)
Just be a bit more careful next time. :D
Post by: Sue T on December 28, 2010, 04:31:27 pm
does it matter wether th reagent used for the hydrolysis of an amide is said as :
(A) aq acid/alkali and reflux
or
(B) H3O+ or H+ or OH-?
does it differ if we're talkin bout a protein or normal amide -- cuz in th past pprs sometimes it says smthin n others ts smthin else. i ignored it at first but its gettin on my nerves now...
ny ideas?
n also they cnt decide wether they want us 2 say reflux or simply heat >:(
Post by: Amelia on December 31, 2010, 08:39:21 pm
A2 question:
does it matter wether th reagent used for the hydrolysis of an amide is said as :
(A) aq acid/alkali and reflux
or
(B) H3O+ or H+ or OH-?
No, it doesnt matters. They act. mean the same thing. ::)
Although, an equation for this reaction is simply written as (for eg:-)
Ch3CONHCH3 +H2O +H+(aq)----------> CH3COOH + CH3NH2
Quote
does it differ if we're talkin bout a protein or normal amide -- cuz in th past pprs sometimes it says smthin n others ts smthin else. i ignored it at first but its gettin on my nerves now...
ny ideas?
Have a look at this ;) http://www.chemguide.co.uk/organicprops/aminoacids/proteinhydrolysis.html
Quote
n also they cnt decide wether they want us 2 say reflux or simply heat >:(
Amides may be hydrolysed by heating under the reflux with either an acid or an alkali. :-X
Post by: Sue T on January 02, 2011, 03:15:09 pm
Post by: Amelia on January 02, 2011, 07:04:24 pm
thanks amelia!
Glad to be of help, mate. :)
Post by: Vin on January 08, 2011, 01:19:54 pm
Like for eg, this question :
A sample of chlorine containing isotopes of mass numbers 35 and 37 was analysed in a
mass-spectrometer.
How many peaks corresponding to Cl2+ were recorded?
A 2 B 3 C 4 D 5
Could anybody fill me in with some sense as to what the ques means in English? and everything in and around this topic? Thanks.
Post by: Amii on January 08, 2011, 01:25:48 pm
Does anybody have notes for mass- spectroscopy?ans will be B
Like for eg, this question :
A sample of chlorine containing isotopes of mass numbers 35 and 37 was analysed in a
mass-spectrometer.
How many peaks corresponding to Cl2+ were recorded?
A 2 B 3 C 4 D 5
Could anybody fill me in with some sense as to what the ques means in English? and everything in and around this topic? Thanks.
1 peak for Cl 35 and Cl 35
1 peak for Cl 35 and Cl 37
1 peak for Cl 37 and Cl 37
I am in a rush - have to get ready for tuitions - I will come back and try to make the explanation proper
Take care :)
Post by: Amelia on January 08, 2011, 04:15:59 pm
This topic is about (M) (M+2) (M+3) peaks. ::)
Both bromine and chlorine naturally occur as mixtures of 2 isotopes, with the relative abundances as shown :-
Chlorine -
35 - Cl relative abundance - 75.8%
37-Cl relative abundance - 22.4%
approximate ratio - 3:1
Bromine -
79 - Br relative abundance - 50.5%
81 - Br relative abundance - 49.5%
approx. ratio - 1:1
The mass spectrum of a compound containing one of these elements should therefore show two molecular ions, one with an m/e value two mass units higher than the other. If the molecule contains two chlorine atome, we should expect to see three molecular ions, at m/e values of M, M+2, M+4. The ratio of the M/M+2 peak should reflect the natural abundances. (i.e 3:1 for chlorine, 1:1 for bromine).
You can just go through this : http://www.chemguide.co.uk/analysis/masspecmenu.html :P
P.S - Sorry, esha. :-X
Post by: Amii on January 08, 2011, 04:57:28 pm
Amii is right. :PLOL why sorry? :-\ :D
This topic is about (M) (M+2) (M+3) peaks. ::)
Both bromine and chlorine naturally occur as mixtures of 2 isotopes, with the relative abundances as shown :-
Chlorine -
35 - Cl relative abundance - 75.8%
37-Cl relative abundance - 22.4%
approximate ratio - 3:1
Bromine -
79 - Br relative abundance - 50.5%
81 - Br relative abundance - 49.5%
approx. ratio - 1:1
The mass spectrum of a compound containing one of these elements should therefore show two molecular ions, one with an m/e value two mass units higher than the other. If the molecule contains two chlorine atome, we should expect to see three molecular ions, at m/e values of M, M+2, M+4. The ratio of the M/M+2 peak should reflect the natural abundances. (i.e 3:1 for chlorine, 1:1 for bromine).
You can just go through this : http://www.chemguide.co.uk/analysis/masspecmenu.html :P
P.S - Sorry, esha. :-X
Thank you so much ;D
I was getting late so I had to run :P
Post by: astarmathsandphysics on January 08, 2011, 10:28:57 pm
chlorine has two isotopes (I think) mass numbers 37 and 36 so Cl2 has three masses
2*37=74, 37+36=73,36+36=72
Hence three peaks, one for each mass
Post by: Vin on January 17, 2011, 07:47:09 pm
Post by: Amelia on January 17, 2011, 07:53:35 pm
THANK YOUU SO MUCHH. :D I shall make sense out of it after my term exams. Thanks a lot, again!
You are welcome. :D
Post by: Sue T on January 19, 2011, 07:17:42 am
so we have this eqn:
CuCO3.Cu(OH)2.x H2O(s) ------> 2CuO(s) + CO2(g) + (x+1)H2O(g)
we're tryin 2 find x
wat i got is:
1) the mass of the basic copper carbonate (LHS) and;
2) loss in mass which is basically CO2 and H2O
there wer 2 expressions in the m.s. tht im supposed 2 use:
(221 + 18x); 44 + 18(x + 1);
ny has ny idea wat 2 do 2 find x ??
p.s. i hate paper 5. >:(
Post by: HUSH1994 on January 21, 2011, 03:58:19 pm
Post by: Amelia on January 22, 2011, 03:51:29 pm
what do u need to know about isomers for AS and is knowledge about geometrical isomers needed?
According to the 2011 syllabus, you only need to know about the Optical Isomers.
Optical Isomers
When drawing a pair of optical isomers, candidates should indicate the three-dimensional structures
according to the convention used in the example below.
mirror plane
(http://www.sexandphilosophy.co.uk/pe_20_21_chiral_alanine.gif)
Post by: Amelia on January 22, 2011, 04:58:44 pm
i have an A2 ppr 5 ques
so we have this eqn:
CuCO3.Cu(OH)2.x H2O(s) ------> 2CuO(s) + CO2(g) + (x+1)H2O(g)
we're tryin 2 find x
wat i got is:
1) the mass of the basic copper carbonate (LHS) and;
2) loss in mass which is basically CO2 and H2O
there wer 2 expressions in the m.s. tht im supposed 2 use:
(221 + 18x); 44 + 18(x + 1);
ny has ny idea wat 2 do 2 find x ??
p.s. i hate paper 5. >:(
eg :- mass of empty boiling-tube - 10.32g
mass of boiling-tube and basic carbonate before heating - 11.19g
mass of boiling-tube and residue after heating - 10.92 g.
After the heating the CO2 and H2O boil away and only CuO remains.
(11.19 - 10.92 g)(mol / 159.09 g) = 1.70e-3 mol rxn
(1.70e-3 mol)(221.1138 + 18.0148x) = 11.19 - 10.32 g
x = 16 (rounding to 2 s.d.)
P.S - I hope to be right. ::)
Post by: Sue T on January 24, 2011, 04:47:11 pm
IF there is no bother - pls elaborate just a lil bit more - im sure im jus missin smthin somewere
please and thank you :-\
Post by: Amelia on January 24, 2011, 05:44:03 pm
After the heating the CO2 and H2O boil away and only CuO remains.
(11.19 - 10.92 g)(mol / 159.09 g) = 1.70e-3 mol rxn
11.19 - 10.92 = 0.27g (mass of residue)
CuO - (63.546+16)x2
Find the moles.
The moles of the residue - 0.87/((63.546+16)x2) = 1.70e-3 mol rxn
Moles x RMM = Mass.
Therefore, use this eqn. to find X.
(1.70e-3 mol)(221.1138 + 18.0148x) = 11.19 - 10.32 g
Post by: Sue T on January 25, 2011, 07:47:32 am
Post by: thecandydoll on January 25, 2011, 02:35:26 pm
I need resources and help in Chemistry P5 Got no idea.No IDEA how to attempt this paper.
Post by: $H00t!N& $t@r on January 25, 2011, 06:00:00 pm
is the formula products - reactants or is it the other way round? ???
Post by: elemis on January 25, 2011, 06:14:11 pm
I'm having some trouble with calculating enthalpy change.
is the formula products - reactants or is it the other way round? ???
Enthalpy change of reaction = reactants - products
Post by: $H00t!N& $t@r on January 25, 2011, 06:27:34 pm
http://www.science.uwaterloo.ca/~cchieh/cact/c120/heatreac.html (http://www.science.uwaterloo.ca/~cchieh/cact/c120/heatreac.html)
Post by: elemis on January 26, 2011, 02:38:16 am
pls check this website... im getting confused beacuse of this
http://www.science.uwaterloo.ca/~cchieh/cact/c120/heatreac.html (http://www.science.uwaterloo.ca/~cchieh/cact/c120/heatreac.html)
Enthalpy change of reaction and enthalpy change of formation are dfferent.
Post by: Amelia on January 28, 2011, 07:14:19 pm
got it ;) thank you - sorry 4 th trouble :)
No problem. ;D
Heyy!
I need resources and help in Chemistry P5 Got no idea.No IDEA how to attempt this paper.
Practise Past Papers.
Here are some tips - http://answers.yahoo.com/question/index?qid=20081211153457AAmQNJl
Post by: elemis on February 04, 2011, 11:00:59 am
Post by: elemis on February 04, 2011, 12:57:52 pm
Post by: elemis on February 04, 2011, 01:38:40 pm
Post by: Amii on February 04, 2011, 02:04:38 pm
QUestion 9C lowest activation energy
Post by: elemis on February 04, 2011, 02:20:52 pm
C lowest activation energy
Stupid me.
And the rest, please ?
Post by: Amii on February 04, 2011, 02:22:07 pm
Questions 23,26,29 and 30Not AT ALL sure
30) B - Butan-2-ol
On hydration can form But-1-ene,trans-but-2-ene and cis-but-2-ene (3 alkenes)
Butan-1-ol will form only But-1-ene(1 alkene)
2-methylpropan-1-ol and 2-methylpropan-2-ol will form 2-methylpropene (1 akene)I guess :-\
Post by: Amii on February 04, 2011, 02:24:55 pm
Stupid me.I don't take CIE. I called my tuition teacher but she was busy.
And the rest, please ?
So if possible I wil check those rest questions with her later.
Post by: elemis on February 04, 2011, 05:17:26 pm
Shouldnt it be 410*4 +151 + (-299-3*410-299) ??
There are 4 C-H bonds broken but the MS pretends only one is broken.
Post by: Amelia on February 04, 2011, 05:32:53 pm
CH4 + I2 --> CH3I + HI
Only one C-H bond is broken to form H-I.
The rest are still intact -> CH3I
Post by: elemis on February 05, 2011, 11:45:44 am
Am I correct ?
Post by: Amii on February 05, 2011, 12:23:50 pm
Questions 23,26,29 and 30
26) B-CH3CHBRCH=CH2 has identical atoms attached to one of the carbon atoms in the double bond- so no cis-trans in this compound.
A,B,D - exhibits cis-trans isomerism.
But only D has a Carbon atom which has 4 different atoms/groups attached to it - hence only D will can show optical isomerism.
I hope it makes sense. :-[
Post by: Amii on February 05, 2011, 01:03:38 pm
Questions 23,26,29 and 3029)D (CH3)3CBr
-Tertiary halogenoalkane.
-Undergoes hydrolysis by SN1 mechanism.
-Reaction takes place in two steps.
-Since the rate determining step doesn't involve the nucleophile,changing the concentration of OH- doesn't affect the rate of the SN1 reaction.
Post by: Amelia on February 05, 2011, 01:15:05 pm
I've identified the chiral centres by drawing a red square next to them.
Am I correct ?
Yes.
The answer to 23 is C.
Post by: elemis on February 06, 2011, 03:31:43 pm
Reduction is the GAIN of Hydrogen, right ? In step 2 a massive molecule is added that has several hydrogens. Isnt this reduction ?
Can you explain why the answer is C ?
In the second one, I have marked the chiral centre. Is it correct ?
Post by: Amelia on February 07, 2011, 08:23:18 pm
Quote
In the second one, I have marked the chiral centre. Is it correct ?
Yes, it is.
Post by: elemis on February 08, 2011, 02:41:52 pm
Post by: elemis on February 08, 2011, 02:44:27 pm
^ There are three methyl groups in 2,2dimethylbutane. Available for Chlorine radicals to attack on. So free radicals are formed from each Ch3's.
I dont understand. Are you talking about the right question ?
Post by: ashish on February 08, 2011, 02:57:25 pm
In the first one, why isnt the answer D ?
Reduction is the GAIN of Hydrogen, right ? In step 2 a massive molecule is added that has several hydrogens. Isnt this reduction ?
Can you explain why the answer is C ?
In the second one, I have marked the chiral centre. Is it correct ?
it is not reduction as we are nor adding hydrogen directly to the molecule!
the alcohol is acidic (releases H+) creating CH3CH2O- which then attack the CH3C+HO- creating an nucleophile which is then attacked by the proton
Post by: elemis on February 08, 2011, 02:59:17 pm
it is not reduction as we are nor adding hydrogen directly to the molecule!
the alcohol is acidic (releases H+) creating CH3CH2O- which then attack the CH3C+HO- creating an nucleophile which is then attacked by the proton
Thanks.
Could you answer my other questions, please ?
Post by: EMO123 on February 08, 2011, 03:08:39 pm
Post by: Amelia on February 08, 2011, 03:34:52 pm
I know the carbon with the red square is DEFINITELY chiral. Why is the one with the PINK square NOT chiral ?
It actually has three chiral's. The pink one you marked is a chiral as well. And also, the first Carbon. (CH-CH3)
I dont understand. Are you talking about the right question ?
Mistook your question. I apologize.
Post by: elemis on February 17, 2011, 04:37:02 am
I drew the following compound, yet the markscheme shows something else altogether.
Would my answer be accepted ?
Post by: Deadly_king on February 17, 2011, 05:15:36 am
Last question, (e) part.
I drew the following compound, yet the markscheme shows something else altogether.
Would my answer be accepted ?
Sorry for lateness but I had to do the whole number to get the required answer.
So your answer will NOT be accepted. Tests 5 and 6 eliminates your answer completely. Now I'll explain :
For the compound you drew, upon test 5, it would get oxidised to a carbon dioxide and a carboxylic acid as the C=C will break. So the resulting carboxylic acid will give a positive result with the following test.(test 6)
Hence your answer is rejected.
Post by: elemis on February 17, 2011, 05:25:48 am
Sorry for lateness but I had to do the whole number to get the required answer.
So your answer will NOT be accepted. Tests 5 and 6 eliminates your answer completely. Now I'll explain :
For the compound you drew, upon test 5, it would get oxidised to a carbon dioxide and a carboxylic acid as the C=C will break. So the resulting carboxylic acid will give a positive result with the following test.(test 6)
Hence your answer is rejected.
2,4-DNPH gives a positive result only with Aldehydes or Ketones not Carboxylic acids.
Post by: Deadly_king on February 17, 2011, 05:42:10 am
2,4-DNPH gives a positive result only with Aldehydes or Ketones not Carboxylic acids.
That's true but before turning into a carboxylic acid......the compound will first be converted into an aldehyde. We need to oxidise much more for it to get converted to carboxylic acid.
Post by: elemis on February 17, 2011, 06:13:10 am
That's true but before turning into a carboxylic acid......the compound will first be converted into an aldehyde. We need to oxidise much more for it to get converted to carboxylic acid.
The question says it is oxidised completely THEN it is tested with DNPH
Post by: Deadly_king on February 17, 2011, 06:23:37 am
The question says it is oxidised completely THEN it is tested with DNPH
Yupz........but note that in test 5, the reaction says warm with acidified dichromate ions. This is a weak reagent which is not sufficiently strong to convert the aldehyde formed to carboxylic acid readily.
If manganate ions had been used, then no aldehyde would have been present since all of them would have readily been converted to carboxylic acid.
This is the difference, your answer would have been okay had manganate ions been used instead of dichromate. ;)
Post by: narnia on February 19, 2011, 11:33:22 am
Also,why acid-base character change as we go down group4?
Post by: Amelia on February 19, 2011, 04:57:00 pm
Why is SiO2 insoluble in water buh dissolves in concentrated alkali ???
Silicon dioxide is a giant molecule (macromolecular structure due to extensive network of covalent bonds in the crystalline lattice structure that extends to infinity). since the solubility of a substance is related to the similarity in bond strength in the solvent (water in this case) AND the solute (silicon dioxide in this case), and the extensive network of covalent bonds is much stronger than the water-water interactions (i.e. hydrogen bonds), therefore silicon dioxide CANNOT be hydrated by water molecules (i.e. water cannot break down the giant crystal lattice structure of the macromolecule SiO2) to form aqueous ions and therefore it remains insoluble in water. (silicon dioxide only dissolves by heating with strong alkali. in this case water is not alkaline enough to hydrolyse silion dioxide to give sillicate ions).
Also Silicon dioxide is an acidic compound.
In one line, Silicon dioxide doesn't react with water, because of the difficulty of breaking up the giant covalent structure.
Quote
Also,why acid-base character change as we go down group4?
http://www.thestudentroom.co.uk/wiki/Revision:Acid-Base_Character_of_Group_4_Oxides
Post by: Nera_egypt on February 20, 2011, 07:58:14 am
In section 1.3 – Genetic information. I couldn't solve SAQ 11 part c on page 41 :( .
I would be really glad if someone helps me work it out :)
P.S: I attached the question
Post by: tmisterr on February 20, 2011, 04:39:03 pm
I have a question in the application booklet related to Biochemistry
In section 1.3 – Genetic information. I couldn't solve SAQ 11 part c on page 41 :( .
I would be really glad if someone helps me work it out :)
P.S: I attached the question
from the peptide fragment, we see -Ala-Ala-: two same amino acids one after another. so we would expect that we see two same codons following each other on the fragment of mRNA. the first time this happens is -GCU-GCU- so we can say that the codon for Ala is GCU. we can confirm this by skipping the next two codons (-GAA-GGA-), which we can suspect to be for the -Glu-Gly- section of the peptide and we see that the next codon is again -GCU- confirming that GCU is the codon for Ala since is coincides with the polypeptide chain, that it two Ala, followed by two other amino acids then Ala again. once you know this, you can easily get the rest ;)
(ii) deciphering gives the codon for tyrosine as UAC. this is for the mRNA so the DNA triplet code from which it was transcribed from is ATG (I hope you know how the bases match). since mRNA is always synthesised from the 5' end to the 3'end the direction of this codon is 5'-ATG-3'
well then, I hope this helps
Post by: tmisterr on February 20, 2011, 04:46:54 pm
Post by: Nera_egypt on February 21, 2011, 07:11:12 am
from the peptide fragment, we see -Ala-Ala-: two same amino acids one after another. so we would expect that we see two same codons following each other on the fragment of mRNA. the first time this happens is -GCU-GCU- so we can say that the codon for Ala is GCU. we can confirm this by skipping the next two codons (-GAA-GGA-), which we can suspect to be for the -Glu-Gly- section of the peptide and we see that the next codon is again -GCU- confirming that GCU is the codon for Ala since is coincides with the polypeptide chain, that it two Ala, followed by two other amino acids then Ala again. once you know this, you can easily get the rest ;)
(ii) deciphering gives the codon for tyrosine as UAC. this is for the mRNA so the DNA triplet code from which it was transcribed from is ATG (I hope you know how the bases match). since mRNA is always synthesised from the 5' end to the 3'end the direction of this codon is 5'-ATG-3'
well then, I hope this helps
Thanks alot, this was the best explanation given to me :)
Post by: narnia on February 21, 2011, 04:45:50 pm
Silicon dioxide is a giant molecule (macromolecular structure due to extensive network of covalent bonds in the crystalline lattice structure that extends to infinity). since the solubility of a substance is related to the similarity in bond strength in the solvent (water in this case) AND the solute (silicon dioxide in this case), and the extensive network of covalent bonds is much stronger than the water-water interactions (i.e. hydrogen bonds), therefore silicon dioxide CANNOT be hydrated by water molecules (i.e. water cannot break down the giant crystal lattice structure of the macromolecule SiO2) to form aqueous ions and therefore it remains insoluble in water. (silicon dioxide only dissolves by heating with strong alkali. in this case water is not alkaline enough to hydrolyse silion dioxide to give sillicate ions).Thanks tons!
Also Silicon dioxide is an acidic compound.
In one line, Silicon dioxide doesn't react with water, because of the difficulty of breaking up the giant covalent structure.
http://www.thestudentroom.co.uk/wiki/Revision:Acid-Base_Character_of_Group_4_Oxides
Post by: EMO123 on February 21, 2011, 06:08:03 pm
Post by: Amelia on February 21, 2011, 06:35:50 pm
What are the Friedel-Crafts Catalyst?
They are the catalysts used in the Friedel-Crafts reactions.
Here are examples of the reactions - http://www.chemguide.co.uk/organicprops/arenes/fc.html
Thanks tons!
Welcome:)
Post by: EMO123 on February 22, 2011, 05:57:01 pm
They are the catalysts used in the Friedel-Crafts reactions.
Here are examples of the reactions - http://www.chemguide.co.uk/organicprops/arenes/fc.html
Welcome:)
thanxxs
Post by: TheLonelyIsland on March 07, 2011, 01:37:04 am
"Which of these samples of gas contains the same number of atoms as 1g of hydrogen
(Mr : H2, 2)?
A 22 g of carbon dioxide (Mr: CO2, 44)
B 8 g of methane (Mr: CH4, 16)
C 20 g of neon (Mr: Ne, 20)
D 8 g of ozone (Mr: O3, 48) "
According to ms, the correct answer is C...How do you do it correctly? Arghh, so frustrating when you don't know, lol...
Thx in advance,
Take care
Post by: Deadly_king on March 07, 2011, 10:36:03 am
Hi, can someone PLZ help me with this question? I'm so stupid, it's really easy, but I keep getting the wrong answer :o :
"Which of these samples of gas contains the same number of atoms as 1g of hydrogen
(Mr : H2, 2)?
A 22 g of carbon dioxide (Mr: CO2, 44)
B 8 g of methane (Mr: CH4, 16)
C 20 g of neon (Mr: Ne, 20)
D 8 g of ozone (Mr: O3, 48) "
According to ms, the correct answer is C...How do you do it correctly? Arghh, so frustrating when you don't know, lol...
Thx in advance,
Take care
Hmm..........it's not very difficult. ;)
The most important thing that you need to know is that 1 mole of any substance contains 6.02 x 1024 number of atoms. ;)
So you've been asked to find a sample with the same number of atoms as 1g of hydrogen. 1 gram of Hydrogen as you should be knowing is equal to 1 mole of hydrogen (Ar of H = 1).
In other words, now you need to look for a sample representing one mole. You can use the following formula to do do so.
No of moles = Mass / Mr
A ----> No of moles = 22/44 =0.5
B ----> No of moles = 8/16 = 0.5
C ----> No of moles = 20/20 = 1
D ----> No of moles = 8/48 = 0.17
You'll note that only C offers this possibility. ;)
Hope it helps :)
Post by: TheLonelyIsland on March 07, 2011, 12:59:14 pm
Post by: HUSH1994 on March 08, 2011, 03:38:40 am
C8H18 + 12.5 O2 ------- 8CO2 + 9 H2O
I need full detailed explanation please,i know the defenition but how to calculate it
Post by: elemis on March 08, 2011, 11:13:58 am
how do u calculate the enthalpy change of combustion, and what do u use the values given at the data booklet for,i need help in:
C8H18 + 12.5 O2 ------- 8CO2 + 9 H2O
I need full detailed explanation please,i know the defenition but how to calculate it
Calculate the energy needed to break the bonds of the reactants and subtract the energy needed to form the products.
Just use the bond dissociation energies given in the Data Booklet.
Post by: Deadly_king on March 08, 2011, 11:55:18 am
thx a lot, i knew it was something really easy... What confused me was that Mr H2=2... Dumb, I know! Thx, again!
Don't worry dear........it happens. You'll get used to it with time and more practice ;)
Post by: xlane on March 08, 2011, 11:57:44 am
Post by: Amelia on March 08, 2011, 02:22:12 pm
i need explanation of partition coefficient and two immiscible layers...its for p5.......so i need the complete explanation.....some 1 help me out please....and does any1 know any p5 question based on this two topics ?
- The word 'partition' means a substance X is distributed between two phases in a dynamic equilibrium.
- It is a heterogeneous equilibrium since the 'solute' is distributed between two distinct phases.
- The two phases may be a gas and liquid-solution or, more likely, two immiscible liquids.
The basic expression is:
Kpartition = [X(phase 1)] / [X(phase 2)]
Here the K is called the partition coefficient or distribution coefficient. If it involves two immiscible liquids, K has no units.
The partition will involve the distribution of a solute between two immiscible liquid phases, which is a more likely and simpler situation to deal with.
If the solute is in the same molecular state in both liquid-phases, the following simple partition equilibrium expression will apply:
Kpartition =
[X(liquid 1)]
----------------
[X(liquid 2)]
K is called the partition/distribution coefficient and has no units and is temperature dependent.
Both concentrations must be in the same units e.g. molarity mol dm-3, g dm-3, mg cm-3 or whatever.
If a substance is added to a mixture which is soluble to a greater or lesser extent in both immiscible liquids, on shaking and then allowing the mixture to settle, the concentrations in each layer become constant. However, there is continual interchange of solute between the liquid layers via the interface i.e. a dynamic equilibrium is formed.
If more of the substance X is added to the system, the solute will distribute itself between the immiscible liquids so that the ratio of the solute concentrations remains the same at constant temperature independently of the total quantity of X in the same molecular state, and that is essentially the partition equilibrium law.
Post by: Amelia on March 08, 2011, 02:33:12 pm
(a) Calculate the partition coefficient Kp for butanedioic acid between ether/water.
Kp = [BDA(ether)] / [BDA(water)] = 0.024 / 0.16 = 0.15
(b) If 10g of BDA had been shaken with 50 cm3 of each solvent at 25oC, what value would expect for Kp if the layers were again analysed?
0.15, since the partition law states the ratio of concentrations remains constant at constant temperature.
(c) If 10g of butanedioic acid was dissolved in 50 cm3 of ether at 25oC, calculate how much of the acid can be extracted with 50 cm3 of water.
If we call x the mass in g extracted into water, V the liquid volumes, and expressing the concentrations in gcm-3, substitution into the partition expression gives ...
Kp =
[(10-x)/Vether] (10-x)
---------------------- = ------- (since V's cancel out)
[x/Vwater] x
Rearranging: 0.15 x = 10 - x, so 1.15x = 10
therefore x = 10/1.15 = 8.7g of BDA extracted.
--- Check the past papers for more questions... or perhaps, the application booklet.
Post by: iluvme on March 09, 2011, 11:42:48 am
Post by: Master_Key on March 09, 2011, 01:08:47 pm
Please explain why Kc does not change with a change in concentration.
If you are changing the concentration of both the reactants then it will change.
If you are changing the concentration of one substance then it seems obvious that one of the reactant is in excess and does not react.
Post by: iluvme on March 12, 2011, 12:32:24 pm
If you are changing the concentration of both the reactants then it will change.
If you are changing the concentration of one substance then it seems obvious that one of the reactant is in excess and does not react.
Thanks :)
Post by: Master_Key on March 12, 2011, 01:08:50 pm
Thanks :)You are welcome.
Post by: iluvme on March 20, 2011, 03:07:38 pm
When 1 mole of HI is allowed to dissociate, in a 1.0 dm-3 vessel at 4500C, only 0.78 mol of HI are present at equilibrium. What is the Kc value for this reaction?
Post by: Amii on March 20, 2011, 03:28:26 pm
<-----------
Intial moles 1 0 0
Change (1-0.78) (0.5X0.22) (0.5X0.22)
-0.22 +0.11 +0.11
Moles at equilibrium 0.78 0.11 0.11
Conc at equilibrium (0.78/1 ) (0.11/1) (0.11/1)
mol dm-3 0.78 0.11 0.11
Kc = [H2][I2]
---------
[HI]2
Kc= 0.11
---------------------------
(0.78
=0.0199 no units
I hope it's right
Post by: iluvme on March 20, 2011, 03:49:02 pm
Post by: Amii on March 20, 2011, 03:52:42 pm
Thank You! <3No problem :D
Check it again.I modified it.
Post by: iluvme on March 20, 2011, 04:14:53 pm
No problem :D
Check it again.I modified it.
Thanks again.
Didn't notice anything wrong. Was just going through the steps. :)
Post by: narnia on April 04, 2011, 07:41:52 pm
Post by: elemis on April 05, 2011, 05:21:52 am
M/J 02, p1, ques.13,21,27....I hv the ans., but how??
Question 13
Al2O3 is amphoteric and will react with the alkaline solution.
SO2 disolves in NaOH and is neutralised.
SiO2 reacts with NaOH to form Na2SiO3 and water.
By process of elimination MgO is the only possible answer.
Question 21
I'm not going to do this for you. Draw each and every possibility out on a sheet of paper and think of all possible placements of the Cl.
Question 27
Ethane is a gas so it is eliminated.
Potassium manganate(VII) would oxidise Ethanol to a carbonyl compound, so it is eliminated.
Ethanoic acid has a permanent dipole allowing for hydrogen bonding between H2O and its molecules. Hence, it mixes well with water. Thus, it is eliminated.
Ethyl Ethanoate is the answer as per process of elimination.
Post by: narnia on April 05, 2011, 06:02:52 am
for question 21,I got confused by the formula given, so asked.
But I really appreciate & thank you for your consideration. =)
Post by: elemis on April 05, 2011, 07:26:03 am
Excuse me please, the answer to question 13 is MgO & not Al2O3.
for question 21,I got confused by the formula given, so asked.
But I really appreciate & thank you for your consideration. =)
My bad, I misread the question. I though we were talking about dissolving in WATER. It seems we have to consider NaOH as the solvent.
Post by: narnia on April 06, 2011, 09:47:33 am
My bad, I misread the question. I though we were talking about dissolving in WATER. It seems we have to consider NaOH as the solvent.no problems.
how can ch3ch2ch2ch3 can result from mixing ethane n cl2 in uv light?? =/
also, when temp increases, the boltzman distrbtn curve move to d right or left?
bcz as far i know, it moves to right n bcums broader n shorter, but in oct04 p1 q12, the corect ans is b...where it movd to d left...im sorry i cant attach the ppr since im on from a mobile device.
Post by: elemis on April 06, 2011, 09:54:35 am
no problems.
how can ch3ch2ch2ch3 can result from mixing ethane n cl2 in uv light?? =/
You know that in free radical substitution of ethane an ethyl radical is formed, correct ?
Thus, in a resulting a termination step :
CH3CH2* + CH3CH2* ---> CH3CH2CH2CH3
Post by: narnia on April 06, 2011, 12:44:28 pm
You know that in free radical substitution of ethane an ethyl radical is formed, correct ?godamn i just overlooked tht part...
Thus, in a resulting a termination step :
CH3CH2* + CH3CH2* ---> CH3CH2CH2CH3
anyways, can u please answer the second part in my pevious post as well if possible?
Post by: HUSH1994 on April 14, 2011, 05:38:50 am
In question 10 i know that the Kp is the pressure of the products over the reactants,which is p(NO2)^2
----------
p(N2O4)
so how do you get the partial pressures for each
Post by: elemis on April 14, 2011, 05:47:43 am
MJ 2006 Paper 1 Q3 & 10 please
In question 10 i know that the Kp is the pressure of the products over the reactants,which is p(NO2)^2
----------
p(N2O4)
so how do you get the partial pressures for each
Question 3
Nitrogen receives 3 electrons from Gallium to form an N3- ion. Can you determine the answer by yourself now ?
Question 10
Lets say initially there were 2 moles of N2O4
50% of this dissociates, hence, 1 mole dissociates giving 2 moles of NO2 (use the molar ratio).
Partial pressure of molecule = no. of moles of molecule present at equilibrium/ TOTAL no. of moles at equilibrium
Can you do it now ?
Post by: HUSH1994 on April 14, 2011, 06:04:02 am
Post by: HUSH1994 on April 14, 2011, 07:19:32 am
Post by: elemis on April 14, 2011, 07:26:48 am
Thanks man i got it,sorry i meant Q2 not 3,3 is too easy
The isotopes of Cl are 35 and 37.
Hence, you can have :
(35-37)+
(37-37)+
(35-35)+
Thus, only three.
Post by: elemis on April 14, 2011, 07:29:25 am
MJ 2006 Paper 1 Q25 & 28 please
Question 25
Conisder the fact that the C-F bond is harder to break than a C-Cl bond.
Now do it for yourself.
Question 28
Think about the reaction between 2,4-Dinitrophenylhydrazine and carbonyl compounds.
Post by: HUSH1994 on April 14, 2011, 07:44:41 am
Post by: elemis on April 14, 2011, 07:49:07 am
Thanks man,i really appreciate your help +rep,and By the way are you doing any AS/A2 exams?
Yeah, I'm answering May/June of this year.
Post by: HUSH1994 on April 14, 2011, 08:23:44 am
Post by: Master_Key on April 14, 2011, 08:53:27 am
Post by: Aadeez || Zafar on April 14, 2011, 10:43:01 am
in fact the reaction mechanism of all the organic reaction categorised by functional group
Post by: red_911 on April 14, 2011, 02:42:30 pm
cause it has only one electron
Post by: Arthur Bon Zavi on April 14, 2011, 02:45:45 pm
• Helium has the highest first ionisation energy why not hydrogen ? ???
cause it has only one electron
It is a noble/inert gas. Does not react and easily form ion.
Post by: Master_Key on April 14, 2011, 03:14:17 pm
• Helium has the highest first ionisation energy why not hydrogen ? ???
cause it has only one electron
Ionization energies increase from left to right. This is a trend in periodic table.
Post by: xlane on April 14, 2011, 10:41:41 pm
Post by: Vin on April 14, 2011, 10:52:46 pm
how to find the oxidation state of the element in the hydride n chloride ?? h2S , SCl2....
You know the oxi. state of hydrogen, i,e +1
Take oxi. state of S = x
so,
2 x (+1) + x = 0
x = -2
as oxi. state of all compounds = 0
Do the same for the other compound. Oxi. state of Cl = -1.
Post by: HUSH1994 on April 15, 2011, 05:44:11 am
Post by: elemis on April 15, 2011, 06:03:20 am
ON 2008 Q2,i tried it many times but its always wrong,can anyone help please?
Show me your workings.
Post by: HUSH1994 on April 15, 2011, 06:05:32 am
Post by: Master_Key on April 15, 2011, 12:18:03 pm
qp12
Question 22.
Please someone help and can you give me notes on reactions of aldehydes.
Post by: elemis on April 15, 2011, 12:46:16 pm
CIE Chemistry 9701 - O/N 2010
qp12
Question 22.
Please someone help and can you give me notes on reactions of aldehydes.
In the first scenario you will EVENTUALLY get a 2-hydroxy carboxylic compound. Think in terms of nucleophilic adittion of a CN- ion and hence you should be able to determine the structure.
The second scenario is straight forward oxidation of an aldehyde to a carboxylic acid.
Post by: Master_Key on April 15, 2011, 02:17:58 pm
In the first scenario you will EVENTUALLY get a 2-hydroxy carboxylic compound. Think in terms of nucleophilic adittion of a CN- ion and hence you should be able to determine the structure.Ari, I am getting the difference as 14. How could it be 16?
The second scenario is straight forward oxidation of an aldehyde to a carboxylic acid.
Can you plz explain.
Post by: elemis on April 15, 2011, 02:36:58 pm
Ari, I am getting the difference as 14. How could it be 16?
Can you plz explain.
When you react an ethanal with HCN you get 2-hydroxy propanenitrile. Hydrolysis of this gives you 2- hydroxypropanoic acid (RMM = 90)
Oxidation of propanal gives propanoic acid (RMM =74)
Do the subtraction....
Post by: Arthur Bon Zavi on April 15, 2011, 03:45:09 pm
Post by: elemis on April 15, 2011, 04:23:44 pm
Ignored the hydrolysis part. ::)
Hmm ?
Post by: Arthur Bon Zavi on April 15, 2011, 04:30:33 pm
Hmm ?
The previous question which you answered. So was getting false difference. Thanks though.
Post by: elemis on April 15, 2011, 04:41:02 pm
The previous question which you answered. So was getting false difference. Thanks though.
You mean Master_key forgot about the hydrolysis ?
Post by: Arthur Bon Zavi on April 15, 2011, 04:54:11 pm
You mean Master_key forgot about the hydrolysis ?
We both were solving in the class.
Post by: Master_Key on April 16, 2011, 01:47:15 pm
CH3CHO + HCN ---> CH3CHOCN
CH3CHOCN + H2O ---> CH3CHOHCOOH (RFM=90).
Post by: TJ-56 on April 18, 2011, 01:08:11 pm
Can someone help me with the following Paper 4 questions?
May/June 2002 Question 3 (c) (ii)
May/June 2003 Questions 2 (b) (iii) & 3 (b) (i) & (c)
Any help is greatly appreciatied, thanx in advance!
Post by: EMO123 on April 18, 2011, 01:30:07 pm
Hey.but which paper?
Can someone help me with the following:
May/June 2002 Question 3 (c) (ii)
May/June 2003 Questions 2 (b) (iii) & 3 (b) (i) & (c)
Any help is greatly appreciatied, thanx in advance!
Post by: TJ-56 on April 18, 2011, 02:08:29 pm
but which paper?
Sorry, didnt notice, Paper 4.
Post by: EMO123 on April 18, 2011, 02:29:26 pm
CO2 + 2NaOH ? Na2CO3 + H2O
SnO2 + 2NaOH ? Na2SnO3 + H2O
Post by: TJ-56 on April 18, 2011, 03:45:18 pm
3(b)(i)
CO2 + 2NaOH ? Na2CO3 + H2O
SnO2 + 2NaOH ? Na2SnO3 + H2O
Thanx.
Regarding the third reaction though, SiO2, silicon (iv) dioxide, on the markscheme gives a similar reaction to that of SnO2, but I have read that SiO2 and NaOH dont react. So i dont know if the book is incorrect or the mark scheme. Do they react?
Post by: elemis on April 18, 2011, 04:21:43 pm
Thanx.
Regarding the third reaction though, SiO2, silicon (iv) dioxide, on the markscheme gives a similar reaction to that of SnO2, but I have read that SiO2 and NaOH dont react. So i dont know if the book is incorrect or the mark scheme. Do they react?
It does react. You would get Na2SiO3 and water
Post by: TJ-56 on April 18, 2011, 05:41:36 pm
It does react. You would get Na2SiO3 and water
oh ok thanx
Post by: EMO123 on April 19, 2011, 12:48:07 pm
oh ok thanxby mistake i wrote it Na2SnO3
it is written right by ari
Post by: thecandydoll on April 19, 2011, 02:17:33 pm
Post by: elemis on April 19, 2011, 02:32:35 pm
A few of the organic questions in A2 Paper 4 are in the form of skeletal forumla.How do you do it?
What do you mean by 'how do you do it ?'.
You have to be more specific if you want any constructive help or people will be simply put off by the ambiguous nature of your questions.
Post by: red_911 on April 19, 2011, 06:03:28 pm
Post by: Amelia on April 19, 2011, 06:30:01 pm
water has which types of intermolecular forces ??Hydrogen Bonds.
Quote
n which has higher boiling point HF or h2O ? explain with reason plzz ???
Water has two hydrogens while HF has only one. So, obviously the water will make multiple bonds (three) and when together then form a greater intermolecular force than that of HF, since HF can form hydrogen bonds with only two other atoms.
Post by: thecandydoll on April 20, 2011, 09:16:37 am
What do those lines depict?
i haveseen you in tsr
Post by: Master_Key on April 20, 2011, 01:18:46 pm
How do you UNDERSTAND THE SKELETAL formula...
What do those lines depict?
i haveseen you in tsr
The Points on line depict Carbon atoms. If it is just a change of line means Carbon-Hydrogen atoms. If Chlorine or smth else is there then it must be specified.
Post by: EMO123 on April 24, 2011, 09:08:51 am
Post by: HUSH1994 on April 24, 2011, 12:56:04 pm
http://www.patentstorm.us/patents/5344988/description.html
Post by: red_911 on April 25, 2011, 09:21:30 pm
Post by: EMO123 on April 26, 2011, 03:47:23 am
Post by: SkyPilotage on April 26, 2011, 08:01:39 pm
which one of these have higher boiling point: alkanes, alkenes, carboxylic acids or halogenoalkanes ... explain all a little please ???
It depends on the intermolecular forces between the molecules, which are Van-der-waals, permanent dipole-dipole and hydrogen bonds( type of Perm. Dipole-Dipole).
Strength of Hydrogen bonds>permanent dipoles>van-der-waals.
As electrons increase, van-der-waals increase since they are actually induced dipoles from the movement of e's.
Hydrogen bonds form between N,O,F lone pair and H.
So if you are talking abt corresponding alkanes<alkenes<alchohols<carboxylic acids. Ofcourse as Carbon atoms increase m.p and b.p increases!
Post by: SkyPilotage on April 26, 2011, 08:06:54 pm
Hydrogen Bonds.
Water has two hydrogens while HF has only one. So, obviously the water will make multiple bonds (three) and when together then form a greater intermolecular force than that of HF, since HF can form hydrogen bonds with only two other atoms.
Doesnt water have all three? van der wall + perm dipoles + hydrogen bonds? I think all molecules have Van-der-waals...except ideal gases (anthr story)
Post by: HUSH1994 on April 27, 2011, 07:41:07 am
Post by: elemis on April 27, 2011, 08:41:53 am
ON 2003 P1 Q20,how do u solve this type of question,is there another way other than drawing the structures?
Nope, there's no other way.
Post by: $H00t!N& $t@r on April 28, 2011, 07:51:39 pm
Can someone please explain q2 d (ii) 2010 O/N var 23 in detail
Thanks
Post by: Vin on April 28, 2011, 08:09:50 pm
Post by: $H00t!N& $t@r on April 28, 2011, 08:27:28 pm
Can you please post the paper?
Here it is: http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w10_qp_23.pdf (http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w10_qp_23.pdf)
Post by: Vin on April 28, 2011, 09:03:09 pm
Let oxi. no. of S be x
H2S => +2 + x = 0
sum of all oxidation states of atoms 0
so x = -2
SO2 => x + 2(-2)
x = +4
S => for any atom its 0
SO2 is reduced from +4 to 0
^This is what I did when i first saw the ques. After checking the ms I read the question again, however it clearly says "the oxidation numbers of each of the sulfur-containing substances in this reaction? "
which should be 0
and not oxi. no. of Sulfur in these compounds :/ If you get what I mean.
Sorry to complicate things all over again, I'm confused myself. Lets wait for someone smarter to conform
Post by: $H00t!N& $t@r on April 28, 2011, 09:15:14 pm
2 H2S + 3O2 -> 2SO2 + 2H2O
Let oxi. no. of S be x
H2S => +2 + x = 0
sum of all oxidation states of atoms 0
so x = -2
SO2 => x + 2(-2)
x = +4
S => for any atom its 0
SO2 is reduced from +4 to 0
^This is what I did when i first saw the ques. After checking the ms I read the question again, however it clearly says "the oxidation numbers of each of the sulfur-containing substances in this reaction? "
which should be 0
and not oxi. no. of Sulfur in these compounds :/ If you get what I mean.
Sorry to complicate things all over again, I'm confused myself. Lets wait for someone smarter to conform
I understand what you mean... they want us to find out the oxidation number for the element attached to sulphur and not sulphur itself, but I thought that the oxidation number for S is -2 cuz it's right below oxygen.. :-\ and for the second one, SO2, why did you find the oxidation number for sulphur and not oxygen, the way you did for the first one, H2S... the last one was pretty simple :)
Post by: Sue T on April 30, 2011, 04:53:38 pm
3(c) The molecule dichlorocarbene, CCl2, can be produced under certain conditions. It is
highly unstable, reacting with water to produce carbon monoxide and a strongly acidic
solution.
(i) Suggest the electron arrangement in CCl2 and draw a dot-and-cross diagram
showing this. Predict the shape of the molecule.
th answer was
.. ..
Cl:C:Cl or Cl=C–Cl
+ –
bent or non-linear or angle = 100–140o
how do we arrive at this structure????
Post by: Amelia on April 30, 2011, 05:20:28 pm
i have a ques from paper 41 mj 10
3(c) The molecule dichlorocarbene, CCl2, can be produced under certain conditions. It is
highly unstable, reacting with water to produce carbon monoxide and a strongly acidic
solution.
(i) Suggest the electron arrangement in CCl2 and draw a dot-and-cross diagram
showing this. Predict the shape of the molecule.
th answer was
.. ..
Cl:C:Cl or Cl=C–Cl
+ –
bent or non-linear or angle = 100–140o
how do we arrive at this structure????
One lone pair, two bond pairs. Repulsion between lone pairs and bond pairs.
http://upload.wikimedia.org/wikipedia/commons/8/8c/Dichlorocarbene-from-MW-2001-2D.png
Post by: Vin on April 30, 2011, 08:19:21 pm
I understand what you mean... they want us to find out the oxidation number for the element attached to sulphur and not sulphur itself, but I thought that the oxidation number for S is -2 cuz it's right below oxygen.. :-\ and for the second one, SO2, why did you find the oxidation number for sulphur and not oxygen, the way you did for the first one, H2S... the last one was pretty simple :)
Okay, S8 is the molecule for sulfur, and is written as S in chemical eqaution, in the atomic form right? Hence, S is 0.
Ok, we know the oxidation state of oxygen, O, which is -2.
So, S + O2 = 0
x + 2 (-2) = 0
x = -4
Cuz to find the oxidation no. for any atom X in a compound you're supposed to know the other as well.
Post by: iluvme on May 01, 2011, 03:13:58 pm
Q4 f and g
Q5 f ii)
Thank You :)
Post by: EMO123 on May 01, 2011, 04:13:32 pm
November 2006,i am sory but can you tell us which paper
Q4 f and g
Q5 f ii)
Thank You :)
Post by: iluvme on May 01, 2011, 04:14:56 pm
i am sory but can you tell us which paper
Oh sorry Paper 2
Post by: SkyPilotage on May 01, 2011, 04:50:49 pm
November 2006,
Q4 f and g
Q5 f ii)
Thank You :)
Q4 f) First of all cracking alkanes forms alkanes or alkenes, since Z is unsaturated then its an alkene CnH2n.
You found the molecular mass in ii) as 42 g so you can do it by trial and error or : 12n+(1)x(2n)=42 ---> 14n=42 ---> n=3 so its C3H6 (propene)
g) whenever u show the monomer or a polymer, the molecule should be shown in the form of "ethene" so just show 2 molecules of propene (without double bond ofcourse) as the polymer....
Q5 f)ii) solubility of water relates to hydrogen bonds... So the product must have formed hydrogen bonds with water...(usully from the hydroxyl group OH-)
Post by: iluvme on May 01, 2011, 04:52:14 pm
Q4 f) First of all cracking alkanes forms alkanes or alkenes, since Z is unsaturated then its an alkene CnH2n.
You found the molecular mass in ii) as 42 g so you can do it by trial and error or : 12n+(1)x(2n)=42 ---> 14n=42 ---> n=3 so its C3H6 (propene)
g) whenever u show the monomer or a polymer, the molecule should be shown in the form of "ethene" so just show 2 molecules of propene (without double bond ofcourse) as the polymer....
Q5 f)ii) solubility of water relates to hydrogen bonds... So the product must have formed hydrogen bonds with water...(usully from the hydroxyl group OH-)
Q5 It doesn't have to do anything with water being polar? :-\
Post by: SkyPilotage on May 01, 2011, 04:55:23 pm
Q5 It doesn't have to do anything with water being polar? :-\Hydrogen bonds are types of permanent dipole-dipole forces which exists only if the molecule is polar!!!!! But a polar molecule may have Permanent dipole forces but no hydrogen bonds...So u must mention tht hydrogen bonds form!
Post by: iluvme on May 01, 2011, 05:16:10 pm
Besides I've got more doubts, :-[
November 2007 Paper 2
Q1 b, Q2 e, Q3 d, Q5 e and d
Post by: EMO123 on May 01, 2011, 05:40:53 pm
covalent bond is due to the attraction between the nuclei and electrons and there is an attraction electrostatic forces
Post by: EMO123 on May 01, 2011, 05:51:49 pm
460K is Al2Cl6
1150K is AlCl3
2 e)ii)
in the attachment
2 e)iii)
in the attachment
Post by: EMO123 on May 01, 2011, 06:10:04 pm
and
Reduction Product is H
Post by: EMO123 on May 01, 2011, 06:11:25 pm
there is no question 5 e :D
Besides I've got more doubts, :-[
November 2007 Paper 2
Q1 b, Q2 e, Q3 d, Q5 e and d
Post by: EMO123 on May 01, 2011, 06:24:12 pm
Step I
Reagent :- PI3
Condition :- Room Temperature
Step II
Reagent :- KCN in aqueous Ethanol
Condition :- Heat Under Reflux
Step III
Reagent :- HCl
Condition :- Heat Applied
Post by: NidZ- Hero on May 03, 2011, 06:39:45 am
Post by: SkyPilotage on May 03, 2011, 10:15:56 am
hey how is SO2 a gud reducing agentI dont get what you mean by how but SO2 reduces KMnO4 and potassium dichromate too..... while it gets oxidised.
Post by: kulsoom on May 03, 2011, 11:00:28 am
i needed some help in NMR spectrometry...as in how do u deduce the molecular formula of a compound from the given chemical shifts? i dont understand it at all...
please hellppp!!
Thank uuu!! :) :)
Post by: Arthur Bon Zavi on May 03, 2011, 05:43:45 pm
I dont get what you mean by how but SO2 reduces KMnO4 and potassium dichromate too..... while it gets oxidised.
Reducing agent gets oxidised. Oxidising agent gets reduced.
Post by: Arthur Bon Zavi on May 03, 2011, 05:46:00 pm
hey guys!
i needed some help in NMR spectrometry...as in how do u deduce the molecular formula of a compound from the given chemical shifts? i dont understand it at all...
please hellppp!!
Thank uuu!! :) :)
Do you owe CIE book of Brian Ratcliff, Helen Eccles, et al ? Page number 394.
Post by: red_911 on May 03, 2011, 05:55:21 pm
just one question ... if we compare silver chloride n sodium chloride, sodium chloride in almost ionic while silver chloride has covalent character ... explain it terms of ion polarisation the difference between both ? ???
Post by: Arthur Bon Zavi on May 03, 2011, 06:07:05 pm
hey fellas :D
just one question ... if we compare silver chloride n sodium chloride, sodium chloride in almost ionic while silver chloride has covalent character ... explain it terms of ion polarisation the difference between both ? ???
NaCl :
has ionic bonding;
giant ionic structure;
both elements face greater difference in electonegativies, so will bond more strongly thus structure is rigid.
AgCl :
holds polar covalent bonding;
less difference in electronegativities (recall the trend of electrinegativity across and down the periodic table), so will bond less strongly and hence sometimes covalently.
http://en.wikipedia.org/wiki/Ionic_polarization#Ionic_polarization
Post by: red_911 on May 03, 2011, 06:26:44 pm
NaCl :
has ionic bonding;
giant ionic structure;
both elements face greater difference in electonegativies, so will bond more strongly thus structure is rigid.
AgCl :
holds polar covalent bonding;
less difference in electronegativities (recall the trend of electrinegativity across and down the periodic table), so will bond less strongly and hence sometimes covalently.
http://en.wikipedia.org/wiki/Ionic_polarization#Ionic_polarization
yea u are right fidato .. n Thanks for replying .. but your answer was with reference to electronegativity difference, i want the answer by considering ion polarisation :-\
Post by: SkyPilotage on May 03, 2011, 07:15:55 pm
yea u are right fidato .. n Thanks for replying .. but your answer was with reference to electronegativity difference, i want the answer by considering ion polarisation :-\Well, this is because The silver ion is smaller than the sodium ion and has a higher charge density so is able to attract the electrons from the chloride ion closer to it ( we say the anion has been polarised) hence showing more covalent character than sodium chloride. This is also the same for the bromides or iodides but not the same in the flourides because flourine has a smaller radius so it is much harder to move electrons away from it due to the nuclear attraction to the electrons...
Post by: red_911 on May 03, 2011, 07:28:36 pm
Well, this is because The silver ion is smaller than the sodium ion and has a higher charge density so is able to attract the electrons from the chloride ion closer to it ( we say the anion has been polarised) hence showing more covalent character than sodium chloride. This is also the same for the bromides or iodides but not the same in the flourides because flourine has a smaller radius so it is much harder to move electrons away from it due to the nuclear attraction to the electrons...but i checked on many websites, sodium has a smaller ionic radius than silver :S
Post by: SkyPilotage on May 03, 2011, 08:02:52 pm
but i checked on many websites, sodium has a smaller ionic radius than silver :SI mean inside the compound, silver has a smaller ionic radii than the sodium ion in NaCl....This is only shown in chlorides and bromides..
By the way I think bond polarisation isnt required in As? As only required you to know abt Aluminum?
Post by: Arthur Bon Zavi on May 05, 2011, 10:46:42 am
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w06_qp_1.pdf
Post by: Master_Key on May 05, 2011, 12:15:49 pm
Question 1 please.
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w06_qp_1.pdf
Answer is A.
required 25 cm3 of 1.0 × 10–2 mol dm–3 potassium hydroxide for complete neutralisation.
moles = .025*1.0 × 10–2
=2.5 × 10-4 moles.
Ca2+ = 2H+
= 2.5 × 10-4 * 0.5 moles of Ca2+.
= 1.25 × 10-4 moles of Ca2+ in .05 dm3.
C = m/v
1.25 × 10-4
C = ---------------
0.05
C = 2.5 × 10-3
Post by: Vin on May 05, 2011, 12:26:02 pm
Post by: Arthur Bon Zavi on May 05, 2011, 03:33:47 pm
Answer is A.
required 25 cm3 of 1.0 × 10–2 mol dm–3 potassium hydroxide for complete neutralisation.
moles = .025*1.0 × 10–2
=2.5 × 10-4 moles.
Ca2+ = 2H+
= 2.5 × 10-4 * 0.5 moles of Ca2+.
= 1.25 × 10-4 moles of Ca2+ in .05 dm3.
C = m/v
1.25 × 10-4
C = ---------------
0.05
C = 2.5 × 10-3
I didn't see the ratio as 2 : 1 and so didn't divide. Thanks.
Post by: EMO123 on May 06, 2011, 05:35:15 am
M/J 2002 P1 Q32 :)
Post by: Arthur Bon Zavi on May 06, 2011, 05:53:19 am
i got one fantastic doubt
M/J 2002 P1 Q32 :)
H2O (2 bonding pairs, 2 lone pairs) = 104.5o
CH4 (4 bonding pairs) = 109.5o
NH3 (3 bonding pairs, 1 lone pair) = 107o
SF6 (6 bonding pairs. S has 12 electrons now, after bonding) = 90o
CO2 (linear) = 180o
BF3 (3 bonding pairs. Look at the number of electrons in the outer shell of B after bonding !) = 120o
Do the rest yourself, you may find answer as D.
Post by: EMO123 on May 06, 2011, 08:01:51 am
H2O (2 bonding pairs, 2 lone pairs) = 104.5oare you shure :-\
CH4 (4 bonding pairs) = 109.5o
NH3 (3 bonding pairs, 1 lone pair) = 107o
SF6 (6 bonding pairs. S has 12 electrons now, after bonding) = 90o
CO2 (linear) = 180o
BF3 (3 bonding pairs. Look at the number of electrons in the outer shell of B after bonding !) = 120o
Do the rest yourself, you may find answer as D.
Post by: I like cheese on May 06, 2011, 08:58:09 am
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9701%20-%20Chemistry/&file=9701_w05_qp_1.pdf
(Oct/Nov 2005 paper 1)
qn. 7,12,16,25
-----------------------------------------------------------------------------------------------------------------------------------------
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9701%20-%20Chemistry/&file=9701_w04_ms_1.pdf
(Oct/Nov 2004 paper 1)
Qns. 3,7,10,12,18, and why isnt the answer to Qn 27 Option B ???
-----------------------------------------------------------------------------------------------------------------------------------------
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9701%20-%20Chemistry/&file=9701_s06_qp_1.pdf
(May/June 2006 paper 1)
Qn 2,38
-----------------------------------------------------------------------------------------------------------------------------------------
not asking someone to do it all but if a couple of people could do 1 or 2 qns at a time that would be awesome ;D ;D ;D ;D ;D ;D ;D ;D :P
Post by: Master_Key on May 06, 2011, 01:59:56 pm
So ive done almost all the past papers...and now the qns which i dont know how to do have piled up
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9701%20-%20Chemistry/&file=9701_w05_qp_1.pdf
(Oct/Nov 2005 paper 1)
qn. 7,12,16,25
-----------------------------------------------------------------------------------------------------------------------------------------
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9701%20-%20Chemistry/&file=9701_w04_ms_1.pdf
(Oct/Nov 2004 paper 1)
Qns. 3,7,10,12,18, and why isnt the answer to Qn 27 Option B ???
-----------------------------------------------------------------------------------------------------------------------------------------
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9701%20-%20Chemistry/&file=9701_s06_qp_1.pdf
(May/June 2006 paper 1)
Qn 2,38
-----------------------------------------------------------------------------------------------------------------------------------------
not asking someone to do it all but if a couple of people could do 1 or 2 qns at a time that would be awesome ;D ;D ;D ;D ;D ;D ;D ;D :P
(Oct/Nov 2005 paper 1)
qn. 7,12,16,25
7. Strong acid-strong base will have maximum negative change in enthalpy and Weak-Strong will have lower. So ans is A.
12. MgO = 1:1
Al2O3 = 2:3
SO2= 1:2
Plot your graph and you will get D.
16.Colourless gas = CO2 So a carbonate is sure. MgO and Bao react vigorously.
25. 2,4-dnph reacts with aldehydes.
Post by: I like cheese on May 06, 2011, 02:30:37 pm
Quote
(Oct/Nov 2005 paper 1)
qn. 7,12,16,25
7. Strong acid-strong base will have maximum negative change in enthalpy and Weak-Strong will have lower. So ans is A.
12. MgO = 1:1
Al2O3 = 2:3
SO2= 1:2
Plot your graph and you will get D.
16.Colourless gas = CO2 So a carbonate is sure. MgO and Bao react vigorously.
25. 2,4-dnph reacts with aldehydes.
ok thanks man but ummmm for qn 25, in both options A and D , there is 1 aldehyde......so which one??? and does a carboxylic acid react with PCl5??
Post by: EMO123 on May 06, 2011, 03:21:35 pm
2,4-dinitrophenaylhydrazine also reacts with ketones to give a deep yellpw pr orange precipitate
there fore
y is a aldehyde
so there is only one aldehyde not ketone also present in that combination
:. D is the right answer
hope i helped
Post by: EMO123 on May 06, 2011, 03:44:42 pm
q-3
10cm3 of CH3SH
60cm3 of 3O2
so 70cm3 of total solution
now as water is already water at room temprature
so remove it from it
so you will get 50cm3
hope it works for you
Post by: EMO123 on May 06, 2011, 04:06:33 pm
Post by: EMO123 on May 06, 2011, 04:10:45 pm
q-10
as there are 2 molecules of H
:. it is (0.20*1)/1.1
and 1 molecule of oxygen
:. it is (0.10*1)/1.1
:. answer is D
Post by: EMO123 on May 06, 2011, 04:17:56 pm
q-12
the energy is added more to increase the proportion of molecules
this energy is activation energy
i think it is not right anyone plzz check this answer
Post by: EMO123 on May 06, 2011, 06:31:44 pm
from
(Oct/Nov 2004 paper 1)
q-18,27
(May/June 2006 paper 1)
q-2,38
Post by: Master_Key on May 07, 2011, 07:31:10 am
master_key
2,4-dinitrophenaylhydrazine also reacts with ketones to give a deep yellpw pr orange precipitate
there fore
y is a aldehyde
so there is only one aldehyde not ketone also present in that combination
:. D is the right answer
hope i helped
There are no ketone groups in that question so i mentioned "Aldehyde".
2,4-dnph reacts with Carbonyl compounds.
Post by: EMO123 on May 07, 2011, 10:31:24 am
There are no ketone groups in that question so i mentioned "Aldehyde".ya you are right but it was just for the reference
2,4-dnph reacts with Carbonyl compounds.
Post by: TJ-56 on May 13, 2011, 05:09:53 pm
I plotted the mass of CuO on the y-axis against the mass of the basic carbonate on the x-axis. Now the 2 values to calculate the gradient are 5.70g (basic carbonate, x-axis) and 4.00g (CuO, y-axis). So how exactly do we use these numbers to calculate x ?
Appreciate any help.
Post by: thecandydoll on May 14, 2011, 07:42:59 am
How do you fill in the table?
DO we need to find conc in solvent a and b?
If so how?
Post by: noor.h92 on May 14, 2011, 12:23:48 pm
has anyone done the 2011 biology paper 3. (As-level). In question 1ai- where you have to draw a graph what would be my x-axis and y-axis?
Post by: elemis on May 14, 2011, 12:27:24 pm
Hi,
has anyone done the 2011 biology paper 3. (As-level). In question 1ai- where you have to draw a graph what would be my x-axis and y-axis?
2011.... You from the future ?
Also, does this look like the CIE Biology thread ?
Post by: TJ-56 on May 14, 2011, 01:42:34 pm
How do we calculate the value of x in May June 2008 Paper 5 question 2 (f) ?Anyone?
I plotted the mass of CuO on the y-axis against the mass of the basic carbonate on the x-axis. Now the 2 values to calculate the gradient are 5.70g (basic carbonate, x-axis) and 4.00g (CuO, y-axis). So how exactly do we use these numbers to calculate x ?
Appreciate any help.
I also have a question in the October November 2009 paper: question 1 (d) (table), I found it really confusing, and the ms didnt help a lot. Can someone explain?
Post by: Master_Key on May 14, 2011, 03:52:54 pm
Ionic Bonding Poem - a snippet of chemical poetry
(anon Y11 student, Whitby Community College, Oct 31st 2002)
How do I long for a full outer shell!
being chlorine having seven, is a horrid hell
but my name is sodium and I have one spare!
I want to lose it, can we not share?
No? for are we not a perfect match
chuck it to me, I promise to catch
then we can live our separate ways
and live with full shells to the end of our days!
and so our tale comes to an end
as positive and negative we shall remain friends
Post by: HUSH1994 on May 15, 2011, 06:59:19 am
Post by: EMO123 on May 15, 2011, 07:04:41 am
A short poem.good poem man
Ionic Bonding Poem - a snippet of chemical poetry
(anon Y11 student, Whitby Community College, Oct 31st 2002)
How do I long for a full outer shell!
being chlorine having seven, is a horrid hell
but my name is sodium and I have one spare!
I want to lose it, can we not share?
No? for are we not a perfect match
chuck it to me, I promise to catch
then we can live our separate ways
and live with full shells to the end of our days!
and so our tale comes to an end
as positive and negative we shall remain friends
Post by: Master_Key on May 15, 2011, 08:30:54 am
Nice one master key, +rep :)Thank you.
Post by: EMO123 on May 15, 2011, 10:02:40 am
Quote
Esters are a useful group of compounds due to their distinctive smells. One exampleneed the answer of D)
of an ester is ethyl ethanoate, its formation is shown below.
CH3COOH(aq) + C2H5OH(aq)?CH3COO C2H5(aq)
+ H2O(l)
a) Systems like this are described as being a 'dynamic equilibrium'. Explain
the term 'dynamic equilibrium'
b) Write down the expression for the equilibrium constant, Kc, for this reaction.
c) Calculate the value of Kc for this reaction given the equilibrium concentrations below.
[CH3COOH] = 0.08 moldm-3
[C2H5OH] = 0.08 moldm-3
[CH3COO C2H5] = 0.25 moldm-3
[H2O] = 0.1 moldm-3
d) Concentrated sulphuric acid is added to the reaction mixture as it removes water molecules. What effect would this have on the equilibrium position of this system?
Post by: EMO123 on May 15, 2011, 10:09:32 am
Quote
2. The Haber-Bosch process is used for the large-scale production of ammonia fromi have doubt in B)
nitrogen and hydrogen gas. The reaction is shown below:
N2(g) + 3H2(g)?2NH3(g)
a) Write an expression for Kp for this reaction.
b) What are the units of Kp for this reaction? (Assume pressure is measured in kPa)
c) When the temperature is raised for this process the proportion of NH3(g) in the mixture decreases. Explain this observation.
d) What effect will an increase in the total pressure have on the equilibrium position?
Post by: aneeta on May 15, 2011, 12:33:26 pm
Post by: aldehyde1612 on May 15, 2011, 01:50:46 pm
OMG OMG!!! IM FREAKING OUT ABOUT PAPER 2 TOMORROW!! x_x ORGANIC IS KILLING ME!! ANY SUGGESTIONS?!just relax.. and stop freaking out!
I suggest you dont study any chem for now... just quickly skim over your short notes and stuff if you havent done so already. THEN STOP. listen to music and take a walk .. or something, anything and STOP FREAKING OUT. You'll be fine.
And if tomoro it gets too hectic.. just breaaathe. ;D
Post by: aldehyde1612 on May 15, 2011, 03:23:38 pm
need the answer of D)
Wouldnt it move the position of equilibrium to the right (i.e. favouring the forward reaction), cause H2O is removed and according to LeChat's Principle the system works to counteract that- thus the forward reaction as it produces more H2O.
Post by: EMO123 on May 15, 2011, 03:25:37 pm
Wouldnt it move the position of equilibrium to the right (i.e. favouring the forward reaction), cause H2O is removed and according to LeChat's Principle the system works to counteract that- thus the forward reaction as it produces more H2O.thanxx
Post by: aneeta on May 15, 2011, 04:11:02 pm
just relax.. and stop freaking out!
I suggest you dont study any chem for now... just quickly skim over your short notes and stuff if you havent done so already. THEN STOP. listen to music and take a walk .. or something, anything and STOP FREAKING OUT. You'll be fine.
And if tomoro it gets too hectic.. just breaaathe. ;D
Noooooooooo>< I'm like dying.... Organic :-[ ufff. I need to get like really good in p2 cuz p1 sucked. Well I didn't get enough time to finish everything!! And p1 is going to be bad!! I don't even want to think of that!! -______-
Post by: EMO123 on May 16, 2011, 09:12:08 am
Noooooooooo>< I'm like dying.... Organic :-[ ufff. I need to get like really good in p2 cuz p1 sucked. Well I didn't get enough time to finish everything!! And p1 is going to be bad!! I don't even want to think of that!! -______-https://studentforums.biz/revison-notes/%28notes%29-acash09%27s-chemistry-notes-%28cie-as-level%29-%28notes%29/
hope i helped
Post by: aneeta on May 16, 2011, 12:43:47 pm
Post by: iluvme on May 16, 2011, 12:49:50 pm
How was it for you?
Post by: EMO123 on May 16, 2011, 12:51:29 pm
Post by: aneeta on May 16, 2011, 12:51:52 pm
Went fine not bad as I hoped. :)
How was it for you?
Terrible=( what variant did you get?
Post by: elemis on May 16, 2011, 12:53:04 pm
Went fine not bad as I hoped. :)
How was it for you?
I did pretty darn good. Maximum 4 marks gone down the drain.
I'm pretty happy actually !
Post by: iluvme on May 16, 2011, 12:56:35 pm
Terrible=( what variant did you get?
22.
Quote
I did pretty darn good. Maximum 4 marks gone down the drain.
I'm pretty happy actually !
Made some really silly mistakes. Might lose about 10 or so.
Post by: EMO123 on May 16, 2011, 01:04:19 pm
Post by: Master_Key on May 16, 2011, 02:06:53 pm
Post by: Vin on May 16, 2011, 05:32:54 pm
Post by: Aadeez || Zafar on May 16, 2011, 08:35:46 pm
Post by: TheLonelyIsland on May 17, 2011, 01:31:28 am
expecting threshold to be uphow high are you expecting them?
Post by: Arthur Bon Zavi on May 17, 2011, 12:11:34 pm
Post by: TJ-56 on May 19, 2011, 04:44:43 pm
Thanks in advance and I really appreciate the help.
Post by: EMO123 on May 19, 2011, 04:48:59 pm
Regarding the E(cell) values, how do we know if a reaction can occur or not? For example: mj02, q4(c): (i)no reaction occurs, well how can we decide so? and (ii) a reaction occurs. Is there like a 'least value of an E(cell) of a reaction to occur'?. And also, can someone please explain which value we subtract from which in calculating this E(cell) value? I notice sometimes in every paper its different and I have found it really confusing.which year or which variant
Thanks in advance and I really appreciate the help.
Post by: TJ-56 on May 19, 2011, 04:53:07 pm
which year or which variantThanks for trying to help. Well I just posted an example: May June 2002 Paper 4 Q4 (c), but as I wrote I found the topic a bit confusing.
Post by: EMO123 on May 19, 2011, 04:53:57 pm
Thanks for trying to help. Well I just posted an example: May June 2002 Paper 4 Q4 (c), but as I wrote I found the topic a bit confusing.its k i will try to do that
Post by: aldehyde1612 on May 19, 2011, 06:19:17 pm
can someone explain? .. like how we're supposed to do question 3.d. from May 2010 p42.
ANY helps appreciated.
***Thanks for the overwhelming amount of help you guys gave. lol jokes. i figured it out myself.
Post by: TJ-56 on May 20, 2011, 04:14:47 pm
In October November 2005 Paper 4 Question 3 (c) (i): I got the H2O2 part, but not the KI. How do we know which to use: the K+ or I- half equation? This may be silly, but I am a bit confused :S
Thanx in adv
Post by: da_rockstar on May 21, 2011, 05:09:02 pm
cuz i was going through the application booklet...and i saw in oil slick part that they use fluorinated molecules to repel water and absorb oil in sorbent boom.
but i had doubts over it since as far as i remember fluorine forms the strongest hydrogen bonds doesn't it?
Post by: leebux101 on May 21, 2011, 10:01:10 pm
for those who are doing A2 this year, do we have to revise the AS stuff as well???
Post by: Arthur Bon Zavi on May 21, 2011, 10:11:19 pm
URGENT
for those who are doing A2 this year, do we have to revise the AS stuff as well???
Definitely.
Post by: TJ-56 on May 21, 2011, 10:16:07 pm
Definitely.
All of the topics?
I mean I have solved all papers from 2002 till 2010, but there are many AS topics that are not mentioned or required for the A2.
Post by: leebux101 on May 21, 2011, 10:32:13 pm
All of the topics?yeah i agree with TJ-56, "definitely" is the wrong answer.... for sure.
I mean I have solved all papers from 2002 till 2010, but there are many AS topics that are not mentioned or required for the A2.
theres about 5 marks in total only for AS organic Chem, is anybody studying AS organic then?
Post by: Arthur Bon Zavi on May 21, 2011, 10:38:27 pm
yeah i agree with TJ-56, "definitely" is the wrong answer.... for sure.
theres about 5 marks in total only for AS organic Chem, is anybody studying AS organic then?
He didn't mention all topics, he mentioned AS topics. If I would have been there, I would revise all the AS topics atleast once.
Post by: TJ-56 on May 21, 2011, 11:29:59 pm
He didn't mention all topics, he mentioned AS topics. If I would have been there, I would revise all the AS topics atleast once.
Maybe it's best if we did so, but to be honest, it's easier said than done, plus, paper 4 is just a few days away and hence there's barely enough time to go through the past papers.
Post by: leebux101 on May 21, 2011, 11:37:13 pm
Post by: Arthur Bon Zavi on May 22, 2011, 08:32:07 am
yeah ur ryt fidato, but as Tj-56 said, theres barely time to revise all of A2 chem... i guess i will go over some AS organic chem though...
Yes, and also if you can, properties and reactions of group I, II and VII.
Post by: TJ-56 on May 24, 2011, 09:36:53 am
Thanx in advance
Post by: TJ-56 on May 24, 2011, 07:40:54 pm
Post by: Aadeez || Zafar on May 24, 2011, 07:42:00 pm
I still think its not fair that there are two variants for every paper, and with much different difficulties. Comparing Paper 41 (A threshold 66) and Paper 42 (threshold 54), 42 is suicide. During the exam this difficulty can effect the rest of the 'easy' questions as well due to frustration and the inability to solve the 'harder' questions. God help us tomorrow, and I hope it'll be an easy paper(s), as my preparation is a disaster, despite countless hours of study, this paper is a walk in hell.is a2 harder than as
Post by: TJ-56 on May 24, 2011, 07:44:56 pm
is a2 harder than asYeah, something like that.
Post by: Aadeez || Zafar on May 24, 2011, 07:47:58 pm
Yeah, something like that.is it much harder or a bit
hw much did u scored in AS level chem
Post by: leebux101 on May 24, 2011, 07:48:39 pm
I still think its not fair that there are two variants for every paper, and with much different difficulties. Comparing (MJ10) Paper 41 (A threshold 66) and Paper 42 (threshold 54), 42 is suicide. During the exam this difficulty can effect the rest of the 'easy' questions as well due to frustration and the inability to solve the 'harder' questions. God help us tomorrow, and I hope it'll be an easy paper(s), as my preparation is a disaster, despite countless hours of study, this paper is a walk in hell.lol! when i read this, i thought i was the one writing this...
i sooooo agree with u. tomorows papers is going down the drain... sooooooooooo much left and i am chilling on the laptop... :(
Post by: TJ-56 on May 24, 2011, 07:51:14 pm
is it much harder or a bit
hw much did u scored in AS level chem
To be honest, I'm finding it MUCH harder, but maybe it's due to the fact that I'm self studying, maybe if there was a good A2 teacher here, I would have found it easier.
I scored a B in chemistry, the worst of my AS grades, and hence I must do really really well if I will have to raise it to an A, but I hardly see that coming, in fact I'm praying for a B right now.
Post by: TJ-56 on May 24, 2011, 07:55:06 pm
lol! when i read this, i thought i was the one writing this...Lol, I know! I think this exam is the least exam I have prepared for, in like, well since starting the IGCSE's. That's not the problem, it's that I HAVE prepared, ALOT! But suddenly now everything Chemistry related text I read, It's like pure gibberish.
i sooooo agree with u. tomorows papers is going down the drain... sooooooooooo much left and i am chilling on the laptop... :(
Post by: AN10 on May 24, 2011, 09:12:22 pm
Post by: TJ-56 on May 24, 2011, 09:48:17 pm
Mj 10 41 Q9 c iii, For propanoic acid can we write COOH proton instead of OH proton. Please help I am confused. :-\I'm not sure, it is the 'OH' that is responsible for the peak, so..
I think it's the same, but you never know with CIE, so I think its better to be safe than sorry.
Gd luck.
Post by: leebux101 on May 24, 2011, 11:29:10 pm
Lol, I know! I think this exam is the least exam I have prepared for, in like, well since starting the IGCSE's. That's not the problem, it's that I HAVE prepared, ALOT! But suddenly now everything Chemistry related text I read, It's like pure gibberish.i know right!! .. tell me about it! anyway God help us tomorow.. Inshallah.
Post by: iluvme on May 25, 2011, 10:26:08 am
Please help!
Post by: EMO123 on May 25, 2011, 10:39:51 am
November 2007 Paper1 Q 5, 9, 21 and 32.5) answer is A
Please help!
There is always a hydrogen bonding between to OH groups
so A
Post by: EMO123 on May 25, 2011, 10:43:46 am
November 2007 Paper1 Q 5, 9, 21 and 32.in 32 i understand the 2and three but i dont understand 1
Please help!
so no answer for that i can give
anyone help us
Post by: Amelia on May 25, 2011, 06:04:09 pm
November 2007 Paper1 Q 5, 9, 21 and 32.
Please help!
Q5 - as you know already is A.
Q9 - It's B. The 50 cm3 metal salt reacted with 25 cm3, so the oxidation state will be 1:1. The orginal oxidation state was +3, hence, the new state will be +2.
Q21 - It's A. (Ch3)4Pb, supplies methyl radicals. And we do require methyl radicals to react with chlorine to give off the products.
Q32 - It's A. All three options are right. *Carbon monoxide is a strong reducing agent because it is easily oxidised to carbon dioxide - where the oxidation state is the more thermodynamically stable +4.
Post by: iluvme on May 25, 2011, 06:14:38 pm
Though I yet didn't get Q9 :/
Post by: aneeta on May 27, 2011, 12:21:49 am
Post by: EMO123 on May 27, 2011, 04:13:23 am
Does any one know where I can get chemistry past papers, paper 1 2000?of Cie AS levels is it there
i dont think so
Quote
Thanks Lia and Emo. Smileyohh i also dint get that so not posted
Though I yet didn't get Q9 :/
Post by: iluvme on May 27, 2011, 11:07:01 am
June 2009, Paper 1
10
11
15
18
20
29
30
38 and
40
Thank You :)
Post by: iluvme on May 27, 2011, 11:18:42 am
WOAH ! That's a lot. I'll see if I can provide you with some hints.
:) Thanks.
Post by: elemis on May 27, 2011, 11:31:35 am
A catalyst serves to LOWER the activation energy... does that help ?
Question 40
Alkaline Hydrolysis gives you the Sodium salt and the alcohol
Question 38
To determine if Option 1 is correct, look at the double bonds and the alcohol group at the end. KMnO4 would oxidise the alcohol to what ? An aldehyde or a ketone ? If you cant tell, then look at the no. of carbons the hydroxyl group is joined to.
Next, look at the double bonds. Can you see a place where a secondary alcohol (which upon further oxidation would form a ketone) could forrm as a result of oxidation by KMnO4 ?
Option 2 should be easy to prove/disprove.
Option 3 - Hint: Can you see any carboxylic acid groups ?
Post by: elemis on May 27, 2011, 11:37:07 am
It is a liquid at room temperature and atmospheric pressure.
Ethane is a gas and the others are liquids.
It does not mix completely with water.
To mix with water you need to be able to form hydrogen bonds. This statement implies that the mystery compound DOES NOT form hydrogen bonds.
It does not decolourise acidified potassium manganate(VII).
The mystery compound will not undergo oxidation.
Question 29
There is no easy way to solve these kind of questions. Just draw out the structures and see if it helps.
Post by: iluvme on May 27, 2011, 12:40:28 pm
Question 10
A catalyst serves to LOWER the activation energy... does that help ?
Question 40
Alkaline Hydrolysis gives you the Sodium salt and the alcohol
Question 38
To determine if Option 1 is correct, look at the double bonds and the alcohol group at the end. KMnO4 would oxidise the alcohol to what ? An aldehyde or a ketone ? If you cant tell, then look at the no. of carbons the hydroxyl group is joined to.
Next, look at the double bonds. Can you see a place where a secondary alcohol (which upon further oxidation would form a ketone) could forrm as a result of oxidation by KMnO4 ?
Option 2 should be easy to prove/disprove.
Option 3 - Hint: Can you see any carboxylic acid groups ?
Q 10, Energy is more at Ea1, which means activation energy is more? so uncatalysed reaction. Righto
Q 40, I was confused about the word partial. :/
Q 38, Right right right, Didn't notice that part, about COOH being absent.
Post by: iluvme on May 27, 2011, 12:46:16 pm
Question 30
It is a liquid at room temperature and atmospheric pressure.
Ethane is a gas and the others are liquids.
It does not mix completely with water.
To mix with water you need to be able to form hydrogen bonds. This statement implies that the mystery compound DOES NOT form hydrogen bonds.
It does not decolourise acidified potassium manganate(VII).
The mystery compound will not undergo oxidation.
Question 29
There is no easy way to solve these kind of questions. Just draw out the structures and see if it helps.
Right, got both. :)
Though Q 29, I got mixed up with the (CH3)2C(OH) part.
Thanks Ari. :)
Post by: HUSH1994 on May 27, 2011, 01:31:52 pm
Question 10+rep,man your explanations are really amazing :)
A catalyst serves to LOWER the activation energy... does that help ?
Question 40
Alkaline Hydrolysis gives you the Sodium salt and the alcohol
Question 38
To determine if Option 1 is correct, look at the double bonds and the alcohol group at the end. KMnO4 would oxidise the alcohol to what ? An aldehyde or a ketone ? If you cant tell, then look at the no. of carbons the hydroxyl group is joined to.
Next, look at the double bonds. Can you see a place where a secondary alcohol (which upon further oxidation would form a ketone) could forrm as a result of oxidation by KMnO4 ?
Option 2 should be easy to prove/disprove.
Option 3 - Hint: Can you see any carboxylic acid groups ?
Post by: EMO123 on May 27, 2011, 01:48:01 pm
I've got more doubts, Can somebody please explain?Question 11
June 2009, Paper 1
10
11
15
18
20
29
30
38 and
40
Thank You :)
3N2 is not really much unreactive gas but in this case they are unreactive gas
but my doubt is from the last two equations which is correct
plzz explain
i also need help in Q 15
in Question 18
i got the residue as Fe(OH)2
but i dint get the gas as NH3 i got it as SO2
and also need help in Q 20
Post by: HUSH1994 on May 27, 2011, 06:30:46 pm
Q20 : the compound has 3 double bonds,therefore you use 2 to the power of 3 which is equal to 8.
Post by: EMO123 on May 28, 2011, 06:59:32 am
Question 11in Q no. 11
3N2 is not really much unreactive gas but in this case they are unreactive gas
but my doubt is from the last two equations which is correct
plzz explain
i also need help in Q 15
in Question 18
i got the residue as Fe(OH)2
but i dint get the gas as NH3 i got it as SO2
and also need help in Q 20
as said that only the heat is evolved and and an unreactive gas i produced
that means that it just forms an unreactive gas as it is the same o the option C is the correct answer
in Q no. 15
at some extend you are right only hush
and one more thing in it is that ions are always in aqueous forms so they are soluble so the answer for this is A
in Q no. 18
it is that as it is given that and alkaline solution is formed so it can be only NH3
and for the solid residue is Fe(OH)2 because it is given that one cation is Fe2+ and the other thing given is it is soluble in water so it will form a hydroxide so the answer for it is C.
hope it helps
and well triad Hush
Post by: EMO123 on May 28, 2011, 07:04:34 am
may be this helps
http://www.chemguide.co.uk/basicorg/isomerism/geometric.html#top
Post by: EMO123 on May 28, 2011, 07:53:31 am
in Q15,the AgCl is insoluble,so a reaction with a more reactive metal causes Ag's displacement(from AgCl to NaCl).now i got you thanxx Hush
Q20 : the compound has 3 double bonds,therefore you use 2 to the power of 3 which is equal to 8.
after i tryed it all the ways i got you what you were meaning so thanxx hush
Post by: HUSH1994 on May 28, 2011, 09:36:33 am
Post by: Dania on May 28, 2011, 04:26:45 pm
a. 2.91
b. 3.86
c. 4.55
d. 4.74
e. 4.94
Post by: iluvme on May 29, 2011, 07:18:55 pm
Which alcohol gives only one oxidation product when warmed with dilute acidified potassium
dichromate(VI)?
A butan-1-ol
B butan-2-ol
C 2-methylpropan-1-ol
D 2-methylpropan-2-ol
The answer is B
Post by: The Golden Girl =D on May 29, 2011, 07:33:39 pm
Could answer this question and explain it for me please?
Which alcohol gives only one oxidation product when warmed with dilute acidified potassium
dichromate(VI)?
A butan-1-ol
B butan-2-ol
C 2-methylpropan-1-ol
D 2-methylpropan-2-ol
The answer is B
I guess what they mean by one oxidation product is only one product from this oxidation ...if it's a primary Alcohol A , then it has two products => Aldehyde and carboxylic acid ...but if it's a secondary Alcohol B then it will give ONLY a Ketone which is why It's B
I hope I helped :)
Post by: iluvme on May 29, 2011, 07:57:25 pm
I guess what they mean by one oxidation product is only one product from this oxidation ...if it's a primary Alcohol A , then it has two products => Aldehyde and carboxylic acid ...but if it's a secondary Alcohol B then it will give ONLY a Ketone which is why It's B
I hope I helped :)
Got it. Thanks. :)
Post by: The Golden Girl =D on May 29, 2011, 08:00:56 pm
Post by: $H00t!N& $t@r on May 30, 2011, 02:09:06 pm
M/J 2008 q 6, 27 and 29
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_s08_qp_1.pdf (http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_s08_qp_1.pdf)
Post by: $H00t!N& $t@r on May 30, 2011, 03:05:55 pm
Post by: The Golden Girl =D on May 30, 2011, 03:15:29 pm
Post by: elemis on May 30, 2011, 03:21:18 pm
EDIT : Answering it now.
Post by: elemis on May 30, 2011, 03:28:19 pm
Using the formula density = mass/volume you should be able to determine the mass of ice heated.
Next, you must realise that ice is basically water... Can you calculate the no. of moles of water in 1 gram of ice ? Should be easy.
Now, use the formula PV=nRT.
Rearrange for Volume. Remember the volume you calculate will be in m3 NOT dm3.
Post by: elemis on May 30, 2011, 03:31:46 pm
Look at the double bonds in each option... that is where the water molecule was removed from.
Try drawing each structure with an added hydrogen on one of the carbons in the double bond and a OH on the other carbon.
Do not forget this point in the question :
is unreactive towards mild oxidising agents.
Its a hint... Think about it.
Post by: elemis on May 30, 2011, 03:33:37 pm
In this scenario, look for the carboxylic acid group.... COOH. Also, look for carbons with four different structures attached to them.
Post by: $H00t!N& $t@r on May 30, 2011, 04:16:23 pm
Dehydration results in the removal of a water molecule.
Look at the double bonds in each option... that is where the water molecule was removed from.
Try drawing each structure with an added hydrogen on one of the carbons in the double bond and a OH on the other carbon.
Do not forget this point in the question :
is unreactive towards mild oxidising agents.
Its a hint... Think about it.
Oh i see.. so its D because it is the result of dehydrating tertiary alcohol which is unreactive towards mild oxidising agents. ;D
Post by: $H00t!N& $t@r on May 30, 2011, 04:24:21 pm
The question says 1cm3 of ice is heated.
Using the formula density = mass/volume you should be able to determine the mass of ice heated.
Next, you must realise that ice is basically water... Can you calculate the no. of moles of water in 1 gram of ice ? Should be easy.
Now, use the formula PV=nRT.
Rearrange for Volume. Remember the volume you calculate will be in m3 NOT dm3.
For this one, I got 2.72 instead of 2.67 :-[
Post by: $H00t!N& $t@r on May 30, 2011, 04:25:27 pm
You should be able to eliminate two of the options straight away.
In this scenario, look for the carboxylic acid group.... COOH. Also, look for carbons with four different structures attached to them.
I understand it now. :)
Post by: elemis on May 30, 2011, 04:34:00 pm
For this one, I got 2.72 instead of 2.67 :-[
Yes, that's what I got too. The reason they put 2.67 is to probably test your confidence.
Post by: $H00t!N& $t@r on May 30, 2011, 04:37:32 pm
Yes, that's what I got too. The reason they put 2.67 is to probably test your confidence.
ah i see.. :D
Post by: SkyPilotage on May 30, 2011, 04:47:05 pm
Yes, that's what I got too. The reason they put 2.67 is to probably test your confidence.Hmm...Well There is another way to solve it...
-Its given that there are 24 dm3 in 298K
So whats the volume in 596K? Cross multiply to get 48 dm3...
-1 mole-------------->48 dm3 ( at 596K)
-X mole-------------->Y dm3 (at 596K)
To find out the number of moles of ice..We have the mass as 1g and molar mass as 18g..So cross-multiply to get 2.666666666666 dm3.....
I think using the PV=NRT doesnt give the exact answer since steam is NOT an ideal gas. :)
Post by: $H00t!N& $t@r on May 30, 2011, 08:11:28 pm
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w09_qp_12.pdf (http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w09_qp_12.pdf)
Post by: elemis on May 31, 2011, 06:01:05 am
i have another question.. Q28 October/November 2009 variant 12
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w09_qp_12.pdf (http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w09_qp_12.pdf)
Option B is correct since, when you draw it out (see below).
It is a rearrangement of the target molecule in the question. Just substitute the Br with a COOH group.
Post by: $H00t!N& $t@r on May 31, 2011, 08:37:57 am
Option B is correct since, when you draw it out (see below).
It is a rearrangement of the target molecule in the question. Just substitute the Br with a COOH group.
Thanks I get it :)
Post by: $H00t!N& $t@r on May 31, 2011, 08:42:27 am
October/November 2010 variant 12
Q 11, 12 and 40
http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w10_qp_12.pdf (http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9701%20-%20Chemistry/9701_w10_qp_12.pdf)
Post by: elemis on May 31, 2011, 12:16:29 pm
Y is a product. Initially there is no Y present. What does this tell you about the concentration at t=0 ?
Similarily, what concentration of X is present at t=0 ? You should now be able to eliminate 2 out the 4 options.
When equilibrium has been reached [X(aq)] has fallen to 0.25mol dm–3
Using that statement can you not calculate the moles of X and Y present at equilibrium using the given equation ?
Now calculate the Kc... It should all come together.
Question 12
This is something simple that you should be well aware of. Read your textbook.
Question 40
I'll give you a hint. Think about SN1 reactions and the formation of a carbocation....
Post by: Vin on May 31, 2011, 04:14:32 pm
Question:
A mass spectrum of magnesium shows three peaks at mass numbers 24, 25 and 26. The relative heights of these peaks are 60, 8 and 10 respectively. What is the mean atomic mass of Mg?
Response:
Wrong
24.1
Should have chosen
24.8
25.4
25.0
Post by: elemis on May 31, 2011, 04:55:54 pm
Post by: Vin on May 31, 2011, 06:09:31 pm
^Its a waste of time By the way. Qs are too easy
Post by: SkyPilotage on May 31, 2011, 08:37:44 pm
Score 0 of 1Well it should be done by 24x(60/78)+25x(8/78)+26x(10/78) But the answer is 24.4 :( Info should be enough to solve .... This question is similar to May June 2007 question 1....
Question:
A mass spectrum of magnesium shows three peaks at mass numbers 24, 25 and 26. The relative heights of these peaks are 60, 8 and 10 respectively. What is the mean atomic mass of Mg?
Response:
Wrong
24.1
Should have chosen
24.8
25.4
25.0
Post by: Aadeez || Zafar on May 31, 2011, 08:45:08 pm
Well it should be done by 24x(60/78)+25x(8/78)+26x(10/78) But the answer is 24.4 :( Info should be enough to solve .... This question is similar to May June 2007 question 1....i got it 24.35 and in the periodic tabel it is 24.3 , i think choosing the closest one would be best
and thus the ans would have been 24.1 as it it is the closest to 24.35, see the examinaer REPORT, somtinh would be mentioned there
Post by: SkyPilotage on May 31, 2011, 08:46:37 pm
i got it 24.35 and in the periodic tabel it is 24.3 , i think choosing the closest one would be bestlol I checked the website.... And the answer is 24.8...Definitely a mistake... :) Its none of the above Vin :P
and thus the ans would have been 24.1 as it it is the closest to 24.35, see the examinaer REPORT, somtinh would be mentioned there
Post by: Aadeez || Zafar on May 31, 2011, 08:49:26 pm
using the dat
i get atomic mass of Mg to be= 1900/78
no mistake in MS
Post by: SkyPilotage on June 01, 2011, 09:47:28 am
hw cn it be 24.8, we have to calculate the ans from data provide and not frmthe websitedude am talking abt the answer on the website is 24.8..I know its 24.35..Check my working above..
using the dat
i get atomic mass of Mg to be= 1900/78
no mistake in MS
I meant there is a mistake on the website answers thts all ;D
Post by: SkyPilotage on June 01, 2011, 12:22:02 pm
Question 21 : Confused between A and C....
Post by: EMO123 on June 01, 2011, 12:35:13 pm
November 2009 V1..Question 19but which paper
Question 21 : Confused between A and C....
Post by: SkyPilotage on June 01, 2011, 12:48:27 pm
but which paperwhat do you think? :P Paper 1 dude :)
Post by: angel_ak on June 01, 2011, 01:46:42 pm
NH4NO3 ----------------> N2O + 2H2O
what are the changes in the oxidation num of the two N atoms in NH4NO3 when this reaction proceeds?
1. -2, -4
2. +2, +6
3. +4, -6
4. +4, -4
Post by: SkyPilotage on June 01, 2011, 01:57:25 pm
ammonium nitrate NH4NO3, can decompose explosively when heated.Do it stepwise...
NH4NO3 ----------------> N2O + 2H2O
what are the changes in the oxidation num of the two N atoms in NH4NO3 when this reaction proceeds?
1. -2, -4
2. +2, +6
3. +4, -6
4. +4, -4
-Oxidation number of N in NH4+ is -3 because 4H have oxidation number of +4 ANd the overall charge is 1+...
-Oxidation Number of N in NO3- is +5 because 3O have oxidation number of -6 And the overall charge is 1-....
-Oxidation number of N in N20 is +1 because O has an oxidation number of -2 And the overall charge is 0...
So from -3 to +1 is +4 and from +5 to +1 is -4..Therefore the answer is D... :)
Post by: aloha32 on June 01, 2011, 03:23:37 pm
Often the first onez always seem to be wierd calculations onez as compared to the rest of the paper
Thank you! :)
Post by: angel_ak on June 01, 2011, 05:17:22 pm
0.200 mol of a hydrocarbon undergo complete combustion to give 35.2g of carbon dioxide and 14.4g of water as the only products. what is the molecular formula of the hydrocarbon?
1. C2H4
2. C2H6
3. C4H4
4. C4H8
Post by: EMO123 on June 01, 2011, 05:28:30 pm
THANKEW!! ;Dfor this we get a ratio of 1:2
0.200 mol of a hydrocarbon undergo complete combustion to give 35.2g of carbon dioxide and 14.4g of water as the only products. what is the molecular formula of the hydrocarbon?
1. C2H4
2. C2H6
3. C4H4
4. C4H8
but there are two options of 1:2 haydrocarbons
Post by: SkyPilotage on June 01, 2011, 05:33:50 pm
for this we get a ratio of 1:21:4 ratio...
but there are two options of 1:2 haydrocarbons
N of Co2 = 35.2/44 = 0.8
N of H20 = 14.4/18 = 0.8
So ratio of CxHy:CO2:H2O
1 : 4 : 4
So there are 4 C and 8 H and 12O so the Hydrocarbon iS C4H8...
Post by: Vin on June 01, 2011, 05:38:38 pm
What is the empirical formula of this fertiliser?
A NO3 B NHO3 C NH2O D N2H4O3
D. How and why? :/
Post by: SkyPilotage on June 01, 2011, 05:56:19 pm
18 A solid nitrate fertiliser reacts with an alkali to produce a gas which turns damp pH paper blue.Ammonium salts + alkalis -----> ammonia gas...
What is the empirical formula of this fertiliser?
A NO3 B NHO3 C NH2O D N2H4O3
D. How and why? :/
So its NH4NO3 which has the molecular formule N2H4O3...
Post by: EMO123 on June 01, 2011, 06:07:20 pm
1:4 ratio...i am sorry to say but if the ratio is 1:4 then it will be C4H16
N of Co2 = 35.2/44 = 0.8
N of H20 = 14.4/18 = 0.8
So ratio of CxHy:CO2:H2O
1 : 4 : 4
So there are 4 C and 8 H and 12O so the Hydrocarbon iS C4H8...
Post by: SkyPilotage on June 01, 2011, 06:08:44 pm
i am sorry to say but if the ratio is 1:4 then it will be C4H161:4 ratio As in 1 mole of CxHy to 4 mole of CO2 to 4 mole of H2O
Post by: EMO123 on June 01, 2011, 06:13:12 pm
1:4 ratio As in 1 mole of CxHy to 4 mole of CO2 to 4 mole of H2Ok now i got that you meant
4 mole of C a
and
4 mole of H2
right so it is 4(CH2)
i.e.C4H8
thanxx buddy
Post by: SkyPilotage on June 01, 2011, 06:14:15 pm
k now i got that you meantYou got it :D
4 mole of C a
and
4 mole of H2
right so it is 4(CH2)
i.e.C4H8
thanxx buddy
Post by: EMO123 on June 01, 2011, 06:17:35 pm
You got it :Dya i go it now
Post by: Vin on June 01, 2011, 08:03:12 pm
Ammonium salts + alkalis -----> ammonia gas...
So its NH4NO3 which has the molecular formule N2H4O3...
Oh yea the other options cant be salts :/ thanks man
another one. forgot how to do these. got it by common sense but i wanna know how to solve them (see attc.)
Post by: SkyPilotage on June 01, 2011, 08:18:01 pm
Oh yea the other options cant be salts :/ thanks manInitial N(moles) 4 ---> 0 : 0
another one. forgot how to do these. got it by common sense but i wanna know how to solve them (see attc.)
Final N(moles) 4-X ---> X : 0.5X
*PS: Since the ratio of 2NO2 and 2NO is 1:1 So Moles of NO increase by same amount as NO2 decreases....And O2 increases by half of the amount that NO2 decreases.
-Since its given that O2 is 0.8(final moles) 0.5X=0.8----> that means that NO2 decreases by 1.6 and NO increases by 1.6...Thats it. :)
Post by: Vin on June 01, 2011, 09:32:11 pm
Initial N(moles) 4 ---> 0 : 0
Final N(moles) 4-X ---> X : 0.5X
*PS: Since the ratio of 2NO2 and 2NO is 1:1 So Moles of NO increase by same amount as NO2 decreases....And O2 increases by half of the amount that NO2 decreases.
-Since its given that O2 is 0.8(final moles) 0.5X=0.8----> that means that NO2 decreases by 1.6 and NO increases by 1.6...Thats it. :)
Dude, I owe you my life. Respect. Ok i have many doubts. sorry. i still have a lot of papers to do so im not wasting my time on thinking too much on something i dont get. so here..
Post by: SkyPilotage on June 01, 2011, 09:48:25 pm
Dude, I owe you my life. Respect. Ok i have many doubts. sorry. i still have a lot of papers to do so im not wasting my time on thinking too much on something i dont get. so here..QUESTION 1:
-Its given that there are 24 dm3 in 298K
So whats the volume in 596K? Cross multiply to get 48 dm3...
-1 mole-------------->48 dm3 ( at 596K)
-X mole-------------->Y dm3 (at 596K)
To find out the number of moles of ice..We have the mass as 1g and molar mass as 18g..So cross-multiply to get 2.666666666666 dm3.....
I think using the PV=NRT doesnt give the exact answer since steam is NOT an ideal gas.
Question 2:
Find the percentage composition of N in the fertiliser which is 15grams of N in 100 grams of fertiliser-----> 0.15 (15%)
Find the mass Of N in 14 g of fertiliser -----> 0.15 x 14g = 2.1 grams
Find the number of moles of nitrogen ----> 2.1/14= 0.15 moles
Find the concentration (n/V) --------------> 0.15/5=0.03 moldm-3
I am so much glad to help :D
Post by: Vin on June 01, 2011, 10:15:28 pm
Q2 I missed 100 g :S Thanks again !
Thank you so much for taking out your time!
Post by: bulono on June 02, 2011, 03:15:39 am
TiO2+2c=Ti+2CO
enthalpy change of formation of TiO is -940 and for CO it is -110
why will we take both -940 and -110 as positive???
Post by: bulono on June 02, 2011, 03:24:54 am
Post by: Master_Key on June 02, 2011, 07:43:14 am
HELPP!!!IT is not taken positive. It is taken to reference of enthalpy change of formation.
TiO2+2c=Ti+2CO
enthalpy change of formation of TiO is -940 and for CO it is -110
why will we take both -940 and -110 as positive???
Post by: SkyPilotage on June 02, 2011, 09:38:11 am
Q1 I was using pv = nrt. :O You're right. thank you !Dude I am very glad to help Soo You DONT owe me anything!!! ;D
Q2 I missed 100 g :S Thanks again !
Thank you so much for taking out your time!
And By the way I keep wasting my time anyways So You are actually helping me to make some use of my time :P
Just be confiident and kill The MCQ paper tmrw and Gd luck to you and everyone else (and me :P) ;D
Post by: EMO123 on June 02, 2011, 10:08:37 am
but if serious than it was white and printed with black ink ;)
Post by: SkyPilotage on June 02, 2011, 10:41:36 am
the paper was a bit tricky and cooltricky and coool? That doesnt go along together :P
but if serious than it was white and printed with black ink ;)
I have mine in 2 hours :D
How did you do?
Post by: xlane on June 02, 2011, 03:13:57 pm
Post by: The Golden Girl =D on June 02, 2011, 05:12:27 pm
emo 123 what u can do is...write sth like ...the paper was quite easy and u all can sole it.....give them hope.
This is your first warning.
Respecting other members is one of the site's rules.
Post by: xlane on June 02, 2011, 09:22:24 pm
By the way u moderators should help students ......and demoralizing students does not fall in the category of helping.........so i suggest u do sth about it.
Post by: EMO123 on June 04, 2011, 10:50:45 am
i understand.....but doesnt demoralizing students get a warning......i guess no....because its not there is SF rule right ??i think you should wait for 24 hours after your paper or our paper because it is a rule you may have not read the rules or something
By the way u moderators should help students ......and demoralizing students does not fall in the category of helping.........so i suggest u do sth about it.
Post by: xlane on June 04, 2011, 02:58:16 pm
Post by: EMO123 on June 05, 2011, 06:45:49 pm
abbe yaaaar.......... :o :o...hw does sayin the paper was easy effect the 24 hour rule....lets say an A caliber student finds the paper easy where as a C caliber student finds the paper hard.....so sayin the paper is easy or hard does not effect any rules or regulations.......if u dont understand dat much..den :-\ :-\ :-\ :-\ :-\ :-\ya that not effection by that but no disscussion of Q/A as such it is over so we should our dissussion
Post by: Hissa on June 06, 2011, 03:12:04 pm
what exaaaaaaaaactly do we need to know from the AS stuff?
The papers were so easy, im just worried they might get extra AS stuff :-\
Post by: elemis on June 06, 2011, 03:27:47 pm
Tomorrow is the A2 practical
what exaaaaaaaaactly do we need to know from the AS stuff?
The papers were so easy, im just worried they might get extra AS stuff :-\
There's a CIE Chemistry A2 Practical ? :S
Post by: EMO123 on June 16, 2011, 03:42:14 pm
Paper 04
Q-2 c)
Q-3 b)
Q-4
more doubts coming from me soon
Post by: EMO123 on June 18, 2011, 03:48:28 pm
May/June 2007somebody do this
Paper 04
Q-2 c)
Q-3 b)
Q-4
more doubts coming from me soon
Post by: samriddh on June 18, 2011, 04:24:40 pm
hello all =]
just been two days i started chem A level (both AS & A2) and any doubts i find ill be postin here...
so my first doubt,
can someone please clearly explain me what orbitals are ???
accordin to the definition, it says tht they are space around the nucleus where there is max. probablity of findin the electron within its orbit.
for example when n=3; l=0,1,2
so by 0,1,2 does it mean 0 from the nucleus (like coverin all the orbits in between as well> n=1 and n=2) ???
well orbitals are simply spaces where electrons most likely travel to while moving around the nucleus...and the number like u said n=1,2,3 etc are generally shells numbers which contain several orbitals...
it's like shells( like K,L,M,N...etc) contain subshells(1s, 2s etc where 1s means s subshell form shell no. one) now s-subshell contains 2 orbitals....
well it's my first post in this site so i dont know if u will find this much helpful but nonetheless i thought you could use some help so i posted..:)
Post by: Romeesa-Chan on June 23, 2011, 07:23:53 pm
May/June 2007
Paper 04
Q-2 c)
Q-3 b)
Q-4
more doubts coming from me soon
Ans 2.c- part i:
Gas A:
16 = O
17 = OH
18 = H2O
Therefore, it'll be H2O
Gas B:
14 = N
16 = O
28 = N2 (14+14)
30 = NO (14+16)
44 = N2O (14+14+16)
Therefore, it'll be N2O
part ii: NH4NO3 ? N2O + 2H2
Ans3.b- Not so sure.
(i) Pb is in two oxidation states: +2 and +4
The ratio will be 1:2
(ii) Pb3O4 ? 3PbO + ½O2
(iii) Pb3O4 + 4HNO3 ? 2Pb(NO3)2 + PbO2 + 2H2O
(iv) Pb2+ more basic than Pb4+
Ans4- The d-orbital will be split in two ways: along the axis and in between the axes.
(ii) Ligands have lone pair of electrons which is how they form a covalent bond with the transition metal ion. The electrons in the orbitals pointing towards the ligands have higher energy.
c(i)- Accoring to graph and data given:
C= red and D=blue
c(ii)- More energy less wavelength. Therefore, at C.
Hope that's cleared. :D
Post by: EMO123 on June 23, 2011, 07:40:28 pm
Ans 2.c- part i:thanx dear it really helped i can +rep after about 1 hour so i will do it in the morning now i have gtg
Gas A:
16 = O
17 = OH
18 = H2O
Therefore, it'll be H2O
Gas B:
14 = N
16 = O
28 = N2 (14+14)
30 = NO (14+16)
44 = N2O (14+14+16)
Therefore, it'll be N2O
part ii: NH4NO3 ? N2O + 2H2
Ans3.b- Not so sure.
(i) Pb is in two oxidation states: +2 and +4
The ratio will be 1:2
(ii) Pb3O4 ? 3PbO + ½O2
(iii) Pb3O4 + 4HNO3 ? 2Pb(NO3)2 + PbO2 + 2H2O
(iv) Pb2+ more basic than Pb4+
Ans4- The d-orbital will be split in two ways: along the axis and in between the axes.
(ii) Ligands have lone pair of electrons which is how they form a covalent bond with the transition metal ion. The electrons in the orbitals pointing towards the ligands have higher energy.
c(i)- Accoring to graph and data given:
C= red and D=blue
c(ii)- More energy less wavelength. Therefore, at C.
Hope that's cleared. :D
Post by: Romeesa-Chan on June 23, 2011, 07:41:21 pm
thanx dear it really helped i can +rep after about 1 hour so i will do it in the morning now i have gtg
Welcome. :D Glad to help! :D
Pray for my exam! (:
Post by: EMO123 on June 24, 2011, 03:00:04 am
Welcome. :D Glad to help! :Dbest of lucksssssssss
Pray for my exam! (:
do your level best
Post by: ahmedigcse4eva on September 14, 2011, 08:20:56 pm
Would you please tell me what are the topics in AS chem cambridge, and if there notes for the AS chem.
Thanks alot
Post by: Most UniQue™ on September 14, 2011, 10:13:06 pm
Hey guys,
Would you please tell me what are the topics in AS chem cambridge, and if there notes for the AS chem.
Thanks alot
Here's the syllabus, read it and you'll know what are included:
http://www.cie.org.uk/qualifications/academic/uppersec/alevel/subject?assdef_id=736
And notes, you can find it here:
Notes Here (https://studentforums.biz/reference-material-83/aas-level-notes-for-physics-economics-computing-applied-ict-maths/)
Notes Here (https://studentforums.biz/reference-material-83/ultimate-resource-guide-for-a-level-bio-chem-and-phy-!!/)
GOOD LUCK Mate!
Post by: ahmedigcse4eva on September 18, 2011, 02:30:55 pm
Post by: MKL on September 24, 2011, 03:14:35 am
Post by: Becca on October 22, 2011, 03:28:58 am
It's +2 because the cold dilute acidified KMnO4 produces a diol by adding a hydroxyl group to each of the two carbon atoms in the carbon-carbon double bond within the molecule. This creates two chiral carbon atoms for those two carbon atoms... On the ather hand, when hot concentrated acidified KMnO4 is used, one chiral caron atom is lost because the hydroxyl group within the original cholesterol molecule is oxidised to form a ketone functional group, so the carbon atom attached to that hydroxyl group ceases to be a chiral carbon atom.
Hope you understand my explanation! :)
Becca
Post by: donhassan on October 27, 2011, 02:42:08 pm
if we have a cell made up of
AG ion + e=ag E=0.80v
CU ion + 2 e=cu E=0.34
if we add water to cu ion/cu beaker and let AG content remain same will the EMF increase or decrease and WHY ?
Post by: tmisterr on October 27, 2011, 06:05:50 pm
Post by: The Golden Girl =D on October 30, 2011, 03:28:23 pm
According to the 2011 syllabus, you only need to know about the Optical Isomers.
Optical Isomers
When drawing a pair of optical isomers, candidates should indicate the three-dimensional structures
according to the convention used in the example below.
mirror plane
(http://www.sexandphilosophy.co.uk/pe_20_21_chiral_alanine.gif)
i ain't a CIE Student but can you tell me how to distinguish between Optical Isomers :-\
EDIT
Thanks =D
Post by: dlehddud on October 31, 2011, 03:10:04 am
Post by: The Golden Girl =D on October 31, 2011, 12:37:27 pm
Post by: The Golden Girl =D on October 31, 2011, 03:06:20 pm
How can Optical Isomers be separated Chemically?
Post by: dlehddud on October 31, 2011, 11:13:07 pm
Post by: dlehddud on October 31, 2011, 11:14:36 pm
Post by: The Golden Girl =D on November 01, 2011, 02:43:55 pm
I don't even think it's in the Syllabus (EDEXCEL)
Post by: dlehddud on November 03, 2011, 05:38:17 am
Post by: SZM on November 03, 2011, 11:37:14 am
Hve u done Alevel? if so which board? edexcel or cambridge?
by the way if u hv done, cn u tel me ur experience towards alevel??? Becoz i am doing alevel privately not schooling, thaty. so pls tel me ur experience towards alevel??
waitn for ur reply.
need ur reply urgenlty.
Post by: dlehddud on November 05, 2011, 12:00:02 pm
Post by: dlehddud on November 08, 2011, 04:50:17 am
Could someone please explain why the E anode for O2 production becomes more negative?? 0.0
Post by: angel_ak on November 29, 2011, 01:55:11 pm
Post by: Deadly_king on January 02, 2012, 11:06:13 am
does silicon reacts with water??
Try these links, it may help ;)
http://www.chemguide.co.uk/inorganic/period3/elementsreact.html
http://www.lenntech.com/periodic/water/silicon/silicon-and-water.htm
Post by: Deadly_king on January 02, 2012, 11:08:31 am
2010 oct/nov paper 43 q5 c)i)
Could someone please explain why the E anode for O2 production becomes more negative?? 0.0
DO you have any link to question?
I don't have the paper, so it might be helpful to me if you could wire the full question or give me a link ;)
Post by: emzie on March 02, 2012, 05:28:02 pm
I just have this doubt in equilibrium
Is the Kp value affect by change in pressure at constant temperature, if not then why?
Thank in advance!
Post by: Romeesa-Chan on March 02, 2012, 06:13:47 pm
Hello! Hope you all doing great :)
I just have this doubt in equilibrium
Is the Kp value affect by change in pressure at constant temperature, if not then why?
Thank in advance!
It doesn't change the Kp value; it only affects the equilibrium position.
Correct me if I am wrong. :)
Post by: Most UniQue™ on March 02, 2012, 07:03:25 pm
It doesn't change the Kp value; it only affects the equilibrium position.
Correct me if I am wrong. :)
Seconded
Post by: raamishstudent on April 19, 2012, 07:25:28 am
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s10_qp_11.pdf
qs 8,17,16,24,38,31
Post by: Arthur Bon Zavi on April 19, 2012, 02:31:31 pm
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s10_qp_11.pdf
qs 8,17,16,24,38,31
[8] Strontium (atomic weight - 88)
ONE mole of group 2 element needs ONE mole of chlorine molecule to completely react.
Number of moles of group 2 element (in this case, Strontium)= (2.92)/(88) = 0.033
Number of moles of chlorine = (5.287 - 2.920)/(35.5 X 2) = 0.033
[16] Silicon, Phosphorus, Sulfur
Regarding ionization energy, Phosphorus is an exception as it disturbs the increasing trend towards the right of the periodic table. It satisfies the other condition of lowest melting point.
[17] Second ionization energy
Ionization energy decreases as we go down the group 2 elements as removing outer electron becomes easier because reactivity increases as we descend down the periodic table for group 2.
[24] Propan-2-ol
NaBH4 reduces propanone.
REMEMBER THIS : Secondary alcohol on oxidation gives a ketone, so in return while on reduction, ketone gives a secondary alcohol.
[31] 1 and 2
1 is diamond and 2 is graphite. These show giant molecular structure while 3 forms ionic lattice.
[38] 2 and 3
Numbering of a hydrocarbon chain starts from only one side and cannot start from both sides simultaneously. This is obeyed by 2 and 3.
Post by: raamishstudent on April 20, 2012, 06:10:04 am
PPr: http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s10_qp_11.pdf
Ms:http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s10_ms_11.pdf
Post by: LekhB on May 04, 2012, 06:12:11 am
Post by: Physics09 on May 04, 2012, 07:00:52 pm
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w07_qp_1.pdf
Q: 30 and 21
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s08_qp_1.pdf
Q:23 and 27
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w08_qp_1.pdf
Q: 28
Post by: Deadly_king on May 05, 2012, 12:18:20 am
Explain why the carbon chlorine bond photodissociates and the carbon fluorine bond does not?
Carbon-fluorine bond is an extremely strong bond that requires very high amount of energy in order to dissociate ;)
http://en.wikipedia.org/wiki/Carbon%E2%80%93fluorine_bond
Carbon-chlorine bond however is comparatively weaker and thus easy to dissociate ;)
Post by: Deadly_king on May 05, 2012, 12:41:16 am
Exam in three days! Urgent help please!
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w07_qp_1.pdf
Q: 30 and 21
21. C-Pb bond is a very weak bond...........So tetra methyl lead acts a source of methyl radicals ;)
30. Upon breaking the ester bonds, we can have only two possible acids formed namely CH3CH2CH2CO2H or CH3CO2H.
Hence answer is B ;)
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s08_qp_1.pdf
Q:23 and 27
23. Answer is D due to high molecular weight which will prevent the hydrocarbon from rising upon water ;)
27. The formula C10H14O indicates that it should be an alcohol and the fact that it is unreactive to mild oxidising agents, it cannot be a primary one ;)
So you just have to eliminate all the answers which upon hydration would give a primary alcohol ;) This leaves us with B, C and D! But only D gives a tertiary alcohol which we all know is not affected by mild oxidising agents ;)
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w08_qp_1.pdf
Q: 28
28. The reaction is one forming an ester ;) So to form an ester we usually need and alcohol and a carboxylic acid ;)
If you break the ester bond, you'll get as products CH3CO2H and CH3CH2CH(CG3)CH2OH, this is because the second carbon atom of 2methyl-butyl methanoate contains a double bond with O which indicated that it should be the acidic carbon atom ;) The rests just fits in ;)
Check the answers, if they are good, then my explanation are correct ;) IF not, then am afraid I can't help :/
AM afraid I don't currently have enough time to check the marking schemes or my notes........I just replied according to my knowledge :/
Hope it helps :)
Post by: NidZ- Hero on May 05, 2012, 03:16:48 pm
- Q3
- Q5
- Q8
- Q11
- Q26
http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FIGCSE%2FChemistry%2FCIE%2F2002+Nov/
paper 1
Post by: Ghost Of Highbury on May 05, 2012, 08:48:48 pm
can Someone please Explain me in details the answerrs of the Following ?
- Q3
- Q5
- Q8
- Q11
- Q26
http://www.freeexampapers.com/past_papers.php?l=Past_Papers%2FIGCSE%2FChemistry%2FCIE%2F2002+Nov/
paper 1
First I.E + Second I.E is same for Co and Al (757 + 1640 for Co) and (577+1820) = 2397. Use data booklet for IEs
A will have a dipole because Chlorine atoms and Fluorine atoms can be places in different carbons, and due to different electronegativities,
a net dipole remains. Same reason for C and D. The answer, C2Cl4 doesn't have a net dipole because Cl atoms have same Electronegativity and thus
net dipole cancels and is zero.
A 2 g sample of hydrogen at temperature T and of volume V exerts a pressure p. p = nrT/V = 2RT/V (moles of H2 = 2/1 = 2)
check which option gives same pressure. For e.g A gives p = nRT/V = RT/V (n = 2/2 = 1). B gives p = 4RT/V
C gives correct answer because n = 1 + 1 = 2.
2HI <-> H2 + I2
Initially b 0 0
Equilibrium b-x b/2 b/2
(b^2/4)/(1-b)^2
26. First break the double bonds, add the -OH groups and make the diols. NOw, for oxidation to ketone, 2 hydrogen atoms should be lost, one from the -OH group, one from the same carbon atom. And remember u need a diketone, so 2 ketone groups.
It isnt possible to get a diketone in A and D because you'll only get one ketone group in each. From B and C, B gives u an aldehyde after losing th 2 h-atoms, C is the answer.
Post by: Twinkle Charms on May 06, 2012, 07:16:29 am
If flasks are connected at constant temperature, what is the final pressure??
Post by: Arthur Bon Zavi on May 06, 2012, 08:28:08 am
Flask X contains 1dm3 of helium at 2kPa pressure and flask Y contains 2dm3 of Neon at 1kPa pressure.
If flasks are connected at constant temperature, what is the final pressure??
Total volume = 1dm3 + 2dm3 = 3dm3
PV = nRT
As temperature is constant, PV is constant, so P1V1 = P2V2
Flask X : (2)(1) = P2(3)
P2 = 2/3 kPa
Flask Y : (1)(2) = P2(3)
P2 = 2/3 kPa
Total pressure = sum of partial pressures
Total pressure = 2/3 + 2/3 = 4/3 kPa
Post by: perky on May 07, 2012, 09:16:51 am
Post by: Ghost Of Highbury on May 08, 2012, 04:04:27 am
oct/nov 2009 ppr 11 q 28,29,30
27 : 2,4 DNPH reacts with aldehydes and ketones. This molecule has a ketone group, so it will react.
HBr will get added to the C=C
NaBH4 will reduce the C=O group.
Fehling's reagent only reacts with aldehydes, no -CHO group so no reaction.
28 : This one's easy. Start with the general aldehyde molecule R-CHO. The carbon being partially positive, gets attacked by the nucleophile
-CH2CO2CH3. So the final product is R-CH(OH)CH2CO2CH3. Now R can be any alkyl group, the answer is C, where R is the ethyl group (CH3CH2)
29. B. Nucleophilic substitution. CN- ions replace Br. CN gets hydrolysed to COOH. Do these for B and u get the required product.
Post by: raamishstudent on May 08, 2012, 06:07:19 am
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w05_qp_1.pdf
Qs 31 How to do it?
Qs 34) WHy is C not the answer
Qs 20) MY isomers are becoming 4 but the answer is 3.
Qs 5) How to do it?
Please help the exam is tommorrow
Post by: NidZ- Hero on May 08, 2012, 02:32:55 pm
First I.E + Second I.E is same for Co and Al (757 + 1640 for Co) and (577+1820) = 2397. Use data booklet for IEs
A will have a dipole because Chlorine atoms and Fluorine atoms can be places in different carbons, and due to different electronegativities,
a net dipole remains. Same reason for C and D. The answer, C2Cl4 doesn't have a net dipole because Cl atoms have same Electronegativity and thus
net dipole cancels and is zero.
A 2 g sample of hydrogen at temperature T and of volume V exerts a pressure p. p = nrT/V = 2RT/V (moles of H2 = 2/1 = 2)
check which option gives same pressure. For e.g A gives p = nRT/V = RT/V (n = 2/2 = 1). B gives p = 4RT/V
C gives correct answer because n = 1 + 1 = 2.
2HI <-> H2 + I2
Initially b 0 0
Equilibrium b-x b/2 b/2
(b^2/4)/(1-b)^2
26. First break the double bonds, add the -OH groups and make the diols. NOw, for oxidation to ketone, 2 hydrogen atoms should be lost, one from the -OH group, one from the same carbon atom. And remember u need a diketone, so 2 ketone groups.
It isnt possible to get a diketone in A and D because you'll only get one ketone group in each. From B and C, B gives u an aldehyde after losing th 2 h-atoms, C is the answer.
Thankkkks Alllot
Post by: Ghost Of Highbury on May 08, 2012, 03:01:09 pm
Hey guys just need some1 to clear some doubts of mine.
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w05_qp_1.pdf
Qs 31 How to do it?
Qs 34) WHy is C not the answer
Qs 20) MY isomers are becoming 4 but the answer is 3.
Qs 5) How to do it?
Please help the exam is tommorrow
Q31) The relative molecular mass is the average of all the isotopic mass. So some atoms might have different number of neutrons, hence different nucleon number. The different number of neutrons won't affect the radius. So 1 is correct, 2 is wrong, hence D.
34) 1 is correct, because H+ reacts with OH- ions of the first equilibrium and equilibrium moves to the right. So number of hydroxyapatite molecules decrease. 2 is correct because H+ will react with the PO4(3-) ions and decrease its concentration again moving the equilibrium to the right. 3 is not correct, because Calicium ions won't react with the acid. I'm not sure about the explanation to that, if someone could throw some light on that, it'd be great.
20) 1,1 dichloro ethene. cis 1,2 dichloro ethene and trans 1,2 dichloro ethene. 3 isomers.
5) He and Nitrogen will only have VDW's forces between them, so they can behave like an ideal gas compared to the others. HCl will have VDW's forces stronger than that in CH4 because more number of electrons. It will also have hydrogen bonding, so that's the answer.
Post by: NidZ- Hero on May 08, 2012, 03:09:56 pm
http://www.freeexampapers.com/get_past_papers.php?l=Past_Papers%2FA+Level%2FChemistry%2FCIE%2F2002+Nov%2F9701_w02_qp_1.pdf
Questions : 8 ,29, 39
http://www.freeexampapers.com/get_past_papers.php?l=Past_Papers%2FA+Level%2FChemistry%2FCIE%2F2006+Nov%2F9701_w06_qp_1.pdf
Question : 4 , 5 , 6 , 9 , 11 , 14 , 23 , 27 , 37.
http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s07_qp_1.pdf
Question : 9 ,26
ASAP
Thanku
Post by: Ghost Of Highbury on May 08, 2012, 08:00:02 pm
I have to sleep in a bit, so i'll do w06 questions very quickly. Hope someone can help u with the other 5
4) Construct balanced equations of the combustion. Find out how much volume of O2 hasn't reacted or is in excess. Find out the volume of CO2 given off using mole concept, add 'em up. Do this for all the alkanes.
5) B, sideways overlap. Weak bond. The rest are head to head
6) Ammonia has a lot of different types of bonding between the molecules, hence it deviates the most from the ideal gas. H2 is the closest to an ideal gas, hence straight line. D
9) Use this hess's cycle.
H2 + CO + 0.5O2 ---(-283)---> CO2 + H2
\ /
\(-286+44) /
\ /
V V
CO + H2O(g)
11) acid alchol ester water
1 1 0 0
1-x 1-x x x
(x^2)/(1-x)^2 = 4
Solve for x
14) Calcium nitrate -> calcium oxide + NO2 + O2
Balance, convert to moles, find out the mass of CaO remaining.
23) B - Different groups attached to each Carbon will produce the cis trans.
27) B - Tertiary alchol, so doesn't get oxidized. No reaction. Forms ONa and gives H2 gas.
37) This is a tough one. Can't draw now, but i'll tell u what to do. break the double bonds and put O atoms to each carbon broken from the double bond. See if the 2 products contain a ketone group. One gives an aldehyde group if i'm not wrong. try it
Good luck, I hope you do great tomorrow :)
Night
Post by: perky on May 09, 2012, 03:50:16 pm
Post by: whale95 on November 17, 2012, 05:23:06 pm
Post by: astarmathsandphysics on November 20, 2012, 01:48:52 pm
Post by: rickyponting66 on November 20, 2012, 07:40:15 pm
How you determine the sign?? Ans is D.. How you determine the sign... How you solved it... please write the calculation.
Please reply asap.. thank you.
Post by: astarmathsandphysics on November 20, 2012, 09:02:13 pm
Post by: astarmathsandphysics on November 20, 2012, 09:04:21 pm
Post by: skylinegtrr5 on March 23, 2013, 03:08:55 am
I have a few question which I think I some need explaination. Thank you.
Chemistry 2009 Winter Paper 11- Question 21 and 29
2009 Summer Paper 11- Question 11
Post by: NidZ- Hero on March 31, 2013, 03:43:50 pm
Thanks :)