IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: Phosu on March 19, 2010, 12:17:20 pm
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nov 09 pp1 var 1
que 4, 7, 21(how do we find the double bonds)?, 29 (please explain mechanism),
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attach d paper
i cant download it for sme reason :S:s
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ok here you go
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4) lone pair-lone pair repulsion is maximum, followed by lone pair-bond pair, least is bond pair-bond pair
S had 2 lone pairs, 2 bond pairs,which is 104.5 degrees(x)
C has 4 bond pairs and no lone pairs, which gives rise to angle of 109.5 degrees(y)
N has 3 bond pairs and 1 lone pair, which is 107.5(Z)
so order is y>z>x
which is C
7)
in A ox state of Cr before is +6 and after rxn is also +6
x+ (4X-2)=-2
and later
x+ (3X-2)=0
x=+6
in B it goes from +6 to +6
I'll explain how
before reaction, the compound has a charge of -2
x+ (4X-2)=-2
x=+6
After rxn
2x+(7X-2)=-2
x=+6
In C
x+(2X-2)+(2X-1)=0
x=+6
after rxn
x+(2X-4)=-2
x=+6
In D
x+(2X-2)+(2X-1)=0
x=+6
After rxn
2x+(3X-2)=0
x=+3
Hence reduction in ox no in D
21)Total no of double bonds
3 in the cyclohexane ring
1 in the aldehyde (C=O)
1 is the alkene C=C between C11 and C12
so total of 5
cis isomer has the higher groups(R) on the same side.
Hence A
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29) CN- will replace Br
2-ethyl-3-methylbutanoic acid has a total of 7 carbons.
Carbon from CN contributes to this no.
Hence the reactant has 6 carbons point 1
This eliminates option C and D with 7 carbons
It can't be A...cause CN replaces Br which is on the 3rd carbon in the chain...It won't give an ethyl side chain on the 3rd carbon
Hence it's B check the diagram
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great explanation, Thanks alot :D
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;D
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how do we know which structures will be formed?
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1 n 2
not 3 because when u split 3, u get one ethyl radical and another butyl radical...
It should be a propyl-propyl radical joining together
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i have doubts tooo :-\
nov.06
Q40> Why is no.3 wrong? isnt c-o double bond a ketone?
Q37> PLZ XPLAIN
Q21> PLZ XPLAIN
Q14, Q11, Q6, Q9
THANX IN ADVANCE ;)
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after im done with my school work :)
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after im done with my school work :)
Great work you doing Nid, as always.
+rep. :)
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Q9> da ans. is A, but if so why is da initial rate of da reaction lower than the original exp.??????????
Q.16> plzz xplain
Q.17> mentioned in da que. dat excess of oxygen is available so i thought dat da
CO----> CO2 and da SO2-------> SO3
rrrrrrrrite??? ???
plzzzzzzzzzzz reply
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Q9)
I don't knw how the reaction rate came in here :-\
Anyway I'll explain
Kc= [X2Y]2
[X2]2[Y2]
Kc=2
ratio of X2Y and X2 is the same....hence conc same....
they get cancelled in the eqn above...
therefore 1/[Y2] =2
conc of [Y2]= 1/2=0.5
for reaction two
Ratio of Y2: X2: X2Y
1 : 2 :2
Kc= [0.5]1/2 [1]
[1]
=1/root 2
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Q16) Cl is more electronegative than Iodine. This is because its radius is much smaller...HCl bond is therefore stronger, shorter bond length
Q17) Ok...the ans to that is..the conversion of SO2 to SO3 is a reversible reaction and is exothermic, at high temperature when burning in the air, it is bound not to happen...It require a catalyst too(V2O5 as in contact process)
hope that helps
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Oh & Thanks Mani :D
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oh, am sorry its Q11> not 9
thanx :D
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Q11) Ans is B.
Le chatelier's principle....
When u add more H2O2, it takes longer for the breakdown since the amount of catalyst may not be sufficient...more H2O2 and hence more vol of O2 produced
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Thank You sooooooooooooo much :-* :-*
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i have doubts tooo :-\
nov.06
Q40> Why is no.3 wrong? isnt c-o double bond a ketone?
Q37> PLZ XPLAIN
Q21> PLZ XPLAIN
Q14, Q11, Q6, Q9
THANX IN ADVANCE ;)
cud u plzzz help me in these Que's toooo??? :-[ :-[
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Oh & Thanks Mani :D
You're welcome.
Well deserved. ;D
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1) jun 09 pp1 q40
2) a 2g sample of hydrogen at temperature T and volume V exerts a presssure P. Deuterium 1H2, is an isotope of hydrogen. which of the following would also exert a pressure P at the same temperature T?
a) 2g of deuterium of volume V
b) 4g of deuterium of volume V/2
c) a mixture of 1g of hydrogen and 2g of deuterium of total volume V
d) a mixture of 2g of hydrogen and 1g of deuterium of total volume 2V
3) Which property applies to halogens as group VII is descended?
a)Hydrogen-halogen bond energy decreases
b) intermolecular forces between the elements decrease
c) ions become less strongly reducing
d) thermal stability of the hydrides increases.
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Im doing Edexcel Chem
so i cant help with question 2,
but question 3, the answer is A
a) H-F has the strongest bond enthalpy while H-I has the lowest bond enthalpy because F is more electronegative and it is hard to break the bond
b) Intermolecular forces increase down the group because the size of the atom increases so more chance of an instantaneous dipole-induced dipole because there are more electrons
c) OILRIG - oxidation is loss of electron, reduction is gain.... but here we are considering Reducing Power, or how easily they are oxidised....as mentioned earlier.... Flourine is the most electronegative...so it is most difficult to remove electrons from it... so ions become more strongly reducing.
d) Same explanation as bond enthalpy...
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Q9 nd 25 of da first 2 papers ( specimen 01)
Q21 and 26 of next 2 sheets ( may 01)
thats all !!!
plz reply ASAP ;)
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9) Cl2(g)----> Cl(g) + Cl(g)
atomisation energy...+244kj/mol
Cl(g)--->2Cl- -364 kj/mol
I think the answer should be A
25) A,B and D are incorrect... cuz it's possible to have a gas of this composition, it may be immiscible in water and it can have a mass less than 32...consider HCHO...mass 30
So, C
21)stoichio...
unsaturates...alkenes have formula Cn H2n in this hydrocarbon, there r 10 less hydrogen atoms...5 H2
so 4.1g~0.01 moles of the hydrocarbon will react with 0.05 moles of H2
0.05 moles of H2= 0.05 X2 grams=0.1g
no option? :-\
C is 0.12g and it will react completely with squalene...so C i guess
26) B it's a tertiary alcohol...won't undergo oxidation
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Thank You :-*
for Q9 the ms says C ( dunno why)
da rest of da que's r ok, but can u tell me in Q21 why dont we react the alkene with 6H2 , to compensate for the 10 and 2 more atoms to break the double bond ??? ???
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For question 9, first the Cl2 molecule breaks and the enthalpy change from the data booklet is +244. then you have two Cl atoms. definition of electron affinity is the enthalpy change when one mole of gaseous atoms each gain an electron to have a single negative charge. so since there are two, they will form two Cl-. enthalpy change for one mole is -364 but since there are two, it will be -728, but remember +244 was used in the beginning to to break the Cl-Cl bond so -728+244= -484
for question 26, I believe the answer is A, because A contains a aldehyde group and it can be oxidised further into a carboxylic acid which will change the coLour from orange to green. This can not be done with the compound in the question because it contains a tertiary alcohol and a ketone and both cannot be oxidised further so the colour will remain orange. I do not believe it is B because B has a tertiary alcohol which cannot be oxidised further so again the colour will remain green.
for question 21,
it is a weird question but this is how i did it. Since it is unsaturated, one of the reactions alkenes undergo is hydrogenation to form alkanes in the presence of platinum catalyst, which are the conditions stated in the question. so if C30H50 undergoes full hydrogenation, it will become an alkane C30H62 so u need to a add 12 moles of hydrogen atoms so (4.1/410)*12= 0.12 which is C.
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Thank You sooo much :D
and ya ur ans. for q26 is riteee , i checked the ms ;D
cud u also help me with Q21 of 5th sheet ( nov01)
and Q31 and 32 of last sheet ( nov01)
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q 21, I should be D. A would be formed since it is a reducing atmosphere, reduction is the addition of hydrogen so ethene can be reduced to ethane. B can also be form when methane reacts with ammonia to form methyl-amine. ethanol, C ,can also be formed when ethene reacts with water (steam) the high energy can convert water to steam. D is an acid, which are you usually formed by oxidation reactions, so it can't be formed here.
q 31 D again. the most abundant is the second bar. one sulphur atom has a mass of 32.1 so the second bar must be 2 sulphur atoms S2 since its mass is 64.2, making 1 correct. for 2 there is no way to come to this conclusion. and 3, they are not different isotopes they are just different molecules of sulphur with different number of atoms in each molecule.
q32 I should be A. as you know temperature is the measure of kinetic energy in a substance so even though there be some at very high kinetic energy and some with a very low kinetic energy(think of Boltzmann distribution curve), since they are at a constant temperature, the average kinetic energy should be the same. 2 is true also since in 1 dm3 they have the same number of moles and since P has an Mr which is four time greater it is 4 times heavier. 3 is also true, same number of molecules of each gas because of same number of moles so they contribute equal pressure.
Can you do me a favour and send me these two papers (specimen 01 and oct/nov 01)? I can't find them on xtreme papers or FEP and they seem like they would be good practice.
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q 21, I should be D. A would be formed since it is a reducing atmosphere, reduction is the addition of hydrogen so ethene can be reduced to ethane. B can also be form when methane reacts with ammonia to form methyl-amine. ethanol, C ,can also be formed when ethene reacts with water (steam) the high energy can convert water to steam. D is an acid, which are you usually formed by oxidation reactions, so it can't be formed here.
q 31 D again. the most abundant is the second bar. one sulphur atom has a mass of 32.1 so the second bar must be 2 sulphur atoms S2 since its mass is 64.2, making 1 correct. for 2 there is no way to come to this conclusion. and 3, they are not different isotopes they are just different molecules of sulphur with different number of atoms in each molecule.
q32 I should be A. as you know temperature is the measure of kinetic energy in a substance so even though there be some at very high kinetic energy and some with a very low kinetic energy(think of Boltzmann distribution curve), since they are at a constant temperature, the average kinetic energy should be the same. 2 is true also since in 1 dm3 they have the same number of moles and since P has an Mr which is four time greater it is 4 times heavier. 3 is also true, same number of molecules of each gas because of same number of moles so they contribute equal pressure.
Can you do me a favour and send me these two papers (specimen 01 and oct/nov 01)? I can't find them on xtreme papers or FEP and they seem like they would be good practice.
ya, sure ill sent them as soon as i scan dem 4 u :)
nd 4 Q 31, how do we know if the bars represent isotopes or atoms ????????
6 The density of ice is 1.00 g cm–3.
cud u also xplain this Q ......
Q>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
thanx sooo much for da help, By the way all ur answers r riteeee!!!
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Why is ethanoic acid a stronger acid in liquid ammonia than in aqueous solution?............. hey guys s03 question m confused........
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The strength of an acid is determined by how easily it can donate a proton.
Thus, when in liquid ammonia, there is stronger interaction between the NH3 molecules and the acid, than with water. This causes, protons to 'come out' easier than in water, as the internal bonding forces are more strongly overcome.
Thus in ammonia, it has a greater ability to donate a proton.
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thanks.............yea that makes sense so yes it will more easily dissociate..........however u cant check this wid indicator can u ...............to deternmne its strenght
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can any1 help me out wiv
mj-08 ppr 1
question numbr 6 !?!?
Thanks
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1g= 1/18 mole
1 mole=24dm3 at 298K and 1atm
but temp is 596 K...2 times
so 1 moles at 596K occupies 48dm3
1/18 X48= 2.67
C is the ans
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thanks.............yea that makes sense so yes it will more easily dissociate..........however u cant check this wid indicator can u ...............to deternmne its strenght
Yes, that reactions with this substance will simply be more vigorous.
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hi :) can someone explain to me what is permanent dipole, induced dipole interaction please? and there's one more, dipole-dipole interaction, if i'm not mistaken. can someone explain the differences too? i'm so confused when it comes to this T.T
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ya, sure ill sent them as soon as i scan dem 4 u :)
nd 4 Q 31, how do we know if the bars represent isotopes or atoms ????????
6 The density of ice is 1.00 g cm–3.
cud u also xplain this Q ......
Q>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
thanx sooo much for da help, By the way all ur answers r riteeee!!!
hey, sorry for the late reply, been a bit busy. nid already answered this question for you :). But for question 31 i don't know how to explain, it's just a matter of observing the question let me give a try. from the graph, the masses of each bar increasing by 32.1. As you know, the Ar of sulphur is 32.1 (one mole of sulphur atoms). If two sulphur atoms are to covalently bond forming a diatomic molecule, then you add another 32.1 to get the Mr of a diatomic sulphur molecule S2 (64.2) and you keep of adding till you get to the octa-atomic S8 which has an Mr of 256.8. so each bar is representing a different molecule(not atom), the first is mono-atomic the second diatomic the third tri-atomic the fourth tetra-atomic....u get the point, so they cannot be isotopes.
Thanks, much appreciated, tell me when they are ready.
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its alrite, never mind 8)
ya, got ur point , Thank You :D
sure , i'll inform u when dey r ready :)
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1g= 1/18 mole
1 mole=24dm3 at 298K and 1atm
but temp is 596 K...2 times
so 1 moles at 596K occupies 48dm3
1/18 X48= 2.67
C is the ans
Thanks =D
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Use of the Data Booklet is relevant to this question.
Which properties would be expected from radium, 88Ra, or its compounds?
1 Radium carbonate decomposes only at a very high temperature.
2 Radium hydroxide is very insoluble.
3 Radium does not react with cold water.
hey guys a doubt the answer to this is only 1 is correct but shouldnt 2 also be correct
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Radium hydroxide is soluble, because it is lower down in group 2, and so the bond will not be strong. Genereally, the higher up group 2 u go, the more insoluble stuff gets. Becuase the lower you go the weaker the cation becomes.
Anyways, the reason why 1 is correct is because, the lower down you go the stronger carbonates get, why?? explained here: Thermal stability of carbonates and nitrates (http://www.chemguide.co.uk/inorganic/group2/thermstab.html#top)