ALL CIE CHEMISTRY DOUBTS HERE !!

[Twinkle Charms]

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

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??

[Deadly_king]

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.

[Twinkle Charms]

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.

yo im gettin smarter haha 

thank yooouuu

[Deadly_king]

yo im gettin smarter haha 

thank yooouuu

Hahaha.....very good
keep up the good work 

You are most welcome

[Twinkle Charms]

Q) Tin reacts with iodine in an organic solvent to form a covalent compound SnI_x 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 SnI_x 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 SnI_x ??

[Deadly_king]

Q) Tin reacts with iodine in an organic solvent to form a covalent compound SnI_x 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 SnI_x 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 SnI_x ??

Equation is as follows :

Sn + (x/2)I₂ ----> SnI_x

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 SnI_x

Number of moles of Sn used = Mass/Ar = 0.342/119

Therefore same number of moles of SnI_x 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

[Twinkle Charms]

Equation is as follows :

Sn + (x/2)I₂ ----> SnI_x

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 SnI_x

Number of moles of Sn used = Mass/Ar = 0.342/119

Therefore same number of moles of SnI_x 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

wow thanks =)

[Deadly_king]

wow thanks =)

Mention not

[$!$RatJumper$!$]

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?

[Deadly_king]

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

[$!$RatJumper$!$]

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?

[Arthur Bon Zavi]

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! 

[moon]

can anyone pls explain Q2 (b), Q4 (b)(iii), (iv) & Q7 (b)Nov.2008 ppr4? thanx in advance.

[Deadly_king]

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!

[Deadly_king]

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.