ALL CIE CHEMISTRY DOUBTS HERE !!

[cs]

I have questions here, attached picture, i am sorry if anyone has answer this before, there are so many pages, i will try to look for it

Edit: found my answer for the second one in previous page, so left the 1st and 3rd one.. Thanks

[$!$RatJumper$!$]

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

[elemis]

@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

[elemis]

@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

[$!$RatJumper$!$]

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?

[cs]

@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

What about the 50% thing?

[cs]

@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?

[Deadly_king]

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.

[Deadly_king]

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.

[$!$RatJumper$!$]

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 :/

[$!$RatJumper$!$]

Nov. 09 Varient 1 Q5,21,28,31 Thankx

[Deadly_king]

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

K_c for reaction 1 is given to be equal to 2.

K_c = [X₂Y]² / [X₂]²[Y₂] = 2

For reaction 2 ---> K_c = [X₂][Y₂]^1/2 / [X²Y]

If you square the K_c equation for reaction 2, you'll find that it becomes the reciprocal of the Kc equation for reaction 1.

Hence for reaction 2 ---> K_c² = 1/2.

So K_c will come to be (1/2)^1/2 ----> K_c = 1/(2^1/2)

Answer is A.

Hope it's clear now

[cs]

The 50% indicates that the equilibrium is at dynamic equilibrium.

Since 50% of reactant becomes products while the other 50% remains as reactants.

ok, got it finally, my third question please, anyone.. Thank you!
I have two days to do chem and i'm done.  

[Deadly_king]

ok, got it finally, my third question please, anyone.. Thank you!
I have two days to do chem and i'm done.  

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.

[cs]

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.

understood, i thought of that too, but not so sure.Thanks! My friends are sitting for Bio on Thursday too.. I'm not    Good Luck!