Chemistry P4| A2 only

[shamil]

Does anyone have a summary of everything we need to know for analytical chem? IF so please do post, thank you

 

[sahania]

you deleted the post histephenson007 ....

 

[histephenson007]

you deleted the post histephenson007 ....

There is no Q3 i) ii) ... do u mean Q3. d i) ii) ??? (In that case, the only possible solution we could agree on was to try the trial and error method..... )

For Q4 c) Because we have to find out which one of these oxidation processes are likely to occur, we need to find out the E cell of both these process. The oxidation with the higher E cell is more likely to occur.

These are 2 solution:
-Fe2+ in acid
-Fe(OH)2 in alkaline

Sorry, internet got dced

 

[sahania]

oh yeah , sorry it was 3 d ...

 

[sahania]

And this skeletal formulae thingie is not working for me ..
http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w11_qp_43.pdf
Q5(c)
please help

 

[sahania]

7c too ... mass spectometry and NMR uhhhhhhhhhhh .... i'm unable to attempt these questions at all despite of going through the topics in OCR and application booklet ... it was tedious studying them maybe that's why i couldn't grasp much ...

 

[histephenson007]

And this skeletal formulae thingie is not working for me ..
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_43.pdf
Q5(c)
please help

which part is not working?!

 

[sahania]

check out part c of Q5... they've asked to draw skeletal formulae... i'm unable to attempt it ... all my skeletal formulae went wrong

 

[larina]

Can someone please tell me how to do Nov 2011, p43, question 3(e)(iv)!!
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_43.pdf
im using the equation pH= pKa + log [base/acid] and im gettingg 4.36...

histephenson007 can u answer me this one pls!?

 

[ousamah112]

explain may/june 2007 paper 4 q 4 part a n b
http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_s07_qp_4.pdf

 

[larina]

7c too ... mass spectometry and NMR uhhhhhhhhhhh .... i'm unable to attempt these questions at all despite of going through the topics in OCR and application booklet ... it was tedious studying them maybe that's why i couldn't grasp much ...

to find the no. of carbon atoms, always use this formula---> 100/1.1 * (M+2/M)
and if M:M+2 is 3:1 ==> chlorine atom is present
if M:M+2 is 1:1==> bromine atom is present
and for the determining the molecular structure of the organic compound, u have to use the Data Booklet...

 

[histephenson007]

7c too ... mass spectometry and NMR uhhhhhhhhhhh .... i'm unable to attempt these questions at all despite of going through the topics in OCR and application booklet ... it was tedious studying them maybe that's why i couldn't grasp much ...

For 7.c)i) You'll hav to use this formula : No. of Carbon atoms = [(100/1.1) * (M+1/M)] . So, you'd get 3 Carbon atoms.

ii) There was M+2 peak present which had similar height to the M peak. This sort of peaks can only be formed with Bromine, because in the booklet, it is the only element we learned to have a 1:1 ratio abundance in its isotopes.

iii) Only one because there are only the M and M+2 peaks were in the spectrum ( because the question gave the M:M+1 ratio, we are to assume that the M+1 peak was solely caused by the 13C isotope). If there were more than one Bromine atoms, there could have been the 1:2:1 peak ratio thingies found.

iv)From 3 Carbons and 1 Bromine, we can get the empirical formula : C3H7Br

In the NMR,
There is one peak with 6H, but the there are only 3 Carbon atoms, so it isn't possible for a benzene ring to be there. So, there must be two -CH3 s on either side of a central carbon.
And the other 1H peak must've been formed by the hydrogen on the central Carbon atom.
So, the structure shouldbe CH3CH(Br)CH3

I recommend reading the chemguide version for NMR. It is very very very helpful

 

[sahania]

thanks alotttttttt

 

[histephenson007]

histephenson007 can u answer me this one pls!?

First, we need to get the Ka value from the pKa ..
pKa=-log(Ka)
So, Ka = 10 ^(-4.76)

The acid equation is : CH3COOH <---> (H+) + CH3COO-
So, Ka = ([H+]*[CH3COO-]) / [CH3COOH]
Re-arranging the equation: [H+] = (Ka*[CH3COOH]) / [CH3COO-]

But, we don't know what [CH3COO-] is,
So, we hav to find it with the formula C = {n(H+) - n(OH-)} / V
So, C = (0.1*0.01-0.01*0.25) / 0.02 = 0.075

Then, we can calculate for the H+ with the above reaction. And then pH. My teacher just used some pH = 4.76 + log(0.05/0.075), but I don't understand why.

 

[histephenson007]

check out part c of Q5... they've asked to draw skeletal formulae... i'm unable to attempt it ... all my skeletal formulae went wrong

Think of it as a free-radical substitution reaction and add chlorine on each Carbon atom. Beware with the left ones.

 

[lewis_neu]

Hey frens, has anyone got any notes on plastics!! thermosetting and thermoplastics..................... if so plz post or plz share the link

 

[Mobeen]

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w11_qp_42.pdf
http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w11_ms_42.pdf
Q2 a (i) how did we get the order with respect to H+ ?

 

[DragonCub]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_42.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_ms_42.pdf
Q2 a (i) how did we get the order with respect to H+ ?

By comparing 1 and 2, we get [CH3CHO] is 1st order.
By comparing 2 and 3, we get [CH3OH] is 1st order.
Then by comparing 1 and 4:
[CH3OH] becomes twice large, so does [H+].
We already know that [CH3OH] is 1st order, so it being twice large shall double the rate. But actually, the rate quadruples (4 times large).
The extra doubling effect comes from [H+]. It doubles to give a double effect. Therefore it is 1st order.

 

[vita199]

can sum1 plzzz help in these questionss....i dont get how electrode potentials increases or decreases when u increase conc of of the reactants..these are the papers.q3 of oct/2010/41 and q5c of oct/2010/43..

 

[larina]

First, we need to get the Ka value from the pKa ..
pKa=-log(Ka)
So, Ka = 10 ^(-4.76)

The acid equation is : CH3COOH <---> (H+) + CH3COO-
So, Ka = ([H+]*[CH3COO-]) / [CH3COOH]
Re-arranging the equation: [H+] = (Ka*[CH3COOH]) / [CH3COO-]

But, we don't know what [CH3COO-] is,
So, we hav to find it with the formula C = {n(H+) - n(OH-)} / V
So, C = (0.1*0.01-0.01*0.25) / 0.02 = 0.075

Then, we can calculate for the H+ with the above reaction. And then pH. My teacher just used some pH = 4.76 + log(0.05/0.075), but I don't understand why.

why did u subtract the no.of moles??