AQA A2 CHEM4 &amp;amp; CHEM5 June 2015 [official discussion thread]

Original post by Itsmonique

Is anyone planning on retaking chem1 or chem2?

Yepp chem1 :smile:

Reply 41

No.An Acylation reaction basically is the addiction of an Acyl group RCOCl to a benzene ring where your electrophile is the acylium ion RCO+
I hope that makes sense :smile:

[im not the best at chem4]

Posted from TSR Mobile

Reply 42

Itsmonique

Original post by TheGemini

Yepp chem1 :smile:

yep, me too. I hope to get an A this time. :biggrin:

When do you plan to start revising and doing past papers?

Reply 43

Original post by Itsmonique

yep, me too. I hope to get an A this time. :biggrin:

When do you plan to start revising and doing past papers?

I'm still doing chem4 ,I've got about 4 topics left then I'll get to it .
Im planning on starting papers after Christmas :/ Wbu?
I made a mistake of throwing away my notes :frown:

now I have to start afresh.
What did you get in chem1?
Chem1 is a horrible,it's very hard to master .Ive got an A in chem2 but just a B (70Ums) in Chem1 .I think it's the calculations that let me down :|

Posted from TSR Mobile

Reply 44

Itsmonique

Original post by TheGemini

I'm still doing chem4 ,I've got about 4 topics left then I'll get to it .
Im planning on starting papers after Christmas :/ Wbu?
I made a mistake of throwing away my notes :frown:

now I have to start afresh.
What did you get in chem1?
Chem1 is a horrible,it's very hard to master .Ive got an A in chem2 but just a B (70Ums) in Chem1 .I think it's the calculations that let me down :|

Posted from TSR Mobile

We have just got to finish off 4 more chapters for chem4.
I'm just going over each topic in chem1 thoroughly so that I have a full understanding of the content before I start past papers. I plan to start past papers for chem1 during the christmas holidays, although everyone has a mock for chem1 in decemember.
I got a C(68 UMS) in chem 1, which I wa really upset about because I revised so much and did a lot of exam practice. I hope to get an A this time round. I think balancing the equations let me down.

Reply 45

Hudl

16

How far is everyone into CHEM4

Reply 46

Itsmonique

Original post by Hudl

How far is everyone into CHEM4

Just have acid and bases, and structure determination left to do. We missed quite a few chemistry lessons this term. Wbu? How are you finding it?

Posted from TSR Mobile

Reply 47

Soo confused on biodiesel ,glycerol /Propan-1,2,3-triol stuff :frown:

Help !!

Posted from TSR Mobile

Reply 48

banker-121

aqa jan 11
The acid dissociation constant, Ka, for the weak acid HY has the value1.35 x 10–5 mol dm–3 at 25 °C.
A buffer solution was prepared by dissolving 0.0236 mol of the salt NaY in50.0 cm3 of a 0.428 mol dm–3 solution of the weak acid HY

Calculate the pH of this buffer solution
I don't understand why u don't need to find the excess moles

Reply 49

Original post by dee1995

aqa jan 11
The acid dissociation constant, Ka, for the weak acid HY has the value1.35 x 10–5 mol dm–3 at 25 °C.
A buffer solution was prepared by dissolving 0.0236 mol of the salt NaY in50.0 cm3 of a 0.428 mol dm–3 solution of the weak acid HY

Calculate the pH of this buffer solution
I don't understand why u don't need to find the excess moles

You don't need to find the excess moles for that part of the question, excess moles need to be used in the next part of the question (when NaOH is added).

Have you completed the question? If not tell me and I'll happily go through it with you

Reply 50

Hudl

16

Just had my CHEM4 mock results today ew

Reply 51

banker-121

Original post by Dylann

You don't need to find the excess moles for that part of the question, excess moles need to be used in the next part of the question (when NaOH is added).

Have you completed the question? If not tell me and I'll happily go through it with you

Hey, I understand that in part b you have to find the excess but I don't understand why you don't have to in part A, in other words when don't u have to find out excess moles in buffer questions thanks

Reply 52

Original post by dee1995

Hey, I understand that in part b you have to find the excess but I don't understand why you don't have to in part A, in other words when don't u have to find out excess moles in buffer questions thanks ��

When you dissolve a salt of a weak acid into a solution of the weak acid they do not react (at least you assume they don't). For example, ethanoic acid CH3COOH and Sodium Ethanoate do not necessarily react together in solution. Therefore the moles of both substances remains constant (Ethanoic acid barely dissociates anyway).

You only have to find excess moles when things react. In part (a) nothing is reacting, you just have the moles of the salt (NaY) and the moles of the acid (HY) and the volume. You then can easily work out the concentrations of both, plug into Ka equation, find [H+] and -log it.

Remember in buffer solutions [H+] does not equal [X-] (or in this case [Y-])

Reply 53

banker-121

Original post by Dylann

When you dissolve a salt of a weak acid into a solution of the weak acid they do not react (at least you assume they don't). For example, ethanoic acid CH3COOH and Sodium Ethanoate do not necessarily react together in solution. Therefore the moles of both substances remains constant (Ethanoic acid barely dissociates anyway).

You only have to find excess moles when things react. In part (a) nothing is reacting, you just have the moles of the salt (NaY) and the moles of the acid (HY) and the volume. You then can easily work out the concentrations of both, plug into Ka equation, find [H+] and -log it.

Remember in buffer solutions [H+] does not equal [X-] (or in this case [Y-])

oo yes im getting the hang of this question so the Y from NaY and HY means they are from the same acid/salt so wont react?
by any chance do you have any similar questions where you dont have to find the excess moles? thanks so much :colondollar:

Reply 54

Original post by dee1995

oo yes im getting the hang of this question so the Y from NaY and HY means they are from the same acid/salt so wont react?
by any chance do you have any similar questions where you dont have to find the excess moles? thanks so much :colondollar:

Yes that's right, we assume they don't react.

I can't specifically link you to some (I'll check when I'm free) but www.a-levelchemistry.co.uk is decent for questions. Take note that many buffers questions will involve finding excess moles since the whole purpose of a buffer is to react with whatever is added thus you'll need to find excess moles of [H+] to find pH

Posted from TSR Mobile

Reply 55

Danny786

Does anyone have the CHEM4 June 2014 paper and MS? If so, can you post it here!
Thanks... :smile:

Reply 56

Original post by Danny786

Does anyone have the CHEM4 June 2014 paper and MS? If so, can you post it here!
Thanks... :smile:

It's not available and you shouldn't post it on here because a lot of people are doing their mocks with that paper. If you want it because you're doing a mock I recommend revising properly, it's not a bad paper honestly. If you're not doing a mock then maybe ask your teacher :smile:

Reply 57

FN510

Original post by TheGemini

Soo confused on biodiesel ,glycerol /Propan-1,2,3-triol stuff :frown:

Help !!

Posted from TSR Mobile

Just remember:
Methanol + triglyceride ---> biodiesel + glycerol (glycerol distills off)

R group from the triglyceride joins with the CH3 and the OH joins what remains of the triglyceride to make glycerol

Sodium hydroxide + triglyceride ---> soap + glycerol (glycerol distills off)

R group from the triglyceride joins with the Na and the OH joins what remains of the triglyceride to make glycerol

Reply 58

Danny786

Original post by Dylann

It's not available and you shouldn't post it on here because a lot of people are doing their mocks with that paper. If you want it because you're doing a mock I recommend revising properly, it's not a bad paper honestly. If you're not doing a mock then maybe ask your teacher :smile:

What kind of things come up and are they harder, or equal to previous papers. I have a feeling there will be some very tricky questions!

Reply 59