OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD

Reply 200

76584

8

does anyone have the june 2015 papers? need them for revision

Reply 201

Miminfl

6

Guys please help with question (ii)
The mark scheme doesn't make sense :frown:

been trying to understand it but doesn't make sense

Reply 202

ReeceM1

12

Original post by Miminfl

Guys please help with question (ii)
The mark scheme doesn't make sense :frown:

been trying to understand it but doesn't make sense

I haven't seen the mark scheme for this answer but I'll try to tell you how'd I go about doing it:

With solution A, there is 5x10^-3 moles of acid reacting with 2.5x10^-3 moles of NaOH therefore after the reaction has occured some acid will still remain as it is in excess. The acid will be in equilibrium with its conjugate base (ethanoate ions).

But with solution B, there is 5x10^-3 moles of acid reacting with the same amount of NaOH, therefore there would be full neutralisation due to the 1:1 quantities. There would be no acid remaining as it all would have reacted, therefore no equilibrium will exist.

Reply 203

Miminfl

6

Original post by ReeceM1

I haven't seen the mark scheme for this answer but I'll try to tell you how'd I go about doing it:

With solution A, there is 5x10^-3 moles of acid reacting with 2.5x10^-3 moles of NaOH therefore after the reaction has occured some acid will still remain as it is in excess. The acid will be in equilibrium with its conjugate base (ethanoate ions).

But with solution B, there is 5x10^-3 moles of acid reacting with the same amount of NaOH, therefore there would be full neutralisation due to the 1:1 quantities. There would be no acid remaining as it all would have reacted, therefore no equilibrium will exist.

Thank you, that really helped , I didn't know I had to find out the number of moles. :smile:

Reply 204

Miminfl

6

Original post by Hunnybeebee

In solution A, there is excess acid as there's 5x10^-4 of ethanoic acid, but there's 2.5x10^-4 of NaOH. As there is excess, there will be an equilibrium of CH3COOH, and it's conjugate base of CH3COO- (which was formed by CH3COONa).

In solution B, these is equal amount of mol of each solution, therefore they will completely react:

CH3COOH + NaOh --> CH3COONa + H2O. No equilibrium, therefore cannot be a buffer solution as it isn't a weak acid and its conjugate base.
Thank you :smile:

it makes sense now

Reply 205

ReeceM1

12

Does it mention somewhere further up in the stem of the question anything about diluting something to 1dm3 / or using 1dm3 of sewage water?

If it does then it would be x40 because only 25cm3 is used in the titration, so this would be scaled up 40x to get the values for 1dm3.

If it doesn't mention that then I'm not sure.

Reply 206

bakedbeans247

20

Ah i've seen somehting similar before in a MS, "Where did this number come from?". Funny enough it was also the number 40. Here, on this video, skip to 1 : 38 : 45 https://www.youtube.com/watch?v=VaJBbcXaFBM&index=5&list=PLbJbCQgCpBPs0IQaePXLxLvLnv9Jq3TO1

or here's another thread that answered your q:
http://www.thestudentroom.co.uk/showthread.php?t=1665699

(edited 7 years ago)

Reply 207

rory58824

2

Give this a go, hundreds of questions here:

http://www.a-levelchemistry.co.uk/OCR%20Chemistry%20A/OCR%20Chemistry%20A%20home.html

Reply 208

AqsaMx

5

Anyone able to explain the ozone lysis of the second one? Thank you!

Reply 209

xsashax

3

Original post by AqsaMx

Anyone able to explain the ozone lysis of the second one? Thank you!

Sorry but I haven't even heard of ozone lysis, which board is this for??

Posted from TSR Mobile

Reply 210

supercrazyxo

1

Original post by xsashax

Sorry but I haven't even heard of ozone lysis, which board is this for??

Posted from TSR Mobile

Its from ocr

Reply 211

popo111

10

Original post by xsashax

Sorry but I haven't even heard of ozone lysis, which board is this for??

Posted from TSR Mobile

It's not something we learn, but rather than application based question.

Reply 212

AqsaMx

5

It's an OCR A application question. June 2014 paper F324 :smile:

Reply 213

Clankdroid

1

Original post by AqsaMx

Anyone able to explain the ozone lysis of the second one? Thank you!

for hexa-2,4-diene, it has structural formula of CH3CHCHCHCHCH3, so CH3-CH=CH-CH=CH-CH3,

so the products would be (CH3)CH=O + O=CH-CH=O + O=CH(CH3),

as the C=C double bonds become C=O, for both carbons in the C=C double bond

Reply 214

itsConnor_

11

Jan 2013 F325 is so bloody hard :frown:

Reply 215

xsashax

3

Original post by AqsaMx

It's an OCR A application question. June 2014 paper F324 :smile:

Oh okay thanks

Reply 216

username2097061

2

Really? Wow that's impressive. I have my mock next week so I guess I'll have to wait. However, seeing the statements on here and the fact that it was featured on bbc news, certainly does not excite me for it.

Reply 217

thad33

12

Original post by mil88

Really? Wow that's impressive. I have my mock next week so I guess I'll have to wait. However, seeing the statements on here and the fact that it was featured on bbc news, certainly does not excite me for it.

I sat the exam last year in the summer. I think I got around 5/7 for the working out but not the right structure. I walked out thinking that question was really hard but looking back on it like 5 minutes ago it isn't too bad.
The timing is the biggest stress. If you can nail down the rest of the paper and leave yourself with a good 15-20 minutes for NMR then you should be ok.

Reply 218

username2097061

2

Original post by thad33

I sat the exam last year in the summer. I think I got around 5/7 for the working out but not the right structure. I walked out thinking that question was really hard but looking back on it like 5 minutes ago it isn't too bad.
The timing is the biggest stress. If you can nail down the rest of the paper and leave yourself with a good 15-20 minutes for NMR then you should be ok.

Fair enough, but I heard that the biggest issue was time. So I'm not sure how I'm going to have 15-20 mins left over for that question, when apparently loads of questions were application based, and thus take longer to answer?