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Aqa chem 4/ chem 5 june 2016 thread

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Original post by Signorina
If it doesn't ask you to leave your answer to a specific d.p, does that mean you can round it to any decimal place?? doesn't really say on the mark schemes???


PH is always to 2dp. Everything else, try and make your answers to 3 sig.fig
Original post by hopingmedicinae
Just for Reference guys;

Terylene is a polyester made from 1-4 benzene dicarboxylic acid + ethane-1-2-diol

Kevlar is a polyamide - Benene 1-4 diamine + Benzene 1-4 dicarboxylic acid

Nylon 6-6 - Polyamide - 1-6 diaminohexane + 1--6 hexanedioic acid


do we need to memorise the monomers for each or will they give it to us?
Original post by Lilly1234567890
do we need to memorise the monomers for each or will they give it to us?


Im pretty sure you dont have to remember the monomers, although you need to remember that Terylene is a polyester and Kevar and Nylon6,6 are polyamides etc
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BJA98
4
What mark out of 100 is everyone getting and what grade are they aiming to get? Im just about 80 and aiming for an A.
Original post by Sniperdon227
im so confused you've bypassed my question and contradicted yourself so many times


Original post by Parallex
2 people including myself have tried answering whatever you're asking, I literally don't even know what else you want me to say. There's nothing to it other than that...


I think the main problem here is how you are trying to convey the information. Yes H2SO4 is diprotic so theres double the amount of H+ than you usually see. But you dont say it's 2[H+]. You say that mol [H+] = 2 x mol H2SO4. When you say 2[H+] it looks you are saying that the acid gives 2 lots of H+ and the amount there is, is that amount divided by 2. It can get confusing. Obviously you do not mean that really but the way you are saying it sounds like you are.
Original post by shiney101


The only thing I don't get is why you wouldn't divide the final moles by the volume it gives but this is my working for the rest of it- hope it helps!
13451142_462918260584339_129731565_n.jpg46.4KB
Original post by BJA98
What mark out of 100 is everyone getting and what grade are they aiming to get? Im just about 80 and aiming for an A.


Around 80+. I need an A. I don't really want to scrape it either.
Original post by SirRaza97
I think the main problem here is how you are trying to convey the information. Yes H2SO4 is diprotic so theres double the amount of H+ than you usually see. But you dont say it's 2[H+]. You say that mol [H+] = 2 x mol H2SO4. When you say 2[H+] it looks you are saying that the acid gives 2 lots of H+ and the amount there is, is that amount divided by 2. It can get confusing. Obviously you do not mean that really but the way you are saying it sounds like you are.


First reply: There's 0.02904 moles of H+ (2x the moles of H2SO4) and 0.014112 moles of OH-. Excess moles of H+ = 0.014928. Concentration of H+ = 0.014928/(46*10^-3). -log(10) of this = 0.49 to 2 dp.

Second reply; What do you mean? H2SO4 fully dissociates so the moles of H+ is 2x the moles of H2SO4. That's all it is.

Third reply was talking about an individual H2SO4 molecule, which dissociates to give 2H+.

:s-smilie:
http://filestore.aqa.org.uk/subjects/AQA-CHEM4-W-QP-JAN11.PDF
Hi i need help with 2di and 2dii...

For 2di, how come you're meant to to use moles of NaY in the calculation instead of concentration?
and for 2dii
Could someone perhaps explain what's happening? THANK YOU!!! Anyone :smile:
Original post by Parallex
First reply: There's 0.02904 moles of H+ (2x the moles of H2SO4) and 0.014112 moles of OH-. Excess moles of H+ = 0.014928. Concentration of H+ = 0.014928/(46*10^-3). -log(10) of this = 0.49 to 2 dp.

Second reply; What do you mean? H2SO4 fully dissociates so the moles of H+ is 2x the moles of H2SO4. That's all it is.

Third reply was talking about an individual H2SO4 molecule, which dissociates to give 2H+.

:s-smilie:


The first reply makes perfect sense. I don't know what the second reply is trying to ask. That's exactly what first reply did

"Third reply was talking about an individual H2SO4 molecule, which dissociates to give 2H+"

H2SO4 dissociates to give 2 lots of H+ relative to a monoprotic acid.
(edited 7 years ago)
I'm confused with buffer calculations/calculating the pH when you add two solutions.

When do I need to use the total volume and when do I convert things to concentration? I always mix it up :frown:
Most of the people on this thread are getting confused with buffers! :')

please read this, it helped me so much to understand the theory and also how to carry out the calculations:

https://chemrevise.files.wordpress.com/2014/04/mod-4-revision-guide-3-acid-base-equilibria.pdf

(Scroll down for the buffers bit)
Original post by BJA98
What mark out of 100 is everyone getting and what grade are they aiming to get? Im just about 80 and aiming for an A.


I'm around 84, but some of the boundaries of the paper are so high :frown:

I did the Jan 2010 paper this morning and the grade boundaries were a dream :biggrin: 74% for an A, but I guess that was because it was one of the first papers in the spec.
Original post by lahigueraxxx
Most of the people on this thread are getting confused with buffers! :':wink:

please read this, it helped me so much to understand the theory and also how to carry out the calculations:

https://chemrevise.files.wordpress.com/2014/04/mod-4-revision-guide-3-acid-base-equilibria.pdf

(Scroll down for the buffers bit)


This is wonderful, thank you so much!
Original post by Signorina
http://filestore.aqa.org.uk/subjects/AQA-CHEM4-W-QP-JAN11.PDF
Hi i need help with 2di and 2dii...

For 2di, how come you're meant to to use moles of NaY in the calculation instead of concentration?
and for 2dii
Could someone perhaps explain what's happening? THANK YOU!!! Anyone :smile:


2d(i) You can use the concentration but that would take much longer to do but should give the same answer. You would fond the conc of each in the total volume and use that. Using mol values negates the fact that they are in equal volume so it's much easier.

2d(ii) Adding OH- will get rid of H+ ions. So equilibrium will shift to the right to make up for the loss of H+ thus more of the acid will dissociate. So the amount of HA decreases and the amount of Y- increases. Again the conc of each can be calculated but it would take longer to do.

HA mol - mol of OH added and NaY mol + mol of OH added.
Original post by NotMolly
This is wonderful, thank you so much!


You're welcome! :smile:
What are the conditions for the use of NaBH4 as a reducing agent to reduce aldehydes and ketones?
Original post by Signorina
http://filestore.aqa.org.uk/subjects/AQA-CHEM4-W-QP-JAN11.PDF
Hi i need help with 2di and 2dii...

For 2di, how come you're meant to to use moles of NaY in the calculation instead of concentration?
and for 2dii
Could someone perhaps explain what's happening? THANK YOU!!! Anyone :smile:


You can either use both moles or both concentrations. This works because the number of moles = concentration/volume, so the factor of 1/v cancels on each side of the fraction.
why is it at half neutralisation point that there are equal moles of HA and A-?
Original post by hopingmedicinae
Hope this helps! By the way, number 8 is weirdly back in the benzene ring, I forgot to mark it so had to go back to it!


Ah i see now thankyou! I was thinking 3 and 4 were symmetrical so I was getting 10!

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