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# F334 (19/06/2013 9:00AM) - Salters Chemistry (Chemistry of Materials) Revision Thread watch

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1. (Original post by stealth_writer)
Help with Q 2 (b) (ii) Jan 13 please
Okay to start, 16cm3 of the NaOH reacted therefore using n=cv (16/1000)*0.25=4x10^-3 mol of NaOH reacted, it is a one to one ratio so 4x10^-3 mol of the acid reacted, this is was in 25cm^3 of the diluted acid solution.

From here you can take 2 approaches
1) you could work out the concentration of the diluted solution and then scale it:
using c=n/v c=4x10^-3/(25/1000)=0.16moldm^-3, you know the ratio between the volumes of the diluted and undiluted solutions 250/14=17.85..., so the original solution is 17.85... time more concentrated so the original concentration is 17.85...x0.16=2.86moldm^-3

2) Or another way of doing it is this, you know that number of moles of the acid in 25cm^3 solution 4x10^-3, so in 250cm^3 diluted solution you got the 25cm^3 from must contain 10 times as much moles of acid 4x10-3x10=4x10^-2, this must contain the same number of moles of acid as the diluted solution because no moles of solution is lost, so the original concentration must be this number of moles over the original volume 4x10^-2/(14/1000)=2.86moldm^-3

I'm assuming you can take it from here, give us a shout if you need further help

I got the mark scheme if anyone needs it
Attached Images
3. F334 Jan 2013.pdf (95.1 KB, 80 views)
4. (Original post by Dude Ranch)
Ah thanks so much its so simple now!
(Original post by Tikara)

iv) is your stock enantiomer question - if it was monodentate then it wouldn't have this but due to the bidentate nature if you mirror the image it's nonsuperimposable (not the same when put on top of each other)

my problems in that question were that I put diaminoethane rather than 1,2-diaminoethane
and didn't know that electronegativity is only in a bond grrrr
your welcome and ah i see iv now!
5. (Original post by Mtheodore)

I got the mark scheme if anyone needs it
here you go

That disappearance question is hurting my head arghh
Attached Images
6. F334 Jan 2013 QP.pdf (344.6 KB, 75 views)
7. (Original post by stealth_writer)
and the very final question about rate of Iodine disappearing or what not

cheers!
I think I got the disappearance question guys :3

After some searching, it's just to do with the moles - granted I had to go back a bit to see the molar ratios
rate of disappearance of I- is the rate of production times 2 because I- to I2 are in a 2:1 ratio!

S2O82- + 2I- --> I2 +2SO42-

So just to put things in context the rate of disappearance of S2O82- would be the same as the reaction rate (as the 1:1 ratio) and the rate of production of 2SO42- would be 2 times the rate of iodine production cause of the moles!!

hope that helps (and is correct) it seems to make sense though
8. Can anyone explain to me how you work out the repeating unit? For example 1ai in the january 2013 paper please
9. make sure to get lots of sleep people! I didn't get much last night and feel it affected my exam this morning ( so make sure y'all are ready and make sure you know all of the experimental techniques like reflux and recrystallisation!!

and Good luck for tomorrow!!!!
10. if i draw the polarity signs on unnecessary atoms will i lose marks?

for example on 3b jan 2013, I usually would have drawn those extra ones in red. will I get penalised for that? not sure when to do it on the atom behind since i see it done like that sometimes but not other times
11. (Original post by Kreayshawn)
if i draw the polarity signs on unnecessary atoms will i lose marks?

for example on 3b jan 2013, I usually would have drawn those extra ones in red. will I get penalised for that? not sure when to do it on the atom behind since i see it done like that sometimes but not other times
No you won't get penalised; I would've done the same, as it's best to be safe
12. (Original post by Tikara)
I think I got the disappearance question guys :3

After some searching, it's just to do with the moles - granted I had to go back a bit to see the molar ratios
rate of disappearance of I- is the rate of production times 2 because I- to I2 are in a 2:1 ratio!

S2O82- + 2I- --> I2 +2SO42-

So just to put things in context the rate of disappearance of S2O82- would be the same as the reaction rate (as the 1:1 ratio) and the rate of production of 2SO42- would be 2 times the rate of iodine production cause of the moles!!

hope that helps (and is correct) it seems to make sense though
Yah! I also finally got this today! Such a cheeky question tbh. Cheers for writing it out; I'll laugh if they ask us the same tomorrow
13. What is the charge on an ethanediote ligand????
14. (Original post by hannah1994)
What is the charge on an ethanediote ligand????
2-, and it's ethanedioate
15. (Original post by Dualcore)
2-, and it's ethanedioate
thankyou and ah yeah
16. (Original post by hannah1994)
thankyou and ah yeah
No problem
17. All the best everyone for tomorrow. I'm guessing it's the last exam for the majority, so try your best to end on a massive high! Best of luck x
18. anyone pulling an all-nighter?
19. (Original post by amber109)
anyone pulling an all-nighter?
Half a nighter. Want to go through all the papers.
20. (Original post by EstebanK0)
Half a nighter. Want to go through all the papers.
Me too bro. T_T with you guys on this

-Worst bit is. I have an exam on thurs. And friday.
21. Could someone help me with question Q2biii January 2011 I'm really stuck for some reason
I thought when finding order with respect to one substance the other two have to be constant concentration right? for each row? So confused! Could someone try explain ^^''
22. (Original post by RainieXD)
Me too bro. T_T with you guys on this

-Worst bit is. I have an exam on thurs. And friday.
I have an exam on thurs too :/

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