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ocr a f325 revision thread

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Reply 3640
Avatar for emah123
emah123
OP
5
Original post by tangaroo
i could've sworn it wasn't THAT bad guys?


errr yeah it was.
I'm guessing, There was too many easy (ish ) questions and then a a couple v.v.v.v.v.v hard questions.
A: 75
B: 66
C:59
D:53
E: 47
Anyone agree?
Well that sucked.
Original post by _HabibaH_
If you tell me you enjoyed it, I will cry...real tears. :unimpressed:


It was tolerable is what I would say, but definitely more maths and buffer based really
Reply 3644
Avatar for amin666
amin666
Original post by Brad0440
I just wrote because its concentration doesn't affect the rate. I meant to go back at the end to expand on it, but I ran out of time. :s-smilie:


i wrote becasue its a solution and water is in excess so the change in concentration doesn't affect the rate so its zero
Oh, and the concentration of water was zero because it was in excess and was therefore effectively contant, and thus had no effect on the rate of reaction
Reply 3646
Avatar for Bazuka
Bazuka
Original post by Allotropy
Overall I thought the paper was hard but doable. Looking through the answers people have posted I seem to be the only one who calculated a value of 3.55 for the buffer solution pH, did anyone else get this?

I got that but then I realised I hadn't taken away the moles of HCOOH which react with NaOH to give the HCOO-... I then changed it and got 3.99 :smile:
what did everyone draw for the optical isomers????
Original post by amin666
why order of H2O was zero?


Because it's in such great excess, that any change in concentration is negligible.
Reply 3649
Avatar for Brad0440
Brad0440
On the question where you had to draw the optical isomers, did people put that the =O was bonded to the central metal ion, or the O-?
Does anyone remember what the alkali we had to work pH out from was? I swear it's Ca(OH)2 and I forgot to multiply the concentration by 2 =( Unless it was NaOH then I did it right LOL
alright chapitos? that was an alright paper
Original post by TauMuon
Oh, and the concentration of water was zero because it was in excess and was therefore effectively contant, and thus had no effect on the rate of reaction



Original post by amin666
i wrote becasue its a solution and water is in excess so the change in concentration doesn't affect the rate so its zero


I wrote that... along with it didn;t appear in the rate equation.... ;/

Original post by amin666
i wrote becasue its a solution and water is in excess so the change in concentration doesn't affect the rate so its zero
Original post by keepontrying
what did everyone draw for the optical isomers????


I drew the iron with three of those bidentate ligands around it. I think you could also have taken two of the bidentate ligands and left two waters on there. Not sure really.
Original post by Brad0440
On the question where you had to draw the optical isomers, did people put that the =O was bonded to the central metal ion, or the O-?


I put the O-
Reply 3655
Avatar for amulla9
amulla9
did anyone get NiSO4.7H20 for the last question
Original post by TauMuon
For the redox one I think I did:

2[Fe(H2O)6]^3+ + Al ====> 2[Fe(H2O)6]^2+ + Al^2+

Which seems completely wrong to me. Not sure I should have included the complexes. Oh well. Don't know if that's exactly what I put; can't remember what the Aluminium was oxidised to...


For the ligand substitution one I think I did:

[Fe(H2O)6]^3+ + 6CN^- ====> [Fe(CN)6]^3- + 6H2O



For the use of NH3 in the bidentate ligand question:

NH3 oxidised the ligand (or words to that effect) :s-smilie:


The amount of answers I have no idea are correct or not is too damn high..


I got the same equations as you did, I'm hopefully going for 85+!! :biggrin:
When they say write an equation, is it ok for the ionic equation cause they didn't specify FULL or IONIC
Reply 3658
Avatar for AishaTWx
AishaTWx
i'm hoping the grade boundaries low, safe to say, this is the hardest paper i've ever done
Original post by Sarangtaec
Does anyone remember what the alkali we had to work pH out from was? I swear it's Ca(OH)2 and I forgot to multiply the concentration by 2 =( Unless it was NaOH then I did it right LOL


NaOH - relax!

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