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

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Reply 60
Original post by mechanism
Guys - does anyone know how to do the last question F324 June 2015? I really can't get my head around it.


https://www.youtube.com/watch?v=UuEfvamQrP4
This may help with that
Reply 61
Original post by 28Ikky
Tell me about it!! I have my I S AA tomorrow and its the evluative on identifying 2 weak acids -__- my last attempt to bump my grade up and IDK wth to do :frown: pm somebody :frown:


Any tips on what to revise
Reply 62
You know cos the paper is synoptic with AS, does anybody know how much AS I should revise? Would it be worth me going over a few papers?
Reply 63
Also, I achieved an A at AS, does anybody know what is the lowest grade I can achieve on F326 to still get an A* or an A provided I achieve high marks on the exams??
Reply 64
Original post by AqsaMx
Also, I achieved an A at AS, does anybody know what is the lowest grade I can achieve on F326 to still get an A* or an A provided I achieve high marks on the exams??


480 ums to get an A

Lets say you got 240 at AS, and you get A in f324 and f325, tallying up to 240+120+72 = 432

Youd need 40 ums out of 60 for an A in f326, which is like 33 out of 40 (high C)
Reply 65
Original post by zirak46
480 ums to get an A

Lets say you got 240 at AS, and you get A in f324 and f325, tallying up to 240+120+72 = 432

Youd need 40 ums out of 60 for an A in f326, which is like 33 out of 40 (high C)


33 is a b? But thank you :smile:
Reply 66
Original post by AqsaMx
33 is a b? But thank you :smile:


I think grade boundaries were 37 for A and 34 for B, i may be mistake, best to be safe and think its higher
Original post by Sade_101
Can someone help with this question please:

A 25.0 cm3 portion of a solution containing Fe2+ and Fe3+ ions was acidified and titrated against potassium permanganate (VII) solution. 15.0 cm3 of a 0.0200 mol dm-3 solution of potassium manganate (VII) were required. A second 25.0cm3 portion was reduced with zinc and titrated against the same manganate solution. This time 19.0cm3 of the oxidant was required.
a) calculate the concentrations of Fe2+ and Fe3+ ions in the solution
b) what percentage of the total iron content of the solution was Fe3+

So far I've done:
MnO4- + 8H+ + 5Fe2+ ---> Mn2+ + 4H2O + 5Fe3+
mole of MnO4 = 3x10-4 mol and mole of Fe2+ = 1.5x10-3 mol (1:5 mole ratio)
concentration of Fe2+ = 0.06 moldm-3

I have no idea how to work out the concentration of Fe3+


What is the actual answer? I got Fe3+ = 0.032 moldm-3 for this but could be wrong
Reply 68
Original post by zirak46
I think grade boundaries were 37 for A and 34 for B, i may be mistake, best to be safe and think its higher


It was 33 for B, but if I got 33, I can't get an A* even if I get above 90 on both exams?
has anyone done a qual on transition metals? have it tomorrow :frown:
Does anyone know how strict examiners when asking for the conditions required for a reaction? (i.e. if I were to write H2SO4 instead of concentrated H2SO4, would I then lose credit?)
Reply 71
Original post by rory58824
Does anyone know how strict examiners when asking for the conditions required for a reaction? (i.e. if I were to write H2SO4 instead of concentrated H2SO4, would I then lose credit?)


I would think so, as H2SO4(aq) probably wouldn't work.
Original post by rory58824
Does anyone know how strict examiners when asking for the conditions required for a reaction? (i.e. if I were to write H2SO4 instead of concentrated H2SO4, would I then lose credit?)


I thought they usually say ignore concentrated or dilute but you'd need to say aq for ester hydrolysis? I might be wrong!


Posted from TSR Mobile
Original post by Saywhatyoumean
I thought they usually say ignore concentrated or dilute but you'd need to say aq for ester hydrolysis? I might be wrong!


Posted from TSR Mobile


I've seen some mark schemes which may say (conc.) H2SO4 for example, but like alow said, stating H2SO4 (aq) would be wrong in a reaction such as oxidising an primary alcohol.

I think it might just be best to remember everything and not leave anything out!
Most of the time it says ignore Ref to Concentration on the mark scheme. Although I have seen times where it says conc but I'm not sure I've ever seen dilute?

So when it asks for reagents I just put concentrated and either way it'll normally be right.
Original post by rory58824
Does anyone know how strict examiners when asking for the conditions required for a reaction? (i.e. if I were to write H2SO4 instead of concentrated H2SO4, would I then lose credit?)


Most of the time they ignore it
I've only seen when it's required for esters and for the reduction of nitrobenzene - pretty sure you must state concentrated H2SO4 and concentrated HCl for those ones
Original post by ReeceM1
Most of the time it says ignore Ref to Concentration on the mark scheme. Although I have seen times where it says conc but I'm not sure I've ever seen dilute?

So when it asks for reagents I just put concentrated and either way it'll normally be right.


Well you would have to put dilute acid/alkali for the hydrolysis of esters.

Original post by tcameron
Most of the time they ignore it
I've only seen when it's required for esters and for the reduction of nitrobenzene - pretty sure you must state concentrated H2SO4 and concentrated HCl for those ones


Thanks for that, I might as well try and remember the exact conditions for each reaction, just in case.
Reply 77
image.jpgimage.jpg
I've worked out the precipitate is Fe(OH)3 but could someone explain how to calculate the gas from density PLEASE
Original post by AqsaMx
image.jpgimage.jpg
I've worked out the precipitate is Fe(OH)3 but could someone explain how to calculate the gas from density PLEASE


I assume you mean the next question?
Think about the equation to find the volume of a gas: n=v/24,000 cm3
Assume 1cm3 = 1=333x10^-3 g
So that means 24,000cm3 = 1.333x10^-3 x 24,000
That equals 31.992 gmol^-1
On the periodic table the molecular mass closet to this value must be O2 (oxygen) which is 32gmol^-1
So the gas is O2
(edited 7 years ago)
Reply 79
Original post by tcameron
I assume you mean the next question?
Think about the equation to find the volume of a gas: n=v/24,000 cm3
Assume 1cm3 = 1=333x10^-3 g
So that means 24,000cm3 = 1.333x10^-3 x 24,000
That equals 31.992 gmol^-1
On the periodic table the molecular mass closet to this value must be O2 (oxygen) which is 32gmol^-1
So the gas is O2


But how do we know 1cm3 = 1.333x10-3 and why are you multiplying volume by 24?
Sorry, I'm really confused :frown:

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