OCR A2 Chemistry Unit 5, F325 15th June 2015

Ugh i got 4.3x on the magnesium buffer, don't think that's right, most stuff seem good tho, i only got max 3 marks on the last question, plus i did the last 3 marker in acid conditions. I think the paper was good for most of it tho. Got 8.4x104 for the rate question, 0 for the ph of hydrogen electrode. 56.7/57.6 summert for kc. Starting to see I've lost some easy marks unfortunately 😫 still think I've got a good chance of getting a high B though, if not a low

Reply 3981

Cdp3

2

Original post by Nikeshp79

It can't as it contains no protons to donate

Another stupid mark lost

Reply 3982

HennersPD

3

Original post by ForgottenApple

Dunno lol
Most discriminating mark on paper imo -> Iodine as a solid lmao

haha!

Reply 3983

tewas

2

PLEASE DOES SOMEONE REMEMBER THE VALUES FOR THE DILUTION pH QUESTION??

Reply 3984

crypticdar

Guys is there an unofficial mark scheme?

Reply 3985

prof moriartress

Original post by games211

did anyone get 8.3 for anything?

I got 8.3 recurrent for Kc but everyone else got 8.3 * 10^4

Reply 3986

ECrazyMeC

Did anyone put i2 in solid state when filling in equations for haber cycle?

Reply 3987

hamza3256

8

did anybody put H+ ions for the last question involving Cr?

Reply 3988

klarriel

9

for the adding cobalt or copper to ammonia etc did we just have to put the complex ion examples???

Reply 3989

Hazza_

1

Original post by ECrazyMeC

Did anyone put i2 as solid when filling in equation for haber cycle??

Yeah, I am pretty sure iodine's standard state is solid.

Reply 3990

justfly

2

was CR +3 ON FIRST q?!!?

Reply 3991

tewas

2

would we only lose 1 or 2 marks for doing the equations in acidic conditions?

Reply 3992

kaihiwatari

6

Original post by ECrazyMeC

Did anyone put i2 in solid state when filling in equations for haber cycle?

I did

Reply 3993

plasmaman

18

>oxidation number (1)
>example of catalysis (1)
>bidentate definition (1)
>formula of the bidentate ligand (1)
>structure of the bidentate ligand and how it acts as a ligand (2)
>draw the optical isomers (1)
>reactions of Cu2+ ions (6)

>calculate Kc (6)
>what happens to pressure as equilibrium is approached (1)
>predict the sign and explain effect on yield (2)
>oxygen is now in excess, state and explain effect on Kc (3)

>calculate k (4)
>graphs for two temperatures (2)
>effect on k (1)
>graph for the first order relationship (2)

>acid definitions (1)
>pH of base (2)
>pH of solution (2)
>full equation (1)
>ionic equation (1)
>pH of buffer (1)
>adding NH3 and the effect on equilibrium (2)
>pH when Mg was added (4)

>predict the products when Fe2+ reacts with halogens (3)
>order of negativity (1)
>ph electrode question (1)
>question below this about equilibrium (2)
>conjugate acid-base pair question (1)
>why not use acidic conditions (1)

>born-haber cycle (4)
>lattice enthalpy definition (2)
>calculate lattice enthalpy (2)

>calculate entropy for hydrogen (2)
>calculate delta G and discuss feasibility (2)
>explain the significance of the temperature relating to feasibility (2)

>2 reactions (2)
>percentage purity (6)
>equations (last question) (3)

82 marks

Reply 3994

HeyC97

4

Original post by tewas

Shall i make an unofficial? Send me asnswers and questions

Yes please!! but i cant remember any questions

Reply 3995

hamza3256

8

Original post by tewas

would we only lose 1 or 2 marks for doing the equations in acidic conditions?

that's what i did, but apparently we are meant to use OH- instead of H+ ions

Reply 3996

AndyChow

19

Original post by kaihiwatari

Ugh i got 4.3x on the magnesium buffer, don't think that's right, most stuff seem good tho, i only got max 3 marks on the last question, plus i did the last 3 marker in acid conditions. I think the paper was good for most of it tho. Got 8.4x104 for the rate question, 0 for the ph of hydrogen electrode. 56.7/57.6 summert for kc. Starting to see I've lost some easy marks unfortunately 😫 still think I've got a good chance of getting a high B though, if not a low

Other 2 yes.
Magnesium one is tricky
0.25 mol of Mg means 0.5 mol of acid will react
so [C2H5COO-]=1.5M
[C2H5COOH]=0.5M

pH=-log(1.35E-5) +log(1.5/0.5)
pH=5.35 right?

Reply 3997

annnnnnnn

4