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OCR Chemistry A 2016 unofficial mark scheme 27/05/16

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ums / grade boundaries thoughts?
Reply 21
Original post by X_IDE_sidf
ums / grade boundaries thoughts?


I thought it was an easy paper compared to the old spec, but cause it's a new spec might be slightly lower than usual, so probably similar to previous years imo
Original post by Bosssman
I thought it was an easy paper compared to the old spec, but cause it's a new spec might be slightly lower than usual, so probably similar to previous years imo


So sort of 80% for A
70% B
60% C etc.?
Original post by X_IDE_sidf
ums / grade boundaries thoughts?

I think they will be near average. Fairly high 80% for an A?
Reply 24
Original post by X_IDE_sidf
So sort of 80% for A
70% B
60% C etc.?


Probably
Original post by Chmbiogeog
I think they will be near average. Fairly high 80% for an A?


81% for an a in the atoms and bonds paper jan 2013
Anyone know the answer to the enthalpy change of formation question?
Reply 27
Original post by Apat1326
Anyone know the answer to the enthalpy change of formation question?


It's in the mark scheme, -58.5kJ/mol

Edit: thought you mean the other one, it was -129kJ/mol
Reply 28
Original post by medicapplicant
i think this is what i got for the last question
1st reagent Br2 or Cl2
halogenated compound
2nd reagent NaOh


For the second reagent, I just put OH- ions. Will I lose mark?
*No of atoms in 5.00g coin: 3.98 x 10^22 (3sf) *

For this one, I got 3.97 x 10^22 for using the RAM calculated in the part before?? Did you have to use the periodic table one or the calculated one?
Reply 30
Original post by SGHD26716
For the second reagent, I just put OH- ions. Will I lose mark?


Hard to tell, totally depends on what the mark scheme says, it could say either
Reply 31
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For the non polar molecule question I put E-2,3-dichlorobut-2-ene as this was the only molecule where the resultant dipole was right in the middle of the molecule
Original post by Bosssman
Hard to tell, totally depends on what the mark scheme says, it could say either


How many marks was the copper coin one i put it to 4 sig figs... did you have to write the x10^22 bit or did it already have it there and you just put the 3.whatever in?
Reply 33
Oh wow, I though the exam went awfully! And here everyone's saying it wasn't too bad...

Can anyone remember the marks for any of the non multiple choice questions? I'd love to know how badly I did so I can prepare...
Original post by SGHD26716
For the second reagent, I just put OH- ions. Will I lose mark?


im not sure but i think that should be alright because OH- is the reacting species
I think the answer to the first question on Section B is different - wasn't it an ion, and thus the electron and proton quantities are different?
Reply 36
Original post by medicapplicant
im not sure but i think that should be alright because OH- is the reacting species


I put H2O, didn't even think of a hydroxide! Wow.
Reply 37
Original post by Bosssman
Here's my attempt at an unofficial mark scheme, it is for the OCR Chemistry A, new spec, breadth in chemistry paper, really can only put the maths questions down, but if there are written ones I can remember I will put them down. This is only really for comparison of numbers, and general ideas

Multiple choice:
BCDACBBDCBCBBCBDDBBD (if anyone happened to store their multiple choice answers on there calculator, please post them below to compare)

Same number of protons and electrons, different number of neutrons

Relative atomic mass: 63.62
No of atoms in 5.00g coin: 3.98 x 10^22 (3sf)
Percentage error of water removed mass: 1.72%
Volume of H2SO4 required to neutralise NaOH: 24.3cm^3
Enthalpy change of reaction: -58.5 kJmol^-1
Concentration of KI required: 3.3 moldm^-3
Concentration in equilibrium question: 0.876 moldm^-3

Homologous series: alkenes
General formula: CnH2n

Electron configuration of bromide ion: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6

Equilibrium question: The equilibrium position will shift to the right, as the forward reaction is exothermic, in an attempt to reduce the effect of the temperature decrease. The equilibrium position will also shift to the right as there are fewer gas molecules in the products of the forward reaction, in an attempt to reduce the effect of the pressure increase. This may vary to the actual conditions as low temperature reduces rate of reaction, and it's expensive to have a high pressure.

Ways to reduce percentage uncertainty: use a larger mass of crystals to get larger mass of water removed, reducing % error

Halogen precipitate colours: chloride, white; bromide, cream; iodide, yellow

Last question: step 1 reagent, Br2, step 2 reagent, NaOH

If anyone remembers the written answers, please post and I'll add them,
Any incorrect (although I don't think there are) please tell and I'll correct

Also, post about how you found the exam, as this will give a good idication of grade boundaries


You cannot use Br2 because you will make a dihaloalkane and you cannot make an alcohol from dihaloalkane. You had to use HBr to make haloalkane and then NaOH to make alcohol.
Reply 38
Original post by LordStark
I think the answer to the first question on Section B is different - wasn't it an ion, and thus the electron and proton quantities are different?


What was the question?
Reply 39
Original post by laney1999
How many marks was the copper coin one i put it to 4 sig figs... did you have to write the x10^22 bit or did it already have it there and you just put the 3.whatever in?


No, you had to put the x 10^22, and also I don't think significant figures was specified in the question

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