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

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    ums / grade boundaries thoughts?
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    (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
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    (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.?
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    (Original post by X_IDE_sidf)
    ums / grade boundaries thoughts?
    I think they will be near average. Fairly high 80% for an A?
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    (Original post by X_IDE_sidf)
    So sort of 80% for A
    70% B
    60% C etc.?
    Probably
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    (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
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    Anyone know the answer to the enthalpy change of formation question?
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    (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
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    (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?
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    *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?
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    (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
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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
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    (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?
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    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...
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    (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
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    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?
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    (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.
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    (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.
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    (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?
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    (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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