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

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    (Original post by Aeluna)
    I put H2O, didn't even think of a hydroxide! Wow.
    H20 is one of the right answers i think i remember it being correct on one of the past papers i did
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    (Original post by tamoni4)
    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.
    The end product was a diol...
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    (Original post by Bosssman)
    The end product was a diol...
    There's probably more than one correct method
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    (Original post by tamoni4)
    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.
    24.8 for h2s04 and 0.912 for the kc value
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    (Original post by Bosssman)
    No, you had to put the x 10^22, and also I don't think significant figures was specified in the question
    I think it did say to 3sf?
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    (Original post by 456CJ)
    24.8 for h2s04 and 0.912 for the kc value
    Yes it was 24.8, that has been changed
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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

    I didn't put "electrons" will I lose marks?

    Relative atomic mass: 63.62
    No of atoms in 5.00g coin: 3.98 x 10^22 (3sf)

    do you remember the calculation for the coin question?

    Percentage error of water removed mass: 1.72%

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

    how to make sure all the water's gone: I put reweigh and reheat repeatedly until constant mass is achieved.

    Volume of H2SO4 required to neutralise NaOH: 24.3cm^3
    Enthalpy change of reaction: -58.5 kJmol^-1
    Spoiler:
    Show
    I also got this wrong idk what the **** I did I got 20 something

    Concentration of KI required: 3.3 moldm^-3
    Concentration in equilibrium question: 0.876 moldm^-3
    Spoiler:
    Show
    I got this wrong cause I though kc=25.1


    Homologous series: alkenes
    General formula: CnH2n

    there was a question after this bit I don't remember, all I know is I put: h2, nickel catalyst.
    then there was a diagram: BLANK - H2S04 catalyst and heat - THE GIVEN ALKENE (I can draw it out from memory but idk how to put it on here) - ANOTHER BLANK

    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.

    what happens when you put chlorine gas into bromine ions:I put chlorine is more reactive that bromine, so it displaces bromine that causes whatever change the question was asking for. ionic equation for the equation given with PBII put: PB^2+ (aq) + 2I- (aq) —> PbI2 (s)

    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

    I also remember a random gas question I got 32.7cm^3

    (Original post by Bosssman)
    The end product was a diol...
    it wasnt a haloalkane thoughit was just an alkane with OH groups on either end
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    (Original post by Aeluna)
    There's probably more than one correct method
    Yes I totally agree, however if HBr was used that would result in a single OH group, and the product had two OH groups
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    (Original post by Aeluna)
    What was the question?
    didn't it ask for the similarities and differences between isotopes of the same element???
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    (Original post by alkaline.)
    Same number of protons and electrons, different number of neutrons

    I didn't put "electrons" will I lose marks?

    Relative atomic mass: 63.62
    No of atoms in 5.00g coin: 3.98 x 10^22 (3sf)

    do you remember the calculation for the coin question?

    Percentage error of water removed mass: 1.72%

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

    how to make sure all the water's gone: I put reweigh and reheat repeatedly until constant mass is achieved.

    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

    there was a question after this bit I don't remember, all I know is I put: h2, nickel catalyst.
    then there was a diagram: BLANK - H2S04 catalyst and heat - THE GIVEN ALKENE (I can draw it out from memory but idk how to put it on here) - ANOTHER BLANK

    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.

    what happens when you put chlorine gas into bromine ions:I put chlorine is more reactive that bromine, so it displaces bromine that causes whatever change the question was asking for. ionic equation for the equation given with PBII put: PB^2+ (aq) + 2I- (aq) —> PbI2 (s)

    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

    I also remember a random gas question I got 32.7cm^3

    it wasnt a haloalkane thoughit was just an alkane with OH groups on either end
    Thank you for your input, I'll add to the original
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    (Original post by therealbatman365)
    *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?
    Yeah I want an answer to that too. I used the Relative Atomic Mass from the question before because the questions were c) i) and ii) so I presumed they were connected, can't remember what I got though but it was 3. something x10-22
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    (Original post by Bubblegum2090)
    didn't it ask for the similarities and differences between isotopes of the same element???
    If it was the difference between isotopes then it's the same number of protons and electrons, different neutrons. But I do remember something about a bromide ion and one point.
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    (Original post by Bosssman)
    The end product was a diol...
    Damn, I haven't noticed the second OH.
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    Easy is paper, could have been far worse, this was an above average paper so I'm thinking paper two will be an absolute terror, but that should drop the grade boundaries at least...
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    the kc question may be wrong too as (0.876^2)/2^2+1.2 =1.47 which is not correct as kc was 1.6
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    (Original post by adamwall99)
    Yeah I want an answer to that too. I used the Relative Atomic Mass from the question before because the questions were c) i) and ii) so I presumed they were connected, can't remember what I got though but it was 3. something x10-22
    I used the periodic table value, I'm sure for that question they'll allow either as it didn't specify
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    (Original post by therealbatman365)
    *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?
    I did that, the whole question was talking about the coin and that question said 'one of these coins' or something like that so i used the calculated RAM. I'm not sure if they will penalise you for using the datasheet one though
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    (Original post by 456CJ)
    the kc question may be wrong too as (0.876^2)/2^2+1.2 =1.47 which is not correct as kc was 1.6
    You multiply 2^2 and 1.2, not add them
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    (Original post by ball1st1cpengu1n)
    Easy is paper, could have been far worse, this was an above average paper so I'm thinking paper two will be an absolute terror, but that should drop the grade boundaries at least...
    Very true, there was a lot of the hard stuff left out which does suggest it's concentrated into depth :/
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    (Original post by Bosssman)
    You multiply 2^2 and 1.2, not add them
    for **** sake i completely forgot.
 
 
 
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