AQA CHEM 2 June 2012 paper and unofficial mark scheme

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    Again wondering if anyone can help me with this?

    I did the graphs on the first two pages at the end and was running out of time - stupidly all I did was do a steeper gradient for the first (same volume), shallower gradient for the second (same volume) and steeper gradient (same volume) for the third.

    I think that's two marks for the third, but for the first two, would I get one mark for each for the change in gradient?

    Or would the two-mark allocation be 1) for a change in volume and 2) for a correct sort of size of change in volume? (i.e. 0 marks for me on both)

    Thanks
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    (Original post by rturnbull94)
    That's pretty good to be honest with you
    It's a resit haha, I'm doing A2 but I resat Chem2 to get more UMS.
    I was terrified to resit it, because I didn't think I'd do any better, but hopefully I have!!
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    (Original post by cuckoo99)
    holy **** FFS i didnt see that either OMG OMG OMG OMG EASY MARKS GONE :O:O:O:O
    I missed 8bi! I'm hating life right now. I think i'm lucky to get a D and it wasn't even a bad exam compared to mocks.
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    Do you think you will get penalized if you didn't draw the height exactly double and 2/5th as high on the graphs and just put it higher and lower?
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    (Original post by Scott.M)
    Do you think you will get penalized if you didn't draw the height exactly double and 2/5th as high on the graphs and just put it higher and lower?
    You probably just had to draw them within the correct 'lines' that the graph was on. If you did that, you'd get the mark.
    The other mark is for starting at the same point as the original line.
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    (Original post by knocker)
    UNOFFICIAL MARK SCHEME - please point out any errors (done with paper in front of me and I'm a teacher...)

    1(a) steeper line, ending twice as high (2)
    (b) less steep line, ending 2/5 as high (2)
    (c) steeper line, ending at same level (2)
    (d) concentration of reactants decreases with time, hence frequency of collisions decreases as does the rate (2)
    (e) 2 H2O2 --> 2 H2O + O2 (1)
    (f) The HBr is not used up (1)

    2(a)(i) -49 (3)
    (ii) it is an element (1)
    (b) increases, 4 moles gas --> 2 moles gas, eqilibrium shifts right to oppose the change (3)
    (c) increases, reaction is endothermic, equilibrium shifts right to oppose the change (3)
    (d)(i) no net carbon produced (1)
    (ii) CH3OH + 3/2 O2 --> CO2 + 2 H2O (1)
    (iii) 3 H2 + 3/2 O2 --> 3 H2O (1)
    (e) q = 4389 J, DH = - -399 kJ/mol (3)


    3(a) numbers to add 2, 15, 15, 6 (1)
    (b) P4 = zero, H3PO4 = +5 (2)
    (c) Add up Ar values to show they are different to 5dp (1)
    (d)(i) speeds up rate of reaction without being used up (1)
    (ii) addition of water (1)
    (iii) CH3CH=CH2 + H2O --> CH3CH(OH)CH3, propan-2-ol (2)

    4(a) Forms brittle TiC (1)
    (b)(i) 2 FeTiO3 + 7 Cl2 + 6 C --> 2 FeCl3 + 2 TiCl4 + 6 CO (1)
    (ii) TiCl4 + FeCl3 + 7 Na --> Fe + Ti + 7 NaCl (1)
    (c) Cu2+ + Fe --> Cu + Fe2+, Cu2+ ions reduced as they gain electrons (2)
    (d) 2 O2- --> O2 + 4e (1)

    5(a)(i) Ba + 2 H2O --> Ba(OH)2 + H2 (1)

    (ii) reactivity increases (1)

    (b) Mg(OH)2 (1)

    (c) Barium meal / X-ray tracer, insoluble (2)

    6(a)(i) Cl2 --> 2 Cl.
    CH2Cl2 + Cl. --> .CHCl2 + HCl
    .CHCl2 + Cl2 --> CHCl3 + Cl. (3)
    (ii) UV light, Free-radical substitution (2)
    (b)(i) CHCl3 + Cl2 --> CCl4 + HCl (1)
    (ii) CHCl3 contains C-H bonds, no C-H absorption at 2850-3300 (2)
    (c) CFC's - UV light breaks C-Cl bond, makes Cl. radicals
    Cl. + O3 --> ClO. + O2
    ClO. + O3 --> 2 O2 + Cl. chain reaction
    2 O3 --> 3 O2 (4)
    (d)(i) CH2=C(F)CF3 (1)
    (ii) it contains no C-Cl bonds (1)

    7(a)(i) elimination mechanism (3)
    (ii) E hex-3-ene (H's on opposite sides) and Z-hex-3-ene (H's on same side) (2)
    (iii) Same MF and SF, different arrangement of atoms in space (2)
    (b) H2SO4 electrophilic addition mechanism (4)

    8(a) Safety - NaOH is corrosive, Environmental - Process 2 uses CO2 from local factory, less CO2 emissions/global warming (2)
    (b)(i) Nucleophilic substitution with :OH- (3)
    (ii) BCA (3)
    (c) Fermentation, 35oC, Yeast, aqueous, anerobic (4)
    (d) primary alcohol, acidified potassium dichromate(VI)
    CH2(OH)CH2CH2CH2OH + 4[O] --> HOOCCH2CH2COOH + 2 H2O (3)

    9(a)(i) Iodine/I2, Cl2 + 2 I- --> 2 Cl- + I2, redox (3)
    (ii) silver(I) chloride, Ag+ + Cl- --> AgCl, ppt dissolves (3)
    (b)(i) H+ + Cl- --> HCl, HCl (2)
    (ii) 2 I- --> I2 + 2e-
    H2SO4 + 8 H+ + 8 e- --> H2S + 4 H2O
    H2SO4 = oxidising agent, yellow solid = sulfur/S/S8 (4)
    (iii) NaOH reacts with H+, equilibrium shifts right to make more H+ and opppose the change
    bromine is used in small quantities
    For the last question there was Br- Ions on the rhs aswell Na+ + Br- = NaBr no?
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    (Original post by Picture~Perfect)
    Mark scheme
    A big thanks to knocker for this mark scheme

    Question 1:
    (a) steeper line, ending twice as high (2)
    (b) less steep line, ending 2/5 as high (2)
    (c) steeper line, ending at same level (2)
    (d) concentration of reactants decreases with time, hence frequency of collisions decreases as does the rate (2)
    (e) 2 H2O2 --> 2 H2O + O2 (1)
    (f) The HBr is not used up (1)

    Question 2:
    (a)(i) -49 (3)
    (ii) it is an element (1)
    (b) increases, 4 moles gas --> 2 moles gas, eqilibrium shifts right to oppose the change (3)
    (c) increases, reaction is endothermic, equilibrium shifts right to oppose the change (3)
    (d)(i) no net carbon produced (1)
    (ii) CH3OH + 3/2 O2 --> CO2 + 2 H2O (1)
    (iii) 3 H2 + 3/2 O2 --> 3 H2O (1)
    (e) q = 4389 J, DH = - -399 kJ/mol (3)


    Question 3:
    (a) balancing equation (1)
    (b) P4 = zero, H3PO4 = +5 (2)
    (c) Add up Ar values to show they are different to 5dp (1)
    (d)(i) speeds up rate of reaction without being used up (1)
    (ii) addition of water (1)
    (iii) CH3CH=CH2 + H2O --> CH3CH(OH)CH3, propan-2-ol (2)

    Question 4:
    (a) Forms brittle TiC (1)
    (b)(i) 2 FeTiO3 + 7 Cl2 + 6 C --> 2 FeCl3 + 2 TiCl4 + 6 CO (1)
    (ii) TiCl4 + FeCl3 + 7 Na --> Fe + Ti + 7 NaCl (1)
    (c) Cu2+ + Fe --> Cu + Fe2+, Cu2+ ions reduced as they gain electrons (2)
    (d) 2 O2- --> O2 + 4e (1)

    Question 5:
    (a)(i) Ba + 2 H2O --> Ba(OH)2 + H2 (1)
    (ii) reactivity increases (1)
    (b) Mg(OH)2 (1)
    (c) Barium meal / X-ray tracer, insoluble (2)

    Question 6:
    (a)
    (i) Cl2 --> 2 Cl.
    CH2Cl2 + Cl. --> .CHCl2 + HCl
    .CHCl2 + Cl2 --> CHCl3 + Cl. (3)
    (ii) UV light, Free-radical substitution (2)
    (b)
    (i) CHCl3 + Cl2 --> CCl4 + HCl (1)
    (ii) CHCl3 contains C-H bonds, no C-H absorption at 2850-3300 (2)
    (c) CFC's - UV light breaks C-Cl bond, makes Cl. radicals
    Cl. + O3 --> ClO. + O2
    ClO. + O3 --> 2 O2 + Cl. chain reaction
    2 O3 --> 3 O2 (4)
    (d)
    (i) CH2=C(F)CF3 (1)
    (ii) it contains no C-Cl bonds (1)

    Question 7:
    (a)
    (i) elimination mechanism (3)
    (ii) E hex-3-ene (H's on opposite sides) and Z-hex-3-ene (H's on same side) (2)
    (iii) Same MF and SF, different arrangement of atoms in space (2)
    (b) H2SO4 electrophilic addition mechanism (4)

    Question 8:
    (a) Safety - NaOH is corrosive, Environmental - Process 2 uses CO2 from local factory, less CO2 emissions/global warming (2)
    (b)
    (i) Nucleophilic substitution with :OH- (3)
    (ii) BCA (3)
    (c) Fermentation, 35oC, Yeast, aqueous, anerobic (4)
    (d) primary alcohol, acidified potassium dichromate(VI)
    CH2(OH)CH2CH2CH2OH + 4[O] --> HOOCCH2CH2COOH + 2 H2O (3)

    Question 9:
    (a)
    (i) Iodine/I2, Cl2 + 2 I- --> 2 Cl- + I2, redox (3)
    (ii) silver(I) chloride, Ag+ + Cl- --> AgCl, ppt dissolves (3)
    (b)
    (i) H+ + Cl- --> HCl, HCl (2)
    (ii) 2 I- --> I2 + 2e-
    H2SO4 + 8 H+ + 8 e- --> H2S + 4 H2O
    H2SO4 = oxidising agent, yellow solid = sulfur/S/S8 (4)
    (iii) NaOH reacts with H+, equilibrium shifts right to make more H+ and opppose the change
    bromine is used in small quantities
    Thanks for this! Would I get the mark for doing the half equation like this: 8I- ----> 4I2 + 8e-?
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    CAN SOMEONE PLEASE TELL ME FOR THE FIRST QUESTION!!! I never did anything twice as high or twice as low..I just drew them all from the starting point and either higher or lower???Would that be work 1 mark for each?? i.e. for the 1st i drew it up. 2nd one down and last one i drew same as first ..But i never did it twice or much for any just up or down!!..would i still get 1 mark for each! :/.
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    9.
    (b)
    (i) H+ + Cl- --> HCl, HCl (2)

    For this one I wrote h2so4 +Cl- ---> Cl+ + HSO4
    WOULD THIS BE RIGHT!!
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    (Original post by Ayakashi)
    For the last question there was Br- Ions on the rhs aswell Na+ + Br- = NaBr no?
    As it was in solution the Na+ and Br- would just float around separately as spectator ions and not join to form NaBr.
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    (Original post by joker11)
    Thanks for this! Would I get the mark for doing the half equation like this: 8I- ----> 4I2 + 8e-?
    Fine.
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    (Original post by king101)
    CAN SOMEONE PLEASE TELL ME FOR THE FIRST QUESTION!!! I never did anything twice as high or twice as low..I just drew them all from the starting point and either higher or lower???Would that be work 1 mark for each?? i.e. for the 1st i drew it up. 2nd one down and last one i drew same as first ..But i never did it twice or much for any just up or down!!..would i still get 1 mark for each! :/.
    1 mark for slope, and 1 for correct height. So you would get 1/2 for each one.
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    (Original post by king101)
    9.
    (b)
    (i) H+ + Cl- --> HCl, HCl (2)

    For this one I wrote h2so4 +Cl- ---> Cl+ + HSO4
    WOULD THIS BE RIGHT!!
    Possibly, if you meant H2SO4 + Cl- --> HCl + HSO4- but I think they wanted H+ + Cl- --> HCl as it is an acid-base reaction whose 'usual' ionic equation is just H+ + OH- --> H2O
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    (Original post by berryripple)
    Again wondering if anyone can help me with this?

    I did the graphs on the first two pages at the end and was running out of time - stupidly all I did was do a steeper gradient for the first (same volume), shallower gradient for the second (same volume) and steeper gradient (same volume) for the third.

    I think that's two marks for the third, but for the first two, would I get one mark for each for the change in gradient?

    Or would the two-mark allocation be 1) for a change in volume and 2) for a correct sort of size of change in volume? (i.e. 0 marks for me on both)

    Thanks
    As has been mentioned before - you'll get some marks. 1 for correct steepness and 1 for correct final volume.
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    What do you think the mark for 140 UMS will be? I think it was 93 in January. I just hope the UMS isn't as high as last summer
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    (Original post by Scott.M)
    Do you think you will get penalized if you didn't draw the height exactly double and 2/5th as high on the graphs and just put it higher and lower?
    I don't think they'll be THAT harsh on us :L
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    (Original post by Mojojojo)
    I missed 8bi! I'm hating life right now. I think i'm lucky to get a D and it wasn't even a bad exam compared to mocks.
    we lost same amount of marks then :/
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    (Original post by deej34)
    What do you think the mark for 140 UMS will be? I think it was 93 in January. I just hope the UMS isn't as high as last summer
    An A is always 112 UMS the UMS boundaries don't change, what changes is the raw mark needed to reach that UMS boundary.
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    *fingers crossed* for a lowish boundary, I need an extra few UMS to push up my ISA.
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    i need atleast 110 or so ums , then i will be happy
 
 
 
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