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    Is it not Zn2+|Zn||...
    And the electron configuration of Co -1s2 2s2 2p6 3s2 3p6 3d9 ???



    (Original post by hi-zen-berg)
    The thread is a shambles so just put some solutions in order. If anyone has corrections / knows marks allocated fire away. Cheers. Edit: * THANKS to all who contributed from chem4/5 thread.

    Q1 Periodicity (12 marks)

    Covalent
    P
    P4O10 + 6H2O -> H3PO4
    (3 marks)

    Ionic
    Na
    2Na + 2H2O --> 2NaOH + H2
    (3 marks)

    Al2O3
    Ionic
    Reacts with acids and bases
    3H2O + Al2O3 + 2NaOH -> 2NaAl(OH)4
    Al2O3 + 6HCl -> 2AlCl3 + 3H2O
    (4 marks)

    Q2 BH Cycle (8 marks)
    Thermodynamics Calcs ****
    2K(s) + ½O2(g)
    2K(g) + ½O2(g)
    2K(g) + O(g)
    (3 marks)

    Enthalpy of Lattice dissociation = +2328kJ.mol-1
    (3 marks)
    K+ ions are bigger than Na+ ions, same charge so lower charge density
    Weaker attraction between K+ and O^(2-) than Na+ and O^(2-)
    (2 marks)

    Q3 Enthalpy of solution (6 marks)

    MgCl2 + aq --> Mg^2+(aq)+ 2Cl-(aq) (1 mark)

    Solubility will increase with temperature. The question stated itself that solubility is defined as the amount of solute available to dissolve in a solvent ( in this case water ). Hence, by the reverse reaction due to exothermic enthalpy of solution, more magnesium chloride will be produced, hence increasing its solubility. (3 marks)

    Enthalpy of solution = -155 kJ.mol-1 (2 marks)

    Q4 Electrochemistry (11 marks)

    Weakest oxidising agent is Zn2+ (1 mark)Standard conditions to give EMF of -0.76:100kPa/298K/zero current/1M all solutions (2 marks)Cell giving EMF of 0.48:Zn|Zn2+||Co2+|Co (2 marks)
    SO2 + V2O5 -> SO3 + V2O4 (2 marks)V2O4 + 0.5 O2 -> V2O5
    Q5 Hydrogen cells (9 marks)

    O2 + 4e- + 2H2O ---> 4OH- Postive electrode

    H2 + 2OH- ---> 2H2O + 2e- Negative electrode

    Overall: O2 + 2H2 ---> 2H2O
    (3 marks)

    Increase pressure increase EMF (2 marks)

    Graph = straight horizontal line (1 mark)

    No effect on EMF for increase in surface area of Pt (1 mark)

    Environmental advantage – CO2 not produced, greenhouse gas, global warming (1 mark)

    Why H fuel cells may not be Carbon neutral?
    H2 obtained by electrolysis of H2O
    May use energy from combustion of fossil fuels
    (1 mark)

    Q6 Contact process/enthalpy calcs (14 marks)
    Enthalpy change -196 (3 marks)Entropy -189 (2 marks)
    What does sign show about the product compared to thereactants? Products more ordered/less disordered (1 mark)DeltaG = -135 kJ.mol-1Delta G is less than/= 0 therefore feasible (3 marks)Reactants pure to prevent catalyst poisioning (1 mark)Q6 Chromium/inorganic RXNs (12 marks)
    CrCl3 + 6H2O -> 3Cl- + [Cr(H2O]6]3+ (1 mark)P = [Cr(H2O)3(OH)3]NaOH / any OH-[Cr(H20)6]2+ + 3OH --> [Cr(H2O)3(OH)3]+ 6H20(3 marks)Q = CO2Na2CO3 / any CO3^(2-)2[Cr(H20)6]2+ + 3CO3^(2-) --> 2[Cr(H2O)3(OH)3]+ 3CO2 + 3H2O(3 marks)RXN 4 = [Cr(OH)6]3-Excess NaOH / Excess of any OH-[Cr(H20)6]2+ + 6OH- --> [Cr(OH)6]3-+6H2O(3 marks)Zn + HCLBlue(2 marks)

    Q7 Cobalt Chem (12 marks)
    Co Atom is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7
    Co2+ 1s2 2s2p6 3s2p6d7
    Characteristic features: complex formation, coloured compounds, var. ox state. (5marks)


    [Co(h2o)6]2+ + 3NH2CH2CH2NH2 --> [Co(NH2CH2CH2NH2)3]2+ + 6H2O

    4 mol reactant to 7 mol product therefore deltaS positive

    DeltaH roughly 0 as same type and number of bond broken

    So delta G is negative and rxn is feasible

    Structure of complex ion: Co central metal, 3 NH2CH2CH2NH2’s joined via 6 coord bonds to the N atoms, octahedral shape, 2+ charge aroundentire complex (7 Marks)



    Redox titration (16 marks)
    Autocatalyst = a product of a reaction catalyses that reaction
    i.e.Mn2+
    (2 marks)

    H2SO4 was the acid to tick (1 mark)

    Reaction was slow as requires two negative ions to collide, which repulse each other, so high Ea. (I also just dropped in, little/no Mn2+formed so uncatalysed)
    (3 marks)

    Calculation:Ratio of C2O4^2- to MnO4^1 was 2:5
    There were 26.4cm3 of 0.02 moldm3 MnO4- = 0.000528 mols MnO4 therefore therewas 0.00132 moles C2O4^2- in 25cm3 therefore 0.0132 moles in 250cm3,

    There was 3 C2O4^2- per one mole of the salt therefore there were 4.4x10-3moles salt

    Mr K3(Fe(c2o4)3).3h20 = 491.1, Mass = moles x mr = 4.4x10-3 x 491.1 = 2.16g

    Original mass = 2.29 therefore 2.16/2.22 * 100 = 94.4% purity( 7 Marks) – Common wrong answer of 35% is likely to be 5/7marks

    Colourimetry
    Make solutions with known concs of MnO4- ions Plot calibration curve of concs to lightabsorpted/transmitted
    Measure absorption/transmission of light in unknown soln,
    extrapolate MnO4- conc (3 Marks)

    *** Ongoing discussion as to whether to divide by two. General consensus is NO. It has been suggested that AQA won't give credit for dividing by two. Exam marking is generally standardized anyway, so it may not be a binary right or wrong way to do this question given that quite a few people have interpreted it in a different way

    Preicted grade boundaries from isaisababy:
    Full UMS: 93
    A*: 86
    A: 78
    B: 70
    C: 59?
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    (Original post by hi-zen-berg)
    nice one. u going for A*?
    i think ill just be happy to sit on my AS UMS and get an A. A-levels have tired me out now
    (Original post by hi-zen-berg)
    nice one. u going for A*?
    i think ill just be happy to sit on my AS UMS and get an A. A-levels have tired me out now
    (Original post by Azzer11)
    I was hoping for an A*, although I'm not so sure after that horrible Chem4. And yeah aha I agree, I have just 2 more left and then freedom!
    Physics I presume
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    (Original post by Tokhmehsag)
    Is it not Zn2+|Zn||...
    And the electron configuration of Co -1s2 2s2 2p6 3s2 3p6 3d9 ???
    4s orbital gets filled before 3d
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    (Original post by hi-zen-berg)
    4s orbital gets filled before 3d
    Does 3d not get filled first in all transition metals?
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    (Original post by Tokhmehsag)
    Does 3d not get filled first in all transition metals?
    Afraid not mate
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    what do you think 50.7% as the purity answer would get out of 7?
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    (Original post by Tokhmehsag)
    Is it not Zn2+|Zn||...
    And the electron configuration of Co -1s2 2s2 2p6 3s2 3p6 3d9 ???
    Oxidation on the left hand side right? So I think it was Zn|Zn2+||Co2+|Co
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    (Original post by isaisababy)
    Oxidation on the left hand side right? So I think it was Zn|Zn2+||Co2+|Co
    Yeah but it was
    Zn2+ + 2e- --> Zn
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    For the final part of Chromium question with excess OH-, would [Cr(H2O)2(OH)4]- be acceptable or not? I remember seeing a couple textbooks had this instead of [CR(OH)6]3-
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    Are all the questions included in this markscheme?
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    (Original post by hi-zen-berg)
    Honestly not sure about H2SO4 mate its not the standard answer

    Hmm I also am unsure about the mno4- thing. I think that sounds a bit convoluted for what they asked. Fingers crossed for 2/3 marks if you have written the method and got the principle of a calibration curve and extrapolating
    Hello, Jw how many marks this would be....

    1) Prepare samples of Mn04- over a large range of concentrations
    2) Use calorimeter to measure absorbance- using a filter to give the highest absorbance
    3) Plot a graph of Concentration on X and absorbance on Y- line of best fit
    4 ) measure absorbance of unknown using the same filter - read off the graph by looking at line of best fit.
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    (Original post by hi-zen-berg)
    The thread is a shambles so just put some solutions in order. If anyone has corrections / knows marks allocated fire away. Cheers. Edit: * THANKS to all who contributed from chem4/5 thread.

    Q1 Periodicity (12 marks)

    Covalent
    P
    P4O10 + 6H2O -> H3PO4
    (3 marks)

    Ionic
    Na
    2Na + 2H2O --> 2NaOH + H2
    (4 marks)

    Al2O3
    Ionic
    Reacts with acids and bases
    3H2O + Al2O3 + 2NaOH -> 2NaAl(OH)4
    Al2O3 + 6HCl -> 2AlCl3 + 3H2O
    (5 marks)
    ________________________________ ____________________
    Q2 BH Cycle (8 marks)
    Thermodynamics Calcs ****
    2K(s) + ½O2(g)
    2K(g) + ½O2(g)
    2K(g) + O(g)
    (3 marks)

    Enthalpy of Lattice dissociation = +2328kJ.mol-1
    (3 marks)
    K+ ions are bigger than Na+ ions, same charge so lower charge density
    Weaker attraction between K+ and O^(2-) than Na+ and O^(2-)
    (2 marks)
    ________________________________ ____________________
    Q3 Enthalpy of solution (6 marks)

    MgCl2 + aq --> Mg^2+(aq)+ 2Cl-(aq) (1 mark)

    Solubility will increase with temperature. The question stated itself that solubility is defined as the amount of solute available to dissolve in a solvent ( in this case water ). Hence, by the reverse reaction due to exothermic enthalpy of solution, more magnesium chloride will be produced, hence increasing its solubility. (3 marks)

    Enthalpy of solution = -155 kJ.mol-1 (2 marks)
    ________________________________ ____________________
    Q4 Electrochemistry (11 marks)

    Weakest oxidising agent is Zn2+ (1 mark)
    Standard conditions to give EMF of -0.76:
    100kPa/298K/zero current/1M all solutions (2 marks)

    Cell giving EMF of 0.48
    Zn|Zn2+||Co2+|Co (2 marks)

    SO2 + V2O5 -> SO3 + V2O4
    V2O4 + 0.5 O2 -> V2O5 (2 marks)

    Reactants need purifying to prevent catalyst poisioning (1 mark)
    E(Au+/Au) > E(O2/ H2O) So Gold not oxidised by O2- Alternatively calculate EMF ofthe cell to get -0.45V which proves reaction not feasible (3 marks)
    ________________________________ ____________________
    Q5 Hydrogen cells (9 marks)

    O2 + 4e- + 2H2O ---> 4OH- Postive electrode

    H2 + 2OH- ---> 2H2O + 2e- Negative electrode

    Overall: O2 + 2H2 ---> 2H2O
    (3 marks)

    Increase pressure increase EMF (2 marks)

    Graph = straight horizontal line (1 mark)

    No effect on EMF for increase in surface area of Pt (1 mark)

    Environmental advantage – CO2 not produced, greenhouse gas, global warming (1 mark)

    Why H fuel cells may not be Carbon neutral?
    H2 obtained by electrolysis of H2O
    May use energy from combustion of fossil fuels
    (1 mark)
    ________________________________ ____________________
    Q6 Contact process/enthalpy calcs (14 marks)

    Enthalpy change -196 (3 marks)
    Entropy -189 (2 marks)
    What does sign show about the product compared to the reactants? Products more ordered/less disordered (1 mark)
    DeltaG = -135 kJ.mol-1
    Delta G is less than/= 0 therefore feasible (3 marks)

    Reactants pure to prevent catalyst poisioning (1 mark)
    ________________________________ ____________________
    Q7 Chromium/inorganic RXNs (12 marks)
    CrCl3 + 6H2O -> 3Cl- + [Cr(H2O]6]3+ (1 mark)

    P = [Cr(H2O)3(OH)3]
    NaOH / any OH-
    [Cr(H20)6]2+ + 3OH --> [Cr(H2O)3(OH)3]+ 6H20(3 marks)

    Q = CO2
    Na2CO3 / any CO3^(2-)
    2[Cr(H20)6]2+ + 3CO3^(2-) --> 2[Cr(H2O)3(OH)3]+ 3CO2 + 3H2O
    (3 marks)

    RXN 4 = [Cr(OH)6]3-
    Excess NaOH / Excess of any OH-
    [Cr(H20)6]2+ + 6OH- --> [Cr(OH)6]3-+6H2O(3 marks)

    Zn + HCL
    Blue(2 marks)
    ________________________________ ____________________

    Q8 Cobalt Chem (12 marks)
    Co Atom is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7
    Co2+ 1s2 2s2p6 3s2p6d7
    Characteristic features: complex formation, coloured compounds, var. ox state. (5marks)


    [Co(h2o)6]2+ + 3NH2CH2CH2NH2 --> [Co(NH2CH2CH2NH2)3]2+ + 6H2O

    4 mol reactant to 7 mol product therefore deltaS positive

    DeltaH roughly 0 as same type and number of bond broken

    So delta G is negative and rxn is feasible

    Structure of complex ion: Co central metal, 3 NH2CH2CH2NH2’s joined via 6 coord bonds to the N atoms, octahedral shape, 2+ charge aroundentire complex (7 Marks)
    ________________________________ ____________________


    Q9 Purification calc/ Redox titration (16 marks)
    Autocatalyst = a product of a reaction catalyses that reaction
    i.e.Mn2+
    (2 marks)

    H2SO4 was the acid to tick (1 mark)

    Reaction was slow as requires two negative ions to collide, which repulse each other, so high Ea. (I also just dropped in, little/no Mn2+formed so uncatalysed)
    (3 marks)

    Calculation:Ratio of C2O4^2- to MnO4^1 was 2:5
    There were 26.4cm3 of 0.02 moldm3 MnO4- = 0.000528 mols MnO4 therefore therewas 0.00132 moles C2O4^2- in 25cm3 therefore 0.0132 moles in 250cm3,

    There was 3 C2O4^2- per one mole of the salt therefore there were 4.4x10-3moles salt

    Mr K3(Fe(c2o4)3).3h20 = 491.1, Mass = moles x mr = 4.4x10-3 x 491.1 = 2.16g

    Original mass = 2.29 therefore 2.16/2.22 * 100 = 94.4% purity( 7 Marks) – Common wrong answer of 35% is likely to be 5/7marks

    Colourimetry
    Make solutions with known concs of MnO4- ions Plot calibration curve of concs to lightabsorpted/transmitted
    Measure absorption/transmission of light in unknown soln,
    extrapolate MnO4- conc (3 Marks)

    *** Ongoing discussion as to whether to divide by two. General consensus is NO. It has been suggested that AQA won't give credit for dividing by two. Exam marking is generally standardized anyway, so it may not be a binary right or wrong way to do this question given that quite a few people have interpreted it in a different way

    Preicted grade boundaries from isaisababy:
    Full UMS: 93
    A*: 86
    A: 78
    B: 70
    C: 59?
    Will the one about solubility be CE= 0 if we said decrease but did the correct explanation?
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    For the first question for P i put simple molecular as I wasn't thinking straight instead of covalent will I still get the mark?
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    (Original post by OhMasson)
    For the first question for P i put simple molecular as I wasn't thinking straight instead of covalent will I still get the mark?
    **** I did the same

    If it asked for bonding, probably not
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    Is the bonding in Phosphorous oxide not van Der waals?
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    (Original post by SubwayLover1)
    Will the one about solubility be CE= 0 if we said decrease but did the correct explanation?
    2/3
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    Yeah I Put it would decrease with the right explanation of why it would decrease( well I guess wrong because it was increase), would I lose al marks?

    (Original post by RefusedAccess)
    2/3
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    For the percentage purity I put two calculations one for C2O4 and one for Fe on an extra sheet ,I didn't cross any out ,do I get a few marks or not ???
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    (Original post by RefusedAccess)
    2/3
    Are you sure ?:0
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    (Original post by SubwayLover1)
    Are you sure ?:0
    I can think of two possibilities depending on how AQA will assess this question this year;

    1) you will get 2/3 due to the question asking to explain and then conclude with increase or decrease in one answer box. Typically they have two answer boxes, one for the conclusion, second for the explanation, hence if this the case you should get 2/3

    2) they specify 'If dH negative, then answer must conclude increase in solubility. If state decrease, then CE 0/3.

    I personally think it will be the former, so you should get 2/3, or at least I hope you do, best of luck either way!
 
 
 
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