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    (Original post by laurawoods)
    http://filestore.aqa.org.uk/subjects...W-QP-JAN10.PDF



    7) For that very last question , how does the graph actually look? I have drawn a graph , but i don;t know if that would pick up the marks?

    !

    http://www.chemguide.co.uk/physical/...roduction.html

    the graph looks like the one at the bottom of this page
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    (Original post by saba146)
    http://www.chemguide.co.uk/physical/...roduction.html

    the graph looks like the one at the bottom of this page
    cool cool thanks! wouldn't it touch zero eventaully? would it harm the marks if, in my answer, i made it touch the coordinate x axes after quite some time passed?
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    (Original post by saba146)
    http://www.chemguide.co.uk/physical/...roduction.html

    the graph looks like the one at the bottom of this page
    pls can u help me with the other questions too>?
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    (Original post by laurawoods)
    Ok cool...so are u aiming for an A star grade in the summer?
    Yahuh but I need 112/120 to get an A* which seems preeeeetty unlikely atm :L
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    (Original post by laurawoods)
    http://filestore.aqa.org.uk/subjects...W-QP-JAN10.PDF


    1) In 1ci ) , for the equation 2 that they are asking , would they accept,
    V2O4 + 1/2 O2 --> V2O5
    yep i think they would allow it
    because in the MS they have actually doubled it all up?

    2) In question number 3c ) is my answer , given below, right?

    Pt (s) /H2(g) /H2O (l) // O2 (g) / OH- (aq) / Pt (s)

    Pt|H2(g)|OH-(aq),H2O(l)||O2(g)|H2O(l),OH–( aq)|Pt this is the right answer but you can also have :
    pt (s)| h2(g)|oh-(aq)||o2(g)|(oh-)|pt ( this is without water you dont have to include water)
    yours equation isnt right because the 1st h20 should be oh-



    The MS is confusing me...

    3) For 3h) would they accept my answer: Hydrogen could be produced by the electrolysis of water, and this requires electricity which is generated by burning fossil fuels. This burning of fossil fuels will release CO2 into the envt. think this would be fine but i would also mention that electrolysis isnt carbon neutral

    4) For 6a) would you get the second mark , with a wrong product?
    not sure about this one sorry
    5) Pls can u explain the calculation question in 8c ?
    i explained this in the post above

    6) For 9a) would this get 3 marks for the colours explanation?

    Ti (III) ions are coloured because it has got partially filled 3d orbitals. Therefore, d-d transitions can take place which means electrons are excited from ground state to higher energy level. Light is absorbed from the visible range. Ti (IV) has empty 3d orbitals. So no d-d electron transitions taking place. No light energy from visible range absorbed.
    i think this would get you the marks
    Would this get the marks, because the MS seems so different?

    7) For that very last question , how does the graph actually look? I have drawn a graph , but i don;t know if that would pick up the marks?

    phew and thanks!
    )
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    Can somebody explain what I need to know for thermodynamics please :-)
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    (Original post by saba146)
    equation for the reaction of sodium hydroxide with phosphoric acid:
    H3PO4 + 3NaOH = Na3PO4 + 3H2O

    moles of naoh = 0.5 x ( 21.2/1000) = 0.0106
    moles of h3po4 used in titre = 0.0106/3 =3.53x10-3

    however this is the moles in 25cm3 the tank has 25000 dm3
    25cm3 = 25/1000 =0.0025dm3

    ratio of volume in tank to volum titrated = 25000/0.025 =1x106

    so the moles on h3po4 in tank will be 1x106 times that we used for the tiration
    therefor moles of h3po4 in tank = 3.53x10-3 x 1x106= 3530 moles

    now we need to find the moles of phosphorus oxide
    equation for phosphorus oxide reacting with water to make phosphoric acid is:

    P4O10 + 6H20 = 4H3PO4

    we know the moles of h3po4 is 3530
    moles of p4o10 = 3530/4 = 882.5 moles

    find the mass of p4o10
    mass = moles x mr
    = 882.5 x 248 = 250630g
    = 250630 / 1000 = 250.63kg
    Thankyou!
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    (Original post by saba146)
    )
    Hello for the conventional representation question, I don't understand why it is OH- instead of H2O (i.e. i don't really understand why mine is wrong)? because the H2 is being oxidised to H2O isn't it ?
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    anyone know where I can find the jan 13 chem5 paper and mark scheme?? lease )
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    hey

    new to this
    was wondering how is everyone finding chem 5
    im really struglling on redox and chromium chem alosngside transiiton metals
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    (Original post by saba146)
    )
    hello normally in these mark schemes, I see they say "the CO (for example) bind to the iron / Fe"

    Why do they say iron and Fe why not say Fe2+, if i say Fe2+ would this lead to the deduction of marks? thanks for your extended help!
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    anyone know if there is a thread for physics 5?
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    aqa chem1 / chem 2 / chem 5 june 2011 examiners reports
    Thanks
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    (Original post by saba146)
    )
    A solution of iron(II) sulfate was prepared by dissolving 10.00 g of FeSO4.7H2O (Mr
    = 277.9) in water and making up to 250 cm3 of solution. The solution was left to stand, exposed to air, and some of the iron(II) ions became oxidised to iron(III) ions. A 25.0 cm3
    sample of the partially oxidised solution required 23.70 cm3 of 0.0100 mol dm–3 potassium dichromate(VI) solution for complete reaction in the presence of an excess of dilute sulfuric acid. Calculate the percentage of iron(II) ions that had been oxidised by the air.

    In this pls can someone M5 to me , why do we need to perform the SUBSTRACTION. question 7d from June 2010 question paper.
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    (Original post by laurawoods)
    A solution of iron(II) sulfate was prepared by dissolving 10.00 g of FeSO4.7H2O (Mr
    = 277.9) in water and making up to 250 cm3 of solution. The solution was left to stand, exposed to air, and some of the iron(II) ions became oxidised to iron(III) ions. A 25.0 cm3
    sample of the partially oxidised solution required 23.70 cm3 of 0.0100 mol dm–3 potassium dichromate(VI) solution for complete reaction in the presence of an excess of dilute sulfuric acid. Calculate the percentage of iron(II) ions that had been oxidised by the air.

    In this pls can someone M5 to me , why do we need to perform the SUBSTRACTION. question 7d from June 2010 question paper.
    Because your finding put how much of the Fe2+ has gone, so the difference between the old and the new value that you work out must be the moles of Fe2+ which have been oxidised
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    Could someone please explain to me why exothermic reactions occur more spontaneously than less exothermic reactions (as in not as much heat given out) or endothermic reactions. Thanks
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    (Original post by candyhearts)
    Could someone please explain to me why exothermic reactions occur more spontaneously than less exothermic reactions (as in not as much heat given out) or endothermic reactions. Thanks
    Because then its more likely that gibbs free energy would be negative according to the equation.

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    If the surface area of one of the electrode was doubled what impact would this have on the end?

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    (Original post by samfreak)
    If the surface area of one of the electrode was doubled what impact would this have on the end?

    Posted from TSR Mobile
    Well a greater surface area means more of the reaction can take place (oxidation or reduction) so really it depends which electrode you're on about :L
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    I cannot do redox equilibria for the life of me and i'm having a bit of panic about it.. Thermodynamics isn't particularly nice either.
 
 
 
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