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    (Original post by eggfriedrice)
    First one correct.

    Second one you'd divide by three and times by two which is equivalent to multiplying by 2/3.
    oh yeah i meant multiplying.
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    (Original post by A-New-Start)
    oh yeah i meant multiplying.

    Aha good, you're all set then.
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    jan 2013 the last question, how do you know n = +2 from the molar ratio?? thank u!!
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    is everyone going to be revising on tsr tomorrow?
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    (Original post by _HabibaH_)
    an explanation of June 10 6b would be much appreciated. I just don't understand how the e- could get that way.
    Each Cr is 6+ you are making 2Cr2+, therefore each one needs to gain 4 electrons, hence 4x2=8e-
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    (Original post by galacticfear)
    Each Cr is 6+ you are making 2Cr2+, therefore each one needs to gain 4 electrons, hence 4x2=8e-
    Thankyouuuuuu! Finally, I get it.
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    (Original post by RnTf)
    jan 2013 the last question, how do you know n = +2 from the molar ratio?? thank u!!
    KMnO4 = 2.25x10-2 x 13.2/1000 = 2.97x10-4 mol
    n(v) = 0.126/50.9 = 2.48x10-3 mol
    with factor of 5 you do 2.48x10-3 / 5 = 4.96x10-4
    ratio = vn+/mnO4- = 4.96x10-4/2.97x10-3 i just rounded them to 5:3 then wrote the equation
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    With a grand total of 25 hours spent revising/learning chemistry since March, I can truly say I am extremely confident... I can sense an all nighter coming on -,-
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    (Original post by DavidH20)
    Standard cell potentials can only be used to predict a reaction's feasibility under standard conditions - actual reactions are very rarely at RTP, standard states, unit ion concentrations etc. so may differ from the hypothetical cell reactions.

    Furthermore, the cell potentials say nothing about how fast a reaction will be - it can be hypothetically feasible but so slow it effectively doesn't happen.
    thanks! can also ask how you would tackle a question like 5cii june 2012? And also, that question in another paper about glycolate in hair... ?
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    (Original post by AyshaK)
    KMnO4 = 2.25x10-2 x 13.2/1000 = 2.97x10-4 mol
    n(v) = 0.126/50.9 = 2.48x10-3 mol
    with factor of 5 you do 2.48x10-3 / 5 = 4.96x10-4
    ratio = vn+/mnO4- = 4.96x10-4/2.97x10-3 i just rounded them to 5:3 then wrote the equation
    Thank you, but from the mark scheme, they worked out the value of n just from the molar ratio.:confused::confused:
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    Help!
    "The student diluted 0.015 mol dm-3 nitric acid with an equal volume of water and measured the pH of the diluted acid at 25C.
    Calculate the pH of the diluted acid"

    Explain please.
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    (Original post by RnTf)
    Thank you, but from the mark scheme, they worked out the value of n just from the molar ratio.:confused::confused:
    I think it is comparing oxidation numbers when it is reacted in the ratio 5:3 if I remember correctly. The oxidation numbers have to balance in the overall equation, so you can deduce it from that.
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    MY ORIGINAL POST: http://www.thestudentroom.co.uk/show...7#post43103717

    I understand all that, but if you then compare the 2nd equation to the 1st, then dont we have to divide the moles of Cu(II) by 2 to get the original number of moles in the bronze sample?




    (Original post by Brad0440)
    The moles of thiosulphate are 4.48x10-3. You then use the ratios in the third equation (thiosuphate plus iodine) to get:
    thiosulphate:iodine
    2:1
    4.48x10-3:2.24x10-3

    The moles of iodine formed can then be used in the second equation (as the iodine formed from this reaction are then used in the third reaction which was used above) to get the ratio:
    Iodine:copper(II)
    1:2
    2.24x10-3:4.48x10-3

    Which means that the moles of Cu2+ are 4.48x10-3, like the book says.

    I think that what you did was try to use the ratio of thiosulphate:iodide(2:2) in the third equation, and then link that with the ratio in the second equation (Iodide:copper(II)(4:2)), which you can't do as the iodides in each of the reactions are not the same amounts.

    (Original post by Pride)
    no I think it's one to one as well.

    For every 2 moles of S2O3^2- reacted, 1 mol I2 reacts with it.
    Say all that 1mol was produced by the previous reaction, then 2 moles of Cu2+ produced that.
    Say all 2 moles of that Cu2+ was produced by the previous reaction, that is produced by 2 moles of Cu.

    So it's 1:1 thio : Cu
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    if electrode potential is negative does that mean it can lose electrons and if its positive it can gain electrons? sorry don't know how to word the question
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    (Original post by kuku2013)
    if electrode potential is negative does that mean it can lose electrons and if its positive it can gain electrons? sorry don't know how to word the question
    If you are comparing two electrode potentials, the one that is most negative supplies electrons, whether the overall electrode potential is positive or negative.
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    (Original post by LosMutos)
    ..
    The concentration of the acid has halved
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    (Original post by _sparks)
    MY ORIGINAL POST: http://www.thestudentroom.co.uk/show...7#post43103717

    I understand all that, but if you then compare the 2nd equation to the 1st, then dont we have to divide the moles of Cu(II) by 2 to get the original number of moles in the bronze sample?

    no, because all the Cu2+ formed came from the previous reaction, the oxidation of Cu. That's 1:1.
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    (Original post by kuku2013)
    if electrode potential is negative does that mean it can lose electrons and if its positive it can gain electrons? sorry don't know how to word the question
    The more negative an electrode potential is, the more likely it is to lose electrons. the more positive an electrode potential is, the more likely it is to gain electrons.
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    how can we determine the amount of oxygen


    Potassium manganate (VII)

    how can we work out how many oxygen are present ?
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    (Original post by master y)
    The concentration of the acid has halved
    why?
 
 
 
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