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    Anyone wanna ask some questions?
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    (Original post by minthumbugs)
    imagine it as a number line, the more negative, the more to the left so the reaction goes backwards
    the more positive, the more to the right so the reaction goes forwards

    oxidising agents oxidise other things so are reduced themselves
    oh thankyou thats a good way to remember it!
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    (Original post by otrivine)



    I am giving you a very hard demanding question,

    Aqueous ferrate (VI) ions Fe04 2- decompose in the presence of hydrogen ions, forming Iron(III) ions, oxygen and water. Construct an ionic equation for this reaction
    is it
    3FeO4 2- + 12H+ ------> 2Fe3+ + 6H2O + 3O2
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    (Original post by DudeBoy)
    Anyone wanna ask some questions?
    sure, you start
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    Is it for overall equations, the overall charge must be zero
    Whereas for half equations the Charges either side much just balance ie.
    Fe2+ --> fe3+ + e- (both equal 2)



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    (Original post by AyshaK)
    is it
    3FeO4 2- + 12H+ ------> 2Fe3+ + 6H2O + 3O2
    nearly!


    can you please confirm the electrode with me pleasse
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    (Original post by A-New-Start)
    Fe is being reduced from 6+ to 3+
    whereas Oxygen is oxidised from 2- to 0
    so what do you do with that?
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    (Original post by _HabibaH_)
    That came under Ka and acids in one of the papers.
    Wouldnt it be amount of H+ over initial concentration of acid?
    So for a weak acid
    Ka=[H+]^2/[HA] If you had [HA] and Ka and you worked out [H+], would percentage dissociation be [H+]/[HA] x 100 ?
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    (Original post by bluedate)
    Same
    really? no uni accept general study so whats the point?
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    (Original post by DudeBoy)
    Anyone wanna ask some questions?
    do we need to know all the definitions for the step in born haber
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    (Original post by otrivine)
    sure, you start
    What is the reaction that occurs at each electrode of a hydrogen fuel cell?
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    (Original post by eggfriedrice)
    I changed my answer

    2FeO4 2- + 10H+ -> 2Fe 3+ + 3/2 O2 + 5H2O
    How you did it?
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    On June 2010 paper Q6e)
    Says C is a gas, at RTP C has mass of 1.17g

    I want to know why they did 1.17 x 24
    I thought the formula was n= v/24...

    Please help!!

    Think I'm getting panicky sorry for sounding demanding

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    (Original post by kuku2013)
    do we need to know all the definitions for the step in born haber
    absolutely, specially lattice enthalpy, enthalpy of solution and enthalpy of hydration
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    (Original post by _sparks)
    Okay someone please answer this.
    I cant figure this out.
    Basically, dont we have to divide the moles of thiosulfate by 2 to get moles of Cu
    The book got the moles of 2Cu^2+

    Its page 223 of the OCR book. I attached the image too.
    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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    (Original post by DudeBoy)
    What is the reaction that occurs at each electrode of a hydrogen fuel cell?

    h2+ 2OH- ---> 2H2O + 2e-

    1/2O2 + H2O + 2e- ---> 2OH-



    H2 + 1/2O2 -----> H2O
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    (Original post by amin666)
    really? no uni accept general study so whats the point?
    I know right, and it's 2 hours long, what a pointless waste of time
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    (Original post by bluedate)
    On June 2010 paper Q6e)
    Says C is a gas, at RTP C has mass of 1.17g

    I want to know why they did 1.17 x 24
    I thought the formula was n= v/24... :confused:
    Don't know if this is right or wrong but I did this.

    1.17g per cubic decimetre so 1.17g per dc3 so (1.17/1000)x24000 = 28 which is basically N2.
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    (Original post by _sparks)
    Okay someone please answer this.
    I cant figure this out.
    Basically, dont we have to divide the moles of thiosulfate by 2 to get moles of Cu
    The book got the moles of 2Cu^2+

    Its page 223 of the OCR book. I attached the image too.
    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.
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    (Original post by bluedate)
    On June 2010 paper Q6e)
    Says C is a gas, at RTP C has mass of 1.17g

    I want to know why they did 1.17 x 24
    I thought the formula was n= v/24...

    Please help!!

    Think I'm getting panicky sorry for sounding demanding

    for gases you have to times by 24 as one mole of gas occupies 24dm^3
    im pretty sure thats right
 
 
 
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