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    (Original post by otrivine)
    how can we determine the amount of oxygen


    Potassium manganate (VII)

    how can we work out how many oxygen are present ?
    Which question is this?
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    (Original post by otrivine)
    how can we determine the amount of oxygen


    Potassium manganate (VII)

    how can we work out how many oxygen are present ?
    K is +1, Mn is +7, therefore total charge is +8, therefore you need 4 x -2 oxygen to cancel it out.
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    (Original post by _HabibaH_)
    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.
    thanks
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    (Original post by galacticfear)
    If you are comparing two electrode potentials, the one that is most negative supplies electrons, whether the overall electrode potential is positive or negative.
    thanks
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    (Original post by LosMutos)
    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.
    is it 2.12??
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    (Original post by otrivine)
    why?
    When you add an equal amount of water, it is double the volume but the conc is not changed, so overall the conc halves.. if that makes sense..?
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    (Original post by galacticfear)
    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.
    thank you
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    (Original post by RnTf)
    is it 2.12??
    How did you work that out from just the concentration?
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    (Original post by eggfriedrice)
    How did you work that out from just the concentration?
    It is a strong acid, therefore full dissociation.
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    (Original post by master y)
    When you add an equal amount of water, it is double the volume but the conc is not changed, so overall the conc halves.. if that makes sense..?
    so always halve rigjt
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    (Original post by otrivine)
    so always halve rigjt
    Only when equal amount of volume is added. if double the volume of the acid is added then that would be divided by three right?
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    (Original post by RnTf)
    is it 2.12??
    i got 1.12
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    (Original post by _HabibaH_)
    Only when equal amount of volume is added. if double the volume of the acid is added then that would be divided by three right?
    ??? why three?
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    Stupid question here but would we ever get a question with ka and a strong acid?
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    The answer is -log(0.0075) = 2.125(3dp)

    Yeah you divide the conc. by two
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    fml
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    (Original post by otrivine)
    ??? why three?
    Because if equal volumes = 1+1, double of the existing volume being added would be 1 (existing) +2 (added volume)

    Idk, can someone refute or verify this? I could just be complicating things, I probably am.
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    (Original post by LosMutos)
    The answer is -log(0.075) = 1.125(3dp)
    really? should not be that..
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    (Original post by eggfriedrice)
    Stupid question here but would we ever get a question with ka and a strong acid?
    for sure I seen quite alot
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    (Original post by LosMutos)
    The answer is -log(0.075) = 1.125(3dp)
    Isn't it 0.015 - one zero could make the difference between one pH and another. Each ph differs by ^10
 
 
 
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