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    Can someone please explain the 2 limitations of predictions using standard electrode potentials?
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    (Original post by georgiaaaxo)
    please can someone help me specimen paper q7b I get it all just not why you multiply the conc by 40 as well as 34? where's the 40 from?


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    you have to find UNDILUTED h2o2 therefore you first multiply by 10 to go from 25 to 250, but since undiluted is 25 we then multiply by 40 and then since we have it in mol/dm-3 we need to change it to g/dm-3 so we basically multiply the no of mol. by molar mass ( 34 ) to get mass
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    (Original post by DudeBoy)
    1/4 of the intial rate
    How?
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    (Original post by eggfriedrice)
    Attachment 225848
    I dont see how sodium ethanoate is a salt of that. Anyone care to explain?
    It's not. If you read the examiners comments, it says that the candidate got it wrong.
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    (Original post by A-New-Start)
    you have to find UNDILUTED h2o2 therefore you first multiply by 10 to go from 25 to 250, but since undiluted is 25 we then multiply by 40 and then since we have it in mol/dm-3 we need to change it to g/dm-3 so we basically multiply the no of mol. by molar mass ( 34 ) to get mass
    ok I get that you need to find in diluted, so atm I've got the value for 250cm3 but still don't get where the 40 is from!?


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    (Original post by DudeBoy)
    2 moles of HCHO = intial rate
    In the same reaction 1/2O2 is formed
    So for every one mol of HCHO formed 1/4 moles of O2 is formed
    Do you get it?
    in the specimen paper they asked to calculate the initial rate of fomation of oxygen, why did they get 4 for the dominator
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    (Original post by DudeBoy)
    2 moles of HCHO = intial rate
    In the same reaction 1/2O2 is formed
    So for every one mol of HCHO formed 1/4 moles of O2 is formed
    Do you get it?
    Understood perfectly, thank you! Just needed to know about the moles of HCHO.
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    Can someone help me with question8cii on jan 2013 I have worked out the number of moles of kmnO4 but then I don't knw wat to do next? Plus the equation doesn't seem to balance neither
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    (Original post by A-New-Start)
    you have to find UNDILUTED h2o2 therefore you first multiply by 10 to go from 25 to 250, but since undiluted is 25 we then multiply by 40 and then since we have it in mol/dm-3 we need to change it to g/dm-3 so we basically multiply the no of mol. by molar mass ( 34 ) to get mass
    sorry, autocorrect I meant undiluted on my last reply


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    (Original post by eggfriedrice)
    How?
    2 moles of HCHO = intial rate
    In the same reaction 1/2O2 is formed
    So for every one mol of HCHO formed 1/4 moles of O2 is formed
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    (Original post by eggfriedrice)
    Attachment 225848
    I dont see how sodium ethanoate is a salt of that. Anyone care to explain?
    If you look at what the examiner said they said that sodium ethanoate was wrong because the canditate misunderstood the question

    (Original post by Meado123)
    Can someone please explain the 2 limitations of predictions using standard electrode potentials?
    Is it that non-standard conditions where used and the rate of reaction was too slow?
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    (Original post by Brad0440)
    It's not. If you read the examiners comments, it says that the candidate got it wrong.
    So is the idea that you add NaOH to form a salt of that particular weak acid?
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    Ok i have a question about the enthalpy change of neutralisation in the book. Say the equation was HNO3 + KOH --> KNO3 + 2H20 ( i know that's not balanced, but still ) so then we first scale up to one mole by dividing our answer by 0.05 and then we divide by two, right? so that the the 2mol of water formed can become one.
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    (Original post by meg0001)
    Do you ned to memorise MnO4- and I2/S2O3^2- equations or will they be given in the exam?
    (Original post by Imekolo)
    You need to know them
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    What equations are these?
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    (Original post by beccac94)
    If you look at what the examiner said they said that sodium ethanoate was wrong because the canditate misunderstood the question
    Ah I see. Would be helpful if they included the correct answer xD
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    (Original post by otrivine)
    in the specimen paper they asked to calculate the initial rate of fomation of oxygen, why did they get 4 for the dominator
    its a 0.5: 2 ratio, so youve got the rate of hcoh which is 2 moles, so the rate of oxygen would be a quarter of it
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    (Original post by beccac94)
    If you look at what the examiner said they said that sodium ethanoate was wrong because the canditate misunderstood the question



    Is it that non-standard conditions where used and the rate of reaction was too slow?
    And activation energy is too high
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    (Original post by needtosucceed=))
    its a 0.5: 2 ratio, so youve got the rate of hcoh which is 2 moles, so the rate of oxygen would be a quarter of it
    so is that how they got 4
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    (Original post by eggfriedrice)
    I changed my answer

    2FeO4 2- + 10H+ -> 2Fe 3+ + 3/2 O2 + 5H2O
    a quick question

    for the second half equation what did u put involving hydrogen and water and oxygen/
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    (Original post by _HabibaH_)
    So is the idea that you add NaOH to form a salt of that particular weak acid?
    You would need to put an actual salt of hexylresorcinal to get the mark, so for example, one of the OH groups could become O-Na+ (or something like that). I think that's what the question is looking for.
 
 
 
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