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    (Original post by DudeBoy)
    And activation energy is too high
    Oh I forgot to put that bit, but its sort of assumed that the rate of reaction is too slow due to not as many molecules having successful collisions because the activation energy is too high for the molecules to exceed
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    pleaseeeee can someone tell me where the 40 comes from in 7b June 2010 working out the conc, I get everything else!! would be really appreciated


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    (Original post by otrivine)
    a quick question

    for the second half equation what did u put involving hydrogen and water and oxygen/
    I didn't make a half equation?
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    (Original post by otrivine)
    so is that how they got 4
    yep, you have to divide by 4 to get a quarter of the rate for o2
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    Who was i talking about 3I2 and I2 crap, last night?
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    (Original post by otrivine)
    so is that how they got 4
    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 georgiaaaxo)
    pleaseeeee can someone tell me where the 40 comes from in 7b June 2010 working out the conc, I get everything else!! would be really appreciated


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    What 40??? On the mark scheme I have you times it by 34 which is the mr of H2O2
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    (Original post by beccac94)
    What 40??? On the mark scheme I have you times it by 34 which is the mr of H2O2
    you also have to time by 40 to go from 25cm3 to 1dm3
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    (Original post by Brad0440)
    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.
    Okay thanks, that's fine then
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    (Original post by A-New-Start)
    you also have to time by 40 to go from 25cm3 to 1dm3
    I think the way the mark scheme has done it a different way
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    (Original post by A-New-Start)
    Who was i talking about 3I2 and I2 crap, last night?
    Me xD
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    (Original post by needtosucceed=))
    yep, you have to divide by 4 to get a quarter of the rate for o2
    can u show me so sorry
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    (Original post by A-New-Start)
    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.
    No its the specimen paper
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    (Original post by A-New-Start)
    Who was i talking about 3I2 and I2 crap, last night?
    Me xD
    (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
    The thing is, aren't we assuming the initial rate is for the 2 mols of HCHO?
    I mean we could just as easily say the initial rate for O2 was 1x10-12 and hence the initial rate of HCHO would be four times that?
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    (Original post by eggfriedrice)
    Me xD


    The thing is, aren't we assuming the initial rate is for the 2 mols of HCHO?
    I mean we could just as easily say the initial rate for O2 was 1x10-12 and hence the initial rate of HCHO would be four times that?
    yep we are, so 2 moles of hcoh = rate of 1 x 10-12
    so 1 mole of o2 would be half of that = 5 x 10^-13

    so o.5 moles would be half of that again which = 2.5 x 10^-13

    or you could just divide by 4 to do it all in one step
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    (Original post by beccac94)
    I think the way the mark scheme has done it a different way
    so the x40 is from 25cm3 to 1dm3? how does that work? surely you'd just /25 and then /1000?
    SO CONFUSED


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    (Original post by otrivine)
    can u show me so sorry
    see my post above
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    (Original post by beccac94)

    Is it that non-standard conditions where used and the rate of reaction was too slow?
    I've just had a look through a mark scheme and that was the two points they gave yes
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    (Original post by georgiaaaxo)
    so the x40 is from 25cm3 to 1dm3? how does that work? surely you'd just /25 and then /1000?
    SO CONFUSED


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    multiplying by 40 or dividing by 25/1000 is the same thing
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    (Original post by Meado123)
    I've just had a look through a mark scheme and that was the two points they gave yes
    i think high activation energy is also one?
 
 
 
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