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    (Original post by Kwaks)
    I did:

    192/23 = 8.347826.......
    Ans X 100 = 834.7826....g

    Because you have to work out the mass of air and you already know how much oxygen you have in that air. (I don't know if that is right)


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    I did that and then for the next part did you do 1/74 and times that by 834.7826? 74 because that was the Mr of C4H10O
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    Anyone else put 0.23 for the first calculation, 0.059 for the second and put 0.74 for the Kno3 percentage moles one?

    Please reply.
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    solid im pretty sure, as solubility of group 2 hydroxides decreases up group 2
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    A Grade boundaries? 46?
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    (Original post by Kira Yagami)
    Do you think, if they're nice I put get the mark for the correct equation too?

    Oh man lol what is wrong with me... Like I do the 'hard' part right...
    I would give you the mark tbh. The equation was correct and 'M' could stand for any group 2 metal. And Mg is a group 2 metal. (That's if I was an examiner)


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    (Original post by Kaz_96)
    I did that and then for the next part did you do 1/74 and times that by 834.7826? 74 because that was the Mr of C4H10O
    Yep that's exactly what I did. And I got a mass of 11g or so???


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    (Original post by Average_Andrew)
    solid im pretty sure, as solubility of group 2 hydroxides decreases up group 2
    That's carbonates?


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    (Original post by Kwaks)
    Yep that's exactly what I did. And I got a mass of 11g or so???


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    Yep i got around 11.8
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    (Original post by Average_Andrew)
    solid im pretty sure, as solubility of group 2 hydroxides decreases up group 2
    For this question it was unclear whether the water was in liquid or in gaseous form, Mg reacts with water to form Mg(OH)2 where as it reacts with steam to form MgO. So since nuclear reactors are very hot I made the assumption that the water would be in gaseous form and subsequently probably lost both of the marks for that question.
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    Can anyone remember the working out for the energy density q? I think it was number of moles x the enthalpy of combustion value they gave you?
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    The Uranian decays to Thorium? Is this right
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    I put 0.74 for the Kno3, is this correct?
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    (Original post by Kira Yagami)
    Anyone else put 0.23 for the first calculation, 0.059 for the second and put 0.74 for the Kno3 percentage moles one?

    Please reply.
    I got:
    1st = 835g
    2nd = 11g
    KNO3 = 34%


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    (Original post by JgR1997)
    The Uranian decays to Thorium? Is this right
    I got the same as you


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    (Original post by Kwaks)
    I got:
    1st = 835g
    2nd = 11g
    KNO3 = 34%


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    For the KNO3 it was just mass/Mr for the 3 involved, then do the number of moles of KNO3/ (Moles of KNO3 + the other 2) x 100?
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    Is there an unofficial mark scheme anywhere?
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    (Original post by Kaz_96)
    Can anyone remember the working out for the energy density q? I think it was number of moles x the enthalpy of combustion value they gave you?
    I did:
    1000/Mr
    And X enthalpy change of combustion


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    (Original post by JgR1997)
    The Uranian decays to Thorium? Is this right
    Correct.
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    (Original post by Kaz_96)
    For the KNO3 it was just mass/Mr for the 3 involved, then do the number of moles of KNO3/ (Moles of KNO3 + the other 2) x 100?
    Yeah, I think so... What did you get as the final answer???


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    Does anyone think I'll get a mark for the energy density Q? It was supposed to be near to 30000 or 36000 or something but then I divided this by 1000 so I got 36.something idk

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