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    How many marks out of 100 should I be looking at if I want to get an A* in this paper?
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    (Original post by OhNo_23)
    In the specification it says that we need to be able to define enthalpy of atomisation of an element and of a compound. I have never come across one for a compound :/ Does anyone have any ideas of how this would be defined or have an example of an equation where this is the case?
    Thanks in advance
    Enthalpy of atomisation of an element: the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state.

    Eg. Na(s) --> Na(g)

    Enthalpy of atomisation of a compound: the enthalpy change when one mole of a compound in its standard state is converted into its constituent atoms in gaseous form.

    E.g. CH4(g) --> C(g) + 4H(g)

    The enthalpy of atomisation of a compound is equal to the bond energies of the compound. E.g. in the above reaction atomisation energy is equal to 4(C-H) bonds.
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    has anyone got the jan 11 paper for chemistry unit 5 and the mark scheme? it would be very appreciated, thanks
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    (Original post by jimmy303)
    Enthalpy of atomisation of an element: the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state.

    Eg. Na(s) --> Na(g)

    Enthalpy of atomisation of a compound: the enthalpy change when one mole of a compound in its standard state is converted into its constituent atoms in gaseous form.

    E.g. CH4(g) --> C(g) + 4H(g)

    The enthalpy of atomisation of a compound is equal to the bond energies of the compound. E.g. in the above reaction atomisation energy is equal to 4(C-H) bonds.
    Brilliant, thanks for your help! So is it essentially the same thing as mean bond enthalpy, because I notice that is not listed in the spec?
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    (Original post by OhNo_23)
    Brilliant, thanks for your help! So is it essentially the same thing as mean bond enthalpy, because I notice that is not listed in the spec?
    Ermm.......you just need to know the definition and be able to apply and use it for Born-Haber cycles Think only when you apply it you only need to know its role in the Born-Haber cycle so you can work out the lattice enthalpy etc!
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    (Original post by OhNo_23)
    Brilliant, thanks for your help! So is it essentially the same thing as mean bond enthalpy, because I notice that is not listed in the spec?
    Actually it isn't. It will be similar but a value using mean bond enthalpies will be less accurate.

    It is the same principle as comparing calculations using mean bond enthalpies with calculations using the born haber cycle.

    With mean bond enthalpies the value is the average energy needed to break a covalent bond over a range of different compounds.

    The atomisation of a compound on the other hand will only be looking at the energy needed to break the covalent bonds in that one compound.

    Does that make sense?
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    (Original post by jimmy303)
    Actually it isn't. It will be similar but a value using mean bond enthalpies will be less accurate.

    It is the same principle as comparing calculations using mean bond enthalpies with calculations using the born haber cycle.

    With mean bond enthalpies the value is the average energy needed to break a covalent bond over a range of different compounds.

    The atomisation of a compound on the other hand will only be looking at the energy needed to break the covalent bonds in that one compound.

    Does that make sense?
    That makes perfect sense, thanks so much!
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    (Original post by T-Toe)
    How many marks out of 100 should I be looking at if I want to get an A* in this paper?
    For the last three papers an A* (i.e. 90% ums points) has consistently been around 82-84 marks out of 100.
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    (Original post by laura123)
    Do we need to know all the little details about electrochemical cells? Pleeeeeease say no!
    I don't think we need to know about specific non- / rechargeable cells apart from from I think the hydrogen fuel cell.
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    (Original post by asb15)
    has anyone got the jan 11 paper for chemistry unit 5 and the mark scheme? it would be very appreciated, thanks
    http://www.thestudentroom.co.uk/show....php?t=1611471
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    Hi, can anyone help me with question 8c) on January 2010 paper?? I understand up to calculation moles of H3PO4, but don't know where to go next? I looked at the mark scheme and it said the next mark was for times it by 10^6... I don't understand Any help would be much appreciated.

    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN10.PDF
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    (Original post by BethBeth)
    Hi, can anyone help me with question 8c) on January 2010 paper?? I understand up to calculation moles of H3PO4, but don't know where to go next? I looked at the mark scheme and it said the next mark was for times it by 10^6... I don't understand Any help would be much appreciated.

    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN10.PDF
    You times by 10^6 as right now you're only using a 25cm^3 sample. You need to make this 25000dm^3. times by 1000 to make it dm^3 and then another 1000 to make it 25000.
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    (Original post by BethBeth)
    Hi, can anyone help me with question 8c) on January 2010 paper?? I understand up to calculation moles of H3PO4, but don't know where to go next? I looked at the mark scheme and it said the next mark was for times it by 10^6... I don't understand Any help would be much appreciated.

    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN10.PDF
    Hey, yeah this is a nasty mathy question for 5 marks.

    Firstly H3PO4 + 3NaOH --> Na3PO4 + 3H20
    - 1:3 ratio of H3PO4:NaoH

    n(mole) of NaOH added = (21.2/1000)*0.500 = 0.0106 mol
    => therefore n(mole) H3PO4 in 25cm^3 sample = 0.0106/3 = 3.533333*10^-3 mol

    container is 25000dm^3
    25cm^3 = 0.025dm^3
    => 25000/0.025 = 1000000 25cm^3 in 25000dm^3 - hence where the factor of 1000000 came from

    therefore n(mole) H3PO4 in 25dm^3 = (3.53333*10^-3) * 1000000 = 3533.33333 mol

    P4O10 + 6H2O ---> 4H3PO4
    => therefore n(mole) P4O10 = 3533.33333/4 = 883.3333mol

    n = mass/mr => mass = n*mr
    => mass = (883.333333)*((16*10)+(31*4)) = 250866.6667g
    = 250.8666667kg = 251kg
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    (Original post by starburst92)
    You times by 10^6 as right now you're only using a 25cm^3 sample. You need to make this 25000dm^3. times by 1000 to make it dm^3 and then another 1000 to make it 25000.
    (Original post by Cameronn)
    Hey, yeah this is a nasty mathy question for 5 marks.

    Firstly H3PO4 + 3NaOH --> Na3PO4 + 3H20
    - 1:3 ratio of H3PO4:NaoH

    n(mole) of NaOH added = (21.2/1000)*0.500 = 0.0106 mol
    => therefore n(mole) H3PO4 in 25cm^3 sample = 0.0106/3 = 3.533333*10^-3 mol

    container is 25000dm^3
    25cm^3 = 0.025dm^3
    => 25000/0.025 = 1000000 25cm^3 in 25000dm^3 - hence where the factor of 1000000 came from

    therefore n(mole) H3PO4 in 25dm^3 = (3.53333*10^-3) * 1000000 = 3533.33333 mol

    P4O10 + 6H2O ---> 4H3PO4
    => therefore n(mole) P4O10 = 3533.33333/4 = 883.3333mol

    n = mass/mr => mass = n*mr
    => mass = (883.333333)*((16*10)+(31*4)) = 250866.6667g
    = 250.8666667kg = 251kg
    Thank you! Really appreciate this, think I'll be able to do this if it turns up on friday now Hope your revision is going well x
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    (Original post by ncollier4)
    Thank youuuuuuuuuuuuuuuuuuuuuuu

    But where does the catalyst come in? :confused:
    According to the nelson thornes book it would seem you dont need to know it. After the two equations all it states is the catalyst that's used then starts waffling on about other common catalyst .
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    Hey fellow chemists, do you think it's worth doing these exercises (example below)?

    http://www.a-levelchemistry.co.uk/AQ...ics%20home.htm

    Or just sticking with past papers? I'm pretty content with revision so far so I'm not sure :/
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    I think this is my least favourite unit, it's pretty hard tbh
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    I need about 60/120 for an A in Chemistry overall. As a result I haven't really revised for this paper as much as I should have, hopefully I can pull through.
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    (Original post by Raimu)
    I need about 60/120 for an A in Chemistry overall. As a result I haven't really revised for this paper as much as I should have, hopefully I can pull through.
    I thought both unit 4 and 5 were out of 140...?
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    (Original post by BethBeth)
    I thought both unit 4 and 5 were out of 140...?
    I'm pretty sure both of them are out of 120. 96/120 was 80% on the last paper.
 
 
 
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