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    (Original post by choc1234)
    and me!
    Me too please!
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    (Original post by NRican)
    ahh mindblank, can anyone guide me through writing this representation



    using

    haha i literally just did this now! i don't really understand i myself, so if anyone can help that'd be good.
    do know that on the mark scheme it says you do not really need to put H2O in.
    Also that in acidic ad alkaline conditions the e.m.f. is exactly the same because the overall equation is exactly the same!
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    (Original post by T-Toe)
    The markscheme says otherwise :/
    1
    (d)
    (ii)
    Decreases

    Reaction exothermic/H -ve

    (Equilibrium )shifts to left/backwards (as temperature rises)/ equilibrium opposes the change



    What doyu mean ? The mark scheme does say it's Exothermic
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    (Original post by -jessica)
    The way I did it (not 100% sure if this is the right way!):

    Know two equations for P4O10
    -P4O10 + 12NaOH -> 4Na3PO4
    -P4O10 + 6H2O -> 4H3PO4

    moles NaOH = 0.5 x 21.2x10^-3 = 0.0106 mol
    From equations, moles H3PO4 = 0.0106/3 = 3.53x10^-3 mol

    So moles P4O10 = 1/4 x 3.53x10^-3 = 0.883 mol

    Mr of P4O10 = 284

    mass = moles x Mr = 0.883 x 284 = 250.9 = 251 kg (3 s.f)
    awesome! cheers guys and even more credit to sheldon, absolute saviour
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    (Original post by NRican)
    ahh mindblank, can anyone guide me through writing this representation



    using

    This!!!:confused:
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    (Original post by Zabukar)
    Were can I find the Answers for the Exam questions in the Aqa Chem text book by Nelson thornes ?
    http://www.thestudentroom.co.uk/show....php?t=1137126
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    hey guys you know how cobolt and copper form CXCl4 does it happen with Fe and all the others
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    (Original post by OsmosisJones)
    Learn all the easy ones:

    Fe2+ is generally green
    Fe3+ is purple, precitates go brown
    Cu2+ is generally blue
    Cu and Fe form yellow solution with Cl
    Cr3+ is generally green, remember titration, orange goes green, Cr2O7- ->Cr3+
    MnO4- is deep purple, Mn2+ is light pink (titration)
    Carbonates generally stay the same colour if bonded to 2+ ion, hydroxide formed 3+ ions.


    There are others you are required to learn, but these are the easy ones.

    Others include:
    Co2+ is pink solution, blue prec. in hydroxide, yellow in excess ammonia
    Co3+ is yellow, brown in hydroxide
    Cr3+ is purple in excess ammonia
    Cu2+ only allows 4NH3 around it, deep blue colour.
    Fe3+ is pale violet...

    You cannot generalise all of the colours! For example:

    Copper:
    (Cu(H2O)6)2+ is blue
    (Cu(NH3)4(H2O)4)2+ is blue violet
    (CuCl4)2- is yellow green
    Cu(H20)4(OH)2 is a blue ppt.
    CuCO3 is a blue green ppt.

    Co3+ is brown and brown ppt. in hydroxide.

    I'm sorry guys, but you do have to learn all of the colours.... there isn't really a shortcut.... you can break them down into different central metal ions i suppose.

    Hope this helps
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    (Original post by ben10)
    hey guys you know how cobolt and copper form CXCl4 does it happen with Fe and all the others
    No it doesn't
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    if you add excess ammonia to say copper is the reaction:
    [Cu(h2o)6]2+ + 2NH3 ----- Cu(h2o)4(oh)2 + NH4+

    ? or not?>
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    (Original post by choc1234)
    if you add excess ammonia to say copper is the reaction:
    [Cu(h2o)6]2+ + 2NH3 ----- Cu(h2o)4(oh)2 + NH4+

    ? or not?>
    deep blu soln. of [Cu(NH3)4(H20)2]2+
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    when writing cell diagrams
    when do you NOT need to include water/ H+
    i dont get it :|
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    (Original post by NRican)
    deep blu soln. of [Cu(NH3)4(H20)2]2+
    sorry i meant Fe2+
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    (Original post by Willyg11)
    Fe3+ is pale violet...

    You cannot generalise all of the colours! For example:

    Copper:
    (Cu(H2O)6)2+ is blue
    (Cu(NH3)4(H2O)4)2+ is blue violet
    (CuCl4)2- is yellow green
    Cu(H20)4(OH)2 is a blue ppt.
    CuCO3 is a blue green ppt.

    Co3+ is brown and brown ppt. in hydroxide.

    I'm sorry guys, but you do have to learn all of the colours.... there isn't really a shortcut.... you can break them down into different central metal ions i suppose.

    Hope this helps
    The mark scheme normally lets you put half the colour, with the exception of Cu (otherwise there'd be no point in examining it).
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    Found a good way of remembering, [CuCl4]2- is pronounced, [cYoo see ell 4] and Yooo just reminds me of yellow! And that just leaves [CoCl4]2- to be other (only) other one, and that's blue.
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    with the 2+ ions in excess ammonia, does it form [m(h20)2(nh3)4]2+
    and with 3+ ions [m(nh3)6]3+ or other way around?
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    (Original post by choc1234)
    if you add excess ammonia to say copper is the reaction:
    [Cu(h2o)6]2+ + 2NH3 ----- Cu(h2o)4(oh)2 + NH4+

    ? or not?>
    Little ammonia you are right..... for excess it goes to (Cu(NH3)4(H2O)2)2+
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    (Original post by OsmosisJones)
    The mark scheme normally lets you put half the colour, with the exception of Cu (otherwise there'd be no point in examining it).
    Half the colour?
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    (Original post by NRican)
    ahh mindblank, can anyone guide me through writing this representation



    using

    It's in alkaline conditions so you use the equations with OH- rather than H+.

    The most negative electrode is swapped and put on the right hand side.

    Platinum is put on each side as both half reactions lack a solid.

    Oxidised species in the middle!

    Otherwise you just write out the full half equations from left to right, putting in commas to separate them or | if there is a change in state.

    So:

    H2(g) + 2OH-(aq) ---> 2H2O(l)

    becomes:

    Pt | H2(g) | OH-(aq), H2O(l) ||

    And:

    O2(g) + 2H2O(l) ---> 4OH-(aq)

    becomes:

    || O2(g) | H2O(l), OH-(aq) | Pt

    Overall you get:

    Pt | H2(g) | OH-(aq), H2O(l) || O2(g) | H2O(l), OH-(aq) | Pt

    Hope this is helpful to you all!!

    (Original post by Anon1993)
    x
    (Original post by sophiakrywycz)
    x
    (Original post by choc1234)
    x
    (Original post by Sparkly-Star)
    x
    Edit! The reason why it has to be:

    || O2(g) | H2O(l), OH-(aq) | Pt

    and not

    || O2(g) | OH-(aq), H2O(l) | Pt

    is because the water and the oxygen appear on the same side of the half reaction (O2(g) + 2H2O(l) ---> 4OH-(aq)) so need to be stuck together (in some representations they have their own brackets to separate them).

    || [O2(g) | H2O(l)], [OH-(aq)] | Pt

    Same reason why on

    Pt | H2(g) | OH-(aq), H2O(l) ||

    OH- is next to H2(g)

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    Guys in the specimen paper I thought the decomposition of calcium carbonate is:
    CaCO3 ----> CaO + CO2..
    The markscheme says CaCO3 ---> CaO + CO ??!!!??
 
 
 
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