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    (Original post by igobyursula)
    The issue with the calculations on Unit 5 is that there are different types of calculations that they may ask you. For example, they could ask you about redox titrations, or a back titration.

    In this example, and I have it in front of me now, there are two main equations.

    i) a periodicity equation. The reaction between P4O10 and 6H20 to form 4H3PO4

    ii)acid-base neutralisation between NaOH and Phosphoric Acid.

    You work out the moles of Sodium Hydroxide using the conc. and the volume given. You use the ratios in your Acid-base neutralisation equation between the sodium hydroxide and the phosphoric acid to work out the moles of phosphoric acid. From that you can work out the initial amount of P4O10 using the equation between P4O10 and 6H20.

    There isn't really any universal step by step method on these kind of questions as they may be asking for different things. I just keep practising them.
    Thanks... btw do you know how to write ionic equations from a balanced equation?
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    Spent the day helping my friend with the titration questions. I was a bit rusty so I'm going to do all the textbook questions I can be bothered to do tonight

    Got so much to do as I have biology on Monday as well which I have quite a bit to do for. So unprepared :eek3: Kinda regretting focussing on Chem4 and C4 so much.
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    I just did the June 2011 Chem 5 paper, and I was doing pretty well up until question 7. I still managed to get an A* though :cool:. I thought the grade boundaries were surprisingly low though as the majority of the paper wasn't that difficult.
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    (Original post by brittanna)
    I just did the June 2011 Chem 5 paper, and I was doing pretty well up until question 7. I still managed to get an A* though :cool:. I thought the grade boundaries were surprisingly low though as the majority of the paper wasn't that difficult.
    Did you get the "water hydrolyses aren't reversible" one?
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    (Original post by bananarama2)
    Did you get the "water hydrolyses aren't reversible" one?
    No, I got that one wrong. Question 5 was the other question I didn't do very well on either, although I didn't do as badly as I did on question 7 .
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    (Original post by lifeisgood2012)
    Thanks... btw do you know how to write ionic equations from a balanced equation?
    Yes. I do. Identify what has been oxidised or reduced and what it has been reduced or oxidised to. You can do this by writing oxidation no.s on top of each species. The balance using waters, hydrogens and electrons till the charges and atoms are balanced. When you combine the half equations the electrons should cancel and you will get the initial equation again.

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    (Original post by brittanna)
    I just did the June 2011 Chem 5 paper, and I was doing pretty well up until question 7. I still managed to get an A* though :cool:. I thought the grade boundaries were surprisingly low though as the majority of the paper wasn't that difficult.
    In the examiners report to jun 12 paper for chem 5 (which is the hardest past paper so far on this spec according to my chemistry teacher) they refer to june 2011 stating something like the demanding nature of the June 2012 paper is not unique as June 2011 was equally as demanding. The grade boundaries are low for chem 5 in general, but were especially low for june 11/12 because a lot of people found it challenging.

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    (Original post by igobyursula)
    In the examiners report to jun 12 paper for chem 5 (which is the hardest past paper so far on this spec according to my chemistry teacher) they refer to june 2011 stating something like the demanding nature of the June 2012 paper is not unique as June 2011 was equally as demanding. The grade boundaries are low for chem 5 in general, but were especially low for june 11/12 because a lot of people found it challenging.

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    Dont forget that the examiner from June 2012 onwards changed from all the other papers before them, so difficulty and wording was a little different and it threw candidates in June 2012.
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    Yes, wording is an important factor. That's why I think it is important to look at papers from the older specification as well as the current one, just to be extra prepared.

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    Hey, on the past paper's I've been doing there's a question to do with the colours of complexes which keeps coming up..

    So basically its when [Co(NH3)6]2+ (which I think is straw coloured) is allowed to stand in air.

    So I know that it forms [Co(NH3)6]3+ but I've been taught that this is yellow and the mark scheme says it is dark brown.

    Does anyone know why or if I will still get the marks for putting yellow?

    Thanks
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    (Original post by Lucy-1995)
    Hey, on the past paper's I've been doing there's a question to do with the colours of complexes which keeps coming up..

    So basically its when [Co(NH3)6]2+ (which I think is straw coloured) is allowed to stand in air.

    So I know that it forms [Co(NH3)6]3+ but I've been taught that this is yellow and the mark scheme says it is dark brown.

    Does anyone know why or if I will still get the marks for putting yellow?

    Thanks
    I think when it gets oxidised it turns brown, whether you get a mark or yellow I'm not sure.

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    (Original post by Lucy-1995)
    Hey, on the past paper's I've been doing there's a question to do with the colours of complexes which keeps coming up..

    So basically its when [Co(NH3)6]2+ (which I think is straw coloured) is allowed to stand in air.

    So I know that it forms [Co(NH3)6]3+ but I've been taught that this is yellow and the mark scheme says it is dark brown.

    Does anyone know why or if I will still get the marks for putting yellow?

    Thanks
    Almost all of the mark schemes I've come across have said yellow/straw oxidised to a much darker, rusty brown. I really do doubt they would accept yellow for [Co(H2O)6]3+ as it's never usually mentioned on the right hand side of the MS for BOTD answers, so suggest taking and using was the mark schemes say, if in doubt always copy the mark scheme
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    (Original post by igobyursula)
    Hiya...

    Sorry if someone else has already answered this question for you.

    With 2+ hexa-aqua ions, basically (well my chemistry teacher told me because there is no distinction in the book(s))

    Its 3en which forms [M(en)3]2+ + 6H2O

    With 3+ hexa aqua ions it does something different.

    en is H2NCH2CH2NH2

    This is the following equation

    2[M(H2O)6 ]3+ + 3H2NCH2CH2NH2 ----> 2[M(H2O)3(OH)3] + 3H3NCH2CH2NH3

    This is because 3+ ions are slightly more acidic and can donate H+ more easily. Think of the reactions between the hexa aqua ions and carbonate ions.

    Hope this helps. I was completely muddled by it all initially. There is not actually any distinction in the book (I use Nelson Thornes) at all, but I think you are supposed to apply knowledge of the differences between the reactions betweeen carbonate with 2+ hexa aqua ions and 3+ hexa aqua ions. en is a bidentate ligand, as is the carbonate ion.
    thanks alot, the NT is really vague on this topic
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    can someone explain 6) d) i)

    http://filestore.aqa.org.uk/subjects...W-MS-JUN12.PDF

    dont understand why the ratio is 3:5 isnt it 2:5
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    (Original post by Dalts)
    Almost all of the mark schemes I've come across have said yellow/straw oxidised to a much darker, rusty brown. I really do doubt they would accept yellow for [Co(H2O)6]3+ as it's never usually mentioned on the right hand side of the MS for BOTD answers, so suggest taking and using was the mark schemes say, if in doubt always copy the mark scheme
    Haha its the motto to live by with AQA! Cheers
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    (Original post by JSN)
    can someone explain 6) d) i)

    http://filestore.aqa.org.uk/subjects...W-MS-JUN12.PDF

    dont understand why the ratio is 3:5 isnt it 2:5
    You have to combine the bottom two half equations first

    Fe2+ \rightarrow Fe3+ + e-
    C2O42- \rightarrow 2CO2 + 2e-

    FeC2O4 \rightarrow Fe3+ + 2CO2 + 3e-

    Then combine this equation with the top equation

    3MnO4- + 5FeC2O4 + 24H+ \rightarrow 3Mn2+ + 5Fe3+ + 10CO2 + 12H2O

    So for every 3 moles of MnO4- which react, 5 moles of FeC2O4 react i.e. a 3:5 ratio
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    (Original post by JSN)
    thanks alot, the NT is really vague on this topic
    I think NT are a bit rubbish in some places. I am eternally grateful to my lovely teacher.:yes:
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    Does the value of emf increase if more electrons are released?
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    (Original post by crc290)
    You have to combine the bottom two half equations first

    Fe2+ \rightarrow Fe3+ + e-
    C2O42- \rightarrow 2CO2 + 2e-

    FeC2O4 \rightarrow Fe3+ + 2CO2 + 3e-

    Then combine this equation with the top equation

    3MnO4- + 5FeC2O4 + 24H+ \rightarrow 3Mn2+ + 5Fe3+ + 10CO2 + 12H2O

    So for every 3 moles of MnO4- which react, 5 moles of FeC2O4 react i.e. a 3:5 ratio
    thanks alot, cant rep you


    (Original post by igobyursula)
    I think NT are a bit rubbish in some places. I am eternally grateful to my lovely teacher.:yes:
    i usually live and die by the NT books but its let me down in chem 4 and 5 haha
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    (Original post by lifeisgood2012)
    Does the value of emf increase if more electrons are released?
    it gets more -ve i think but someone confirm this
 
 
 
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