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    (Original post by lifeisgood2012)
    hiyaaaa guysssss...... does any of you know where i can find practice/ old legacy papers for each of the topics?
    Google tom reds blog and go to chemistry

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    i know this ain't the correct thread but the ppl here are probably very god with chem4.... so can some explain nmr to me and hw to deduce the structure of something using nmr (splitting, integration trace, functional groups)... i really dnt get this topic so if someone can give me their not or explain it to be i will be very grateful
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    Any predictions people
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    (Original post by loknath)
    Any predictions people
    Idk but u1 resit was aaaaaawful so I'm hoping this exam is nicer

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    (Original post by lifeisgood2012)
    i know this ain't the correct thread but the ppl here are probably very god with chem4.... so can some explain nmr to me and hw to deduce the structure of something using nmr (splitting, integration trace, functional groups)... i really dnt get this topic so if someone can give me their not or explain it to be i will be very grateful
    Okay so the way I remember it is..
    no of peaks= number of hydrogen environments
    integration ratio= how many hydrogen atoms there are in each environment. e.g if there are are 3 different hydrogen environment. In the first environment if there are 3 hydrogen atoms, in the second environment if there are 2 hydrogens and in the third environment there are 2 hydrogen atoms...then the integration ratio will be 3:2:2.
    If you can bring up a question I may be able to help you. I'm no expert in this topic by any means but hope that helped you abit
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    for this module are we require to drawing a hess cycle?
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    (Original post by lifeisgood2012)
    for this module are we require to drawing a hess cycle?
    No, just know how to draw a Born Harber Cycle.
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    (Original post by lifeisgood2012)
    for this module are we require to drawing a hess cycle?
    I don't think so, aslong as you can workout the calculation.

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    (Original post by cheesypuff)
    No, just know how to draw a Born Harber Cycle.
    Thank youu
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    (Original post by Hart1995)
    I don't think so, aslong as you can workout the calculation.

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    Thanks mate
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    why is it that the transition metals have similar properties issit because of the closeness in the s and the d energy sub level?
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    (Original post by lifeisgood2012)
    for this module are we require to drawing a hess cycle?
    No, but it could come up in thermodynamic synoptic, doubt it though!
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    (Original post by lifeisgood2012)
    why is it that the transition metals have similar properties issit because of the closeness in the s and the d energy sub level?
    Yes, there are a couple of Things in common
    1. They can form a stable ion with a partially filled d-Orbital
    2. They have variable oxidation states
    3. They can form complex coloured ions
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    Hello, anyone here dyslexic and really struggling with all the chemical formulas, names, and calculations? I need some advice on what the best learning methods would be.
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    Could someone please explain how the answer is 6 to question 4e(iii) on the jan 2011 paper. I don't understand how the multiple taste can be assumed to form 4 coordinate bonds with the Co.

    Is it 6 because you just assume it's going to form the octahedral shape that most complexes are seen as?

    Thanks
    http://filestore.aqa.org.uk/subjects...W-MS-JAN11.PDF
    http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF
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    (Original post by starfish232)
    Could someone please explain how the answer is 6 to question 4e(iii) on the jan 2011 paper. I don't understand how the multiple taste can be assumed to form 4 coordinate bonds with the Co.

    Is it 6 because you just assume it's going to form the octahedral shape that most complexes are seen as?

    Thanks
    http://filestore.aqa.org.uk/subjects...W-MS-JAN11.PDF
    http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF
    Hello,
    Ok this is a bit hard to explain. In q4e(ii) you have to assume the PR forms 4 CO-ordiante bonds as we know the H20 can only form 1 and there are 2 molecules of h20.

    Hence in next questions there is 1PR(4 co-ordinate bonds), 1 unidentate ligand(1 co-ordinate bond) and 1 Cyanide molecules(1 co-ordinate bonds)
    Hence the coordination number is 4 plus 1 plus 1 =6

    Hope that helps
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    (Original post by starfish232)
    Could someone please explain how the answer is 6 to question 4e(iii) on the jan 2011 paper. I don't understand how the multiple taste can be assumed to form 4 coordinate bonds with the Co.

    Is it 6 because you just assume it's going to form the octahedral shape that most complexes are seen as?

    Thanks
    http://filestore.aqa.org.uk/subjects...W-MS-JAN11.PDF
    http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF
    No, you do not assume it.

    In the reaction above, you have:

    PR(aq) + [Fe(H2O)6]2+(aq) → [FePR(H2O)2]2+(aq) + 4H2O(I)

    As you can see that the Pophyrin ring has replaced 4 water molecules. Thus, it must be a mutlidendate ligand, with 4 lone pairs of electrons.

    The final complex has these compounds acting as ligands
    CN- ( 1 co-ordinate bond)
    A unidentate ligand ( 1 co-ordinate bond)
    PR (4 co-ordinate bonds)

    The complex has 6 co-ordinate bonds, so its co-ordination number must be six. (octahedral). Does that make sense?
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    (Original post by cheesypuff)
    Hello,
    Ok this is a bit hard to explain. In q4e(ii) you have to assume the PR forms 4 CO-ordiante bonds as we know the H20 can only form 1 and there are 2 molecules of h20.

    Hence in next questions there is 1PR(4 co-ordinate bonds), 1 unidentate ligand(1 co-ordinate bond) and 2 Cyanide molecules(2 co-ordinate bonds)
    Hence the coordination number is 1 plus 1 plus 2=4

    Hope that helps
    Your reasoning is correct, but the co-ordination number isn't 4. Look at my post above.
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    (Original post by cheesypuff)
    Hello,
    Ok this is a bit hard to explain. In q4e(ii) you have to assume the PR forms 4 CO-ordiante bonds as we know the H20 can only form 1 and there are 2 molecules of h20.

    Hence in next questions there is 1PR(4 co-ordinate bonds), 1 unidentate ligand(1 co-ordinate bond) and 1 Cyanide molecules(1 co-ordinate bonds)
    Hence the coordination number is 4 plus 1 plus 1 =6

    Hope that helps
    Oh I get it! Didn't look at the part of the question before when i was answering this part. Thanks!
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    (Original post by frogs r everywhere)
    Your reasoning is correct, but the co-ordination number isn't 4. Look at my post above.
    Yes I know that was a typo
 
 
 
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