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Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread watch

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    Thought I may as well set up the thread now since no one else had. Some stuff that people might find useful:

    Chemguide - http://www.chemguide.co.uk/ (high quality in depth notes on most of the topics)

    Chemrevise- http://chemrevise.org/3-edexcel-revision-guides/ (short and concise notes for each topic)

    Specification - http://qualifications.pearson.com/co...ry_Issue_6.pdf

    Papers - http://www.physicsandmathstutor.com/...dexcel-unit-4/

    http://www.physicsandmathstutor.com/...dexcel-unit-5/

    (has all the papers currently on edexcel website, as well as the IAL papers in one place)

    If anyone else has some good links I have missed, please let me know and I will add them to the op.

    Feel free to use this thread for any questions you may have, but please remember thatyou are not allowed to ask for details of practicals and anyone doing so will be reported.
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    Finally this thread has been created, how everyone is finding revision ?
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    Do any of you use examiner reports after solving past papers? It really helps me when the mark scheme fails to.
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    kaboom
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    These notes are very much helpful... Thanks to those who created it..
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    just realizing what a piece of piss redox is!
    feelsgoodman
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    Hi guys, I'm also doing edexcel and am dreading exam period :\

    Quick question:

    In the textbook(blue edexcel textbook), it says if you react [Cr(H2O)6]3+ with excess sodium hydroxide, you get [Cr(H2O)2(OH)4]- on page 187 but on page 181, it says reaction of [Cr(H2O)6]3+ with a base gives [Cr(OH)6]3-. So I am slightly confused about what is produced when Cr 3+ reacts with excess sodium hydroxide here
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    (Original post by setarcos)
    Hi guys, I'm also doing edexcel and am dreading exam period :\

    Quick question:

    In the textbook(blue edexcel textbook), it says if you react [Cr(H2O)6]3+ with excess sodium hydroxide, you get [Cr(H2O)2(OH)4]- on page 187 but on page 181, it says reaction of [Cr(H2O)6]3+ with a base gives [Cr(OH)6]3-. So I am slightly confused about what is produced when Cr 3+ reacts with excess sodium hydroxide here
    It depends of the amount of NaOH you're adding. The first one is a result of adding it drop-wise until a precipitate forms and the second one is until an excess amount of NaOH is added and the precipitate dissolves to form a solution.
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    (Original post by aymanzayedmannan)
    It depends of the amount of NaOH you're adding. The first one is a result of adding it drop-wise until a precipitate forms and the second one is until an excess amount of NaOH is added and the precipitate dissolves to form a solution.
    Hmm, not sure abt that because if you add it drop wise, u get

    [Cr(H2O)3(OH)3] (s) precipitate forming. My question is if u add an excess of sodium hydroxide from that stage, do you get [Cr(H2O)2(OH)4]- or [Cr(OH)6]3- because both of them form solutions.

    or is it simply that if you add NaOH in excess at once, you get one and if u add it dropwise in excess u get the other?
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    (Original post by setarcos)
    Hmm, not sure abt that because if you add it drop wise, u get

    [Cr(H2O)3(OH)3] (s) precipitate forming. My question is if u add an excess of sodium hydroxide from that stage, do you get [Cr(H2O)2(OH)4]- or [Cr(OH)6]3- because both of them form solutions.

    or is it simply that if you add NaOH in excess at once, you get one and if u add it dropwise in excess u get the other?
    Oops, I'm sorry - I didn't read your question properly at first. You're right about the precipitate, that is the correct formula for it. But when you add excess, you get [Cr(OH)6]3- as per my knowledge. I can't see why the Edexcel textbook would disagree with that because the revision guide and George Facer say otherwise.
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    (Original post by aymanzayedmannan)
    Oops, I'm sorry - I didn't read your question properly at first. You're right about the precipitate, that is the correct formula for it. But when you add excess, you get [Cr(OH)6]3- as per my knowledge. I can't see why the Edexcel textbook would disagree with that because the revision guide and George Facer say otherwise.
    Yeah that's what I'm confused about, Chemguide, George Facer and CGP all say that you get [Cr(OH)6]3- but on page 187 of the edexcel table where it gives the table with the transition elements, it says you get [Cr(H2O)2(OH)4]- on the addition of excess NaOH to give you a solution after the precipitate dissolves.

    Sorry to be a bit pedantic about this, just really make sure I know the it properly
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    (Original post by setarcos)
    Yeah that's what I'm confused about, Chemguide, George Facer and CGP all say that you get [Cr(OH)6]3- but on page 187 of the edexcel table where it gives the table with the transition elements, it says you get [Cr(H2O)2(OH)4]- on the addition of excess NaOH to give you a solution after the precipitate dissolves.

    Sorry to be a bit pedantic about this, just really make sure I know the it properly
    [Cr(OH)4(H2O)2]- will be the first complex formed when the ppt is redissolved, maybe that's what they mean?
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    (Original post by setarcos)
    Yeah that's what I'm confused about, Chemguide, George Facer and CGP all say that you get [Cr(OH)6]3- but on page 187 of the edexcel table where it gives the table with the transition elements, it says you get [Cr(H2O)2(OH)4]- on the addition of excess NaOH to give you a solution after the precipitate dissolves.

    Sorry to be a bit pedantic about this, just really make sure I know the it properly
    Oops, I swear I thought I replied to this. After complete deprotonation occurs, I believe they're indicating that only one ligand is exchanged. That's probably it.
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    This question is in the Jan 2015 paper:
    http://prntscr.com/aliwr7

    The change in oxidation numbers is as follows:

    For N, it is a decrease of -8. (reduction)
    For Al, there is an increase of 3. (oxidation)

    how do we relate this to the stiochiometric coefficients?
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    (Original post by therecovery)
    This question is in the Jan 2015 paper:
    http://prntscr.com/aliwr7

    The change in oxidation numbers is as follows:

    For N, it is a decrease of -8. (reduction)
    For Al, there is an increase of 3. (oxidation)

    how do we relate this to the stiochiometric coefficients?
    Think about what the half equations would be
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    Has everyone finished the entire course yet?
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    Posted from TSR Mobile

    We have one more topic to go, the synthesis of organic compounds. Although our teacher barely covered the last 3 chapters.
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    (Original post by dental17)
    Any helpful revision notes for the transition metals and ligand exchange/ deprotonation section? I cannot understand these equations aghhh??
    What about you say what you're really stuck in like what do you don't understand in these equations ?
    Maybe someone will be able to help you

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    (Original post by samb1234)
    Think about what the half equations would be
    got the answer thanks
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    This question is from the jan 2015 paper IAL:http://i.imgur.com/u8Ckk75.pngThe question is quite easy, but it is the next part which is tricky:http://i.imgur.com/KsFXyxn.pngThe answer is ethanoic acid.Can someone explain how ethanoic acid is formed? :/
 
 
 
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