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Edexcel - Chemistry Unit 2 - 4 June 2013 Watch

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    (Original post by ramanan)
    Attachment 222958

    the angle B-A-B is 180 so octraheral has bothe 90 and 180
    Ahhh okay that makes sense Thanks and good luck for this afternoon!
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    (Original post by Gnome :))
    Reduction can be defined as:
    -gain of electrons
    -loss of oxygen
    or
    - gain of hydrogen.

    In this case, the hydrocarbon gains hydrogen, and so is reduced
    Really helpful cheers!

    Right sorry for all these questions:

    When you react CH3I with alcoholic ammonia in a sealed tube, why is it that ethylamine (ch3ch2nh2) can't form but dimethylamine ( (CH3)2NH ) can form? I thought amines had a NH2 at the end so don't get dimethylamine


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    (Original post by Matterhorn)
    Really helpful cheers!

    Right sorry for all these questions:

    When you react CH3I with alcoholic ammonia in a sealed tube, why is it that ethylamine (ch3ch2nh2) can't form but dimethylamine ( (CH3)2NH ) can form? I thought amines had a NH2 at the end so don't get dimethylamine


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    Ok i thought CHNH2 would form along with NH4I
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    ) Concentrated sulfuric acid is added to potassium chloride in a test tube. Steamy
    fumes are given off which react with ammonia to give dense white smoke.
    (i) Name the gas given off in this reaction.

    Why is it HCl? I thought this is only for PCl5
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    PS Reviewer
    Just did the Jan 13 paper and got 74/80, so that's 114 ums according to the UMS converter.

    :woo:
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    (Original post by Matterhorn)
    Really helpful cheers!

    Right sorry for all these questions:

    When you react CH3I with alcoholic ammonia in a sealed tube, why is it that ethylamine (ch3ch2nh2) can't form but dimethylamine ( (CH3)2NH ) can form? I thought amines had a NH2 at the end so don't get dimethylamine


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    Ethylamine can't form because there is no CH2 in iodomethane.

    Dimethylamine can form because the amine formed in the first reaction (CH3NH2) can act as a nucleophile:

    CH3I + NH3 --> CH3NH2 + HI

    CH3NH2 + CH3I --> (CH3)2NH + HI

    Here's the chemguide page: http://www.chemguide.co.uk/mechanism...mines.html#top
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    (Original post by James A)
    Just did the Jan 13 paper and got 74/80, so that's 114 ums according to the UMS converter.

    :woo:
    hi, how did you do q 21 b i and ii? I don't really understand. thanks
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    (Original post by Linked)
    ) Concentrated sulfuric acid is added to potassium chloride in a test tube. Steamy
    fumes are given off which react with ammonia to give dense white smoke.
    (i) Name the gas given off in this reaction.

    Why is it HCl? I thought this is only for PCl5

    H2SO4 + 2KCl --> K2SO4 + HCl
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    (Original post by Linked)
    ) Concentrated sulfuric acid is added to potassium chloride in a test tube. Steamy
    fumes are given off which react with ammonia to give dense white smoke.
    (i) Name the gas given off in this reaction.

    Why is it HCl? I thought this is only for PCl5
    No you need to remeber your Halide reactions with Sulfuric Acid. Sulfuric Acid is a relativly strong oxidising agent. So KF and KCl will react once, KBr two reactions and KI 3 reactions.
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    (Original post by James A)
    Just did the Jan 13 paper and got 74/80, so that's 114 ums according to the UMS converter.

    :woo:
    Congrats
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    (Original post by bhowland1994)
    No you need to remeber your Halide reactions with Sulfuric Acid. Sulfuric Acid is a relativly strong oxidising agent. So KF and KCl will react once, KBr two reactions and KI 3 reactions.

    do i have to remember the actual equations - including state symbols?
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    (Original post by SophieL1996)
    hi, how did you do q 21 b i and ii? I don't really understand. thanks
    The chlorine questions?

    21bi. You should know that disproportionation is where one species is both oxidised and reduced. Then you just have to work out the oxidation numbers of the chlorine.
    Cl2: oxidation number is 0 because there is only one species in the molecule, so they would share the electrons equally
    HCl: oxidation number is -1, because the halogens usually have an oxidation state of -1, and hydrogen is +1 so the Cl needs to balance this out
    HClO: oxidation state of +1 because O is -2 (unless it's a peroxide or F2O), and H is +1, so Cl has to be +1 to balance it out.

    bii. Le Chatelier's Principle; the change in equilibrium will counteract the change in concentration. Therefore the equilibrium will shift to the left, to reduce the concentration of HCl.
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    ppl can someone please tell me all the apparatus i have to know for the xam ?!


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    (Original post by Gnome :))
    The chlorine questions?

    21bi. You should know that disproportionation is where one species is both oxidised and reduced. Then you just have to work out the oxidation numbers of the chlorine.
    Cl2: oxidation number is 0 because there is only one species in the molecule, so they would share the electrons equally
    HCl: oxidation number is -1, because the halogens usually have an oxidation state of -1, and hydrogen is +1 so the Cl needs to balance this out
    HClO: oxidation state of +1 because O is -2 (unless it's a peroxide or F2O), and H is +1, so Cl has to be +1 to balance it out.

    bii. Le Chatelier's Principle; the change in equilibrium will counteract the change in concentration. Therefore the equilibrium will shift to the left, to reduce the concentration of HCl.
    thanks but that's part a I mean the iionic equations. thank you
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    (Original post by Gnome :))
    Ethylamine can't form because there is no CH2 in iodomethane.

    Dimethylamine can form because the amine formed in the first reaction (CH3NH2) can act as a nucleophile:

    CH3I + NH3 --> CH3NH2 + HI

    CH3NH2 + CH3I --> (CH3)2NH + HI

    Here's the chemguide page: http://www.chemguide.co.uk/mechanism...mines.html#top
    Thanks will rep soon


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    (Original post by Linked)
    do i have to remember the actual equations - including state symbols?
    Yes to remembering reactions and with state symbols probabably not, but the products are so simple you will know what they are anyway.
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    (Original post by James A)
    Just did the Jan 13 paper and got 74/80, so that's 114 ums according to the UMS converter.

    :woo:
    can u send me the jan 13 paper please?


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    Name:  ImageUploadedByStudent Room1370330279.967745.jpg
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Size:  158.1 KB

    How do you know what to draw here?


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    (Original post by Matterhorn)
    Name:  ImageUploadedByStudent Room1370330279.967745.jpg
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    How do you know what to draw here?


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    Heat under Reflux apparatus



    (Original post by Mimi85)
    yeah i mean what solution goes in what apparatus in the titration
    At most you'll just need to know this really:

    The liberated iodine reacts with thiosulfate ion (after adding KI to your solution being tested)

    I2 + 2S2O3^2- ----> 2I- + S4O6^2-

    They give you this stuff anyway usually, it's calculations you should be most concerned with
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    (Original post by Matterhorn)
    Name:  ImageUploadedByStudent Room1370330279.967745.jpg
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    How do you know what to draw here?


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    You heat under reflux
 
 
 
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