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    guys, i am slightly confused about benzene. Does its structure have a range of single c-c bonds and c=bond. ? And did Kekulu suggest that the bonds are all of equal length? is that correct.
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    (Original post by Cadherin)
    5ci from what?
    http://filestore.aqa.org.uk/subjects...W-QP-JUN11.PDF
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    (Original post by Lilly1234567890)
    guys, i am slightly confused about benzene. Does its structure have a range of single c-c bonds and c=bond. ? And did Kekulu suggest that the bonds are all of equal length? is that correct.
    it has a "bond length intermediate" that are between the size of double and single bonds. and they're all of this length yes.
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    (Original post by Lilly1234567890)
    guys, i am slightly confused about benzene. Does its structure have a range of single c-c bonds and c=bond. ? And did Kekulu suggest that the bonds are all of equal length? is that correct.
    benzene has intermediate between single and double
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    Because there are no spare carbons - with an anhydride and two alcohol units, the only product of the condensation reaction will be water.
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    Why does water hardly dissociate?
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    (Original post by Signorina)
    Ohhh... Gosh I'd never get that in the exam haha, good thing it's only few marks ! Thank u both!
    No worries haha, my chemistry teacher had to explain it to me!
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    (Original post by Cadherin)
    Because there are no spare carbons - with an anhydride and two alcohol units, the only product of the condensation reaction will be water.
    But normally acid anhydride + ethanol -> ester + carboxylic acid
    right?
    I still don't get why you get H20 in this case but you don't normally.

    Also do you mind helping me with 6b. Why isn't it step 1?
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    (Original post by Super199)
    But normally acid anhydride + ethanol -> ester + carboxylic acid
    right?
    I still don't get why you get H20 in this case but you don't normally.

    Also do you mind helping me with 6b. Why isn't it step 1?
    I'd have the same thought process as that really but when you do the reaction you're gonna find urself left with a H and 2 o's so you just kinda know

    Needs to have 2xN, 2xO and 2xH in the step. X has 2x N and 2xO so when you react X and H2 you get all the things which are in the rate eqn so its rate determining.
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    When doing buffers- how do you know whether to add/subtract the acid from what?
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    (Original post by Super199)
    But normally acid anhydride + ethanol -> ester + carboxylic acid
    right?
    I still don't get why you get H20 in this case but you don't normally.

    Also do you mind helping me with 6b. Why isn't it step 1?
    Draw out the mechanism, that might help you understand.

    Because, in the rate equation, second order with respect to NO, first order with respect to H2.

    Therefore, you want the step that involves them reacting in a 2:1 ratio. X is a compound comprising 2NO molecules, which reacts with 1H2.

    Therefore, it is step 2.
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    (Original post by Rabadon)
    I'd have the same thought process as that really but when you do the reaction you're gonna find urself left with a H and 2 o's so you just kinda know

    Needs to have 2xN, 2xO and 2xH in the step. X has 2x N and 2xO so when you react X and H2 you get all the things which are in the rate eqn so its rate determining.
    oh thats a nice way of thinking about it.
    I swear there are lots of scenarios for this thing.

    There is one where you just say it's not in the rate equation
    One for saying something about the molar ratio
    and then one by looking at what you require.
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    (Original post by bethl303)
    When doing buffers- how do you know whether to add/subtract the acid from what?
    Try to understand it, don't try to rote learn it!

    Adding H+ - this will react with A- to form the acid - therefore adding to HA and taking away from A-.
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    http://filestore.aqa.org.uk/subjects...W-QP-JUN11.PDF
    7bi. Splitting pattern why is it a triplet?
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    Can someone please explain to me how to work out the rate determining step!! I always get it wrong thanks


    Posted from TSR Mobile
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    So can anyone confirm whether we need to include δ+/ δ− signs on the carbons/oxygens/halides and whether the M+1 peak is actually on the AQA spec?

    Many thanks
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    (Original post by Cadherin)
    Try to understand it, don't try to rote learn it!

    Adding H+ - this will react with A- to form the acid - therefore adding to HA and taking away from A-.
    Can anyone please help me with this question on Jan 2012 Unit 4 paper question 5d???
    Attached Images
     
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    (Original post by Aethrell)
    So can anyone confirm whether we need to include δ+/ δ− signs on the carbons/oxygens/halides and whether the M+1 peak is actually on the AQA spec?

    Many thanks
    No you don't need to include it.
    And yep it is on the spec but I haven't seen a question on it yet.
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    (Original post by bethl303)
    When doing buffers- how do you know whether to add/subtract the acid from what?
    CH3COOH <----> CHCCOO + H+
    Think of it as an equilibrium: if a base is added then the OH- react with the H+, which makes eqm to shift to the RHS to replace the H+ lost (so you add on the right )if acid added then eqm shifts to the left to oppose increase of H+ ( so you add on the left)
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    (Original post by cutelady)
    Can someone please explain to me how to work out the rate determining step!! I always get it wrong thanks


    Posted from TSR Mobile
    It is the step that includes and only includes the species in the rate equation
 
 
 
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