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    (Original post by hopingmedicinae)
    What paper and which question is it? I c ant see it properly on my laptop
    Not the same person but this was the paper:
    http://filestore.aqa.org.uk/subjects...4-QP-JAN12.PDF

    Question 6
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    (Original post by SirRaza97)
    A few quick questions:

    1. How do you work out the major peak of a compound?
    2. How does Sn/HCl or Ni/H+ or similar reduce benzene to hexane?
    3. When would you ever use PKa in a paper?
    4. Anhydride to amide?
    1.) Not sure, can someone help?
    2) Just provides Hydrogen as far as I'm aware oto break the intermediate C-C/C=C bonds in benzene and Nickel is the catalyst
    3.) When they give you Pka and you need to figure out the PH of something
    Ka = 1X10^-Pka
    Then use Ka to figure out H+ then -log(H+)
    4.) Ammonia!

    (CH3CO)2O + NH3 ----> CH3COONH2 + CH3COOH
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    (Original post by Sexybadman)
    Still struggling to understand what to do can you show maybe a worked example ?


    Posted from TSR Mobile
    This is a little more complicated than we need to know but its a good example I think https://www.youtube.com/watch?v=YrHx2R72hFk

    So this is the information we know
    -The integrations tell you the NUMBER of H in each chemical shift e.g in 2.6 there are 2 H atoms, in 1.2 chemical shift, there are 6 H atoms.
    - There is an O-H bond present as we learnt from the first part of the question
    - There are four different environments ( we know this bc four different chemical shifts.)


    So take the first part, chemical shift is 2.6, and there are two H atoms
    On the H.n.m.r chemical shift data, 2.6 is R-C=O- CH. Because the integration value is 2, we then know there are two equivalent H atoms so this bit of the structure is R-C=O-CH2


    Next one = 2.2 chemical shift and this time integration value is 3 so 3 equivalent H atoms2.2 is again the same R-C=O-CH , but this time there are 3 H atoms so this part of the structure is CH3 - C=O -

    C1.2 has an integration value of 6 - 6 atoms all equivalent, this should in your mind indicate two Ch3 groups anyway. But even if it doesn't, the value is 1.2 which on the data is R-CH3. <--- This only is 3, and there are 6 of these so the structure of this chemical shift is H3C-R-CH3

    Finally, it asks you to put the whole structure together, we already have the three CH3 groups so you have the two ends of the Molecule.The only thing we are missing is the O-H so this is the bond we need to include.

    Also bear in mind all peaks are singlets, this means that none of the carbons that any equivalent hydrogen atoms are bonded to are bonded to any other hydrogens.

    So if you work with everything you have, start with the "corners" almost like a jigsaw and your final structure is

    ; (draw this out, it's easier to imagine it )

    (CH3)3-C(OH)- CH2-C=O-CH3

    Hope this helps!!Posted from TSR Mobile[/QUOTE]
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    Does anyone know any good resources for carbon 13nmr and proton nmr?
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    A 25.0 cm3 sample of 0.0850 mol dm–3 hydrochloric acid was placed in a beaker and100 cm3 of distilled water were added.Calculate the pH of the new solution formed.Give your answer to 2 decimal places. please help asap
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    (Original post by Sheary13)
    A 25.0 cm3 sample of 0.0850 mol dm–3 hydrochloric acid was placed in a beaker and100 cm3 of distilled water were added.Calculate the pH of the new solution formed.Give your answer to 2 decimal places. please help asap
    Whenever you are given a dilution question for acid/alkali use this equation:

    Old volume/new volume X conc
    (To get you H+ or OH-) depends on the question.
    In this case we have dilution of HCl. So we do this:
    25/125 X 0.085 to get us H+
    We then use -log(H+) to get us pH try it out!
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    Thank you!


    (Original post by SubwayLover1)
    HA disassociates Into H+ and A-

    So if you add H+ = There is a large amount of A- in the buffer due to the salt - SO an increase in H+ will cause it to react with A- to form HA( equillbrium shifts to the left)

    Adding OH- = OH- can react with H+ reducing the amount of H+ so equilibrium shifts to right
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    (Original post by Sheary13)
    A 25.0 cm3 sample of 0.0850 mol dm–3 hydrochloric acid was placed in a beaker and100 cm3 of distilled water were added.Calculate the pH of the new solution formed.Give your answer to 2 decimal places. please help asap
    This is all about dilution so using
    MV/1000=MV/1000
    On one side you have the molarity X volume in cm3 and then the other side is your new volume (with the added water) then solve to find you new molarity (0.0607)
    This is your new hydrogen concentration so pH=1.22
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    (Original post by Mriffin)
    This is all about dilution so using
    MV/1000=MV/1000
    On one side you have the molarity X volume in cm3 and then the other side is your new volume (with the added water) then solve to find you new molarity (0.0607)
    This is your new hydrogen concentration so pH=1.22
    I get 1.77
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    I feel like I'm going to really run out of time in this exam
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    for the reason why NaBH4 isnt used to reduce C=C, would you say the lone pair on the hydride ion is repelled by the high electron density in the C=C bond?
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    (Original post by ravichauhan11)
    for the reason why NaBH4 isnt used to reduce C=C, would you say the lone pair on the hydride ion is repelled by the high electron density in the C=C bond?
    idk cus surely you could say that about the C=O bond? I'd maybe go with it not being polar?
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    Name:  image.jpg
Views: 204
Size:  489.9 KB Can someone please explain how this has 11 peaks in its Carbon 13 nmr?!?
    Thanks in advance
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    (Original post by Rabadon)
    idk cus surely you could say that about the C=O bond? I'd maybe go with it not being polar?
    oh no im certain my explanation on the whole is correct ( its explained in the book page 70 ) , just wondering whether you have to specify lone pair on hyride ion. yeah i was thinking non polar but i dont think that will be enough for an explanation.
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    The two carbons on nitrogen count as one, the three above it are separate so that's 4 so far. After that it is symmetrical so 7 for the symmetrical part which equals 11
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    Can anyone think of anything from AS we need to go over? Like do we need to know like the UV thingy and shapes and stuff like that?


    Posted from TSR Mobile
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    http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF

    2dii, If anyone doesn't mind. I don't quite get what's happening.
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    Can someone pls explain question 4b from the june 2010 paper of chem4. Thanks (:

    Paper : http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF

    Mark scheme : http://filestore.aqa.org.uk/subjects...W-MS-JUN10.PDF
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    (Original post by Super199)
    http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF

    2dii, If anyone doesn't mind. I don't quite get what's happening.
    Okay so because you're adding OH- ions, they react with the H+, meaning there's a reduction in H+ in the equilibrium, so that means more HY is used up and more Y- is made to make the H+ back to normal.

    So what you do is you get the moles of OH (5x10^-4) and TAKE AWAY from MOLES of HY and ADD the moles of OH to moles of Y-

    Then, you make them back into concentrations and put them back into Kc to get the new concentration of H+ and then put it into your pH equation.

    I can write it out for you if you want?


    Posted from TSR Mobile
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    (Original post by Gezza_O'Brien)
    Okay so because you're adding OH- ions, they react with the H+, meaning there's a reduction in H+ in the equilibrium, so that means more HY is used up and more Y- is made to make the H+ back to normal.

    So what you do is you get the moles of OH (5x10^-4) and TAKE AWAY from MOLES of HY and ADD the moles of OH to moles of Y-

    Then, you make them back into concentrations and put them back into Kc to get the new concentration of H+ and then put it into your pH equation.

    I can write it out for you if you want?


    Posted from TSR Mobile
    Yh if you don't mind, I've not seen a qs like this before
 
 
 
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