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    (Original post by Mathematicus65)
    Can someone explain how on earth the answer to this is 1.35g not 0.216g...?!Attachment 549265
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    This is my working - hope it helps 😊


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    (Original post by Mathematicus65)
    Can someone explain how on earth the answer to this is 1.35g not 0.216g...?!Attachment 549265
    You can even check after you have got the answer if it's right or not by putting it into the % yield calculation

    Number of moles of D = 1.73/346 = 0.005mols

    Times this by the Mr of the compound you want to find = 0.005 x 108 = 0.54g
    but since it is only 40% yield you do 0.54 / 100 then x 40 = 1.35g
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    (Original post by ranz)
    oh, r u suree?


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    Yes if it is just the zwitterion - only the NH2 and COOH attached to the same carbon are changed - the R groups are kept the same even if there are equal numbers of multiples of each
    Only if you add OH- or H+ are the R groups protonated/remove H
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    (Original post by lai812matthew)
    difference between simple and complex triglyceride
    I don't know
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    someone pls clarifyyy! if a zwitteriom contained 2 nh2 and 2 cooh would they all protonate? or just one each?


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    (Original post by tcameron)
    Yes if it is just the zwitterion - only the NH2 and COOH attached to the same carbon are changed - the R groups are kept the same even if there are equal numbers of multiples of each
    Only if you add OH- or H+ are the R groups protonated/remove H
    i keeep hearing different things from people


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    (Original post by itsConnor_)
    I don't know
    simple is when both three fatty acids used in the triglyceride reaction is the same, complex is when the fatty acids are different to each other. p.s. i just checked the spec it does not specify clearly you should know this, same with LDP and HDP......
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    (Original post by ranz)
    someone pls clarifyyy! if a zwitteriom contained 2 nh2 and 2 cooh would they all protonate? or just one each?


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    they will all protonate
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    (Original post by Mathematicus65)
    Can someone explain how on earth the answer to this is 1.35g not 0.216g...?!Attachment 549265

    Method 1:
    ----------------------------------------------------------
    Take 1.73g as Actual

    Use Actual/Theoretical x 100 = 40%

    Rearrange to find theoretical grams or X

    Convert theoretical grams to moles, 1:1 mole ratio, therefore moles of D = Moles of 1,3 Diaminobenzene

    Use (moles 1,3 diaminobenzene) x 108 to get 1.35g


    . I think you have put 0.005 moles as your theoretical when 0.005 is the actual yield. You have to calculate the theoretical.

    Method 2 (your method)
    -----------------------------------------------------------------
    Calculate moles of D to be 0.005

    So using the moles in the (A/T x 100= 40%). Rearranged to calculate T, it would be: 0.005/40% x 100 = 0.0125 'theoretical moles' of D.

    1:1 ratio therefore moles of diaminobenzene is 0.0125.

    Then 0.0125 x 108 = 1.35.
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    (Original post by ranz)
    someone pls clarifyyy! if a zwitteriom contained 2 nh2 and 2 cooh would they all protonate? or just one each?


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    if in pH below isoelectric point (pH at which amino acid exists as a zwitterion), all COOH are COOH and all NH2 are NH3+
    if in pH above isoelectric point, all COOH are COO- and all NH2 are NH2

    Hope this makes any sense
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    (Original post by ranz)
    i keeep hearing different things from people


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    There has been past papers testing this in the past - I'm 100% sure I'm right
    Attached Images
     
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    (Original post by itsConnor_)
    1. Glycerol and propane-1,2,3-triol (came up in june 13 tho )
    2. Colour change from orange to green (oxidation using acidified K2Cr2O7, H2SO4 catalyst), heat under reflux
    3. CH2OHCH(OH)CH2OH + 5[O] --> COOHCOCOOH + 3H2O
    4. ?

    EDIT: for 4 is it cos no ester linkage or amide linkage?
    Ooops, number 4 was typo; should be suggest why organic product in q3 cannot be hydrolyses.
    1, 2 and 3 three correct.
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    (Original post by ranz)
    i keeep hearing different things from people


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    (Original post by itsConnor_)
    if in pH below isoelectric point (pH at which amino acid exists as a zwitterion), all COOH are COOH and all NH2 are NH3+
    if in pH above isoelectric point, all COOH are COO- and all NH2 are NH2

    Hope this makes any sense
    (Original post by lai812matthew)
    they will all protonate
    They don't all protonate… I did a past paper that had a 'zwitterion' with an extra cooh side group and the mark scheme said "reject if both cooh groups are shown to have donated a proton" because only groups directly involved in being an amino acid (eg RCH(NH2)COOH) actually protonate
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    (Original post by itsConnor_)
    if in pH below isoelectric point (pH at which amino acid exists as a zwitterion), all COOH are COOH and all NH2 are NH3+
    if in pH above isoelectric point, all COOH are COO- and all NH2 are NH2

    Hope this makes any sense
    what about in standard isolectric point


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    (Original post by ranz)
    someone pls clarifyyy! if a zwitteriom contained 2 nh2 and 2 cooh would they all protonate? or just one each?


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    just one each because the definition of a zwitterion is the diploar ionic form of an amino acid that is formed by the donation of a H ion from the carboxyl group to the amino group. As both charges are present there is no overall charge.

    Both = 2
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    What is a simple repeat unit?
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    (Original post by itsConnor_)
    if in pH below isoelectric point (pH at which amino acid exists as a zwitterion), all COOH are COOH and all NH2 are NH3+
    if in pH above isoelectric point, all COOH are COO- and all NH2 are NH2

    Hope this makes any sense
    the isolectric point is when it exists as a zwitterion, not below the isoelectric point
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    Sorry I know people have asked a couple of times in this thread for the 2015 F324 mark scheme but I can't find it, please can someone post it again?
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    (Original post by embem10)
    Sorry I know people have asked a couple of times in this thread for the 2015 F324 mark scheme but I can't find it, please can someone post it again?
    http://stsmith.co.uk/data/documents/F324_MS_June15.pdf
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    Name:  Zwitter ions.PNG
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Size:  23.5 KBName:  Zwitter ion mark scheme.PNG
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Size:  37.7 KB This is what the mark scheme says on Zwitter ions for this question i was confused too so emailed my teacher, and this was her reply but I am still confused:
    The NH2 group that is part of the R group is not affected (ionised) at the zwitterion pH, this is because the pH is adjusted to take this group into account. E.g the pH of this amino acid will be above pH 7 as it has a basic group.The NH2 group will be ionised if the amino acid is placed into a solution with a pH less than the zwitterion i.e. in acidic conditionsHope that helpsMiss P
 
 
 
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