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

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    (Original post by Wunderbarr)
    There are some few ways of drawing it from my travels.

    But generally I've seen the N having a positive charge and then single bond O with a negative charge.
    (Original post by Don Pedro K.)
    Look it up on Google; the N has a + charge and the O has a - charge apparently!

    Eurgh I don't like that, it just feels wrong. Thanks though! Beg it comes up tomorrow now I know how to do it
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    (Original post by MacroniCheese)
    Chromium (II) Ethanoate has two Chromium atoms so the moles of MnO4- needed would be 4x2 = 8. You should then get a titre value double what you got
    oh I see ! thank you very much
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    Does have notes on pratical side of organic chemistry. Like the synthesis parts
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    I can't figure any of these out


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    [QUOTE=RUNSran;65990843] x /QUOTE]

    I think you can just look at Cr2+ oxidising straight to Cr6+.

    If you look at the oxidation of Cr2+ by MnO4-, then you can formulate some equations with:
    Cr2+ -> Cr6+
    MnO4- -> Mn2+
    then balance these by balancing the electrons and add them together (for now I'm just ignoring all the other junk that actually appear in the equations).

    We know MnO4- goes to Mn2+ and Cr2+ goes to Cr6+:

    MnO4- + 5e- -> Mn2+ + stuff
    Cr2+ -> Cr6+ + 4e-

    Therefore: 4MnO4- + 5Cr2+ -> 4Mn2+ + 5Cr6+
    So mole ratio of MnO4- to Cr2+ is 4:5 or the moles of MnO4- will be 4/5 times the moles of Cr2+.

    You were right with dividing the moles given by 10 to get 2.66x10-4 but then you need to multiply this number by 8/5 to get the moles of MnO4-.

    Why 8/5 instead of the just mentioned 4/5? Because there are twice as many Cr2+ to oxidise.

    Finish it for me baby.
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    (Original post by Don Pedro K.)
    Someone explain why it's C please?

    Attachment 554513
    Most negative electrode on left hand side
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    I love how this thread is so populated now that it's the night before the exam
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    (Original post by C0balt)
    I love how this thread is so populated now that it's the night before the exam
    Hey where the flip does the x10 x40 come from in this??

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    (Original post by Franckenstar)
    Most negative electrode on left hand side
    Okay so you have +0.27V on the LHS and +0.34V on the RHS. When you do RHS - LHS, don't you get +0.07V instead of -0.07V?
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    [QUOTE=Wunderbarr;65991421]
    (Original post by RUNSran)
    x /QUOTE]

    I think you can just look at Cr2+ oxidising straight to Cr6+.

    If you look at the oxidation of Cr2+ by MnO4-, then you can formulate some equations with:
    Cr2+ -> Cr6+
    MnO4- -> Mn2+
    then balance these by balancing the electrons and add them together (for now I'm just ignoring all the other junk that actually appear in the equations).

    We know MnO4- goes to Mn2+ and Cr2+ goes to Cr6+:

    MnO4- + 5e- -> Mn2+ + stuff
    Cr2+ -> Cr6+ + 4e-

    Therefore: 4MnO4- + 5Cr2+ -> 4Mn2+ + 5Cr6+
    So mole ratio of MnO4- to Cr2+ is 4:5 or the moles of MnO4- will be 4/5 times the moles of Cr2+.

    You were right with dividing the moles given by 10 to get 2.66x10-4 but then you need to multiply this number by 8/5 to get the moles of MnO4-.

    Why 8/5 instead of the just mentioned 4/5? Because there are twice as many Cr2+ to oxidise.

    Finish it for me baby.
    you guys are amazing I fully get it now. Thank you very much
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    (Original post by Don Pedro K.)
    Okay so you have +0.27V on the LHS and +0.34V on the RHS. When you do RHS - LHS, don't you get +0.07V instead of -0.07V?
    E calomel - O.34 = -0.07 so E calomel = 0.34 - 0.07 = + 0,27 V
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    (Original post by Don Pedro K.)
    Hey where the flip does the x10 x40 come from in this??

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    Yeah I wondered too when I did that paper. I got the right answer without 10*40 thing iirc though. Which paper was this?
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    (Original post by samb1234)
    I'm fairly sure it is A because every nitration is electrophilic in nature. If I did the first nitration, and the NO2 group was e- donating, increasing the negativity of the benzene ring, then at the same temperature I would get further nitration going on (as essentially it's the same situation as before but now it is even more susceptible to attack by electrophiles). The fact I have to increase the temp for this to happen shows that the opposite happens, i.e. with each attack the resulting ring becomes less susceptible to attack by electrophiles hence why you need an increased temp for this to happen.
    Thanks a lot...makes sense. I have googled this and found out that nitro groups are electron withdrawing and alkyl groups are electron donating.

    BTW, I'm a girl :P
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    How much do we need to know about solvent extraction
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    Just to clear some confusion about signs of Ecell. The R-L convention is a convention, there is no science behind it. If the overall Ecell is > 0 then this is the reaction that is spontaneous at standard conditions. If Ecell < 0 then the reverse is spontaneous (this is because Ecell is actually proportional to -G -> G=-nFE). So if you are told that the spontaneous reaction is ___ then this overall reaction, once you do R-L, must have the positive Ecell. Both reactions (forward and reverse) have the same magnitude of Ecell, it is the sign that differentiates them (and gives you the spontaneous direction).
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    Hi, would appreciate if anyone helps me in this question.

    10 cm3 of a gaseous hydrocarbon was mixed with excess oxygen and ignited. The gas volumes were measured at room temperature and pressure before and after combustion and it was found that the total gas volume contracted by 20 cm3. Given that combustion was complete, the formula of the hydrocarbon was:

    A C4H4
    B C4H6
    C C4H8
    D C4H10 The answer is A
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    (Original post by Rahatara Sadique)
    Thanks a lot...makes sense. I have googled this and found out that nitro groups are electron withdrawing and alkyl groups are electron donating.

    BTW, I'm a girl :P
    Yes, nitro groups are one of the most powerful electron withdrawing groups in aromatic chemistry (called ring deactivators). There is an easy way to see why it is withdrawing and that's to spot the fact the nitrogen in the NO2 group doesn't have any lone pairs and so cannot be e donating.
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    (Original post by Supermanxxxxxx)
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    I can't figure any of these out


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    (Original post by Don Pedro K.)
    Do all amino acids exist as zwitter ions in pH 7? The MC in June 14 asked what alanine would look like in pH 7...unless we're meant to know that the isoelectric point of alanine is pH 7?
    Perhaps it may be better to say all zwitterions have a pH of 7, due to the overall neutral charge?

    And then alanine at pH 7 should just be its zwitterion? That would be my "guess" (unfortunately it's just that).
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    (Original post by C0balt)
    Yeah I wondered too when I did that paper. I got the right answer without 10*40 thing iirc though. Which paper was this?
    June 14 :/ I did crap in this one lol...

    (Original post by Franckenstar)
    E calomel - O.34 = -0.07 so E calomel = 0.34 - 0.07 = + 0,27 V
    Ahhh okay thanks !
 
 
 
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