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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 ramadeen)
    i usually just say at 5 lol
    I would too but doing some questions and it says 0-10 in the markscheme


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    (Original post by ramadeen)
    exactly! are they meant to be harder or what? it made me feel so unprepared.
    I have no clue tbh. I just did January 15 and started feeling worse. That paper was ugly!
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    (Original post by Supermanxxxxxx)
    I would too but doing some questions and it says 0-10 in the markscheme


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    It is 0-10 degrees. You must always mention it as a range.
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    I hate green chemistry and experimental uncertainty questions, I'm all for Jan 15 difficulty if it doesn't include those two lol
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    (Original post by C0balt)
    I hate green chemistry and experimental uncertainty questions, I'm all for Jan 15 difficulty if it doesn't include those two lol
    DON'T JINX IT OMD I WOULD DIE IF IT WAS AS HARD AS JAN 15 XD (I suspect all the stuff I just typed is going to not be in caps like it is right now, which will be very disappointing...)

    EDIT: It is in caps! yay.
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    How come one of the methods has a cl included and which one is the correct one to use


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    (Original post by Don Pedro K.)
    DON'T JINX IT OMD I WOULD DIE IF IT WAS AS HARD AS JAN 15 XD (I suspect all the stuff I just typed is going to not be in caps like it is right now, which will be very disappointing...)

    EDIT: It is in caps! yay.
    Lol the gb will be super low tho, I want very low 120 UMS mark XD
    (Original post by Supermanxxxxxx)
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    How come one of the methods has a cl included and which one is the correct one to use


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    Cl is a spectator ion so both are correct, one is a full equation and the other is ionic
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    (Original post by C0balt)
    OH- doesn't matter you just have to consider the oxidation states

    N goes from +5 to - 3 so 8 electrons on the LHS
    Al goes from 0 to +3 so 3 electrons on the RHS
    Common multiple is 24 so multiply first half eq by 3 and second by 8 to cancel the e- therefore D
    (Original post by samb1234)
    Nah you can ignore the OH-. Looking at ox states, OH is same, Nitrogen is -8 and Al is +3. So Al loses 3 electrons, and nitrogen gains 8 so by lowest common multiple electrons transferred are 24, so 8 Al and 3 nitrogen. Physics 4 is fine atm, currently on my third paper for it
    Cheers, guys! Sorry if my questions seem a little basic - I'm basically on my own for all my A levels since we don't receive very good teaching at school. The large majority of my friends already gave up with this module, but I'm still trying.
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    What is the actual definition of "standard electrode potential" and "standard hydrogen electrode" it seems like all the mark schemes and textbooks have different definitions?
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    Also, how much do we need to know about batteries? I can't see it on the specification
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    These are four successive ionization energies (in kJ mol−1) of four different elements.
    Which could be those of a transition element?
    A 658 1310 2653 4175 9573
    B 578 1817 2745 11578 14831
    C 738 1451 7733 10541 13629
    D 496 4563 6913 9544 13352

    The answer is A but how can you tell?
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    (Original post by Ayman!)
    Cheers, guys! Sorry if my questions seem a little basic - I'm basically on my own for all my A levels since we don't receive very good teaching at school. The large majority of my friends already gave up with this module, but I'm still trying.
    Honestly its no trouble at all, im always happy to help. Thats great that you havent given up, youve done so well to get to where you are now with little support so just keep it going for another few days. Best of luck with everything
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    Just started my first IAL paper and I already feel like I'm completely in the dark. Plus anything on organic synthesis is horrible since I've not done any experiments for over a year ( Gap year student) I can't remember any techniques in a lot of detail 😭
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    (Original post by Z.e.e)
    These are four successive ionization energies (in kJ mol−1) of four different elements.
    Which could be those of a transition element?
    A 658 1310 2653 4175 9573
    B 578 1817 2745 11578 14831
    C 738 1451 7733 10541 13629
    D 496 4563 6913 9544 13352

    The answer is A but how can you tell?
    I think because there are no BIG jumps between ionisation energies for the ionisation energies of A, which indicates you are still in one shell. Whereas for B, C and D there are BIG jumps which indicate you are entering an inner shell.

    What does this mean? Well if you count the number of electrons being pinged off, A has the most which could correspond to the d-subshell (since it can hold 10 electrons) and the others show a a maximum of 3 being pinged off before entering an inner shell.
    But then again, it asks which could be a transition element.

    Hopefully that helped in some way
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    (Original post by Ayman!)
    Anyone done IAL January 2016? I liked the paper although the boundaries are super low...
    Where did you find it? It's locked on the edexcel website?
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    (Original post by flipper9182)
    Where did you find it? It's locked on the edexcel website?
    I've attached it a few posts ago, please take a look

    (Original post by Franckenstar)
    If you check the data book on the ionization energy page you'll see that B is Titanium.
    It's better to understand the trend rather than use the data booklet, but yours is very good exam technique
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    definition of standard electrode potential anyone???
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    Unit 5 is sooooooo hard omg ((((((((((((((((
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    (Original post by Z.e.e)
    These are four successive ionization energies (in kJ mol−1) of four different elements.
    Which could be those of a transition element?
    A 658 1310 2653 4175 9573
    B 578 1817 2745 11578 14831
    C 738 1451 7733 10541 13629
    D 496 4563 6913 9544 13352

    The answer is A but how can you tell?
    Look at the ionization energies section on the data book and it shows you that A is for titanium

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    Hi guys, please can someone explain why for question 21 a part iii) you use the Mr of CaCO3 but for part b iii) you are supposed to use the Mr of Calcium ethanedioate?? I don't get it! Surely you use the same for both (CaCO3) especially as in b iii) they have provided the moles of calcium carbonate as 0.15mol, rather than that of calcium ethandioate..

    Thanks
 
 
 
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