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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 Don Pedro K.)
    Haha thought so xD I want to understand how to do it, feel like we could get one like this after that flipping unit 4...Which paper is it from?
    The June 2014 IAL paper was really nasty - a* was 64 I believe
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    (Original post by gabby07)
    The June 2014 IAL paper was really nasty - a* was 64 I believe
    What the heck...I don't think I want to kill my confidence like that so close to the exam xD
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    Hey guys, I also just did the June 2014 IAL but I don't quite get these questions:

    How do you know that a platinum electrode is needed rather than the tin electrode??




    And can someone explain this one, I just have no idea...



    Thank youu
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    Hello, in the edexcel textbook its says [Cr(OH)3(H2O)3] reacts with acids to give [Cr(NH3)6]3+ and with bases to give [Cr(OH)6]3- ?? is this right because ive seen different things in different books
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    (Original post by Funky_Giraffe)
    Hey guys, I also just did the June 2014 IAL but I don't quite get these questions:

    How do you know that a platinum electrode is needed rather than the tin electrode??




    And can someone explain this one, I just have no idea...



    Thank youu
    Because if you put tin electrode you'd get Sn/Sn4+ and Sn/Sn2+ potential set up, rather than only Sn4+/Sn2+. Platinum is inert so it won't set up a potential.
    And if you react amine with dilute acid you get an ionic salt which remains as a white solid upon evaporation of the liquid
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    (Original post by C0balt)
    Because if you put tin electrode you'd get Sn/Sn4+ and Sn/Sn2+ potential set up, rather than only Sn4+/Sn2+. Platinum is inert so it won't set up a potential.
    And if you react amine with dilute acid you get an ionic salt which remains as a white solid upon evaporation of the liquid
    Does the fact that the lone e- pair being less available for donation in phenylamine not affect this at all?
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    C0balt Does it matter if the acid is in excess or not?
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    (Original post by Wunderbarr)
    Does the fact that the lone e- pair being less available for donation in phenylamine not affect this at all?
    It is a weaker base but it still reacts the same way with sulfuric acid
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    (Original post by kyungsoo)
    Hello, in the edexcel textbook its says [Cr(OH)3(H2O)3] reacts with acids to give [Cr(NH3)6]3+ and with bases to give [Cr(OH)6]3- ?? is this right because ive seen different things in different books
    With a strong enough base yes (say excess NaOH), otherwise, weaker bases will only deprotonate once or twice.
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    (Original post by Don Pedro K.)
    C0balt Does it matter if the acid is in excess or not?
    No the amount of acid just change the amount of phenylamine reacting
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    https://a5c076379da85d2c0b501a60f1a4...0Chemistry.pdf

    q5 is the answer D and not C bc the 4s fills before 3d?
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    (Original post by Wunderbarr)
    Does the fact that the lone e- pair being less available for donation in phenylamine not affect this at all?
    I mean you'll get less of the salt than a stronger base like say for example Et3N, the relative strength of the base will just affect the relative position of equilibrium. With PhNH2 you'll still get a good amount of salt, the only thing competing for protonation is water which is a weaker base than PhNH2, so you'd expect protonation primarily on the PhNH2.
    Spoiler:
    Show
    pKa H3O+ = -1.7, PhNH3+ = 4.9
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    (Original post by Protoxylic)
    With a strong enough base yes (say excess NaOH), otherwise, weaker bases will only deprotonate once or twice.
    And by once or twice you mean maybe a couple of hundred times?
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    (Original post by Don Pedro K.)
    And by once or twice you mean maybe a couple of hundred times?
    No, I mean once or twice for one complex ion. [also it'd be a couple billion billion of times for the whole vessel, think of the order of magnitude of avagadro's constant, assuming a mole of ions]
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    (Original post by imnoteinstein)
    https://a5c076379da85d2c0b501a60f1a4...0Chemistry.pdf

    q5 is the answer D and not C bc the 4s fills before 3d?
    A transition metal ion is formed by the loss of 4s (and maybe 3d depending on which ion forms) electrons. A colourless transition metal ion is one with completely filled or completely empty 3d orbitals. The maximum number of electrons that be held within the 3d subshell is 10 (each of the 5 d orbitals can be occupied by 2 electrons), so the answer is D because 3d10 means that all of the d orbitals are filled, and therefore there is no possibility for electrons to transition from a lower to a higher energy orbital by absorbing wavelengths of light. All of the white light passing through the solution is hence transmitted, so the solution is colourless.
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    (Original post by Protoxylic)
    No, I mean once or twice for one complex ion. [also it'd be a couple billion billion of times for the whole vessel, think of the order of magnitude of avagadro's constant, assuming a mole of ions]
    I see you didn't get the song reference...xD
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    (Original post by imnoteinstein)
    https://a5c076379da85d2c0b501a60f1a4...0Chemistry.pdf

    q5 is the answer D and not C bc the 4s fills before 3d?
    Ion. Transition metal ions don't have their 4s orbitals filled as once the 4s is filled, the added nuclear charge (effective) pushes the 3d orbital lower than the 4s in energy (due to quantum mechanical effects such as 4s penetration). So when ionising, the 4s electrons are the first to go.
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    (Original post by Supermanxxxxxx)
    Attachment 552876
    What would z be


    Posted from TSR Mobile
    i think its hydrolysis, so naoh followed by strong acid i think...someone correct me if im wrong.
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    (Original post by Don Pedro K.)
    I see you didn't get the song reference...xD
    Nope
    • Thread Starter
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    (Original post by Protoxylic)
    Nope
    I don't think you want to tbh if it's from where I think it is
 
 
 
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