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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)
    Which of the following reagents would enable you to separate iron(III) hydroxide from amixture of iron(III) hydroxide and copper(II) hydroxide?
    A Dilute hydrochloric acid
    B Aqueous ammonia
    C Dilute nitric acid
    D Sodium hydroxide solution
    im confused, help please!!!
    Is it B? As Fe3+ is insoluble in excess ammonia, whereas copper is soluble in excess ammonia as it forms a deep blue solution.
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    (Original post by ramadeen)
    Which of the following reagents would enable you to separate iron(III) hydroxide from amixture of iron(III) hydroxide and copper(II) hydroxide?
    A Dilute hydrochloric acid
    B Aqueous ammonia
    C Dilute nitric acid
    D Sodium hydroxide solution
    im confused, help please!!!
    Cu(OH)2 dissolves in excess ammonia but Fe(OH)3 won't so you can then filter

    HCl and nitric acid would only protonate them which won't separate
    They both don't dissolve in excess NaOH
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    (Original post by Peppercrunch)
    Is it B? As Fe3+ is insoluble in excess ammonia, whereas copper is soluble in excess ammonia as it forms a deep blue solution.
    (Original post by C0balt)
    Cu(OH)2 dissolves in excess ammonia but Fe(OH)3 won't so you can then filter

    HCl and nitric acid would only protonate them which won't separate
    They both don't dissolve in excess NaOH
    you are right, thanks for your help! Im so stupid, I thought it was Fe(iii) and Fe(ii) didn't even read copper I hope i remember how to read tomorrow
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    (Original post by Bertie Paterson)
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    (Original post by C0balt)
    I got that wrong but I think it's because the nitrogen has a lone pair which can be delocalised into the ring and making the ring more susceptible to further electrophilic attack
    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.
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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.
    Oh I read it wrong, I initially thought it was A but the guy said it was B at the end so I thought it was B lol much confusion
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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.
    And it makes sense come to think of it because nitro group has that resonance **** going on because of N-O
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    PLEASE HELP!
    could anyone explain the method involved in these questions?
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    Does anyone know how to change from CrO4 - to Cr2O7 2-? Is it zinc and HCl or zinc and Sulfuric acid?


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    (Original post by sayshay)
    Does anyone know how to change from CrO4 - to Cr2O7 2-? Is it zinc and HCl or zinc and Sulfuric acid?


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    Sulfuric Acid.
    You use zinc and hcl to conver Cr207 2- to [Cr(H20)6]3+
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    (Original post by sayshay)
    Does anyone know how to change from CrO4 - to Cr2O7 2-? Is it zinc and HCl or zinc and Sulfuric acid?


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    They are both +6. CrO4 2- in alkaline and Cr2O7 2- in acid
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    (Original post by Blazyy)
    Sulfuric Acid.
    You use zinc and hcl to conver Cr207 2- to [Cr(H20)6]3+
    Thank you! Say we're reducing cro4 - and we want cr3+, how do we stop it reducing all the way to cr2+?


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    (Original post by sayshay)
    Thank you! Say we're reducing cro4 - and we want cr3+, how do we stop it reducing all the way to cr2+?


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    Make sure to use a reducing agent with a E(standard) value which is more positive than the reduction from Cr3+ to Cr2+ I think!
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    Would anyone kindly give a good but simple (Edexcel) definition of oxidation state/number ? The one with the bit about being an ion?
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    (Original post by sayshay)
    Thank you! Say we're reducing cro4 - and we want cr3+, how do we stop it reducing all the way to cr2+?


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    Normally you'd get just Cr3+ by using the reactants I mentioned.There is no need to do anything, because Cr2+ and Cr3+ are in equilibrium, but the concentration of Cr2+ is very low because it immediately reacts with oxygen to make Cr3+. Although I'm not sure what to do if you want Cr3+ alone.
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    (Original post by Blazyy)
    Sulfuric Acid.
    You use zinc and hcl to conver Cr207 2- to [Cr(H20)6]3+
    Doesn't Cr2O72- reduce all the way to Cr2+ when reacted with Zinc and HCl? That's a stage in the formation of the chromium(II)ethanoate precipitate isn't it?
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    can someone explain to me why the answer to Q3 of JUNE 14R is D.
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  1. File Type: pdf June 2014 (R) QP - Unit 5 Edexcel Chemistry.pdf (611.8 KB, 34 views)
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    (Original post by Don Pedro K.)
    Doesn't Cr2O72- reduce all the way to Cr2+ when reacted with Zinc and HCl? That's a stage in the formation of the chromium(II)ethanoate precipitate isn't it?
    No because you need to keep it under hydrogen atmosphere as oxygen in the air will oxidize cr2+ to cr3+
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    (Original post by Wunderbarr)
    Would anyone kindly give a good but simple (Edexcel) definition of oxidation state/number ? The one with the bit about being an ion?
    The answer to this question was C so I'd say 'the charge an element's ion would have if its bonding electrons were transferred completely' or something less clunky


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    Q. When X is refluxed with concentrated hydrochloric acid for several hours, cooledand neutralized, there is only one organic product, Y (2- aminoethanoic acid/ glycine). Give the structural formula for X.

    the answer is H2NCH2CONHCH2CO2H.

    I have no idea how we get that - could someone please explain this?
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    (Original post by :D :))
    can someone explain to me why the answer to Q3 of JUNE 14R is D.
    RHS-LHS=emf
 
 
 
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