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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 Ayman!)
    Anyone done IAL January 2016? I liked the paper although the boundaries are super low...
    I just did the paper somehow got 66/90 which I think is 108 ums xD
    Super pleased!!! Why are the gb so low??... I mean there were some really weird questions
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    (Original post by gabby07)
    How are these two displayed complexes different?
    These two compounds are enantiomers to each other. So they would rotate plane polarised light in different directions.

    It might seem like they are the same but as long as you can draw a mirror plane inbetween them, they will be chiral.
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    An organic compound X is much more soluble in dilute hydrochloric acid than in water.Compound X forms a coloured complex with aqueous copper(II) ions.Compound X could be
    A C6H5COOH
    B C6H5NO2
    C C6H5NH2
    D C6H5OH

    could someone please explain why the answers C?
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    (Original post by imnoteinstein)
    An organic compound X is much more soluble in dilute hydrochloric acid than in water.Compound X forms a coloured complex with aqueous copper(II) ions.Compound X could be
    A C6H5COOH
    B C6H5NO2
    C C6H5NH2
    D C6H5OH

    could someone please explain why the answers C?
    C0balt samb1234 Ayman!
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    (Original post by imnoteinstein)
    An organic compound X is much more soluble in dilute hydrochloric acid than in water.Compound X forms a coloured complex with aqueous copper(II) ions.Compound X could be
    A C6H5COOH
    B C6H5NO2
    C C6H5NH2
    D C6H5OH

    could someone please explain why the answers C?
    Amines form blue (green for phenylamine) complex with copper (II) ions. Addition of Cu2+(aq) is a functional group test for amine too.
    Phenylamine is more soluble in HCl than in pure water, I presume, is because amine is protonated and becomes charged which can then form ion-permanent dipole attraction with water molecules.
    (Original post by Don Pedro K.)
    C0balt samb1234 Ayman!
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    To measure the standard electrode potential for the Ag+(aq)|Ag(s) electrode, the mostsuitable chemical for the solution in a salt bridge to connect the two half cells is
    A potassium chloride.
    B potassium iodide.
    C potassium nitrate.
    D potassium sulfate.

    why is the answer C?
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    (Original post by imnoteinstein)
    To measure the standard electrode potential for the Ag+(aq)|Ag(s) electrode, the mostsuitable chemical for the solution in a salt bridge to connect the two half cells is
    A potassium chloride.
    B potassium iodide.
    C potassium nitrate.
    D potassium sulfate.

    why is the answer C?
    I got told it was because it would be very unlikely for any of the other compounds, with the exception of KNO3 to react with anything within the half cells.
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    Hey sorry to bring up unit 4 again but does anyone remember anything from section C in between the first page with the flow chart where u had to fill in reagents like PCl5 and the last double page with the NMR spectra and the infrared/TMS peak questions? Its been bugging me for a while I just cant remember if there was/what was on the page in between them!
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    Can someone explain why the answer is D ?

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    Also why is this one A?

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    Can someone explain solvent extraction please? My teacher left before we finished the course and I can't really understand the textbook's explanation of it
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    (Original post by LilacFlorence)
    Can someone explain solvent extraction please? My teacher left before we finished the course and I can't really understand the textbook's explanation of it
    I believe solvent extraction involves dissolving your product in a certain solvent (in which it is soluble) to separate the product from the rest of the mixture? You shake the product with fresh solvent several times to get as much of the product out of the mixture as possible.
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    (Original post by Don Pedro K.)
    Can someone explain why the answer is D ?

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    Also why is this one A?

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    D because you can see Sulfur is reduced from +7 to +6 and that means Fe2+ is oxidised to Fe3+ for this to happen.
    Similarly, Iodide ions are being oxidised to Iodine so Fe3+ is reduced to Fe2+ for this to happen.

    A because it's the same idea as adding Hydrogen under a Raney Nickel catalyst to Benzene, except replacing hydrogen with chlorine.
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    (Original post by Wunderbarr)
    D because you can see Sulfur is reduced from +7 to +6 and that means Fe2+ is oxidised to Fe3+ for this to happen.
    Similarly, Iodide ions are being oxidised to Iodine so Fe3+ is reduced to Fe2+ for this to happen.

    A because it's the same idea as adding Hydrogen under a Raney Nickel catalyst to Benzene, except replacing hydrogen with chlorine.
    Ahhhh right, thanks !
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    (Original post by Don Pedro K.)
    I believe solvent extraction involves dissolving your product in a certain solvent (in which it is soluble) to separate the product from the rest of the mixture? You shake the product with fresh solvent several times to get as much of the product out of the mixture as possible.
    Thank you
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    Really don't understand these two questions....

    Why is this D
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    And also, how do you answer this question? As in, how many peaks would you normally see and why??
    Attachment 553522553528


    Thank you!
    Attached Images
     
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    https://a5c076379da85d2c0b501a60f1a4...0Chemistry.pdf
    Can someone explain Question7 please?
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    A solution of potassium manganate(VII) was used to determine the concentration ofiron(II) ions in solution by titration in the presence of excess dilute sulfuric acid
    (b) If insufficient acid is added, the titre value is
    A low and a brown precipitate forms.
    B low and a green precipitate forms.
    C high and a brown precipitate forms.
    D high and a green precipitate forms.
    could someone please explain this andd what the role of the acid is in this titration?
    • Thread Starter
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    (Original post by Lili1998)
    https://a5c076379da85d2c0b501a60f1a4...0Chemistry.pdf
    Can someone explain Question7 please?
    You want the max amount of the complex, which has one metal ion and 2 two ligands. Therefore the max is going to be when the ratio of metal to ligand is 1:2, so B (as for example if I have 50/50 then I can only have 2.5cm^3 of the metal reacting with ligand)
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    (Original post by imnoteinstein)
    A solution of potassium manganate(VII) was used to determine the concentration ofiron(II) ions in solution by titration in the presence of excess dilute sulfuric acid
    (b) If insufficient acid is added, the titre value is
    A low and a brown precipitate forms.
    B low and a green precipitate forms.
    C high and a brown precipitate forms.
    D high and a green precipitate forms.
    could someone please explain this andd what the role of the acid is in this titration?
    In my CGP, it says that the acid is added to ensure that there are plenty of H+ ions to allow te oxidising agent (potassium manganate (VII)) to be reduced (look at the half equation for MnO4- --> Mn2+ and you will understand why this is).

    So, based on this, if not enough acid is added, the oxidising agent won't be sufficiently reduced, so I'm guessing the iron (II) ions won't be oxidised properly into Fe3+, so you'll mostly have Fe2+ ions, which are green in colour. I would guess the answer is hence B.

    Is that right?
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    (Original post by samb1234)
    You want the max amount of the complex, which has one metal ion and 2 two ligands. Therefore the max is going to be when the ratio of metal to ligand is 1:2, so B (as for example if I have 50/50 then I can only have 2.5cm^3 of the metal reacting with ligand)
    Ohhh I get it now... thanks
 
 
 

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