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    Please could someone help me to understand why the reaction is first order for [C2H4] in experiments 2&3 in the first question? Do you just add the amounts which [O3] and [C2H4] have increased by seeing as the rate increases at the same time?

    And also how to work out the initial rate of formation of O2 in the specimen paper? I don't understand where the 4 has come from in the mark scheme :s


    http://www.ocr.org.uk/download/asses..._unit_f325.pdf
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    Okay, I'm trying to create a list of colours/ shapes / angles we need to memorise. So far I have :

    Colours:
    Adding NaoH
    Co(2+) ---> Pink solution. Forms a blue precipitate that can turn beige in the air.
    Fe (2+) ---> Pale green solution. Forms green precipitate that an turn rusty brown in air
    Fe (3+) ---> Pale yellow solution. Forms a rusty brown precipitate

    Ligand substituion
    Adding Nh3 to [Cu(h20)6] ---> Pale blue precipitate forms. Sultion turns deep blue

    Bond angles and shapes (' = degrees)
    6 ligands is generally octahedral 90'
    4 ligands is generally tetrahedral 109.5'
    If a 4 ligand complex ion is Platinum or Nickel, it is usually square planar 90'



    Have I missed anything?
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    after flopping biology, i now have to ace this or bye bye future
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    (Original post by Vidja)
    Okay, I'm trying to create a list of colours/ shapes / angles we need to memorise. So far I have :

    Colours:
    Adding NaoH
    Co(2+) ---> Pink solution. Forms a blue precipitate that can turn beige in the air.
    Fe (2+) ---> Pale green solution. Forms green precipitate that an turn rusty brown in air
    Fe (3+) ---> Pale yellow solution. Forms a rusty brown precipitate

    Ligand substituion
    Adding Nh3 to [Cu(h20)6] ---> Pale blue precipitate forms. Sultion turns deep blue

    Bond angles and shapes (' = degrees)
    6 ligands is generally octahedral 90'
    4 ligands is generally tetrahedral 109.5'
    If a 4 ligand complex ion is Platinum or Nickel, it is usually square planar 90'



    Have I missed anything?
    With NaOH:

    Cu2+ --> pale blue solution goes to a blue precipitate.

    Ligand substitution:

    [Cu(H2O)6]2+ + 4Cl- --> [CuCl4]2- + 6H2O
    going from pale blue solution to a yellow solution

    [Co(H2O)6]2+ + 4Cl- --> [CoCl4]2- + 6H2O
    going from pink solution to blue solution

    [Cu(H2O)6]2+ + 4NH3 --> [Cu(H2O)2(NH3)4]2+ + 4H2O
    going from pale blue solution to pale blue precipitate with dark blue solution (know you had that but need to know how to balance it)

    [Co(H2O)6]2+ + 6NH3 --> [Cu(NH3)6]2+ + 6H2O
    going from pink solution to a brown solution


    Not entirely sure about that last one since it depends on the text book whether it's there (it's in the CGP book).
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    At least we know for this that most things will come up, whereas for Biology, OCR were c*nts
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    OCR F215 biology was an absolute nightmare, i have a 2hr 30min geog exam tomorrow and chem on wednesday. Motivation has reached an all time low
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    (Original post by fahimak)
    just had the ocr biology F215 exam and it went really bad
    cant motivate myself to even try and revise for chemistry

    (Original post by 786girl)
    after flopping biology, i now have to ace this or bye bye future

    (Original post by Wor Carroll)
    At least we know for this that most things will come up, whereas for Biology, OCR were c*nts

    (Original post by NWA)
    OCR F215 biology was an absolute nightmare, i have a 2hr 30min geog exam tomorrow and chem on wednesday. Motivation has reached an all time low
    You OCR biologists sound like us Edexcel SNAB biologists in January, you'll really be surprised with the results. Just by looking at the general consensus here it was a nightmare, so try to pick yourselves up for Chem! Best advice I can give.
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    (Original post by haydyb123)
    You OCR biologists sound like us Edexcel SNAB biologists in January, you'll really be surprised with the results. Just by looking at the general consensus here it was a nightmare, so try to pick yourselves up for Chem! Best advice I can give.
    Thanks
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    I got PMed about question 6d on Jan '11 paper, this is my best attempt at trying to explain, excuse the poor arrow drawing.
    Attached Images
     
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    (Original post by haydyb123)
    You OCR biologists sound like us Edexcel SNAB biologists in January, you'll really be surprised with the results. Just by looking at the general consensus here it was a nightmare, so try to pick yourselves up for Chem! Best advice I can give.
    Completely agree :lol: although it was a nightmare and it stayed that way for me .
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    (Original post by haydyb123)
    You OCR biologists sound like us Edexcel SNAB biologists in January, you'll really be surprised with the results. Just by looking at the general consensus here it was a nightmare, so try to pick yourselves up for Chem! Best advice I can give.
    Cheers! I only need around 75 UMS for an A in Biology, but after that paper I still don't feel that confident that I got enough
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    (Original post by blush.ox)
    Completely agree :lol: although it was a nightmare and it stayed that way for me .
    Damn I'm sure you'll destroy the resit though
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    (Original post by Wor Carroll)
    Cheers! I only need around 75 UMS for an A in Biology, but after that paper I still don't feel that confident that I got enough
    You always tend to look on the bad side of exams, so you never know, could be pleasantly surprised!
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    Can anyone help on this question (3c i) on this legacy paper on how the sharp rise is at approx 33.3cm3?
    Attached Images
  1. File Type: pdf 2816_01June10.pdf (202.7 KB, 196 views)
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    (Original post by haydyb123)
    Damn I'm sure you'll destroy the resit though
    Just sat it today, dont think i did :teehee:
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    (Original post by blush.ox)
    Just sat it today, dont think i did :teehee:
    how was it? What came up?
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    (Original post by thesimpsons)
    Can anyone help on this question (3c i) on this legacy paper on how the sharp rise is at approx 33.3cm3?
    I'm not entirely sure on this either I normally would draw the sharp rise at 25cm3 lol.. I think it's to do with the different in moles of the weak acid and strong base. If you find out let me know :P
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    its 0.025dm3 of acid of 0.02 mol dm3 so thats 0.02x0.025moles of acid, and then you're adding the base, so you divide whatever number you get for the moles of acid by the concentration of the base, to give you the volume of base required, which equals 33.3cm3 which is why the sharp rise is there.
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    (Original post by thesimpsons)
    Can anyone help on this question (3c i) on this legacy paper on how the sharp rise is at approx 33.3cm3?
    Ok so the the 33.3 is when pH has gone through 7 and neutralisation has happened
    We do 0.02 x (25/1000) = 5x10^-4 = moles of ethanoic acid
    As the mole ratio of the reaction has to be 1:1 for ethanoic acid and sodium hydroxide there also has to be 5x10^-4 moles of NaOH

    We do 5x10^-4 / 0.015 = 0.0333333 dm3

    0.033dm3 x 1000 = 33.333333 cm3
    So we need 33.3 cm3 of NaOH to neutralise the ethanoic acid, this is why the graph shoots up at 33.3
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    Does anybody have a good resource for these
    Transition metal colours in varying oxidation states
    Summary of all reaction of transition metals including colour changes
    Summary of acid - alkali, acid - metal, acid - carbonate reactions

    I just want to print this stuff off and stick it in my room
 
 
 
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