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AQA - Unit 5 - Energetics, Redox and Inorganic Chemistry watch

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    (Original post by Tiger_Lilly92)
    Hey just wandering why isn't EDTA poisonous? It is used as an antidote to metal poisoning. But i don't understand why it wouldn't substitute the nitrogens and oxygen in haemoglobin as they are unidentate ligands, so substituting for EDTA (a multidentate) would have an increase in entropy??
    EDTA isnt naturally occuring in the environment.. stuff like CO are thats why they are poisionous .. EDTA is used when the body is poisned by transition metals.
    the EDTA would then be extracted in the urine
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    OK, got 85/100 in the Jan 11. Made some stupid mistakes :-(
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    Could somebody just make a post stating which of the equations in the book that need to be remebered
    And what colours need to be remebered

    (I mean just tell me what equations need to be remebered - Not the actual equations)

    Cheers

    (Quote Me)
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    (Original post by Sheldon)
    Not that many really a few old style ones, but tehy are cluttered with unit 4 and 2 questions in them
    Just been going over notes.
    If any one wants anything explained please ask it would help me as well
    have a go on this que.

    how do you identify a reducing agent and write half equation for this reaction

    5S2O82- + Br2 + 6H2O ---> 2BrO3- + 12H+ + 10SO42-
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    (Original post by al_habib)
    have a go on this que.

    how do you identify a reducing agent and write half equation for this reaction

    5S2O82- + Br2 + 6H2O ---> 2BrO3- + 12H+ + 10SO42-
    a reducing agent is one that is itself oxidised .. so its ox state will increase the reducing agent is Br2 as in Br03- the ox state is not 0 as it was in the reactant

    Br2 + 6H20 -------- 2BrO3- + 12H+ + 6e-


    quote me if im wrong
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    Well did jan11 cos everyone was hahah

    Only revision I did was I learnt the transition metal colours and equations

    Got 70/100 ...Thats not being able to do the last 2 or so questions because i couldnt remember the hydrogen peroxide equation with manganete haha

    So.. easy peasy!

    If I had to give one bit of advice to someone doing cramming for the chem5 exam.. It would be to spend your time learning transition metals without doubt!


    Could someone very kindly post the reagant equations and how you tell between them?
    Bit to ask, but would be vair much apreciated!
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    (Original post by ben10)
    a reducing agent is one that is itself oxidised .. so its ox state will increase the reducing agent is Br2 as in Br03- the ox state is not 0 as it was in the reactant

    Br2 + 6H20 -------- 2BrO3- + 12H+ + 6e-


    quote me if im wrong
    you are right, am not sure why the no. of electrons is 10 :confused:

    Br2 + 6H2O ---> 2BrO3- + 12H+ + 10e–
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    (Original post by al_habib)
    you are right, am not sure why the no. of electrons is 10 :confused:

    Br2 + 6H2O ---> 2BrO3- + 12H+ + 10e–
    ahh yes sorry it is ten as on the right hand side the 2Br03- has a 2- charge and the 12+ is there so 12 minus 2 is 10

    i just saw the BrO3 differently cuz its in typing format sorrry
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    (Original post by ben10)
    ahh yes sorry it is ten as on the right hand side the 2Br03- has a 2- charge and the 12+ is there so 12 minus 2 is 10

    i just saw the BrO3 differently cuz its in typing format sorrry
    (12H+) minus 2 (of Br03-) = 10e is that what you mean? sorry a bit :confused: coz the way i use is to get the difference of oxidation no. of Br and then plug in the no. of e- in the equation.

    tx alot for your help
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    (Original post by M.A.)
    I dont get how you get this one either! (Na2SiO3).... unless we just have to remember it?! :confused:
    Think about it logically:

    Na is group 1, yeah? 1 outer electron.

    SiO3 has a charge on 2-.


    See where I'm going with this?


    Same with Mg(OH)2 etc...
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    (Original post by al_habib)
    (12H+) minus 2 (of Br03-) = 10e is that what you mean? sorry a bit :confused: coz the way i use is to get the difference of oxidation no. of Br and then plug in the no. of e- in the equation.

    tx alot for your help
    yes so 10 electrons is correct
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    (Original post by ben10)
    hey guys is it possible for me to get an A star in chemistry
    i got 104 UMS in chem 4
    assuming i got an A in my EMPA ?


    You'd need 112(min)/120 UMS, so you'd need to be looking at around 94+ out of 100

    I hope you get it
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    (Original post by T-Toe)
    You'd need 112(min)/120 UMS, so you'd need to be looking at around 94+ out of 100

    I hope you get it
    That kid in your forum sig is so creepy :'(
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    (Original post by LimeTree.)
    That kid in your forum sig is so creepy :'(
    Looool!! I've been told :rofl:
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    (Original post by T-Toe)
    You'd need 112(min)/120 UMS, so you'd need to be looking at around 94+ out of 100

    I hope you get it
    ad never get that :/ LOL
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    hey, could anyone help me with this question, i know the answer i just dont know why lol:

    Write a half-equation for the formation of hydrogen and NH2 minus ions from ammonia.
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    (Original post by adam1232)
    hey, could anyone help me with this question, i know the answer i just dont know why lol:

    Write a half-equation for the formation of hydrogen and NH2 minus ions from ammonia.
    I think its
    NH3 + e- ---> 1/2 H2(g) + NH2-
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    (Original post by ben10)
    ad never get that :/ LOL
    You'd need 88/100
    because 74/100 = A = 96UMS
    84/100 = A* = 108 UMS

    you need 112UMS = 87.33/100= 88/100
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    hey guys just thought I would post a list of some TM reactions that I made
    If you find them useful I have other notes I made like that for periodicity and some other parts.
    Attached Files
  1. File Type: docx TM Complexes.docx (34.6 KB, 481 views)
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    anyone got any decent ways of remembering transition metal colours? worried about those!

    although the january paper seemed to be all transition metals (all of section B except the last question) so i hope that means its all electrode potentials because i love that!!!!!!
 
 
 
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