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    (Original post by Science_help)
    I got a notif that you tagged me in something ?...
    Tbh I don't really remember :/
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    Does anyone know the actual experiments that require reflux because they change from textbook to textbook???
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    h ttp://www.chatzy.com/27882973356288 please join guys, can quickfire questions for f324 tomorrow
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    I posted this as a separate thread (unanswered).... might as well delete that one and ask here?

    CH3 COOH(l) + CH3 CH2 OH(l) <reversible react.>CH3 COOCH2 CH3 (l) + H2 O(l)

    Initial concentration
    -------------------------------
    CH3COOH - 0.4 mol
    CH3CH2OH - 0.3 mol
    CH3COOCH2CH3 - 0.00 mol
    H20 - 0.15

    Equilibrium concentration
    ----------------------------------
    CH3COOH - 0.2 mol
    CH3CH2OH - UNKNOWN
    CH3COOCH2CH3 - UNKNOWN
    H20 - UNKNOWN

    Q1) Find the unknown values


    EDIT: Does anyone know how to delete a thread? I cannot find the button :/
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    Sorry for bad quality an messy writing but could anyone let me know why the answer i gabe was wrong. Why doesnt it split to produce the fragment that i suggested?
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    Hello! Can anyone clarify what we need to know about transesterification to produce margarine in edexcel chem 4?
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    A moments silence please, for our fallen brothers and sisters, who gave their all today. Never before have so few students, been owed so much by the exam boards. RIP.
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    (Original post by NimbleNeil)
    A moments silence please, for our fallen brothers and sisters, who gave their all today. Never before have so few students, been owed so much by the exam boards. RIP.
    Sack up.
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    Do you know any good books for Edexcel AS?A level Chemistry? The one by Ann Fullick and Bob McDuell is apparently not that great!
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    (Original post by ihaspotato)
    I posted this as a separate thread (unanswered).... might as well delete that one and ask here?

    CH3 COOH(l) + CH3 CH2 OH(l) <reversible react.>CH3 COOCH2 CH3 (l) + H2 O(l)

    Initial concentration
    -------------------------------
    CH3COOH - 0.4 mol
    CH3CH2OH - 0.3 mol
    CH3COOCH2CH3 - 0.00 mol
    H20 - 0.15

    Equilibrium concentration
    ----------------------------------
    CH3COOH - 0.2 mol
    CH3CH2OH - UNKNOWN
    CH3COOCH2CH3 - UNKNOWN
    H20 - UNKNOWN

    Q1) Find the unknown values


    EDIT: Does anyone know how to delete a thread? I cannot find the button :/
    Are you sure that H20 isn't zero? It should be I think? As it is a produc and the initial number of moles should be zero...
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    (Original post by alow)
    Sack up.
    What does that mean :rolleyes:
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    (Original post by NimbleNeil)
    What does that mean :rolleyes:
    http://www.urbandictionary.com/defin...term=sack%20up
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    Why is the electrode postive at the site of reduction in an electrochemical cell?

    I'm not sure id be able to answer that well if they asked that in an exam :/
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    (Original post by wwwwwwwwwwwf)
    Why is the electrode postive at the site of reduction in an electrochemical cell?

    I'm not sure id be able to answer that well if they asked that in an exam :/
    Reduction is gain... so I'm guessing all the negative species are attracted by the positive electrode to become oxidised?!
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    (Original post by K3001N)
    Reduction is gain... so I'm guessing all the negative species are attracted by the positive electrode to become oxidised?!

    Name:  ec.png
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    This might make it a bit more clear, reduction is happening at the Cu electrode (Cu2+ > Cu)

    Cu electrode is postive, so the electrons are flowing from the Fe to the Cu electrode, but I just don't really understand why the Cu electrode is the postive electrode
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    (Original post by wwwwwwwwwwwf)
    Name:  ec.png
Views: 156
Size:  34.7 KB

    This might make it a bit more clear, reduction is happening at the Cu electrode (Cu2+ > Cu)

    Cu electrode is postive, so the electrons are flowing from the Fe to the Cu electrode, but I just don't really understand why the Cu electrode is the postive electrode
    Opposites attract.
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    (Original post by wwwwwwwwwwwf)
    Name:  ec.png
Views: 156
Size:  34.7 KB

    This might make it a bit more clear, reduction is happening at the Cu electrode (Cu2+ > Cu)

    Cu electrode is positive, so the electrons are flowing from the Fe to the Cu electrode, but I just don't really understand why the Cu electrode is the positive electrode
    Ahhhh, it's to do with EMF! Cu has a more positive EMF, this means it gains a positive electrode potential and its reduced. Where as Fe has a negative EMF, this means that it gains a negative electrode potential. So it donates electrons via the external circuit to the Cu.
    basically the Fe reduces the Cu and the Fe gets oxidised.

    More negative the emf the more it donates electrons (becomes oxidised)
    the more positive the emf the more it accepts electrons (becomes reduced)

    I think this should help!
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    This is so nice!
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    Does anyone know why with [Co(H2O)6]2+ NH3 is able to fully displaced (by ligand substitution) all the H2O molecules however with [Cu(H2O)6]2+ only 4 H2O molecules will be displaced by NH3?+
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    Posted from TSR Mobile

    Hey guys could you help me interpret proton NMR spectra?
 
 
 
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