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    (Original post by ryanb97)
    when you increase the temprature, equilibrium tries to oppose it, thus favours the side of the reaction which is endothermic, to reduce the amount of heat,to rreduce the temprature

    when you decrease the temprature, equilibrium tries to oppose it, thus favours the side of the reaction which is exothermic,to produce more heat and even it out

    With presure you need to look at the number of moles, of each Gas on both sides of the equation

    If you increase the pressure, you increase the rate of the side of the equation that has more moles of gas .... so equilibrium shifts to the other side,where there are less moles,and also more of that reactant/product


    Since it is a reversible reaction.... it can go both ways - so you want more product and so you change the conditions (temprature,pressure,concentrati on) to make more product!
    Thanks!
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    In weak acids why does the equilibrium lie to the left?? Not sure if I've got this confused. :confused:
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    (Original post by Turtlefushsia)
    In weak acids why does the equilibrium lie to the left?? Not sure if I've got this confused. :confused:
    BEcause there are less H+ protons produced

    so the higher concentration is on the left, so equilibrium also lies on the left

    Ryan
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    (Original post by andersson)
    Woo Geography over! Time to spend my whole day on Chemistry.
    How was it ?
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    (Original post by ryanb97)
    How was it ?
    Surprisingly good, even though I didn't revise that much for it.
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    (Original post by ryanb97)
    BEcause there are less H+ protons produced

    so the higher concentration is on the left, so equilibrium also lies on the left

    Ryan
    Oh yeah thanks Ryan that makes sense :rolleyes:
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    (Original post by L'Evil Fish)
    You should know them, just look at the periodic table and check which group.
    For instance because Br is in group 7 it has a charge of -?

    Therefore:
    H+
    Li+
    Be2+
    B3+
    C4-
    N3-
    O2-
    F-
    N.....?
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    How do you work out the percentage by mass?
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    (Original post by benwalters1996)
    For instance because Br is in group 7 it has a charge of -?

    Therefore:
    H+
    Li+
    Be2+
    B3+
    C4-
    N3-
    O2-
    F-
    N.....?
    C wouldn't lose electrons, too much energy. That's why it bonds covalently.

    It'll be group 1,2,6,7 but yeah, that's the gist.

    N is group ? So it is N ??
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    (Original post by anamkhalid97)
    How do you work out the percentage by mass?
    what?
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    I'm so looking forward to it, I hope I get an A* in it especially since i got 139 in c1c2c3 and 45/48 in coursework.
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    (Original post by andersson)
    Surprisingly good, even though I didn't revise that much for it.
    nice

    are you starting with C4 first...or diving into the deep-end with C6 ? :P

    ryan
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    (Original post by ThenAaliyahSaid)
    I'm so looking forward to it, I hope I get an A* in it especially since i got 139 in c1c2c3 and 45/48 in coursework.
    You only need an A to get your A* ... this time round
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    (Original post by L'Evil Fish)
    C wouldn't lose electrons, too much energy. That's why it bonds covalently.

    It'll be group 1,2,6,7 but yeah, that's the gist.

    N is group ? So it is N ??
    N is group 0 so i guess it doesn't have a charge because it...bonds covalently...?
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    (Original post by ryanb97)
    what?
    How would you calculate the percentage by mass of nitrogen in ammonium sulfate (NH4)2 SO4
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    Hello everybody,
    I have a question about fuel cells. In the textbook I have it has both acidic and alkaline conditions for reactions, however the Oxbox textbook only says about the acid reaction. I have read the spec for this, and it doesn't seem to shed any light on which one should be learnt. Which one is it that needs to be learnt?
    Cheers.
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    (Original post by anamkhalid97)
    How would you calculate the percentage by mass of nitrogen in ammonium sulfate (NH4)2 SO4
    Oh ,, that makes a bit more sense

    you need to work out the mass of nitrogen ,,,then of ammonium sulfate..or its given

    then : (nitrogen/ammonium sufate) x 100 = your % ....

    i think that is it..

    Ryan
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    (Original post by ryanb97)
    nice

    are you starting with C4 first...or diving into the deep-end with C6 ? :P

    ryan
    Neither, I've already done C4. The only thing I need to work on in that module is water purity.

    So I'm doing the calculations side of Chemistry (start of C5) now.
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    (Original post by ryanb97)
    Oh ,, that makes a bit more sense

    you need to work out the mass of nitrogen ,,,then of ammonium sulfate..or its given

    then : (nitrogen/ammonium sufate) x 100 = your % ....

    i think that is it..

    Ryan
    Okay, thanks!
    • Thread Starter
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    (Original post by benwalters1996)
    N is group 0 so i guess it doesn't have a charge because it...bonds covalently...?
    Well, it won't bond ionically
 
 
 
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