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    [CHEM4 AQA Equilibria] but i guess other people can do it too.

    Kc = [A]^2[B]/[C][D]^3

    FiND the units of Kc.

    So, if you do it this way (adding the powers and subtracting), you get -1. (2+1) - (3+1). So (mol dm^-3)^-1. So, are the units mol-1 dm^3 :lolwut:

    Or if you just cancel out the powers, don't you get 1/mol dm^-3

    Or am I just plain confused?
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    lool you are just plain confused

    btw im clueless so don't quote me
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    (Original post by Ayostunner)
    lool you are just plain confused

    btw im clueless so don't quote me
    I know, my fellow T.I. lover, I know. :sigh:
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    (Original post by Get Cape.Wear Cape.Fly.)
    [CHEM4 AQA Equilibria] but i guess other people can do it too.

    Kc = [A]^2[B]/[C][D]^3

    FiND the units of Kc.

    So, if you do it this way (adding the powers and subtracting), you get -1. (2+1) - (3+1). So (mol dm^-3)^-1. So, are the units mol-1 dm^3 :lolwut:

    Or if you just cancel out the powers, don't you get 1/mol dm^-3

    Or am I just plain confused?
    each concentration is measured in mol dm-3

    so you can go cancelling down the powers at the top and the bottom remembering that [A] means mol dm[<sup]-3[/sup] raised to the power of 1.

    So your equation:

    [A]^2[B]/[C][D]^3

    is raised to the power of 3 at the top and 4 at the bottom so overall it is raised to the power of -1

    Hence the units are mol-1 dm3
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    (Original post by Get Cape.Wear Cape.Fly.)
    [CHEM4 AQA Equilibria] but i guess other people can do it too.

    Kc = [A]^2[B]/[C][D]^3

    FiND the units of Kc.

    So, if you do it this way (adding the powers and subtracting), you get -1. (2+1) - (3+1). So (mol dm^-3)^-1. So, are the units mol-1 dm^3 :lolwut:

    Or if you just cancel out the powers, don't you get 1/mol dm^-3

    Or am I just plain confused?
    Yeah you are right, just look in terms of powers, that power of 1's will cancel. The 2 will cancel the 2 of 3 from the [d]^3 leaving with 1/mol dm^-3 = dm^3 mol^-1
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    I get the same from both way, mol^-1 dm^3.
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    (Original post by charco)
    each concentration is measured in mol dm-3

    so you can go cancelling down the powers at the top and the bottom remembering that [A] means mol dm[<sup]-3[/sup] raised to the power of 1.

    So your equation:

    [A]^2[B]/[C][D]^3

    is raised to the power of 3 at the top and 4 at the bottom so overall it is raised to the power of -1

    Hence the units are mol-1 dm3
    That's what I thought, but the answer says mol dm^-3. A book mistake? :o:
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    (Original post by Preasure)
    I get the same from both way, mol^-1 dm^3.
    Ok goood, maybe the book is wrong :p:...again :shifty:
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    (Original post by Get Cape.Wear Cape.Fly.)
    That's what I thought, but the answer says mol dm^-3. A book mistake? :o:
    yup, book typo
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    (Original post by Get Cape.Wear Cape.Fly.)
    That's what I thought, but the answer says mol dm^-3. A book mistake? :o:
    Yeah, the book has a made a mistake which isn't surprising to me :p:
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    i don't know why we even bother with that book tbh...

    The damn thing seems to think that N2H2 is Ammonia :\

    (p94 if you're sad enough to want to check )
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    (Original post by Bslforever)
    i don't know why we even bother with that book tbh...

    The damn thing seems to think that N2H2 is Ammonia :\

    (p94 if you're sad enough to want to check )
    haha! I noticed that too! I actually sat there thinking about it for about 3 minutes :sigh:
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    (Original post by Get Cape.Wear Cape.Fly.)
    haha! I noticed that too! I actually sat there thinking about it for about 3 minutes :sigh:
    I just love the part at the back of the book above the blurb where it says

    Written by the experts
    Checked by the examiners


    I honestly hope the examiners show a little more care when checking my paper tbh...
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    (Original post by Bslforever)
    I just love the part at the back of the book above the blurb where it says

    Written by the experts
    Checked by the examiners


    I honestly hope the examiners show a little more care when checking my paper tbh...
    Agreed :yes:
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    That VERY same mistake was made in the OFFICIAL MARK SCHEME.

    I spotted it a while back, lemme find the thread.

    EDIT: Here
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    mol-1dm3
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    Treat it as indices. The way I do it is the following:

    \frac{[moldm^{-3}] \times [mol^2dm^{-6}]}{[moldm^{-3}] \times [mol^3dm^{-9}]} = mol^{-1}dm^3

    Cancel down as approprate, then convert to mol^xdm^{-y} or \frac{mol^x}{dm^y}

    Of course you coould just multiply in the 1st step and do \frac{mol^3dm^{-9}}{mol^4dm^{-12}}

    Hope that helps!
 
 
 
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