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Draw a labelled diagram to show H-bonding between 2 molecules of liquid ammonia

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    Liquid ammonia, NH3, and water, H20, both show H-Bonding.


    So basically, i know how to draw it.. but erm could someone please clarify whether it's a or b:

    a) only link two NH3 via h-bonding...h20 molecule not relevant - so don't include.

    or

    b) link two NH3 using a water molecule.
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    (Original post by FoOtYdUdE)
    Liquid ammonia, NH3, and water, H20, both show H-Bonding.


    So basically, i know how to draw it.. but erm could someone please clarify whether it's a or b:

    a) only link two NH3 via h-bonding...h20 molecule not relevant - so don't include.

    or

    b) link two NH3 using a water molecule.
    Two NH3.
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    (Original post by Kinkerz)
    Two NH3.
    so basically...a??

    aah, cheers.

    bdway,

    how do i suggest the formula for sodium chlorate (vii) ?? :confused:
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    (Original post by FoOtYdUdE)
    so basically...a??
    :yep:

    (Original post by FoOtYdUdE)
    bdway,

    how do i suggest the formula for sodium chlorate (vii) ?? :confused:
    Not sure.
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    (Original post by Kinkerz)
    :yep:


    I'd go with NaClO3, but I may be wrong.
    it's NaCl04 butt i dunno y. oh nvm. thnxxx.
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    (Original post by FoOtYdUdE)
    it's NaCl04 butt i dunno y. oh nvm. thnxxx.
    At this point, do you need to know why?
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    If you're using oxidation states:

    O = -2
    Na = +1
    ---
    4(-2) + 1 = -7

    So the Cl must have an oxidation number of +7, hence your sodium chlorate (vii) ion.
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    (Original post by Kinkerz)
    If you're using oxidation states:

    O = -2
    Na = +1
    ---
    4(-2) + 1 = -7

    So the Cl must have an oxidation number of +7, hence your sodium chlorate (vii) ion.
    genius!

    i get it!
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    so basically if it was sodium chlorate (III) then the Cl would have an ox. no. of +3 ?? aaah and then the formula would be NaCl02 ??
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    (Original post by FoOtYdUdE)
    so basically if it was sodium chlorate (III) then the Cl would have an ox. no. of +3 ?? aaah and then the formula would be NaCl02 ??
    That'd be my approach, yes.

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