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    Stuck on this question! Please help quickly!!! Thank you!! ’If I react soap CH3(CH2)14COO-Na+ with H2SO4 show equation acting as a base?'
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    Split the two reactants into their constituent ions, then go from there.
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    (Original post by knope)
    Split the two reactants into their constituent ions, then go from there.
    CH3
    CH2
    COO-
    Na+
    H2
    SO4

    and then?

    thank you
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    (Original post by Greenapplepear)
    CH3
    CH2
    COO-
    Na+
    H2
    SO4

    and then?

    thank you
    What charge is there on a SO4 ion? This is key to getting the right answer, not to mention the correct ions in H2SO4. Right now, you've not found them.

    Also, does the carbon chain [CH3(CH2)14] have a charge or not? Will it break from its COO- functional group?
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    (Original post by knope)
    What charge is there on a SO4 ion? This is key to getting the right answer, not to mention the correct ions in H2SO4. Right now, you've not found them.

    Also, does the carbon chain [CH3(CH2)14] have a charge or not? Will it break from its COO- functional group?
    the charge on SO4 is 2-
    H2 is 2+
    Na is +
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    (Original post by Greenapplepear)
    the charge on SO4 is 2-
    H2 is 2+
    Na is +
    H2 is not an ion, it's a molecule. So it has 0 charge - but H2SO4 actually splits up into 2H+ and SO42-.

    Having said that, I suggest you attempt the solution again now. If you're left with more than four ions to react then look back at my last question to you (which you didn't answer).
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    (Original post by knope)
    H2 is not an ion, it's a molecule. So it has 0 charge - but H2SO4 actually splits up into 2H+ and SO42-.

    Having said that, I suggest you attempt the solution again now. If you're left with more than four ions to react then look back at my last question to you (which you didn't answer).
    so is it Na2SO4 + CH3(CH2)14cooh?

    Thank you very much
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    (Original post by Greenapplepear)
    so is it Na2SO4 + CH3(CH2)14cooh?

    Thank you very much
    Yep, perfect!
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    (Original post by knope)
    Yep, perfect!
    Thank you so much, I really appreciate your help
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    (Original post by Greenapplepear)
    Thank you so much, I really appreciate your help
    No worries, glad I could help at all
 
 
 
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