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N2+2H2=N2H4 What is the concentration of H2 at equalibrium when (N2)=.033 and (N2H2)= .042 M? Keq=5.29

  • chemistry -

    ..........N2 + 2H2 ==> N2H4
    aquil..0.033....x......0.042

    Keq = (N2H2)/(N2)(H2)^2
    Substitute and solve for x.

  • chemistry -

    .04905 M

  • chemistry -

    0.4905 M

  • chemistry -

    Yes but if your prof is picky about the number of significant figures s/he may not give you full credit for that answer.

  • chemistry -

    ok thanks!

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