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Consider a tube that is 1.0 meter long. If NH3 gas is introduce at the 0-m mark at the same time HCl gas is introduced at the 1-m mark, how far down the tube will the ring of gas occur?

  • chem -

    0...x.........|..1-x..|<< 1meter

    distance NH3 travels = x
    distance HCl travels = 1-x
    (x/1-x) = sqrt(molar mass HCl/molar mass NH3)
    Solve for x and 1-x.
    I estimated x = about 60 cm and 1-x about 40 cm.

  • chem -

    thank you

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