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Posted by on Tuesday, July 20, 2010 at 2:04pm.

A sample of sodium-24 with an activity of 12 mCi is used to study the rate of blow flow in the circulatory system. If sodium -24 has a half-life of 15 hours, what is the activity of the sodium after 2.5 d ?

  • Chemistry - , Tuesday, July 20, 2010 at 2:44pm

    You can do this two ways.
    No/N = 2n
    12/N = 22.5*24/15
    12/N = 16
    N = 12/16 = 0.75 mCi after 2.5 days.

    Other way.
    Calculate k from 0.69315/15 = 0.04621 and substitute into the following:
    ln(No/N) = kt
    No = 12
    N = ??
    k = from above.
    t = 2.5 days in hours is 2.5 x 24 = 60
    ln(12/N) = 0.69315/60
    Solve for N = 0.75 mCi.

  • Chemistry - , Saturday, January 28, 2012 at 2:44pm

    1.0 mCi is the freakin answer.

  • Chemistry - , Friday, August 7, 2015 at 11:17am

    Ln(Ao/A)=kt/2.30
    ln(12/A)=0.0462*2.5/2.30
    A=21.47

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