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Cesium-137 has a half-life of 30 years. Suppose a lab stored a 30-mCi sample n 1973. How much will be left in 2063?

  • Chemistry/math -

    Use the half life to calculate the k in
    k = 0.693/(t1/2).
    Then use k in the following equation.
    ln(Co/C) = kt
    Co = 30 mCi
    Solve for C. The answer will be in mCi.
    k = from above
    t = years from dates listed.
    Post your work if you get stuck.

  • Chemistry/math -

    what does k stand for?

  • Chemistry/math -

    k is the proportionality constant for the equation:
    ln(Co/C) = kt and it is evaluated from k = 0.693/t(1/2). In this case, its
    k=0.693/30 years = ??

  • Chemistry/math -


  • Chemistry/math -

    Yes, k = 0.0231 years^-1. Now put that into the ln formula I provided and solve for C following the instructions I gave you.

  • Chemistry/math -

    why is it to the -1?

  • Chemistry/math -

    Are you putting me on?
    1/x = x^-1.
    so 0.693/30 = 0.0231 years^-1. It's just a unit and the -1 means years is in the denominator. It's the same thing as saying wavenumber = 1/5 cm = 0.2 cm^-1 or in words it is 0.2 reciprocal centimeters.

  • Chemistry/math -

    ok thanks

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