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November 28, 2014

November 28, 2014

Posted by **ronny** on Wednesday, October 15, 2008 at 8:19pm.

plz tell me what type of equation to use.

- chemistry math -
**GK**, Friday, October 17, 2008 at 2:06pmln(M/Mo) = -kt

ln = natural log

M = mass left (not needed)

Mo = original mass (not needed)

(M/Mo) = 0.809

k = 0.693/T(half)

T(half) = 5.73X10^3 years.

- chemistry math -
**GK**, Friday, October 17, 2008 at 2:08pmln(M/Mo) = -kt

ln = natural log

M = mass left (not needed)

Mo = original mass (not needed)

(M/Mo) = 0.809

k = 0.693/T(half)

T(half) = 5.73X10^3 years.

- chemistry math -
**GK**, Friday, October 17, 2008 at 2:11pmln(M/Mo) = -kt

ln = natural log

M = mass left (not needed)

Mo = original mass (not needed)

(M/Mo) = 0.809

k = 0.693/T(half)

T(half) = 5.73X10^3 years.

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