chemistry
posted by Anonymous .
What percent of the original sample of strontium90 (halflife = 28.9 years) will remain after the following time periods?after 18 years?
after 47.5 years?
after 1008 years?

k = 0.693/t_{1/2}
ln(No/N) = kt
Substitute into equation 1 and solve for k, the substitute into equation 2 and solve for N. No will be 18, 47.5, 1008 years. 
.....it's actually asking for the % left after so many years

So N = amount left.
% = (N/No)*100 = ?
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