posted by lisa .
a 7.500-mg sample of chromium-55 is analyzed after 14.0 min and found to contain 0.469 mg of chromium-55. what is the half-life of chromium-55
ln(No/N) = kt
Substitute No = 7.500 mg
N = 0.469 mg
t = 14 min
Solve for k. Then substitute k into the following.
k = 0.693/t1/2 and solve for t1/2