chemistry
posted by lisa .
a 7.500mg sample of chromium55 is analyzed after 14.0 min and found to contain 0.469 mg of chromium55. what is the halflife of chromium55

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/t_{1/2} and solve for t_{1/2}