What percentage of sodium-24 with a half life of 15.0 hr would remain after 60hours?

Answer 1

k = 0.693/t_1/2. Calculate k and substitute into the equation below.

ln(No/N) = kt

I would start with a convenient number (say 100) for No, calculate N (which is the number remaining after 60 hours). Then (N/No)*100 = percent remaining.