1. 2.34 = x (1/2)^(36 / 8.14) ... log(2.34) - [(36 / 8.14) log(1/2)] = log(x)
2. .23 = (1/2)^(t / 64.9) ... log(.23) = (t / 64.9) log(1/2)
3. x = 4000 * 5^(48 / 6)
1. Iodine-131 is used to find leaks in water pipes it has a half life of 8.14 days. If 36 days have passed and there is now 2.34g left.
What is the initial amount?
2. Sr-85 is used in bone scans and has a half life of 64.9 days. If there is 23 percent of the sample remaining how much time has passed?
3. A bacteria increases fivefold every 6 hours how much bacteria will there be if the initial amount is 4000 and 48 hours have passed?
1 answer
1 answer