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What is the halflife of sodium 25 if 1.00 gram of a 16.00 gram sample of sodium 25 remains unchanged after 237 seconds?
Wiki User
∙ 12y ago
59.25 seconds
After 1 half-life: 16*(1/2) = 8 g remains
After 2 half-lives: 8*(1/2) = 4g remains
After 3 half-lvies: 4*(1/2) = 2g remains
After 4 half-lives: 2*(1/2) = 1g remains
So after 4 half-lves you have 1 gram of Na25 left. This is also the amount remaining after 237 seconds. Since 4 half-lives have elapsed over 237 seconds, you can divide 237 seconds by 4 to find the half-life of Na25 is 59.25 seconds.
You can also figure it out using the rate of decay formula:
At = Ao*e-kt
where Ao is the initial amount, At is the amount left after some time t. k is the decay constant which is k = ln 2 / t1/2 where t1/2 is the half-life.
In this case Ao = 16g, At = 1g, t=237 sec
Substitute in the formula to solve for k, then take the answer for k and use it in the other formula to solve for the half-life (t1/2):
1 = 16*e-237k
ln (1/16) = ln (e-237k)
-2.7726 = -237k
k = -2.7726/-237 = 0.011699 sec-1
t1/2 = ln 2/k = 0.693147/0.011699 sec-1
t1/2 = 59.25 sec
Wiki User
∙ 12y ago
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