K = 2.303/t log[A0]/[A]
t = 33.6 hours
[A0] = 100
[A] = 12.5
K = 2.303/33.6 log(100/12.5)
= 2.303*0.9030/33.6 = 0.0619 hour^-1
t1/2 = 0.693/K
= 0.693/0.0619 = 11.2hour
second method
(1/2)^n = 0.125
(0.5)^n = (0.5)^3
n = 3
n = total time/half life time
3 = 33.6/ half life time
half life time = 33.6/3 = 11.2 hours >>>>answer
what is the half life of an isotope that decays to 12.5% of its original activity...
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