Solution.
t(1/2)=ln(2)kt(1/2) = \frac{ln(2)}{k}t(1/2)=kln(2)
k=ln(2)t(1/2)k = \frac{ln(2)}{t(1/2)}k=t(1/2)ln(2)
k=1tln(C0/C)k = \frac{1}{t} ln(C0/C)k=t1ln(C0/C)
t=ln(C0C)kt = \frac{ln(\frac{C0}{C})}{k}t=kln(CC0)
C0 = 1
C = 0.6*C0
k=3.38×10−5seck = 3.38 \times 10^{-5} seck=3.38×10−5sec
t = 15113 sec = 4.2 hours
Answer:
t = 4.2 hours
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