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`Ararr` Product, `[A]_(0) = 2M`. After `10 min` reaction is `10%` completed. If `(d[A])/(dt) = k[A]`, then `t_(1//2)` is approximatelyA. `0.693 min`B. `69.3 min`C. `66.0 min`D. `0.0693 min` |
Answer» Correct Answer - C `k(2.3030=)/(10) log.(100)/(90)=0.001054 min^(-1)` `t_(1//2)=(2.3030)/(k)=(2.303)/(0.001054)~~66min` |
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