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A first reaction takes `40 mi n ` for `30%` decomposition. Calculate `t_(1//2)`. |
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Answer» Correct Answer - `77.7 mi n` Use direct relation, `(t_(1//2))/(t_(30%))=(3.0)/(log((100)/(100-30)))` `:.t_(1//2)=40 mi nxx(0.3)/(log((10)/(7)))` `=(40mi nxx0.3)/((1-0.845))[log7=0.845]` `~~77.7mi n` Alternatively `x=30, a =100` `k=(2.303)/(t)log (a)/(a-x)` `=(2.303)/(40 mi n ) log((100)/(100-30))` `=(2.303)/(40)xx0.155mi n^(-1)` `=8.918xx10^(-3)mi n^(-1)` `t_(1//2)=(0.693)/(K)=(0.693)/(8.918xx10^(-3))~~77.7mi n` |
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