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Find the two third life (t_(2//3)) of a first order reaction in which k=5.48xx10^(-14)sec^(-1), (log3=0.4771,log2=0.3010)

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Solution :`t_(2//3)=(2.303)/(K)LOG""(a)/(a-2a//3)=(2.303)/(5.48xx10^(-14)s^(-1))log3=2xx10^(13)s.`


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