1.

A first order reaction is completed by 20% 2 minutes. How much further time is required for 64% of the initial concentration of the reactants to remain.

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SOLUTION :`(Kxx2)/(Kxxt)=(LOG(100/80))/(log(100/64)),2/t=(log1.25)/(log1.5625)=0.0969/0.19382=t=(2xx0.19382)/(0.0969~=4`


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