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(C) For a first-order reaction,the rate constant $K$ is given by: $K = \frac{2.303}{t} \log \frac{a}{a-x}$.For the first part,$a = 100$,$x = 90$,and $

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$30$

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(C) For a first-order reaction,the rate constant $K$ is given by: $K = \frac{2.303}{t} \log \frac{a}{a-x}$.For the first part,$a = 100$,$x = 90$,and $t = 10 \ hours$:$K = \frac{2.303}{10} \log \frac{100}{100-90} = \frac{2.303}{10} \log 10 = \frac{2.303}{10} 	imes 1 = 0.2303 \ h^{-1}$.For the second part,$a = 100$,$x = 99.9$,and we need to find $t$:$t = \frac{2.303}{K} \log \frac{100}{100-99.9} = \frac{2.303}{0.2303} \log \frac{100}{0.1} = \frac{2.303}{0.2303} \log 1000 = 10 	imes 3 = 30 \ hours$.

$A$ first-order reaction takes $10 \ hours$ to complete $90\%$ of the reaction. How much time will it take to complete $99.9\%$ of the reaction (in $hours$)?

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