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(A) The rate of reaction is given by: $	ext{Rate} = k[A]^x$. The unit of rate is $	ext{mol} \ L^{-1} \ sec^{-1}$. The unit of concentration $[A]$ is

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$mol^{1-x} \ L^{x-1} \ sec^{-1}$

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(A) The rate of reaction is given by: $	ext{Rate} = k[A]^x$. The unit of rate is $	ext{mol} \ L^{-1} \ sec^{-1}$. The unit of concentration $[A]$ is $	ext{mol} \ L^{-1}$. Substituting these into the rate law equation: $	ext{mol} \ L^{-1} \ sec^{-1} = k 	imes (	ext{mol} \ L^{-1})^x$. Solving for $k$: $k = \frac{	ext{mol} \ L^{-1} \ sec^{-1}}{(	ext{mol} \ L^{-1})^x} = (	ext{mol} \ L^{-1})^{1-x} \ sec^{-1}$. Expanding this,we get: $k = 	ext{mol}^{1-x} \ L^{-(1-x)} \ sec^{-1} = 	ext{mol}^{1-x} \ L^{x-1} \ sec^{-1}$.

If the rate law of a reaction $n A \rightarrow B$ is expressed as Rate $= -\frac{1}{n} \frac{d[A]}{dt} = \frac{d[B]}{dt} = k[A]^{x}$,the unit of rate constant $k$ will be:

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