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(C) The rate law for the reaction is given by $r = k[A]^x$,where $x$ is the order of the reaction.Initially,$r = k[A]^x$ $(1)$.When the concentration

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$2$ and $3$

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(C) The rate law for the reaction is given by $r = k[A]^x$,where $x$ is the order of the reaction.Initially,$r = k[A]^x$ $(1)$.When the concentration is doubled,the new rate becomes $7r$,so $7r = k[2A]^x$ $(2)$.Dividing equation $(2)$ by equation $(1)$: $\frac{7r}{r} = \frac{k[2A]^x}{k[A]^x}$.$7 = 2^x$.Taking the logarithm on both sides: $\log(7) = x \log(2)$.$x = \frac{\log(7)}{\log(2)} \approx \frac{0.845}{0.301} \approx 2.81$.Since $2 < 2.81 < 3$,the order of the reaction is between $2$ and $3$.

When the concentration of a reactant,$A$,in a reaction $A \to \text{Products}$ is doubled,the rate of reaction increases seven times. The order of the reaction is between:

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