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(C) The rate law for the reaction $xA 	o yB$ is given by: $\frac{1}{x} (-\frac{d[A]}{dt}) = \frac{1}{y} (+\frac{d[B]}{dt})$.Given equation: $0.48 + \

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$5:3$

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(C) The rate law for the reaction $xA 	o yB$ is given by: $\frac{1}{x} (-\frac{d[A]}{dt}) = \frac{1}{y} (+\frac{d[B]}{dt})$.Given equation: $0.48 + \log \{-\frac{d[A]}{dt}\} = \log \{+\frac{d[B]}{dt}\} + 0.7$.Using $\log 3 \approx 0.477 \approx 0.48$ and $\log 5 \approx 0.699 \approx 0.7$,we rewrite the equation:$\log 3 + \log \{-\frac{d[A]}{dt}\} = \log \{+\frac{d[B]}{dt}\} + \log 5$.$\log \{3 	imes (-\frac{d[A]}{dt})\} = \log \{5 	imes (+\frac{d[B]}{dt})\}$.$3 (-\frac{d[A]}{dt}) = 5 (+\frac{d[B]}{dt})$.Rearranging gives: $\frac{-\frac{d[A]}{dt}}{5} = \frac{+\frac{d[B]}{dt}}{3}$.Comparing this with the stoichiometric rate expression $\frac{-\frac{d[A]}{dt}}{x} = \frac{+\frac{d[B]}{dt}}{y}$,we get $x = 5$ and $y = 3$.Therefore,$x : y = 5 : 3$.

In the reaction $xA \to yB,$
$0.48 + \log \left\{ -\frac{d[A]}{dt} \right\} = \log \left\{ +\frac{d[B]}{dt} \right\} + 0.7$
Then,$x : y$ is

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In the reaction $xA \to yB,$$0.48 + \log \left\{ -\frac{d[A]}{dt} \right\} = \log \left\{ +\frac{d[B]}{dt} \right\} + 0.7$Then,$x : y$ is