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Question

(A) The rate law for the reaction is given by $r = k[A]^1[B]^1$.For the first experiment: $0.10 = k(20)(0.5)$ $\Rightarrow 0.10 = 10k$ $\Rightarrow k

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$80$ and $2$

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(A) The rate law for the reaction is given by $r = k[A]^1[B]^1$.For the first experiment: $0.10 = k(20)(0.5)$ $\Rightarrow 0.10 = 10k$ $\Rightarrow k = 0.01 \ L \ mol^{-1} \ s^{-1}$.For the second experiment: $0.40 = k(x)(0.5)$. Substituting $k = 0.01$: $0.40 = 0.01(x)(0.5)$ $\Rightarrow 0.40 = 0.005x$ $\Rightarrow x = 80 \ mol \ L^{-1}$.For the third experiment: $0.80 = k(40)(y)$. Substituting $k = 0.01$: $0.80 = 0.01(40)(y)$ $\Rightarrow 0.80 = 0.4y$ $\Rightarrow y = 2 \ mol \ L^{-1}$.Therefore,the values are $x = 80$ and $y = 2$.

Rate $(mol \ L^{-1} \ s^{-1})$	$[A]$ $(mol \ L^{-1})$	$[B]$ $(mol \ L^{-1})$
$0.10$	$20$	$0.5$
$0.40$	$x$	$0.5$
$0.80$	$40$	$y$

For a chemical reaction $A + B \rightarrow \text{Product}$,the order is $1$ with respect to $A$ and $B$.
Rate $(mol \ L^{-1} \ s^{-1})$ $[A]$ $(mol \ L^{-1})$ $[B]$ $(mol \ L^{-1})$
$0.10$ $20$ $0.5$
$0.40$ $x$ $0.5$
$0.80$ $40$ $y$

What is the value of $x$ and $y$?

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For a chemical reaction $A + B \rightarrow \text{Product}$,the order is $1$ with respect to $A$ and $B$.Rate $(mol \ L^{-1} \ s^{-1})$$[A]$ $(mol \ L^{-1})$$[B]$ $(mol \ L^{-1})$$0.10$$20$$0.5$$0.40$$x$$0.5$$0.80$$40$$y$What is the value of $x$ and $y$?