(C) For the reaction $2 A + B \longrightarrow C$,the rate of reaction is given by:Rate $= -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{d[C]

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(C) For the reaction $2 A + B \longrightarrow C$,the rate of reaction is given by:Rate $= -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{d[C]}{dt}$Given that the rate of formation of $C$ is $\frac{d[C]}{dt} = 2.2 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$.Equating the terms for $A$ and $C$:$-\frac{1}{2} \frac{d[A]}{dt} = \frac{d[C]}{dt}$$-\frac{d[A]}{dt} = 2 \times \frac{d[C]}{dt}$$-\frac{d[A]}{dt} = 2 \times (2.2 \times 10^{-3}) = 4.4 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$.

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Observe the following reaction: $2 A + B \longrightarrow C$. The rate of formation of $C$ is $2.2 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$. What is the value of $-\frac{d[A]}{dt}$ (in $mol \ L^{-1} \ min^{-1}$)?

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A
$2.2 \times 10^{-3}$
B
$1.1 \times 10^{-3}$
C
$4.4 \times 10^{-3}$
D
$5.5 \times 10^{-3}$

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Observe the following reaction: $2 A + B \longrightarrow C$. The rate of formation of $C$ is $2.2 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$. What is the value of $-\frac{d[A]}{dt}$ (in $mol \ L^{-1} \ min^{-1}$)?

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$A$ mass $M$ attached to a horizontal spring executes $S.H.M.$ of amplitude $A_1$. When the mass $M$ passes through its mean position,a smaller mass $m$ is placed over it and both of them move together with amplitude $A_2$. The ratio $\left(\frac{A_1}{A_2}\right)$ is

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