If the rate of disappearance of $N_2O_5$ in the following reaction is $1.2 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$,the rate of production of $NO_2$ in $mol \ L^{-1} \ s^{-1}$ is:
$2N_2O_{5(g)} \longrightarrow 4NO_{2(g)} + O_{2(g)}$

If the rate of reaction is given as $\frac{1}{3} \frac{d[X]}{dt} = -\frac{1}{2} \frac{d[Y]}{dt} = -\frac{d[Z]}{dt}$,the reaction can be represented as

For the reaction,$2N_2O_5 \to 4NO_2 + O_2$,the rate of reaction is

For the reaction,$5Br^-{_{\text{(aq)}}} + BrO_3^-{_{\text{(aq)}}} + 6H^+{_{\text{(aq)}}} \rightarrow 3Br_{2\text{(aq)}} + 3H_2O_{\text{(l)}}$,if $-\frac{\Delta[Br^{-}]}{\Delta t} = 0.05 \ mol \ L^{-1} \ min^{-1}$,then the value of $-\frac{\Delta[BrO_3^{-}]}{\Delta t}$ in $mol \ L^{-1} \ min^{-1}$ is:

$NO_2$ required for a reaction is produced by decomposition of $N_2 O_5$ in $CCl_4$ as per the equation $2 \, N_2 O_{5(g)} \rightarrow 4 NO_{2(g)} + O_{2(g)}$. The initial concentration of $N_2 O_5$ is $3 \, mol \, L^{-1}$ and it is $2.75 \, mol \, L^{-1}$ after $30 \, minutes$. The rate of formation of $NO_2$ is $x \times 10^{-3} \, mol \, L^{-1} \, min^{-1}$. The value of $x$ is . . . . . . .

In the following reaction; $x A \longrightarrow y B$
$\log_{10} \left[ -\frac{d[A]}{dt} \right] = \log_{10} \left[ \frac{d[B]}{dt} \right] + 0.3010$
$'A'$ and $'B'$ respectively can be