For the reaction,$CH_3Br_{(aq)} + OH_{(aq)}^{-} \rightarrow CH_3OH_{(aq)} + Br_{(aq)}^{-}$,the rate law is $\text{rate} = k[CH_3Br][OH^{-}]$. What is the change in the rate of reaction if the concentration of both reactants is doubled?

Consider the reaction:
$Cl_{2(aq)} + H_2S_{(aq)} \rightarrow S_{(s)} + 2H^{+}_{(aq)} + 2Cl^{-}_{(aq)}$
The rate equation for this reaction is:
$\text{rate} = k[Cl_2][H_2S]$
Which of these mechanisms is/are consistent with this rate equation?
$A.$ $Cl_2 + H_2S \rightarrow H^{+} + Cl^{-} + Cl^{+} + HS^{-}$ (slow)
$Cl^{+} + HS^{-} \rightarrow H^{+} + Cl^{-} + S$ (fast)
$B.$ $H_2S \rightleftharpoons H^{+} + HS^{-}$ (fast equilibrium)
$Cl_2 + HS^{-} \rightarrow 2Cl^{-} + H^{+} + S$ (slow)

........ of a reaction cannot be determined experimentally.

For a hypothetical reaction $A + B + C \rightarrow \text{Product}$,the rate is given by $r = -\frac{d[A]}{dt} = K[A]^{1/2}[B]^{1/3}[C]^{1/4}$. The order of the reaction is:

Half-life periods for a reaction at initial concentrations of $0.1 \ M$ and $0.01 \ M$ are $5$ and $50$ minutes,respectively. The order of reaction is

For the reaction $2N_2O_5 \to 4NO_2 + O_2$,the rate of reaction and rate constant are $1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$ and $3.4 \times 10^{-5} \ s^{-1}$ respectively. The concentration of $N_2O_5$ at that time will be: