$A$ reaction was found to be $2nd$ order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled,with everything else kept the same,the rate of reaction will

The following graph shows the relationship between $(a-x)^{-1}$ and time $t$ for a second-order reaction. If $\theta = \tan^{-1}(1/2)$ and $OA = 2 \ L \ mol^{-1}$,then the rate at the start of the reaction will be (in $mol \ L^{-1} \ min^{-1}$):

When the concentration of a reactant,$A$,in a reaction $A \to \text{Products}$ is doubled,the rate of reaction increases seven times. The order of the reaction is between:

For the reaction $A + 2B \rightarrow \text{Product } (P)$,the rate law is given by $\frac{d[P]}{dt} = K[A]^2[B]$. If $[A]$ is taken in large excess,what will be the order of the reaction?

For the reaction $2A + 2B \rightarrow 2C + D$,the rate law is expressed as $\text{rate} = k[A]^2[B]$. Calculate the rate constant if the rate of reaction is $0.24 \ mol \ dm^{-3} \ s^{-1}$ where $[A] = 0.5 \ M$ and $[B] = 0.2 \ M$.

For the reaction $A + B \rightarrow C$,select the appropriate rate law based on the following data:
$1$. $[A] = 0.012, [B] = 0.035 \rightarrow \text{Initial Rate} = 0.10$
$2$. $[A] = 0.024, [B] = 0.070 \rightarrow \text{Initial Rate} = 1.6$
$3$. $[A] = 0.024, [B] = 0.035 \rightarrow \text{Initial Rate} = 0.20$
$4$. $[A] = 0.012, [B] = 0.070 \rightarrow \text{Initial Rate} = 0.80$