If a reaction has the experimental rate expression $\text{rate} = K [A]^2[B]$,what happens to the reaction rate if the concentration of $A$ is doubled and the concentration of $B$ is halved?

For a reaction,$A + 2B \longrightarrow C$,the rate is given by $+\frac{d[C]}{dt} = k[A][B]$. Hence,the order of the reaction is:

For the reaction: $2A + B \to A_2B$; the rate $= K[A][B]^2$ with $K = 2.0 \times 10^{-6} \ L^2 \ mol^{-2} \ s^{-1}$. Initial concentrations of $A$ and $B$ are $[A]_0 = 0.2 \ mol/L$ and $[B]_0 = 0.4 \ mol/L$ respectively. Calculate the rate of reaction after $[A]$ is reduced to $0.12 \ mol/L$.

The rate of the reaction becomes twice when the concentration of reactant becomes $8$ times,then the order of the reaction is:

The rate of a certain reaction depends on concentration according to the equation $\frac{-dc}{dt} = \frac{K_1 C}{1 + K_2 C}$. What is the order of the reaction when the concentration $(C)$ is very high?

The rate constant for a reaction is $10.8 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$. The order of the reaction is .......