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

For conversion of compound $A \rightarrow B$, the rate constant of the reaction was found to be $4.6 \times 10^{-5}\,L\, mol ^{-1}\, s ^{-1}$. The order of the reaction is $..........$ 

The half-life of $2 $ sample are $0.1 $ and $ 0.4 $ seconds. Their respective concentration are $200 $ and  $ 50 $ respectively. What is the order of the reaction

For a chemical reaction,$ A + 2B \to C + D,$ the rate of reaction increases three times, when concentration of $A$ only is increased nine times. While when concentration of $B$ only is increased $2\,times,$ then rate of reaction also increases $2\,times$. The order of this reaction is

If reaction between $A$ and $B$ to give $C$shows first order kinetics in $A$ and second order in $B$, the rate equation can be written as

The rate constant k, for the reaction ${N_2}{O_5}(g) \to $ $2N{O_2}(g) + \frac{1}{2}{0_2}(g)$ is $2.3 \times {10^{ - 2}}\,{s^{ - 1}}$. Which equation given below describes the change of $[{N_2}{O_5}]$ with time? ${[{N_2}{O_5}]_0}$ and ${[{N_2}{O_5}]_t}$ correspond to concentration of ${N_2}{O_5}$ initially and at time $t$.