Calculate the half-life of a first-order reaction if the rate constant of the reaction is $2.772 \times 10^{-3} \ s^{-1}$. (in $s$)

Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with $t_{1/2} = 3 \text{ hours}$. The percentage of sucrose remaining after $6 \text{ hours}$ is . . . . . . . (Nearest integer) (Given: $\log 2 = 0.3010$ and $\log 3 = 0.4771$)

$A$ $20 \%$ first-order reaction is completed in $32 \ \text{min}$. How much time (in $\text{min}$) will it take for $60 \%$ of the reaction to be completed (in $.00$)?

For a first order reaction $A \to B$,the reaction rate at reactant concentration of $0.01 \ M$ is found to be $2.0 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$. The half-life period of the reaction is .......... $sec$.

In a first order reaction $A \to B,$ if $k$ is the rate constant and the initial concentration of the reactant $A$ is $0.5 \ M,$ then the half-life is:

Thermal decomposition of $HCOOH$ is a first order reaction and the rate constant at $T(K)$ is $4.606 \times 10^{-3} \ s^{-1}$. The time required to decompose $90 \%$ of initial quantity of $HCOOH$ at $T(K)$ in seconds is