$A$ first order reaction is $10 \%$ completed in $20 \, \text{minutes}$. What will be the time for $19 \%$ completion of the reaction? (in $\text{min}$)

The rate constant of a reaction is $0.69 \times 10^{-1} \ min^{-1}$ and the initial concentration is $0.2 \ mol \ L^{-1}.$ The half-life period is ........ $\sec$

The half-life of a first order reaction having rate constant $K = 1.7 \times 10^{-5} \ s^{-1}$ is ........ $hr$.

For two first-order reactions:
$A \rightarrow \text{Products}$
$B \rightarrow \text{Products}$
If at the same time,$50\% \, B$ has reacted and $94\% \, A$ has reacted,calculate the ratio $K_1/K_2$.

$A$ and $B$ decompose via first order kinetics with half-lives $54.0 \, min$ and $18.0 \, min$ respectively. Starting from an equimolar non-reactive mixture of $A$ and $B$,the time taken for the concentration of $A$ to become $16$ times that of $B$ is ...... $min.$ (Round off to the Nearest Integer).

$A$ first order reaction has the rate constant,$k = 4.6 \times 10^{-3} \ s^{-1}$. The number of correct statement/s from the following is/are
Given : $\log 3 = 0.48$
$A.$ Reaction completes in $1000 \ s$.
$B.$ The reaction has a half-life of $500 \ s$.
$C.$ The time required for $10 \ \%$ completion is $25$ times the time required for $90 \ \%$ completion.
$D.$ The degree of dissociation is equal to $(1 - e^{-kt})$.
$E.$ The rate and the rate constant have the same unit.