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

For the $1^{st}$ order reaction,the half-life is $5 \ minutes$ when $[A] = 0.1 \ M$. If the concentration of $[A]$ becomes twice,then the half-life becomes:

Calculate the rate constant of a first-order reaction if the concentration of the reactant decreases by $90 \%$ in $30 \ minutes$.

$A$ 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.

What is the time required for $75 \%$ completion of a first order reaction if the rate constant is $23.03 \text{ minute}^{-1}$ (in $\text{ s}$)?

What is the numerical value of the rate constant of a first-order reaction that is $20 \%$ completed in $10 \ min$ (in $min^{-1}$)?