What is the rate constant of a first-order reaction if $60 \%$ of the reactant decomposes in $45 \ minutes$?

$A$ flask contains a mixture of compounds $A$ and $B.$ Both compounds decompose by first-order kinetics. The half-lives for $A$ and $B$ are $300 \ s$ and $180 \ s,$ respectively. If the concentrations of $A$ and $B$ are equal initially,the time required for the concentration of $A$ to be four times that of $B$ (in $s$): (Use $\ln 2 = 0.693$)

In a first order decomposition reaction,the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are $t_1$ and $t_2 \ s$,respectively. The ratio $t_1 / t_2$ will be:

The thermal decomposition of a compound is of first order. If a sample of the compound decomposes $50\%$ in $120 \, minutes$,in what time will it undergo $90\%$ decomposition?

The rate of a first order reaction is $1.5 \times 10^{-2} \ mol \ L^{-1} \ \min^{-1}$ at $0.5 \ M$ concentration of the reactant. The half life of the reaction is ....... $\min$

If the rate constant for a first order reaction is $k$,then find the time required for completion of $80 \%$ of the reaction.