$20 \%$ of a first order reaction was found to be completed at $10:00 \ am$. At $11:30 \ am$ on the same day,$20 \%$ of the reaction was found to be remaining. The half-life period in minutes of the reaction is

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:

Initial concentration of reactant in a first order reaction is $0.08 \text{ mol dm}^{-3}$. What concentration would remain after $40 \text{ minutes}$? (Given $\frac{[A]_0}{[A]_t} = 5.00$)

The time required for $90\%$ completion of a certain first order reaction is $1 \text{ hour}$. Calculate the time required for $99.9\%$ completion of the same reaction.

For a first order reaction $(A) \rightarrow$ products,the concentration of $A$ changes from $0.1 \ M$ to $0.025 \ M$ in $40 \ min$.
The rate of reaction when the concentration of $A$ is $0.01 \ M$ is ............$ \times 10^{-4} \ M/min$.

The half-life of a first-order reaction is $20 \text{ min}$. What is the time taken to reduce the initial concentration of the reactant to $\frac{1}{10}$th of its original value (in $\text{ min}$)?