Half life of a first order reaction is $20 \ min$. The time taken to reduce the initial concentration of reactant to $(1/10)^{th}$ is $\qquad$ (in $min$)

The reaction $N_{2}O_{5} \longrightarrow 2NO_{2} + \frac{1}{2}O_{2}$ is first order in $N_{2}O_{5}$ having rate constant $6.2 \times 10^{-4} \ s^{-1}$. What is the value of rate of reaction when concentration of $N_{2}O_{5}$ is $1.25 \ mol \ L^{-1}$?

For the first-order reaction $N_2O_5 \to 2NO_2 + \frac{1}{2}O_2$,the half-life at $30 \ ^\circ C$ is $24 \ \text{hours}$. Starting with $10 \ g$ of $N_2O_5$,how many grams of $N_2O_5$ will remain after $96 \ \text{hours}$ (in $g$)?

An organic compound undergoes first order decomposition. The time taken for decomposition to $\left(\frac{1}{8}\right)^{\text{th}}$ and $\left(\frac{1}{10}\right)^{\text{th}}$ of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. What is the value of $\frac{t_{1/8}}{t_{1/10}} \times 10$? (Given: $\log 2 = 0.3$)

The initial rate for a first order reaction is $0.6932 \times 10^{-2} \ mol \ L^{-1} \ min^{-1}$ and the initial concentration of the reactant is $0.1 \ M$. Then $t_{1/2}$ is equal to ...... $min$.

For a first order reaction,to obtain a positive slope,we need to plot $...$ where $[A]$ is the concentration of reactant $A$.