$A$ reaction which is of first order with respect to reactant $A$,has a rate constant $6 \, min^{-1}$. If we start with $[A] = 0.5 \, mol \, L^{-1}$,when would $[A]$ reach the value of $0.05 \, mol \, L^{-1}$?

The half-life period of a first-order chemical reaction is $6.93 \, \text{min}$. The time required for the completion of $99 \%$ of the chemical reaction will be ........ $\text{min}$. $(\log 2 = 0.301)$

$A$ first-order reaction decomposes such that the time taken for $1/8$ and $1/10$ of the initial concentration to decompose are $t_{1/8}$ and $t_{1/10}$ respectively. Find the value of $\frac{t_{1/8}}{t_{1/10}} \times 10$. (Given: $\log_{10} 2 = 0.3$)

The half-life of a first-order reaction is $850 \text{ s}$. The initial concentration of the reactant is $[R]_0 = 0.06 \text{ mol dm}^{-3}$. What concentration $[R]$ would remain after $t = 1200 \text{ s}$?

If the decomposition reaction $A_{(g)} \longrightarrow B_{(g)}$ follows first-order kinetics,then the graph of the rate of formation of $B$,denoted by $R$,against time $t$ will be:

Decomposition of nitrogen pentoxide is known to be a first order reaction. $75\%$ of the oxide had decomposed in the first $24 \ minutes$. At the end of an hour,after the start of the reaction,the amount of oxide left will be