If the rate constant is $1.155 \times 10^{-3} \ s^{-1}$,after how many seconds will the concentration of the reactant be reduced to half in a first-order reaction?

The rate constant of a first order reaction is $1.15 \times 10^{-3} \ s^{-1}$. How long will $5 \ g$ of reactant take to reduce to $3 \ g$ (in $s$)?

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

Slope of the graph between $\log \frac{[A]_0}{[A]_t}$ ($y$-axis) and time ($x$-axis) for a first-order reaction is equal to:

For a first order reaction $A_{(g)} \rightarrow 2B_{(g)} + C_{(g)}$ at constant volume and $300 \ K$,the total pressure at the beginning $(t=0)$ and at time $t$ are $P_0$ and $P_t$,respectively. Initially,only $A$ is present with concentration $[A]_0$,and $t_{1/3}$ is the time required for the partial pressure of $A$ to reach $1/3^{rd}$ of its initial value. The correct option$(s)$ is (are) (Assume that all these gases behave as ideal gases)

The rate constant of the reaction $2H_2O_{2(aq)} \to 2H_2O_{(l)} + O_{2(g)}$ is $3 \times 10^{-3} \ min^{-1}$. At what concentration of $H_2O_2$,the rate of reaction will be $2 \times 10^{-4} \ M \ s^{-1}$? ............ $M$