After how many seconds will the concentration of the reactants in a first order reaction be halved,if the decay constant is $1.155 \times 10^{-3} \, \sec^{-1}$?

If the rate constant of a first order reaction is $4.606 \times 10^{-3} \ s^{-1}$,then find the time required for $400 \ g$ of the reactant to reduce to $50 \ g$. (in $min$)

Consider the following first order gas phase reaction at constant temperature:
$A_{(g)} \rightarrow 2B_{(g)} + C_{(g)}$
If the total pressure of the gases is found to be $200 \ torr$ after $23 \ s$ and $300 \ torr$ upon the complete decomposition of $A$ after a very long time,then the rate constant of the given reaction is . . . . . . $\times 10^{-2} \ s^{-1}$ (nearest integer).
[Given: $\log_{10}(2) = 0.301$]

For a first-order reaction,if the value of $a/(a - x)$ is $8$ after $10 \text{ minutes}$,what will be the rate constant $K$?

Azo-isopropane decomposes according to the following equation:
$((CH_3)_2CHN = NCH(CH_3)_2)_{(g)} \xrightarrow{250 - 290 ^oC} (N_2)_{(g)} + (C_6H_{14})_{(g)}$
It is a first-order reaction. If the initial pressure is $P_o$ and the total pressure of the mixture at time $t$ is $P_t$,find the rate constant $K$.

$A$ 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: $\log_{10} 2 = 0.3$)