what percentage of reactant will remain after 120 seconds if the half - life is 60 seconds for a first…

Explanation:

Step1: Calculate the number of half - lives

The time elapsed $t = 120$ s and the half - life $t_{1/2}=60$ s. The number of half - lives $n=\frac{t}{t_{1/2}}$. So, $n=\frac{120}{60}=2$.

Step2: Use the half - life formula for first - order reactions

The fraction of reactant remaining $N/N_0=\left(\frac{1}{2}\right)^n$, where $N$ is the amount of reactant at time $t$ and $N_0$ is the initial amount of reactant. Substituting $n = 2$ into the formula, we get $\frac{N}{N_0}=\left(\frac{1}{2}\right)^2=\frac{1}{4}$.

Step3: Convert the fraction to a percentage

To convert the fraction to a percentage, we multiply by 100. So the percentage of reactant remaining is $\frac{N}{N_0}\times100=\frac{1}{4}\times100 = 25%$.

Answer:

25%