Question

When radon decays, it becomes polonium and yields a first-order rate law. If the rate constant is 0.175 days^-1

Answer 1

This problem is providing us with the rate constant and initial mass for radon's decay. Thus, several questions regarding how much it is left over after specific times are proposed and solved along this.

Radioactive decay

In chemistry, radioactive decay is a process whereby an unstable atomic nucleus loses energy by radiation, and therefore its mass decreases as the time goes by.

In such a way, one can use the following equation for the required questions:

$A=Ao*exp(-t*k)$

Where A is how much is left over, Ao the initial amount, t the elapsed time and k the rate constant. Thus, we can proceed as follows:

(a) we just plug in the 2.96 days:

$A=8.64g*exp(-2.96day*0.175day^{-1})\\\\A=5.15g$

(b) we need to convert 46.4 hours to days:

$A=8.64g*exp(-46.4hr*\frac{1day}{24hr} *0.175day^{-1})\\\\A=6.16g$

(c) we assume this is a first-order reaction so we use its definition of half-life as:

$t_{1/2}=\frac{ln(2)}{k} =\frac{ln(2)}{0.175day^{-1}}=3.96day$

(d) if five half-lives have taken place, the radon which is left over turns out to be:

$A=8.64g*(\frac{1}{2} )^5\\\\A=0.270g$

Learn more about radioactive decay: brainly.com/question/1439925