Question

If a substance has a half-life of 55.6 s, and if 230.0 g of the substance are present initially, how many grams will remain after 10.0 minutes?

Answer 1

`m=0.127g`

Step-by-step explanation:

Hello,

In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:

`k=(ln(2))/(t_(1/2)) =(ln(2))/(55.6s)=0.0125s^(-1)`

In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:

`m=m_0*exp(-kt)=230.0g*exp(-0.0125s^(-1)*600s)\\\\m=0.127g`

Best regards.