What is the answer for question 9)?

http://postimg.org/image/5gm0wsol3/

### 1 Answer | Add Yours

We are given that at time t=0 there is 16.32g of oxygen-19 and at time t=600sec there is .964g of oxygen-19. We are asked to find the half-life of oxygen-19:

Let n be the half-life. Then `16.32(1/2)^(600/n)=.964`

** Note that at time n the formula gives 1/2 of 16.32 as required.**

`16.32(1/2)^(600/n)=.964`

`(1/2)^(600/n)=.964/16.32`

`ln((1/2)^(600/n))=ln(.962/16.32)`

`600/n ln(1/2)=ln(.962/16.32)`

`600/n=(ln(.962/16.32))/(ln(1/2))`

`n/600=ln(1/2)/ln(.962/16.32)`

`n=600(ln(1/2)/ln(.962/16.32))`

`n~~147"sec"`

The half-life is approximately 147 seconds.

The graph:

(The red line is when there is 8.16g of oxygen-19; about 147 seconds)

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes