Question

f a substance has a half-life of 8.10 hr, how many hours will it take for 75.0 g of the substance to be depleted to 3.90 g?

Answer 1

35 hrs

Step-by-step explanation:

half life of the substance
`t_(1/2 )` = 8.1 hr

initial amount
`N_(0)` = 75 g

The final amount
`N` = 3.9 g

The time elapsed
`t` = ?

we use the relationship

`N` =
`N_(0)`
`((1)/(2) )^{(t)/(t_(1/2) ) }`

substituting values, we have

3.9 = 75 x
`(1)/(2)^{(t)/(8.1) }`

0.052 =
`(1)/(2)^{(t)/(8.1) }`

take the log of both side

log 0.052 = log
`(1)/(2)^{(t)/(8.1) }`

log 0.052 =
`(t)/(8.1)` log 1/2

-1.284 =
`(t)/(8.1)` x -0.301

1.284 = 0.301t/8.1 =

1.284 = 0.0372t

t = 1.284/0.037 = 34.5 ≅ 35 hrs