Question

Bromine- is radioactive and has a half life of seconds. What percentage of a sample would be left after seconds? Round your answer to significant digits.

Answer 1

Corrected question:

Bromine-88 is radioactive and has a half life of 16.3seconds. What percentage of a sample would be left after 35seconds? Round your answer to 2 significant digits.

Answer:

23% (3 significant digits)

Step-by-step explanation:

`M_(R) =(M_(O) )/(2^(n) )`

where
`M_(R)` is mass remaining

`M_(O)` is original mass

`n=(t)/(t1/2)`

t1/2= 16.3s , t=35s

n =35/16.3 =2.147

`M_(R) =(M_(O) )/(2^(2.147) )`

`M_(R)` =
`0.2258M_(O)`

0.2258*100% =22.5%

≈23%