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)
Explanation:
[tex]M_{R} =\frac{M_{O} }{2^{n} }[/tex]
where [tex]M_{R}[/tex] is mass remaining
[tex]M_{O}[/tex] is original mass
[tex]n=\frac{t}{t1/2}[/tex]
t1/2= 16.3s , t=35s
n =35/16.3 =2.147
[tex]M_{R} =\frac{M_{O} }{2^{2.147} }[/tex]
[tex]M_{R}[/tex] = [tex]0.2258M_{O}[/tex]
0.2258*100% =22.5%
≈23%
