Bromine-88 is radioactive and has a half life of 16.3 seconds. what percentage of a sample would be left after 35.0 seconds? round your answer to 2 significant digits .

Percentage of bromine-88 left after 35.0 s = 23%

Explanation:

Radioactive decay follows first order kinetics.

Given:

Half life of Br-88 = 16.3 s

time = 35.0 s

where,

= Initial concentration of radioactive substance

a = Amount left after time 't'

K = rate constant

Half-life = 0.693/K

now, substitute the value of rate constant and t (35.0 s) in the formula,

Let a0 (initial concentration of bromine-88) be 100.

[a] = 22.6% = 23%

Percentage of bromine-88 left after 35.0 s = 23%

b

explanation:

mitosis

the answer to your question is c