You have 88g of a radioisotope. if the half-life of this radioisotope is 4 days, how many grams will remain after 12 days?

11.0 g.

Explanation:

kt = ln [A₀]/[A]

k is the rate constant = 0.1733 day⁻¹.

t is the time = 12.0 days.

[A₀] is the initial percentage of radioisotope = 88.0 g.

[A] is the remaining percentage of radioisotope = ??? g.

∵ kt = ln [Ao]/[A]

∴ (0.1733 day⁻¹)(12.0 days) = ln (88.0 g)/[A]

2.079 = ln (88.0 g)/[A]

7.996 = (88.0 g)/[A]

∴ [A] = (88.0 g)/(7.996) = 11.0 g.

most likely he is cheating on you

pie chart because it is in percents ( % )