Carbon-11 is used in medical imaging. the half-life of this radioisotope is 20.4 min. what percentage of a sample remains after 60.0 min? 34.013.05.2871.22.94

Percentage of a sample remains after 60.0 min is 13.03%.

Explanation:

∴ k = 0.693/(t1/2) = 0.693/(20.4 min) = 0.03397 min⁻¹.

kt = ln[A₀]/[A],

where, k is the rate constant (k = 0.03397 min⁻¹).

t is the time of the reaction (t = 60.0 min).

[A₀] is the initial concentration of C-11 ([A₀] = 100.0 %).

[A] is the remaining concentration of C-11 ([A] = ???%).

∵ kt = ln[A₀]/[A]

∴ (0.03397 min⁻¹)(60.0 min) = ln(100%)/[A]

∴ 2.038 = ln(100%)/[A]

∴ 7.677 = (100%)/[A]

∴ [A] = (100%)/(7.677) = 13.03%.

So, percentage of a sample remains after 60.0 min is 13.03%.

a polar bond is formed when electrons are shared unevenly between atoms

the right answer is c i just did the question.