If carbon-15 has a half-life of 2.5 seconds, % of the sample would still be
carbon-15 after 5.0 seconds. (assume you are starting with 100% pure carbon-
15)
(2 points)

After 5 second 25% C-15 will remain.

Explanation:

Given data:

Half life of C-15 = 2.5 sec

Original amount = 100%

Sample remain after 5 sec = ?

Solution:

Number of half lives = T elapsed / half life

Number of half lives = 5 sec / 2.5 sec

Number of half lives = 2

At time zero = 100%

At first half life = 100%/2 = 50%

At second half life = 50%/2 = 25%

Thus after 5 second 25% C-15 will remain.

icc, international association for cereal chemistry