How many half-lives have passed if the amount of parent isotope is 12.5%?

Three half lives.

Explanation:

Given data:

Number of half lives passed = ?

Amount of parent isotope left = 12.5%

Solution:

At time zero = 100%

At 1st half life = 100%/2 = 50%

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

At third half life = 25%/2 = 12.5%

It means three half lives would passed.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2