Sucrose decomposes by first order kinetics with t1/2 = 3.00 hours. What fraction of the sucrose sample remains after 6 hours?

1. For first order, half-life is constant at 3.00 hours. 2. In 6 hours there are $6/3 = 2$ half-lives. 3. Fraction remaining $= \left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$. 4. Distractor A is one half-life; distractor D is three half-lives. _Source: syllabus Chemistry Ch 3 "Chemical Kinetics", p.26_