Answer:
294 grams
Explanation:
The amount of radioactive material left after t hours given that the half-life is to hours is
[tex]A=P(0.5)^{\frac{t}{t_0}}[/tex]Now, in our case t0 = 3, t = 9 and P = 2352 g; therefore, the above equation gives
[tex]A=2352(0.5)^{9/3}[/tex][tex]A=294g[/tex]which is our answer!
Hence, the amount of radioactive copper left after 9 hours is 294 grams.
