Answer:
294 grams
Step-by-step explanation:
The amount of radioactive material left after t hours given that the half-life is to hours is
![A=P(0.5)^{(t)/(t_0)}](https://img.qammunity.org/2023/formulas/mathematics/college/f4itg32foiwy4ryfhg0ezijk4utlx540ur.png)
Now, in our case t0 = 3, t = 9 and P = 2352 g; therefore, the above equation gives
![A=2352(0.5)^(9/3)](https://img.qammunity.org/2023/formulas/mathematics/college/knx6yn29hc9muxt00bgrd0tre58q20gmaq.png)
![A=294g](https://img.qammunity.org/2023/formulas/mathematics/college/nhd18u3k1pdrmsd9nnj1aaom2fv9a3rnqr.png)
which is our answer!
Hence, the amount of radioactive copper left after 9 hours is 294 grams.