1. **Problem statement:** We are given two wastes with ultimate BOD ($BOD_u$) of 280.0 mg/L and rate constants $k_1=0.0800$ d$^{-1}$ and $k_2=0.120$ d$^{-1}$. We need to find the 7-day BOD ($BOD_7$) for each at 15°C. 2. **Formula used:** The BOD exerted at time $t$ is given by $$BOD_t = BOD_u \left(1 - e^{-kt}\right)$$ where $BOD_u$ is the ultimate BOD, $k$ is the rate constant, and $t$ is time in days. 3. **Calculate $BOD_7$ for waste 1:** $$BOD_7 = 280.0 \times \left(1 - e^{-0.0800 \times 7}\right)$$ Calculate the exponent: $$-0.0800 \times 7 = -0.56$$ Calculate $e^{-0.56}$: $$e^{-0.56} \approx 0.5712$$ So, $$BOD_7 = 280.0 \times (1 - 0.5712) = 280.0 \times 0.4288 = 120.06$$ 4. **Calculate $BOD_7$ for waste 2:** $$BOD_7 = 280.0 \times \left(1 - e^{-0.120 \times 7}\right)$$ Calculate the exponent: $$-0.120 \times 7 = -0.84$$ Calculate $e^{-0.84}$: $$e^{-0.84} \approx 0.4317$$ So, $$BOD_7 = 280.0 \times (1 - 0.4317) = 280.0 \times 0.5683 = 159.12$$ 5. **Final answers:** - 7-day BOD for waste 1: **120.06 mg/L** - 7-day BOD for waste 2: **159.12 mg/L** These values represent the BOD exerted in 7 days at 15°C for each waste given their respective rate constants.