The problem asks us to determine the rate of pollutant released in pounds per day, given a concentration of 10 mg/L and a discharge of 0.3 million gallons per day (MGD).

Step-by-step solution:

1. Convert the discharge from MGD to liters per day: $1 \text{ gallon} = 3.78541 \text{ liters}$ So, $0.3 \text{ MGD}$ (million gallons per day) is: $0.3 \times 10^6 \text{ gallons/day} \times 3.78541 \text{ liters/gallon} = 1,135,623 \text{ liters/day}$

2. Determine the mass of the pollutant in milligrams per day: Since the concentration is $10 \text{ mg/L}$, the mass of pollutant released per day is: $10 \text{ mg/L} \times 1,135,623 \text{ L/day} = 11,356,230 \text{ mg/day}$

3. Convert milligrams to pounds: To convert milligrams to pounds, use the conversion factor: $1 \text{ pound} = 453,592.37 \text{ mg}$ So: $\frac{11,356,230 \text{ mg/day}}{453,592.37 \text{ mg/pound}} \approx 25.04 \text{ pounds/day}$

Final Answer:

The rate of pollutant released is approximately 25.04 pounds per day.

Would you like any more details or explanations on this? Here are some related questions:

1. How do you convert different units of volume, such as gallons to liters?
2. Why is it important to use correct conversion factors in environmental calculations?
3. How can this method be adapted for different pollutants or discharge rates?
4. What factors might affect the concentration of pollutants in sewage treatment plants?
5. How do regulatory agencies use calculations like this to monitor pollution levels?

Tip: Always double-check unit conversions to ensure consistency throughout environmental calculations.