Answer:
95.9 kg
Explanation:
First we convert 15.0 mi² to m²:
- 15.0 mi² * ([tex]\frac{1609.34 m}{1mi}[/tex])² = 3.88x10⁷ m²
Then we convert 27.0 ft to m:
- 27.0 ft * [tex]\frac{0.3048m}{1ft}[/tex] = 8.23 m
Now we calculate the total volume of the lake:
- 3.88x10⁷ m² * 8.23 m = 3.20x10⁸ m³
Converting 3.20x10⁸ m³ to L:
- 3.20x10⁸ m³ * [tex]\frac{1000L}{1m^3}[/tex] = 3.20x10¹¹ L
Now we calculate the total mass of mercury in the lake, using the given concentration:
- 0.300 μg / L * 3.20x10¹¹ L = 9.59x10¹⁰ μg
Finally we convert μg to kg:
- 9.59x10¹⁰ μg * [tex]\frac{1kg}{1x10^9ug}[/tex] = 95.9 kg