Answer:
The total mass of mercury in the lake is 631,542.7 kg
Step-by-step explanation:
Question: The given dimensions of the lake as obtained from a similar question posted online are;
The surface area of the lake, A = 100 mi²
The lake's average depth, d = 20 ft.
The concentration of the mercury, C = 0.4 μg Hg/mL = 0.4 × 10⁻⁶ kg Hg/L
Therefore, we have;
The volume of water mixture in the lake, V = A × d
∴ V = 100 mi² × 20 ft. = 2,787,840,000 ft.² × 20 ft. = 55,756,800,000 ft.³
1 ft³ = 28.31685 L
∴ 55,756,800,000 ft.³ = 55,756,800,000 ft.³ × 28.31685 L/ft.³ = 1.57885675 × 10¹² L
The total mass of mercury in the lake, m = C × V
∴ m = 0.4 × 10⁻⁶ kg Hg/L × 1.57885675 × 10¹² L = 631,542.7 kg
The total mass of mercury in the lake, m = 631,542.7 kg.