Question

a polluted lake is 0.300 μg (micrograms) per liter of water, what is the total mass of mercury in the lake, in kilograms, if the lake has a surface area of 15.0 square miles and an average depth of 27.0 feet?

Answer 1

Answer:

95.9 kg

Explanation:

First we convert 15.0 mi² to m²:

• 15.0 mi² * ($\frac{1609.34 m}{1mi}$)² = 3.88x10⁷ m²

Then we convert 27.0 ft to m:

• 27.0 ft * $\frac{0.3048m}{1ft}$ = 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 * $\frac{1000L}{1m^3}$ = 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 * $\frac{1kg}{1x10^9ug}$ = 95.9 kg