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

95.9 kg

Step-by-step explanation:

First we convert 15.0 mi² to m²:

• 15.0 mi² * (
  `(1609.34 m)/(1mi)`)² = 3.88x10⁷ m²

Then we convert 27.0 ft to m:

• 27.0 ft *
  `(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 *
  `(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 *
  `(1kg)/(1x10^9ug)` = 95.9 kg