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aleksandrvk [35]
3 years ago
14

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?
Chemistry
1 answer:
Aloiza [94]3 years ago
7 0

Answer:

95.9 kg

Explanation:

First we <u>convert 15.0 mi² to m²</u>:

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

Then we <u>convert 27.0 ft to m</u>:

  • 27.0 ft * \frac{0.3048m}{1ft} = 8.23 m

Now we <u>calculate the total volume of the lake</u>:

  • 3.88x10⁷ m² * 8.23 m = 3.20x10⁸ m³

<u>Converting 3.20x10⁸ m³ to L</u>:

  • 3.20x10⁸ m³ * \frac{1000L}{1m^3} = 3.20x10¹¹ L

Now we <u>calculate the total mass of mercury in the lake</u>, using the <em>given concentration</em>:

  • 0.300 μg / L * 3.20x10¹¹ L = 9.59x10¹⁰ μg

Finally we<u> convert μg to kg</u>:

  • 9.59x10¹⁰ μg * \frac{1kg}{1x10^9ug} = 95.9 kg
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