linda2016
Answered

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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?

Sagot :

maacastrobr

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
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