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

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?

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Khang Minh 3 years 2021-07-25T15:00:31+00:00 1 Answers 9 views 0

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    2021-07-25T15:01:53+00:00

    Answer:

    95.9 kg

    Explanation:

    First we convert 15.0 mi² to m²:

    • 15.0 mi² * (\frac{1609.34 m}{1mi})² = 3.88×10⁷ 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:

    • 3.88×10⁷ m² * 8.23 m = 3.20×10⁸ m³

    Converting 3.20×10⁸ m³ to L:

    • 3.20×10⁸ m³ * \frac{1000L}{1m^3} = 3.20×10¹¹ L

    Now we calculate the total mass of mercury in the lake, using the given concentration:

    • 0.300 μg / L * 3.20×10¹¹ L = 9.59×10¹⁰ μg

    Finally we convert μg to kg:

    • 9.59×10¹⁰ μg * \frac{1kg}{1x10^9ug} = 95.9 kg

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