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Nuetrik [128]
3 years ago
5

If the concentration of mercury in the water of a polluted lake is 0.250 μg (micrograms) per liter of water, what is the total m

ass of mercury in the lake, in kilograms, if the lake has a surface area of 18.5 square miles and an average depth of 39.0 feet?
Chemistry
1 answer:
aleksandr82 [10.1K]3 years ago
8 0

Answer:

142.36 kg

Explanation:

volume of water in the lake = surface area x depth

= 18.5 x ( 1760 x 3 )² x 39 ft³

= 2.011 x 10¹⁰ ft³

= 2.011 x 10¹⁰ x 28.3168 liter .

= 56.945 x 10¹⁰ liter .

concentration of mercury = .25 x 10⁻⁶ g / liter

= 25 x 10⁻⁸ g / liter

= 25 x 10⁻¹¹ kg / liter

mass of mercury in the water of lake

= 25 x 10⁻¹¹ x 56.945 x 10¹⁰ kg

= 142.36 kg .

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Explanation:

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The given balanced equilibrium reaction is,

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At eqm. conc.     (0.14-x) M        (x) M        (x) M

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7 0
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Explanation:

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