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stich3 [128]
3 years ago
11

If the concentration of mercury in the water of a polluted lake is 0.300 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 13.0 square miles and an average depth of 49.0 feet?
Mathematics
1 answer:
Reika [66]3 years ago
6 0

Answer: 151.305 kg

Step-by-step explanation:

0.300μg = 0.3x10⁻⁶ g/l

mass in kg?

Volume of lake = surface area * depth

surface area = 13mi²

depth = 49 ft

<u>Feet to miles</u>

49 ft → mi = 0.00928 mi

(1 mi = 5280 ft)

V = 13*0.00928 = 0.121 mi³

<u>cubic Miles to cubic meters</u>

1mi³ = 4168182000 m³

0.121 mi³ = 504350022 m³

<u>Cubic meters to liters</u>

1L = 1000 m³

Lake has 5.04350022x10¹¹ L

Concentration = 0.3x10⁻⁶ g/l

Lake has 151305 g of Hg = 151.305 kg of Hg

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