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?
1 answer:
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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