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

Question

3 years ago

11

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:

Reika [66] · Answer 1 · 2022-04-26T02:28:54+03:00

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

Feet to miles

49 ft → mi = 0.00928 mi

(1 mi = 5280 ft)

V = 13*0.00928 = 0.121 mi³

cubic Miles to cubic meters

1mi³ = 4168182000 m³

0.121 mi³ = 504350022 m³

Cubic meters to liters

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