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 mass 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?

Answer: 151.305 kg Step-by-step explanation:0.300μg = 0.3x10⁻⁶ g/lmass in kg?Volume of lake = surface area * depthsurface area = 13mi²depth = 49 ftFeet to miles49 ft → mi = 0.00928 mi(1 mi = 5280 ft)V = 13*0.00928 = 0.121 mi³cubic Miles to cubic meters1mi³ = 4168182000 m³0.121 mi³ = 504350022 m³Cubic meters to liters1L = 1000 m³Lake has 5.04350022x10¹¹ LConcentration = 0.3x10⁻⁶ g/lLake has 151305 g of Hg = 151.305 kg of Hg