Answer:
The additional concentration of nitrogen in the river water due to the farmer's fertilizer is 0.0629 milligrams per liter
Explanation:
The mass of the fertilizer the farmer applies = 1705 kg
The percentage of nitrogen contained in the fertilizer = 10%
Therefore;
The mass of nitrogen contained in the fertilizer = 10% of 1705 kg = 10/100 × 1705 kg
The mass of nitrogen contained in the fertilizer = 10/100 × 1705 kg = 170.5 kg
The percentage of the fertilizer that washes into the river that runs through the farm = 15.0%
The mass of the fertilizer that washes into the river = 15/100 × 170.5 = 25.575 kg = 25575 g = 25575000 mg
The volume of water that flows through the farm = 0.455 cubic feet per second
The volume of water that flows through the farm in a year = 0.455 ft³ × 60 × 60 × 24 × 365 = 14348880 ft³ = 406,315.03 m³ = 406,315,034 l
The number of moles of nitrogen = Mass of nitrogen/(Molar mass of nitrogen)
Molar mass of nitrogen =14.0067 g/mol
Therefore;
The number of moles of nitrogen = 25575 kg/(14.0067 g/mol) = 1.825.912 moles
The additional concentration of the nitrogen in moles = 1.825.912 moles/(406,315,034 l) = 4.49 × 10⁻⁶ mol/liter
The additional concentration of the nitrogen in milligrams of nitrogen per liter of water = 25575000/406,315,034 = 0.0629 milligrams of nitrogen per liter.
Therefore, the additional concentration of nitrogen in the river water due to the farmer's fertilizer = 0.0629 milligrams per liter
