Answer:
the additional concentration of nitrogen in the water = 0.1396 mg/L
Explanation:
Given that a farmer uses 1455 kg of fertilizer → 10% nitrogen
15% of the fertilizer is said to be washed into the river,
i.e.
(15/100) × 1455 = 218.25 kg
The amount of nitrogen molecule in the washed fertilizer is :
10% of 218.25 kg
= (10/100) × 218.25 kg
= 0.1 × 218.25 kg
= 21.825 kg in a year.
The river flows at an avg. rate of 0.175 ft³/sec
Then the amount of river water in a year = [tex]0.175 \dfrac{ft^3}{sec}\times 3.155 \times 10^7 \ sec[/tex]
since 1 ft³ = 28.3168 L
the amount of river water in a year = [tex]0.175(28.3168)\times 3.155 \times 10^7 \ L[/tex]
the amount of river water in a year = 156344132
The concentration of nitrogen molecule= mass/volume
= [tex]\dfrac{ 21.825 \times 10^6 \ mg}{ 156344132 \ \ L }[/tex]
= 0.1396 mg/L
Thus, the additional concentration of nitrogen in the water = 0.1396 mg/L
