Question

A farmer applies 1455 kg of a fertilizer that contains 10.0% nitrogen to his fields each year. Fifteen percent (15.0%) of the fertilizer washes into a river that runs through the farm.If the river flows at an average rate of 0.175 cubic feet per second, what is the additional concentration of nitrogen (expressed in milligrams of nitrogen per liter) in the river water due to the farmer's fertilizer each year?

Answer 1

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 = $0.175 \dfrac{ft^3}{sec}\times 3.155 \times 10^7 \ sec$

since 1 ft³ = 28.3168 L

the amount of river water in a year = $0.175(28.3168)\times 3.155 \times 10^7 \ L$

the amount of river water in a year = 156344132

The concentration of nitrogen molecule= mass/volume

= $\dfrac{ 21.825 \times 10^6 \ mg}{ 156344132 \ \ L }$

= 0.1396 mg/L

Thus, the additional concentration of nitrogen in the water = 0.1396 mg/L