Answer:
the additional concentration of nitrogen = 0.04748 mg/L
Explanation:
From the given information:
The mass of the nitrogen added to the field per year = 10% of 1300.0 kg
= [tex]\dfrac{10}{100} \times 1300[/tex]
= 130 kg
The mass of the nitrogen that is being also washed into the river
= 15% of 130 kg
= [tex]\dfrac{15}{100} \times 130[/tex]
= 15 × 1.3 kg
= 19.5 kg /year
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Converting kg/year into milligram per seconds; we have:
[tex]\dfrac{19.5 \times 10^6 \ mg}{365 \times 3600\times24 \ seconds }[/tex]
= 0.61834 mg/seconds
If the river flows at an average rate of 0.460 cubic feet
Let's convert the cubic feet  into liters
we all know that;
1 ft = 0.33 meter
1 cubic feet = 28.31 Â liters
∴  0.460 cubic feet =  0.460 × 28.31
13.0226 liters
Finally, the additional concentration of nitrogen = [tex]\dfrac{0.61834}{13.0226 }[/tex]
the additional concentration of nitrogen = 0.04748 mg/L
