Question

Please explain how the percentage cover contributes to the calculations. And show the units when performing the calculations. Thank you :)

Question:

On the outskirts of the Everglades a monitoring project was started in before large-scale farming came to the area. The different farm sites that were set up use different fertilizers leading to different degrees of nutrient inputs into the system. The researchers at the monitoring project recorded the levels of nutrients in the water at each of 3 sites as well as the % cover of cattails at each site. (16 pts)

Nutrient Levels	Carbon mg/L	Nitrogen mg/L	Phosphorus mg/L	% Cover
Before Farming	500	40	20	30
Farming Site a	800	200	30	32
Farming Site b	600	50	100	60
farming site c	400	300	60	40

1. What is the limiting nutrient for the growth of cattails before farming? Show your work. (Hint: Redfield Ratio)

2. What is the limiting nutrient for the growth of cattails at farming site A? Show your work. (Hint: Redfield Ratio)

3. What is the limiting nutrient for the growth of cattails at farming site B? Show your work. (Hint: Redfield Ratio)

4. What is the limiting nutrient for the growth of cattails at farming site C? Show your work. (Hint: Redfield Ratio)

Answer 1

C) :

• Limiting nutrient at Site B

• we need to determine the no of moles first

• phosphorus = 100/31 =3.2258

  Nitrogen = 50/14 = 3.5714  

  Carbon = 600/12 = 50

• on Re arranging the Ratio or making the No of Moles of phosphorous as 1 . we need to divide throughout by 3.2258

• so C : N : P = 15.500 : 1.108 : 1

• Here only 1.108 no of moles of Nitrogen present , so the no of moles of carbon for 1.108 moles of nitrogen is (106/16) x 1.108 = 7.3458

• but the Amount of carbon in the site B is 15.500

• so In the site B Limiting nutrient is Nitrogen

D) :

• Limiting nutrient at Site C

• we need to determine the no of moles first

• phosphorus = 60/31 = 1.9354

  Nitrogen = 300/14 = 21.4285

  Carbon = 400/12 = 33.333

• on Re arranging the Ratio or making the No of Moles of phosphorous as 1 . we need to divide throughout by 1.9354

• so C : N : P = 17 : 11.0718 : 1

• Here only 17 moles of Carbon present , so the no of moles of carbon for 17 moles of Carbon present n is (106/16) x 11.0718 = 73.350675

• but the Amount of carbon in the site C is 17

• so In the site C Limiting nutrient is Carbon

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