1.
1,000 kg of carbonaceous materials with a C:N ratio of 12:1 are applied to a soilcontaining mixed population of microorganisms. The amount of C in carbonaceous materials is 40%, 2 5% of C in th e sub st ra te is assimila t e d b y micro o rg an isms, an d t he C: N rat io of microorganisms is 8:1. a)Calculate the amount of N available for crop use. b)If the crop requires 60 kg N, how much urea must be applied? (Hint: urea contains 46% N) (10 marks) QUESTION 3 a) NPK content of fertilizer A is 15 - 15 - 15 and sells for RM60 per 50 kg bag; whereas NPK content of fertilizer B is 10 - 8 - 8 and costs RM22.50 per 20 kg bag. Which is the best buy on the basis of nitrogen? (7 marks) b) A liter of liquid fertilizer weighs 1.5 kg. NPK analysis of the fertilizer is 20 - 4 -10. How much N, P, and K are in one liter? (3 marks)Amt of C in carbonaceous materials 0 . 4 x 1 0 0 0 kg = 400 kgAmt of N in carbonaceous materials 400 kg - 12 = 33.3 kg C:N ratio 12:1)Amt of C assimilated by microbes 0.25 x 400 kg = 100 kgAmt of N in carbonaceous materials 100 kg ÷ 8 = 12.5 kg (C:N ratio 8:1)Amt of available N for crop use 3.3 kg – 12.5 kg = 20.8 kgAmt of additional N required by crop 60 k g - 2 0 . 8 kg = 39.2 kgAmt of urea to be applied (1 kg urea= 0.46 kg N) x 39.2 kg N 85.2 kg
2.
Fertilizer A Fertilizer B N content 0.15 x 50 kg 0.10 x 20 kg= 7.5 kg = 2 kg Price of N RM 60 ÷ 7.5 kg RM 22.50 ÷ 2 =RM 8 / kg =RM 11.25 / kg Fertilizer A is the best buy. Amt of NPK in liquid fertilizer 0.20 x 1.5 kg = 0.3 kg N 0.04 x 1.5 kg = 0.06 kg P 0.10 x 1.5 kg = 0.15 kg K
3.
SCHEMANO ANSWERS MARKSla Amt of C in carbonaceous materials 1½ 0 . 4 x 1 0 0 0 kg = 400 kg Amt of N in carbonaceous materials 400 kg - 12 = 33.3 kg C:N ratio 12:1) 1½ Amt of C assimilated by microbes 0.25 x 400 kg = 100 kg 1½ Amt of N in carbonaceous materials 100 kg ÷ 8 = 12.5 kg (C:N ratio 8:1) 1½ Amt of available N for crop use 1 3.3 kg – 12.5 kg = 20.8 kgb Amt of additional N required by crop 60 k g - 2 0 . 8 kg = 39.2 kg 1 Amt of urea to be applied (1 kg urea= 0.46 kg N) x 39.2 kg N 85.2 kg 2
4.
Fertilizer A Fertilizer B 3 N content 0.15 x 3 50 kg 0.10 x 1 20 kg 1 = 7.5 kg = 2 kg Price of N RM 60 ÷ 7.5 kg RM3a 22.50 ÷ 2 =RM 8 / kg =RM 11.25 / kg Fertilizer A is the best buy. Amt of NPK in liquid fertilizer 0.20 x 1.5 kg = 0.3 kg N b 0.04 x 1.5 kg = 0.06 kg P 1 0.10 x 1.5 kg = 0.15 kg K 1
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