a)(i) Give two examples of each of specimens A and B. (2 marks)
(ii) State three advantages of the application of specimen Ain crop production. (3 marks)
(iii) Mention two methods of applying specimen B in crop production. (2 marks)
(b) An experiment was conducted to determine the percentage of water in soil samples C and D.
The results are as shown in the table below:
| Soil Sample | |
| Data Recorded | C D |
| Mass of crucible | 80g 80g |
| Mass of crucible and fresh soil | 100g 140g |
| Mass of crucible and soil after heating | 95g 100g |
(i) Determine the mass of fresh soil for each of soil samples C and D (2 marks)
(ii) Determine the mass of water contained in each of soil samples C and D. (2 marks)
(ii) Calculate the percentage of water in each of soil samples C and D. (2 marks)
(c) What type of relationship exists between the amount of water and the amount of air in soils? (2 marks)
a) Examples of Specimen A and B:
Specimen A (Organic fertilizer):
Specimen B(Inorganic fertilizer:
(ii) Advantages of the application of specimen A (Organic fertilizer) in crop production
(iii) Methods of applying Specimen B (Inorganic fertilizer)
(b)(i) Determination of mass of fresh soil samples C and D
Soil Sample C;
Mass of crucible and fresh soil - Mass of crucible = 100g - 80g = 20g
Soil Sample D;
Mass of crucible and fresh soil - Mass of crucible = 140g - 80g = 60g
(ii) Determination of mass of water contained in soil samples C and D
Soil Sample C; Mass of crucible and fresh soil - Mass of crucible and soil after heating = 100g - 95g = 5g
Soil Sample D; Mass of crucible and fresh soil - Mass of crucible and soil after heating = 140g - 100g = 40g
(ii) Calculation of percentage of water in each soil sample (i.e.C and D)
Mass crucible and fresh soil - Mass of crucible = yg
Mass of crucible and soil after heating - Mass of crucible = zg
yg = mass of fresh soil
zg = mass of soil after heating
% water = \(\frac{y-z}{y}\)
For soil sample C
100g - 80g = 20g \(\to/) y
95g - 80g = 15g \(\to/) z
\(\frac{20g - 15g \times 100}{20g}\) = 25%
For soil sample D;
140g - 80g = 60g \(\to\) y
100g - 80g = 20g \(\to\) z
\(\frac{20g - 15g \times 100}{20g}\) = 25%
(c) Relationship that exists between the amount of water and the amount of air in soils: The amount of air in the soil is inversely proportional to the amount of water in the soil pore spaces which are not occupied by water.
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