Answer:
a. 300 kg of Fertilizer
b. 225 kg of fertilizer
c.400 Kg of fertilizer
d.600 Kg of fertilizer
Explanation:
The percentage composition ratio of Nitrogen, Phosphorus and Potassium bag of the given fertilizer is 40:15:10.
The percentages can be expressed as fractions as follows:
For nitrogen; 40/100 = 0.4
For phosphorus; 15/100 = 0.15
For potassium; 10/100 = 0.1
To find the quantity of fertilizer required to add to a hectare to supply the given amount of nutrients, the amount to be provided is divided by the percentage or fractional compostion of each nutrient.
Quantity of fertilizer required to add to a hectare to supply;
a. Nitrogen at 120 kg/ha = 120/0.4 = 300 Kg of fertilizer
b.. Nitrogen at 90 Kg/ha = 90/0.4 = 225 Kg of fertilizer
c. Phosphorus at 60 kg/ha = 60/0.15 = 400 Kg of fertilizer
d. Potassium at 60 kg/ha = 60/0.1 = 600 Kg of fertilizer
Given:
K = 0.71 = Kp
The reaction of sulphur with oxygen is
S(s) + O2(g) ---> SO2(g)
initial Pressure 6.90 0
Change -x +x
Equilibrium 6.90-x x
Kp = pSO2 / pO2 = 0.71 = x / (6.90-x)
4.899 - 0.71x = x
4.899 = 1.71x
x = 2.86 atm = pressure of SO2 formed
temperature = 950 C = 950 + 273.15 K = 1223.15 K
Volume = 50 L
Let us calculate moles of SO2 formed using ideal gas equation as
PV = nRT
R = gas constant = 0.0821 L atm / mol K
putting other values
n = PV / RT = 2.86 X 50 / 1223.15 X 0.0821 = 1.42 moles
Moles of Sulphur required = 1.42 moles
Mass of sulphur required or consumed = moles X atomic mass of sulphur
mass of S = 1.42 X 32 = 45.57 grams or 0.04557 Kg of sulphur
The average atomic mass of the element is the sum of the products of the percentage abundance of isotope and its mass number. Therefore, for atomic mass equal to 58.933, the most abundant isotope is cobalt-59. Thus, the answer is letter C.
