Ni(OH)₂ ⇄ Ni⁺² + 2 OH⁻
Ksp = [Ni⁺²][OH⁻]² = S (2S)² = 4S³
where S is molar solubility.
at pH = 10
[H⁺] = 10⁻¹⁰
[H⁺][OH⁻] = 10⁻¹⁴
so [OH⁻] = 10⁻⁴ M
Ksp = S [10⁻⁴ + 2S]²
Ksp is very small so the molar solubility of OH⁻ will be very small
so (10⁻⁴ + 2S) is about 10⁻⁴
so Ksp = S x 10⁻⁸
S =
= 6 x 10⁻⁸ M
Answer:
The correct answer is a) 300 Kg b) 225 Kg c) 400 Kg and d) 600 Kg.
Explanation:
Based on the given value, that is. 40-15-10 shows the compositions of nitrogen, phosphorus, and potassium found in the compound fertilizer in the form of percentage.
41 percent nitrogen is equal to 40/100 = 0.4, 15 percent phosphorus is equal to 15/100 = 0.15, and 10 percent potassium is equal to 10/100 = 0.1.
To find the amount of fertilizer, which is needed per hectare to supply the different nutrients per hectare, there is a need to divide the quantity to be supplied by the percentage composition, now the values comes out as:
a) At 120 Kg per ha, the required nitrogen will be 120/0.4 = 300 Kilogram of fertilizer.
b) At 90 kg per ha, the required nitrogen will be 90/0.4 = 225 Kilograms of fertilizer.
c) At 60 kg per ha, the required phosphorus will be 60/0.15 = 400 Kilograms of fertilizer.
d) At 60 Kg per ha, the required potassium will be 60/0.1 = 600 Kilograms of fertilizer.
Answer:
1. 67.2 kJ/mol
Explanation:
Using the derived expression from Arrhenius Equation
Given that:
time 
= 8.3 days = (8.3 × 24 ) hours = 199.2 hours
time 
= 10.6 hours
Temperature 
= 0° C = (0+273 )K = 273 K
Temperature 
= 30° C = (30+ 273) = 303 K
Rate = 8.314 J / mol
Since 
Then we can rewrite the above expression as:
6 protons, 6 neutrons, and 5 electrons
