Hydrogen bonding and dipole-dipole forces.
The correct answer is A.) dilute
Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.
Solution : Given,
Mass of 
= 100 g
Molar mass of = 27 g/mole
Molar mass of 
= 28 g/mole
First we have to calculate moles of .
The given balanced chemical reaction is,
From the given reaction, we conclude that
2 moles of produced from 1 mole of
3.7 moles of produced from 
of
Now we have to calculate the mass of .
Mass of = Moles of × Molar mass of
Mass of = 1.85 mole × 28 g/mole = 51.8 g
Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.
It can possible be you're arteries or also you're intestines with is large and small.
Answer:
a. Ksp = 4s³
b. 5.53 × 10⁴ mol³/dm⁹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
AB₂ dissociates to give
AB₂ ⇄ A²⁺ + 2B⁻
Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as
AB₂ ⇄ A²⁺ + 2B⁻
1 : 1 : 2
Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s
So, we have
AB₂ ⇄ A²⁺ + 2B⁻
[s] [s] [2s]
So, the solubility product Ksp = [A²⁺][B⁻]²
= (s)(2s)²
= s(4s²)
= 4s³
b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³
Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³
Substituting the value of s into the equation, we have
Ksp = 4s³
= 4(2.4 × 10³ mol/dm³)³
= 4(13.824 × 10³ mol³/dm⁹)
= 55.296 × 10³ mol³/dm⁹
= 5.5296 × 10⁴ mol³/dm⁹
≅ 5.53 × 10⁴ mol³/dm⁹
Ksp = 5.53 × 10⁴ mol³/dm⁹
