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
The image of the bonds are missing, so i have attached it.
Answer:
A) - Sigma bond
-Sp³ and Sp³
- None
B) - Sigma and pi bond
- Sp² of C and p of O
- p of C and P of O
Explanation:
A) For compound 1;
- the molecular orbital type is sigma bond due to the end-to-end overlapping.
- Atomic orbitals in the sigma bond will be Sp³ and Sp³
- Atomic orbitals in the pi bond would be nil because there is no pi bond.
B) For compound 2;
- the molecular orbital type is sigma and pi bond
-Atomic orbitals in the sigma bond would be Sp² of C and p of O
- The Atomic orbitals in the pi bond will be; p of C and p of O
Water<span> can </span>dissolve salt<span> because the positive part of </span>water<span>molecules attracts the negative chloride ions and the negative part of </span>water<span> molecules attracts the positive sodium ions.</span>
1. The balanced equation tells us that 2 moles of H2S produce 2 moles of H2O.
8.3 moles H2S x (2 moles H2O / 2 moles H2S) = 8.3 moles H2O = theoretical amount produced
8.3 moles H2O x (18.0 g H2O / 1 mole H2O) = 149 g H2O produced theoretically
% yield = (actual amount produced / theoretical amount) x 100 = (137.1 g / 149 g) x 100 = 91.8 % yield
2. Calculate moles of each reactant.
150.0 g N2 x (1 mole N2 / 28.0 g N2) = 5.36 moles N2
32.1 g H2 x (1 mole H2 / 2.02 g H2) = 15.9 moles H2
The balanced equation tells us that we need 3 moles of H2 to react with every 1 mole of N2.
So if we have 5.36 moles N2, we need 3x that = 16.1 moles H2. Do we have that much available? No, just under at 15.9 moles. So H2 is the limiting reactant. At the end of the reaction there will be a little N2 left over.
Well since i need points I am going to say read the text
