PbSO₄ partially dissociates in water. the balanced equation is;
PbSO₄(s) ⇄ Pb²⁺(aq) + SO₄²⁻(aq)
Initial - -
Change -X +X +X
Equilibrium X X
Ksp = [Pb²⁺(aq)] [SO₄²⁻(aq)]
1.6 x 10⁻⁸ = X * X
1.6 x 10⁻⁸ = X²
X = 1.3 x 10⁻⁴ M
Hence the Pb²⁺ concentration in underground water is 1.3 x 10⁻⁴ M.
[Pb²⁺] = 1.3 x 10⁻⁴ M.
= 1.3 x 10⁻⁴ mol / L x 207 g / mol
= 26.91 ppm
<h3><u>Answer</u>;</h3>
Actual yield = 46.44 g
<h3><u>Explanation;</u></h3>
1 mole of water = 18 g/mol
Therefore;
The experimental yield = 2.58 moles
equivalent to ; 2.58 × 18 = 46.44 g
The theoretical value is 47 g
Percentage yield = 46.44/47 × 100%
= 98.8%
The questions asks for actual yield = 46.44 g
Answer:
The answer to your question is 122.4 g of O₂
Explanation:
Data
mass of O₂ = ?
moles of H₂O = 7.65
Process
1.- Write the balanced chemical reaction
2H₂O ⇒ 2H₂ + O₂
2.- Convert the moles of H₂O to grams
molar mass of H₂O = 2 + 16 = 18 g
18 g of H₂O ---------------- 1 mol
x ----------------- 7.65 moles
x = (7.65 x 18) / 1
x = 137.7 g H₂O
3.- Calculate the grams of O₂
36 g of H₂O -------------------- 32 g of O₂
137.7 g of H₂O ------------------- x
x = (32 x 137.7) / 36
x = 122.4 g of O₂
Answer:
147.2g
Explanation:
The full solution can be found in the image attached. Graham's law was applied to the problem. The rate of diffusion of a gas is inversely proportional to its molar mass or vapour density. Molar mass= 2vapour density
