Write the equilibrium constant: Pb3(PO4)2(s) = 3Pb2+ (aq) + 2PO2 (aq)

Chemistry

1 answer:

kicyunya [14]3 years ago

7 0

Answer:

Kc = [Pb²⁺]³.[PO₄³⁻]²

Explanation:

Let's consider the following reaction at equilibrium.

Pb₃(PO₄)₂(s) ⇄ 3 Pb²⁺(aq) + 2 PO₄³⁻(aq)

The concentration equilibrium constant is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.

Kc = [Pb²⁺]³.[PO₄³⁻]²

This equilibrium constant is known as the solubility product of Pb₃(PO₄)₂.

You might be interested in

Calculate the percent water in iron ii sulfate heptahydrate

liq [111]

According to its formula FeSO4.7H2O
we can get the percent % by the mass of H2O from this formula
%mass of H2O = (mass of water H2O/ mass of the hydrate)x100
when the mass of water = molar mass x 7 = 18 x 7 = 126
and the mass of hydrate (feSO4) = molar mass = 278
So by substitution:
%mass of H2O = (126/278) x 100 = 45%

6 0

3 years ago

Coal power plants burn large amounts of coal, C(s), in an 02(g) atmosphere to generate electricity. The chemical reaction respon

Stella [2.4K]

Answer:

1.85 × 10⁸ L

Explanation:

Coal power plants burn large amounts of coal, C(s), in an O₂(g) atmosphere to generate electricity. The chemical reaction responsible for producing this energy is shown below:

C(s) + O₂(g) → CO₂(g)

Determine the volume of CO₂ in liters produced when 100 metric ton of C(s) is completely burned in an O₂ atmosphere. The density of CO₂ is 1.98 kg/m³ (1 metric ton = 1000 kg: 1 m³ = 1000 L)

We can establish the following relations:

1 metric ton = 1000 kg
1 kg = 1000 g
The molar mass of C(s) is 12.01 g/mol
The molar ratio of C(s) to CO₂(g) is 1:1
The molar mass of CO₂(g) is 44.01 g/mol
1.98 kg of CO₂(g) occupy a volume of 1 m³ (density = 1.98 kg/m³)
1 m³ = 1000 L

The volume of CO₂ produced when 100 metric ton of C(s) react is:

$100metric/tonC.\frac{10^{6}gC}{1metric/tonC} .\frac{1molC}{12.01gC} .\frac{1molCO_{2}}{1molC} .\frac{44.01 \times 10^{-3} kgCO_{2}}{1molCO_{2}} .\frac{1m^{3}CO_{2}}{1.98kgCO_{2}} .\frac{1000LCO_{2}}{1m^{3}CO_{2}} =1.85 \times 10^{8} LCO_{2}$