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4 years ago

After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2 is

Chemistry

1 answer:

Tanzania [10]4 years ago

4 0

Answer:

Ksp = 8.8x10⁻⁵

Explanation:

<em>Full question is:</em>

<em>After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2+ is 2.8 × 10–2 M. What is Ksp for PbCl2?</em>

<em />

When an excess of PbCl₂ is added to water, Pb²⁺ and Cl⁻ ions are produced following Ksp equilibrium:

PbCl₂(s) ⇄ Pb²⁺ + 2Cl⁻

Ksp = [Pb²⁺] [Cl⁻]²

If an excess of PbCl₂ was added, an amount of Pb²⁺ is produced (X) and twice Pb²⁺ is produced as Cl⁻ (2X):

Ksp = [X] [2X]²

Ksp = 4X³

As X is the amount of Pb²⁺ = 2.8x10⁻²M:

Ksp = 4(2.8x10⁻²)³

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3 years ago

2ca(s)+o2(g) → 2cao(s) δh∘rxn= -1269.8 kj; δs∘rxn= -364.6 j/k

Alinara [238K]

Gibbs free energy change for the reaction at 29°C. is equal to -1378.93 KJ.

<h3>What is Gibbs's free energy?</h3>

Gibbs free energy can be described as the enthalpy of the system minus the product of the temperature and entropy.

If the chemical reaction can be carried out under constant temperature ΔT = 0:

ΔG = ΔH – TΔS

The above equation is known as the Gibbs-Helmholtz equation.

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Learn more about Gibbs's free energy, here:

brainly.com/question/13318988

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Your question is incomplete, most probably the complete question was,

2Ca(s)+O₂(g) → 2CaO(s)

ΔH∘rxn= -1269.8 kJ; ΔS∘rxn= -364.6 J/K

For this problem, assume that all reactants and products are in their standard states.

Calculate the free energy change for the reaction at 29°C.

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2 years ago

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Oksanka [162]

Answer:

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Explanation:

Start your equation with what you have been given. Place the units you need in your answer on the right side of the equal sign.

225mg

----------- X ----------- X ------------- = ? g

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225 mg 1 g 25 lb 5.625 g

--------------- X --------------- X ------------- = ---------------

lb 1000 mg 1 1

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3 years ago

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