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3 years ago

Write an equilibrium expression for each chemical equation involving one or more solid or liquid reactants or products.

Chemistry

1 answer:

Alex_Xolod [135]3 years ago

5 0

Answer:

$Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}$

$Keq=[O_2]^3$

$Keq=\frac{[H_3O^+][F^-]}{[HF]}$

$Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}$

Explanation:

Hello there!

In this case, for the attached reactions, it turns out possible for us to write the equilibrium expressions by knowing any liquid or solid would be not-included in the equilibrium expression as shown below, with the general form products/reactants:

$Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}$

$Keq=[O_2]^3$

$Keq=\frac{[H_3O^+][F^-]}{[HF]}$

$Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}$

Regards!

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Answer:

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Explanation:

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Answer:

43.05 moles of Al needed to react with 28.7 moles of FeO.

Explanation:

Given data:

Moles of FeO = 28.7 mol

Moles of Al needed to react with FeO = ?

Solution:

Chemical equation:

2Al + 3FeO → 3Fe + Al₂O₃

Now we will compare the moles of Al with FeO.

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Answer:

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Hello,

From my understanding of the question, we are required to identify the

1) Acid

2) Base

3) conjugate acid

4) conjugate base in the reaction

Acid (BSA) = CH₃COOH

Base (BSB) = H₂O

CA = conjugate acid = H₃O⁺

CB = conjugate base = CH₃COO⁻

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