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2 answers:

5 0

The phrase that correctly defines chemical equilibrium is a state in which the forward and reverse reactions are proceeding at equal rates

<h3>What is a Chemical equilibrium?</h3>

A Chemical equilibrium refers to the state of a chemical system at which a constant concentration of products and reagents is present.

A Substance turnover occurs only on the particle level which is the reason that chemical equilibrium is also referred to as dynamic equilibrium."

Therefore, the Option B is correct.

Read more about Chemical equilibrium

<em>brainly.com/question/5081082</em>

Semenov [28]3 years ago

3 0

Answer:

B. a state in which the forward and reverse reactions are proceeding at equal rates

Explanation:

"Chemical equilibrium is the state of a chemical system at which a constant concentration of products and reagents is present. Reactions, which take place in homogeneous solutions, seem to have come to rest because no changes in concentrations of the participating substances can be determined. Substance turnover occurs only on the particle level, which is why chemical equilibrium is also referred to as dynamic equilibrium."

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finlep [7]

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Explanation:

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4 0

3 years ago

Calculate the molar solubility of PbBr2 (Ksp = 4.67x10-6) in 0.10M NaBr solution.

Softa [21]

Explanation:

$PbBr_{2}$ will dissociate into ions as follows.

$PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)$

Hence, $K_{sp}$ for this reaction will be as follows.

$K_{sp} = [Pb^{2+}][Br^{-}]^{2}$

We take x as the molar solubility of $PbBr_{2}$ when we dissolve x moles of solution per liter.

Hence, ionic molarities in the saturated solution will be as follows.

$[Pb^{2+}]$ = $[Pb^{2+}]_{o}$ + x

$[Br^{-}]^{2}$ = $[Br^{-}]_{o}$ + 2x

So, equilibrium solubility expression will be as follows.

$K_{sp}$ = $([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}$

Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains $[Br^{-}]_{o}$ = 0.10 and there will be no lead ions. So, $[Pb^{2+}]_{o}$ = 0

So, $[Br^{-}]_{o} + 2x$ will approximately equals to $[Br^{-}]_{o}$ .