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velikii [3]
4 years ago
12

consider the following reaction, which is at equilibrium. What effect will reducing the volume of the reaction mixture have? CuS

(s) + O2 (g) ⇋ Cu (s) + SO2 (g) Consider the following reaction, which is at equilibrium. What effect will reducing the volume of the reaction mixture have? CuS (s) + O2 (g) ⇋ Cu (s) + SO2 (g) The reaction will shift to the left in the direction of reactants. The reaction will shift to the right in the direction of products. The equilibrium constant will decrease. No effect will be observed. The equilibrium constant will increase.
Chemistry
1 answer:
marusya05 [52]4 years ago
4 0

Answer:

No effect will be observed.

Explanation:

Let´s consider the following reaction.

CuS(s) + O₂(g) ⇋ Cu(s) + SO₂(g)

<em>What effect will reducing the volume of the reaction mixture have?</em>

We need to take into account Le Chatelier's Principle: if a system at equilibrium suffers a perturbation, it will react to counteract that perturbation. According to Boyle's Law, if the volume is reduced then the pressure increases. According to Le Chatelier's Principle the system will try to reduce pressure by shifting the equilibrium towards where there are fewer moles of gases. In this case, there is 1 mol of gases on the left side and 1 mol of gases on the right side. That´s why no effect will be observed.

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Determine the molar solubility of AgBr in a solution containing 0.150 M NaBr. Ksp (AgBr) = 7.7 × 10-13.
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Answer:

Molar solubility of AgBr = 51.33 × 10⁻¹³

Explanation:

Given:

Amount of NaBr = 0.150 M

Ksp (AgBr) = 7.7 × 10⁻¹³

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Molar solubility of AgBr

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Molar solubility of AgBr = Ksp (AgBr) / Amount of NaBr

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3 years ago
Rank these acids according to their expected pKa values.
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Answer:

According to their expected pKa values, the order of those acids should be:

1- Cl2CHCOOH is the strongest acid and the lowest pKa.

2- ClCH2COOH is a strong acid, but no more than the first. Medium pKa value.

3- ClCH2CH2COOH is a strong acid, but no more than the two previous acids. High pKa value.

4- CH3CH2COOH  is the weakest acid, so the highest pKa value.

Explanation:

The pKa values are the negative logarithm of dissociation constant. It represents the relative strengths of the acids. Stronger acids show smaller pKa values and weak acids present larger pKa value. The stronger the acid, the weaker it's the conjugate base. The larger the pKa of the conjugate base, the stronger the acid. The strength of an acid is inversely related to the strength of its conjugate.

Conjugate bases are the substance that has one less proton than the parent acid. The conjugate base of the acid presented in the problem are:

ClCH2COOH -> ClCH2COO-  + H+

ClCH2CH2COOH -> ClCH2CH2COO- + H+

CH3CH2COOH -> CH3CH2COO- + H+

Cl2CHCOOH -> Cl2CHCOO - + H+

Cl2CHCOOH. The negative charge presented on its conjugate base is by resonance and inductive effect. This is the strongest acid.

ClCH2COOH. A negative charge is stabilized by resonance and electron-withdrawing but only one atom is present. So this acid is less strong than the first one.

ClCH2CH2COOH. The negative charge is stabilized by resonance and electron-withdrawing atom but the effect is less compared to the two acids showed previously.

CH3CH2COOH. The negative charge is stabilized by resonance and destabilized due to CH3 group. This is the weakest acid among the problem.

Stronger acids have smaller pKa values and weak acids have larger pKa values. Due to the information present in this problem, Cl2CHCOOH is the strongest acid and the lowest pKa. CH3CH2COOH is the weakest acid, so the highest pKa value.

Finally, we can conclude that according to their expected pKa values, the order of those acids should be:

1- Cl2CHCOOH is the strongest acid and the lowest pKa.

2- ClCH2COOH is a strong acid, but no more than the first. Medium pKa value.

3- ClCH2CH2COOH is a strong acid, but no more than the two previous acids. High pKa value.

4- CH3CH2COOH  is the weakest acid, so the highest pKa value.

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