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STALIN [3.7K]
3 years ago
12

The insoluble salts below are put into 0.10 M hydrochloric acid solution. Do you expect their solubility to be more, less, or ab

out the same as in a pure water solution?
1. Zinc sulfide
2. Silver chloride
3. Lead iodide
4. Silver hydroxide
Chemistry
1 answer:
steposvetlana [31]3 years ago
5 0

Answer:

1. Zinc sulfide : about the same solubility, no common ion is found.

2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.

3. Lead iodide  : about the same solubility, no common ion is found.

4. Silver hydroxide : about the same solubility, no common ion is found.

Explanation:

Hello,

In this case, we first must remember that adding a common ion (which is related with the dissolving solid) decreases the solubility of the insoluble solid due to the fact Le Chatelier's principle states the reaction will shift leftwards (reactants) to reestablish equilibrium, therefore, we have:

1. Zinc sulfide : about the same solubility, no common ion is found.

2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.

3. Lead iodide  : about the same solubility, no common ion is found.

4. Silver hydroxide : about the same solubility, no common ion is found.

Best regards.

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As we know,

\chi_{Ne} + \chi_{Xe} = 1\\\chi_{Ne}= 1- 0.761\\\chi_{Ne}= 0.239

According to Dalton's Law of partial pressure-

P_i=\chi_i\times P_{total}

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PV=nRT

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Learn more about Dalton's Law of partial pressure here;

brainly.com/question/14119417

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