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JulijaS [17]
3 years ago
9

A solution contains one or more of the following ions: Ag+, Ca2+, and Co2+. Lithium bromide is added to the solution and no prec

ipitate forms. An excess of lithium sulfate is then added to the solution and a precipitate forms. The precipitate is filtered off and lithium phosphate is added to the remaining solution, producing a precipitate.A. Which ions are present in the original solution?Ca2+Ag+Co2+B. Write a net ionic equation for the formation of the precipitate observed after the addition of lithium sulfate. Include physical states.C. Write a net ionic equation for the formation of the precipitate observed after the addition of lithium phosphate. Include physical states.
Chemistry
1 answer:
olchik [2.2K]3 years ago
6 0
<h3>Answer:</h3>

#1. Ca²⁺

# 2. Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)

#3. 3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)

<h3>Explanation:</h3>

The question above concerns solubility of salts or ions in water.

The solution given contains Ag+, Ca2+, and Co2+ ions.

  • In the first case, when Lithium bromide is added to the solution, there is no white precipitate formed.
  • In the second case, the addition of Lithium sulfate results in the formation of a precipitate because of the Ca²⁺ in the solution combined with the SO₃²⁻ from lithium sulfate to form an insoluble CaSO₄.
  • The net ionic equation for the reaction is;

Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)

  • From the solubility rules, all sulfates are soluble except BaSO₄, CaSO₄, and PbSO₄.
  • In the third case, the addition of Lithium phosphate results in the formation of a precipitate because Ag⁺ ions in the solution combine with phosphate ions ( PO₄³⁻) from lithium phosphate to form an insoluble salt, Ag₃PO₄.
  • The net ionic equation for the reaction is;

3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)

  • According to solubility rules, all phosphates are insoluble in water except Na₃PO₄, K₃PO₄, and (NH₄)₃PO₄.
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For all these questions, we want to find the empirical and molecular formulae of various compounds given their percent composition and molar mass. The technique used to answer one of the questions can accordingly be applied to all of them.

Approaching the first question, we treat the percentages of each element as the mass of that element in a 100 g compound (as the percentages add up to 100%). So, our 100 g compound comprises 7.74 g H and 92.26 g C.

Next, we convert these mass quantities into moles. Divide the mass of each element by its molar mass:

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92.26 g C/12.0107 g/mol = 7.681 mol C.

Then, we look for the molar quantity that's the smallest ("smaller," in this case, since there are only two), and we divide all the molar quantities by the smallest one. Here, it's a very close call, but the number of moles of H is slightly smaller than that of C. So, we divide each molar quantity by the number of moles of H:

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The quotients we calculated represent the subscripts of our compound's empirical formula, which should provide the most simplified whole number ratio of the elements. So the empirical formula of our compound is C₁H₁, or just CH.

Here, it just so happens that we obtained whole number quotients. If we end up with a quotient that isn't a whole number (e.g., 1.5), we would multiply all the quotients by a common number that <em>would </em>give us the most simplified whole number ratio (so, if we had gotten 1 and 1.5, we'd multiply both by 2, and the empirical formula would have subscripts 2 and 3).

To find the molecular formula (the actual formula of our compound), we use the molar mass of the compound, 78.1134 g/mol. The molar mass of our "empirical compound," CH, is 13.0186 g/mol. Since our empirical formula represents the most simplified molar ratio of the elements, the molar masses of our "empirical compound" and the actual compound should be multiples of one another. We divide 78.1134 g/mol by 13.0176 g/mol and obtain 6. The subscripts in our molecular formula are equal to the subscripts in our empirical formula multiplied by 6.

Thus, our molecular formula is C₆H₆.

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As mentioned before, all the questions here can be answered following the procedure used to answer the first question above. In any case, I've provided the empirical and molecular formulae for the remaining questions below for your reference.

2. Empirical formula: C₁₃H₁₂O; molecular formula: C₁₃H₁₂O

3. Empirical formula: CH; molecular formula: C₈H₈

4. Empirical formula: C₂HCl; molecular formula: C₆H₃Cl₃

5. Empirical formula: Cl₄K₂Pt; molecular formula: Cl₄K₂Pt

6. Empirical formula: C₂H₄Cl; molecular formula: C₄H₈Cl₂

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