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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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In a heat engine, 700 J of heat enters the system, and the piston does 400 J of work. What is the final internal (thermal) energ
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The first law of thermodynamics characterises the two types of energy transfer, as heat and as thermodynamic. The final internal (thermal) energy of the system is 1,500 J.

<h3>What is internal energy?</h3>

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Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti
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Answer:

1) 2.0 g

2) 0 g

3) 4.17 g

4) 2.57 g

Explanation:

First of all, we need to know the compounds and the reaction. The ion carbonate is CO3^{-2}, and the ion nitrate is NO3^{-}.

Sodium is in group 1, so it must lose one electron to be stable, and be the cation Na^{+}. Silver has only one electron too, so the cation will be Ag^{+}.

To form the chemical compounds, first we put the cation, then the anion, and change their charges without the signal:

Sodium carbonate: Na2CO3

Silver nitrate: AgNO3

Silver carbonate: Ag2CO3

Sodium nitrate: NaNO3

The balanced reaction will be:

Na2CO3 + 2 AgNO3 --> Ag2CO3 + 2 NaNO3

Now, we must check the stoichiometry, which will be 1:2:1:2 (always in number of moles)

The question wants to know the mass value, so we need to know the molar mass of these compounds. Checking the periodic table will see that:

Na = 23 g/mol, C = 12 g/mol, N = 14 g/mol, O = 16 g/mol, Ag = 108 g/mol

So the molar mass of the compounds must be:

Na2CO3 = 106 g/mol (2x23 + 12 + 3x16)

AgNO3 = 170 g/mol (108 + 14 + 3x16)

Ag2CO3 = 276 g/mol (2x108 + 12 + 3x16)

NaNO3 = 85 g/mol

We have a mixture of the reactants, so one probably would be in excess, so, first will need to test. Let's do the stoichiometry calculus using silver nitrate as the limit, so:

1 mol of Na2CO3 ---------- 2 mol of AgNO3

106 g ------------------------------ 2x170 = 340 g

x ------------------------------------ 5.14 g

By a simple direct three rule:

340x = 544.84

x = 1.6 g of Na2CO3

That means that for this reaction, we only need 1.6 g of Na2CO3 to react with 5.14 of AgNO3. How we have 3.60 g of Na2CO3, it is on excess, and all the AgNO3 will be consumed.

1) The mass of Na2CO3 that remains after the reaction will be the initial less the mass that reacted:

m = 3.6 - 1. 6 = 2.0 g

2) All the AgNO3 reacted, so there isn't a mass present after the reaction.

m = 0 g

3) Now, doing the stoichiometry calculus between AgNO3 and Ag2CO3

2 moles of AgNO3 ------------- 1 mol of Ag2CO3

2x170 g ------------------------------- 276 g

5.14 g --------------------------------- x

By a simple direct three rule:

340x = 1418.64

x = 4.17 g of Ag2CO3

4) Now, doing the stoichiometry calculus between AgNO3 and NaNO3

2 moles of AgNO3 ----------------------- 2 moles of NaNO3

2x170 g ---------------------------------------- 2x85 g

5.14 g ------------------------------------------- x

By a simple direct three rule:

340x = 873.8

x = 2.57 g

8 0
3 years ago
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