<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.
Balance the chemical equation for the chemical reaction.
Convert the given information into moles.
Use stoichiometry for each individual reactant to find the mass of product produced.
The reactant that produces a lesser amount of product is the limiting reagent.
The reactant that produces a larger amount of product is the excess reagent.
To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.
The empirical formula CH₂O has a mass [(12 × 1) + (1 × 2) + (16 × 1)] = 30 g/mol
If the empirical formula is 30 g/mol,
and the molecular formula is 60 g/mol
Then the multiple is = 60 g/mol ÷ 30 g/mol
= 2
Therefor the molecular formula is 2(CH₂O) = C₂H₄O₂ (OPTION 2)
The interactions of climate, living organisms, and landscape .
The initial two columns of the periodic table make the s-square, and the components in this square share practically speaking that they have a tendency to lose electrons to pick up soundness.
