Asolution contains one or more of the following ions: ag+, ca2+, and cu2+. when sodium chloride is added to the solution, no precipitate forms. when sodium sulfate is added to the solution, a white precipitate forms. the precipitate is filtered off and sodium carbonate is added to the remaining solution, producing a precipitate. write net ionic equations for the formation of each of the precipitates observed.

Cl⁻ + Ag⁺ → ClAg ↓

CO₃⁻² + Ca²⁺  →  CaCO₃ ↓

CO₃⁻² + Cu²⁺  →  CuCO₃ ↓

Explanation:

The cations in solution are Ag⁺, Ca²⁺, and Cu²⁺

When sodium chloride is added to the solution, no precipitate forms.

FALSE

NaCl → Na⁺ + Cl⁻

Cl⁻ + Ag⁺ → ClAg ↓

When sodium sulfate is added to the solution, a white precipitate forms.

FALSE - No precipitate formed

The precipitate is filtered off and sodium carbonate is added to the remaining solution, producing a precipitate.

When you filtered off the precipitate (we mean AgCl), in solution there's still remain Ca²⁺ and Cu²⁺

Na₂CO₃ → 2Na⁺  +  CO₃⁻²

CO₃⁻² + Ca²⁺  →  CaCO₃ ↓

CO₃⁻² + Cu²⁺  →  CuCO₃ ↓

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

b. is the answer hoped i