When two element combine to form more than one compound i hope this helps you with work have a nice day :)
<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.
Atoms of oxygen are electronegative and attract the shared electrons in their covalent bonds.
Answer:
It will take 5492 seconds to electroplate 0.5 mm of gold on an object .
Explanation:
Mass of gold = m
Volume of gold = v
Surface area on which gold is plated = 
Thickness of the gold plating = h = 0.5 mm = 0.05 cm
1 mm = 0.1 cm
Density of the gold = 
Moles of gold = 
According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :
of electrons
Number of electrons = N =
Charge on single electron = 
Total charge required = Q
Amount of current passes = I = 8 Ampere
Duration of time = T
It will take 5492 seconds to electroplate 0.5 mm of gold on an object .
