Answer:
All three are present
Explanation:
Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble:  .
.
- Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
- Secondly, addition of liquid ammonia would form a precipitate with silver: ![AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq)](https://tex.z-dn.net/?f=AgCl%20%28s%29%20%2B%202%20NH_3%20%28aq%29%20%2B%20H_2O%20%28l%29%5Crightarrow%20%5BAg%28NH_3%29_2%5DOH%20%28s%29%20%2B%20HCl%20%28aq%29) ; Silver hydroxide at higher temperatures decomposes into black silver oxide: ; Silver hydroxide at higher temperatures decomposes into black silver oxide:![2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g)](https://tex.z-dn.net/?f=2%20%5BAg%28NH_3%29_2%5DOH%20%28s%29%5Crightarrow%20Ag_2O%20%28s%29%20%2B%20H_2O%20%28l%29%20%2B%204%20NH_3%20%28g%29) . .
- Thirdly, we also know we have  in the mixture, since addition of potassium chromate produces a yellow precipitate: in the mixture, since addition of potassium chromate produces a yellow precipitate: . The latter precipitate is yellow. . The latter precipitate is yellow.
 
        
             
        
        
        
Answer:
2. Na2O
Reasoning:
Na (soduim) is the only metal out of all the elements in the answer choices, and is bonded with oxygen, a nonmetal, therefore creating an ionic bond.
        
             
        
        
        
Answer:
B
Explanation:
 The most stable carbonation with OH on the adjacent carbon