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: ![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: 
. The latter precipitate is yellow.
A + B → AB: ✔ synthesis
AB → A + B: ✔ decomposition
Hydrocarbon + O2 → CO2 + H2O: <span>✔ combustion</span>
AB + CD → AD + CB: <span>✔ replacement</span>
Answer : The correct option is, 0.21 moles
Solution : Given,
Molarity of the solution = 0.85 M = 0.85 mole/L
Volume of solution = 250 ml = 0.25 L 
Molarity : It is defined as the number of moles of solute present in one liter of the solution.
Formula used :
Now put all the given values in this formula, we get the moles of solute of the solution.
Therefore, the moles of solute is, 0.21 moles
Answer:
12.5 g
Explanation:
<em>12.5 g of the compound would be formed.</em>
First, let us look at the balanced equation of reaction.
3 moles of Mg is required to react with 1 mole of N2 to produce 1 mole of product.
<em>Recall that: mole = mass/molar mass</em>
9.03 g of Mg = 9.03/24.3 = 0.3716 mole
3.48 g of N2 = 3.48/28 = 0.1243 mole
Mole ratio of Mg/N2 = 3:1
<em>Hence, there is no limiting reactant.</em>
3 moles of Mg is required for 1 mole of product.
0.3716 mole of Mg will therefore require:
0.3716 x 1/3 = 0.1239 moles of product.
Molar mass of product 
= 100.9 g/mol
Mass of 0.1239 mole = mole x molar mass
= 0.1239 x 100.9 = 12.5 g
