Leave the answer blank if no precipitate will form. (Express your answer as a chemical formula.) Formula of precipitate ZnCl2(aq) + KOH(aq)→ (s) K2CO3(aq) + AgNO3(aq)→ (s) (NH4)3PO4(aq) + Ni(NO3)2(aq)→ (s) NaCl(aq) + KNO3(aq)→ (s) Na2CO3(aq) + NH4Cl(aq)→

1. Zn(OH)₂ (s)

2. Ag₂CO₃ (s)

3. Ni₃(PO₄)₂(s)

4. No reaction

5. (NH₄)₂CO₃(s)

Explanation:

Let's state the equations and we analyse some solubility and precipitation information:

ZnCl₂(aq) + 2KOH(aq) → Zn(OH)₂ (s)  +  2KCl (aq)

All the salts from the halogens with group 1, are soluble.

The OH⁻ reacts to Zn cation in order to produce a precipitate. This is ok, but if the base is in excess, the Zn(OH)₂ will be soluble

K₂CO₃(aq) + 2AgNO₃(aq) → Ag₂CO₃ (s) ↓ + 2KNO₃(aq)

All salts from nitrate are soluble

All salts from carbonates are insoluble

2(NH₄)₃PO₄(aq) + 3Ni(NO₃)₂(aq) → Ni₃(PO₄)₂(s) ↓ + 6NH₄NO₃(aq)

Salts from phosphates are insoluble

All salts from nitrate are soluble

NaCl(aq) + KNO3(aq) → NO REACTION

All salts from nitrate are soluble

All the salts from the halogens with group 1, are soluble

Na₂CO₃(aq) + 2NH₄Cl(aq) → 2NaCl(aq) + (NH₄)₂CO₃(s) ↓

All salts from carbonates are insoluble

All the salts from the halogens with group 1, are soluble

1) Zn(OH)2(s)

2)Ag2CO3(s)

3)Ni3(PO4)2(s)

4)/

5)(NH4)2CO3(s)

Explanation:

ZnCl2(aq) + 2KOH(aq)→ Zn^2+ +2Cl- + 2K+ + 2OH-

ZnCl2(aq) + 2KOH(aq)→ Zn(OH)2(s) + 2KCl(aq)

K2CO3(aq) + 2AgNO3(aq)→ 2K+ + CO3^2- + 2Ag+ + 2NO3-

K2CO3(aq) + 2AgNO3(aq)→ 2KNO3(aq) + Ag2CO3(s)

2(NH4)3PO4(aq) + 3Ni(NO3)2(aq) → 6NH4+ + 2PO4^3- + 3NI^2+ + 6NO3-

2(NH4)3PO4(aq) + 3Ni(NO3)2(aq) → Ni3(PO4)2(s) + 6NH4NO3(aq)

NaCl(aq) + KNO3(aq)→ Na+ + Cl- + K+ + NO3-

NaCl(aq) + KNO3(aq)→ NaNO3(aq) + KCl(aq)

Na2CO3(aq) + 2NH4Cl(aq)→ 2Na+ + CO3^2- + 2NH4+ + 2Cl-

Na2CO3(aq) + 2NH4Cl(aq)→ 2NaCl(aq) +( NH4)2CO3(s)

the answer is true.

i have no clue hope you find the answer