Here's how to determine the net ionic equation for the reaction of silver nitrate (AgNO₃) and potassium chromate (K₂CrO₄):

1. Write the balanced molecular equation:

2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)

2. Identify the soluble and insoluble compounds:

* Soluble: AgNO₃, K₂CrO₄, and KNO₃ (most nitrates and potassium salts are soluble)

* Insoluble: Ag₂CrO₄ (silver chromate is a precipitate)

3. Write the complete ionic equation:

2Ag⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s) + 2K⁺(aq) + 2NO₃⁻(aq)

4. Identify spectator ions:

Spectator ions are those that appear on both sides of the equation and do not participate in the reaction. In this case, the spectator ions are K⁺ and NO₃⁻.

5. Write the net ionic equation by removing the spectator ions:

2Ag⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s)

Therefore, the net ionic equation for the reaction of silver nitrate and potassium chromate is:

2Ag⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s)