A solution containing an ionic compound is subjected to the anion and cation confirmation tests performed in this experiment, providing the following results: Addition of silver nitrate: a bright yellow precipitate formed. Flame test: a purple color was observed. Identify the ionic compound and write the chemical reaction that occurs when the ionic compound interacts with the silver nitrate. Identify the spectator ions.

The ionic compound is KI and when it reacts with AgNO₃ the spectator ions are K⁺ and NO₃⁻.

A solution that contains a cation and an anion is subjected to 2 experiments to confirm their identities.

In the flame test, a purple color is observed. The cation that gives purple is the flame test is potassium.

Upon the addition of silver nitrate, a bright yellow precipitate is formed. The precipitate is AgI, so the anion is iodide.

The ionic compound formed by K⁺ and I⁻ is KI.

The molecular equation for the reaction between AgNO₃ and KI is:

AgNO₃(aq) + KI(aq) ⇒ KNO₃(aq) + AgI(s)

The net ionic equation is:

Ag⁺(aq) + I⁻(aq) ⇒ AgI(s)

The spectator ions, those that don't appear in the net ionic equation, are K⁺ and NO₃⁻.

The ionic compound is KI and when it reacts with AgNO₃ the spectator ions are K⁺ and NO₃⁻.

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The spectator ions are K⁺ and NO₃⁻, since these ions do not involve in any reaction.

Explanation:

It is given that a bright yellow precipitate will be formed when the given substance reacts with silver nitrate. Silver nitrate reacts with anion iodide (I⁻) to form yellow precipitate.

In the flame test, if we get purple color then the cation must be Potassium (K⁺).

So if we combine the cation and anion we will get the compound as Potassium Iodide (KI).

K⁺ + I⁻ + Ag⁺ + NO₃⁻ → KNO₃ + AgI (s) ↓

Here Silver and Iodide ions combine to form a precipitate and Potassium and nitrate ions are acting as spectator ions, which do not involve in the reaction.

the answer is definitely c.