Respuesta :
2KI(aq) + Pb(NO₃)₂(aq) = PbI₂(s) + 2KNO₃(aq)
Pb²⁺(aq) + 2I⁻(aq) = PbI₂(s)
The lead iodide precipitate is formed in the reaction.
Pb²⁺(aq) + 2I⁻(aq) = PbI₂(s)
The lead iodide precipitate is formed in the reaction.
Answer: to observe both solutions before the reaction
Explanation:
