No, a Bronsted acid should contain at least one polar group which can donate a proton.
Pb(NO3)2 + 2KI ==> PbI2 + 2KNO3 is a precipitation reaction. The PbI2 is insoluble in water and will precipitate. It is NOT neutralization, because you do not have an acid reacting with a base.
Lead(II) iodide (PbI2) is typically a yellow colored solid.
Pb2+ + 2 NO3- + 2H+ + 2I- -> PbI2 (s) + 2HNO3 (aq)
Formula: PbI2
The oxidation number of lead (Pb) in PbI2 is +2. Iodine (I) typically has an oxidation number of -1 in compounds. Since there are two iodine atoms in PbI2, the total negative charge contributed by iodine is -2. This balances with the +2 charge of lead to make PbI2 neutral.
Pb(NO3)2 + 2KI ==> PbI2 + 2KNO3 is a precipitation reaction. The PbI2 is insoluble in water and will precipitate. It is NOT neutralization, because you do not have an acid reacting with a base.
The Bronsted-Lowry theory is that acids and bases are defined by the way they react with each other. Liquid ammonia and acetic acid are examples.
Lead(II) iodide (PbI2) is typically a yellow colored solid.
Pb2+ + 2 NO3- + 2H+ + 2I- -> PbI2 (s) + 2HNO3 (aq)
Formula: PbI2
The oxidation number of lead (Pb) in PbI2 is +2. Iodine (I) typically has an oxidation number of -1 in compounds. Since there are two iodine atoms in PbI2, the total negative charge contributed by iodine is -2. This balances with the +2 charge of lead to make PbI2 neutral.
PBI2, or lead(II) iodide, consists of one lead (Pb) atom and two iodine (I) atoms. Therefore, there are a total of three atoms in one formula unit of PBI2.
Pb(NO3)2(aq) + 2KI)aq) = 2KNO3(aq)+ PbI2(s) PbI2 is a yellow coloured solid and will precipitate out of the solution.
The balanced equation for lead iodide is: PbI2 (s) ↔ Pb2+ (aq) + 2I- (aq)
Formula: PbI2
PbI2
It would be: PbI2