Question

Considering the following precipitation reaction: Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq),

Answer 1

The net ionic equation of Pb(NO₃)₂(aq) and KI(aq) :

Pb²⁺(aq) +  2I⁻ (aq) ⇒ PbI₂(s) and

D) Pb²⁺, I – would not be spectator ions

Further explanation

The electrolyte in the solution produces ions.

The equation of a chemical reaction can be expressed in the equation of the ions

For strong electrolytes (the ionization rate = 1) is written in the form of separate ions, while the weak electrolyte (degree of ionization <1) is still written as an un-ionized molecule

In the ion equation, there is a spectator ion that is the ion which does not change (does not react) because it is present before and after the reaction

When these ions are removed, the ionic equation is called the net ionic equation

For gases and solids including water (H₂O) can be written as an ionized molecule

So only the dissolved compound is ionized ((expressed in symbol aq)

Reactions that occur:

Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq),

PbI₂(s) solid form that does not decompose in the form of ions

So the ionic equation becomes:

Pb²⁺(aq)+ 2NO₃⁻(aq) + 2K⁺ (aq) + 2I⁻ (aq) ⇒ PbI₂(s) + 2K⁺ + 2NO₃⁻

There is an ion spectator that is 2NO₃⁻ and  2K⁺ , so that if it is removed a net ionic equation will be formed:

Pb²⁺(aq) +  2I⁻ (aq) ⇒ PbI₂(s)

Learn more

the net ionic equation

brainly.com/question/8885824

brainly.com/question/11854070

brainly.com/question/10280219

brainly.com/question/9830467

Keywords: the net ionic equation, an ion spectator

Answer 2

Answer: The ions which are not spectator ions are $Pb^{2+}$ and $I^-$

Explanation:

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

The given chemical equation is:

$Pb^{2+}(aq)+NO_3^-(aq)+2K^+(aq)+2I^-(aq)\rightarrow PbI_2(s)+2K^+(aq)+2NO_3^-(aq)$

The ions which are present on both the sides of the equation are $K^+$ and $NO_3^-$ and are spectator ions.

Thus the net ionic equation is:

$Pb^{2+}(aq)+2I^-(aq)\rightarrow PbI_2(s)$

Hence, the correct answer is $Pb^{2+}$ and $I^-$