Respuesta :
Explanation :
In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
(a) The given balanced ionic equation is,
[tex]Ni(NO_3)_2(aq)+Na_2S(aq)\rightarrow NiS(s)+2NaNO_3(aq)[/tex]
The ionic equation in separated aqueous solution will be,
[tex]Ni^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+S^{2-}(aq)\rightarrow NiS(s)+2Na^+(aq)+2NO_3^-(aq)[/tex]
In this equation, [tex]Na^+\text{ and }NO_3^-[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]Ni^{2+}(aq)+S^{2-}(aq)\rightarrow NiS(s)[/tex]
(b) The given balanced ionic equation is,
[tex]NaNO_3(aq)+KBr(aq)\rightarrow KNO_3(s)+NaBr(aq)[/tex]
The ionic equation in separated aqueous solution will be,
[tex]Na^+(aq)+NO_3^-(aq)+K^+(aq)+Br^-(aq)\rightarrow KNO_3(s)+Na^+(aq)+Br^-(aq)[/tex]
In this equation, [tex]Na^+\text{ and }Br^-[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]NO_3^-(aq)+K^+(aq)\rightarrow KNO_3(s)[/tex]
(c) The given balanced ionic equation is,
[tex]Li_2SO_4(aq)+BaCl_2(aq)\rightarrow BaSO_4(s)+2LiCl(aq)[/tex]
The ionic equation in separated aqueous solution will be,
[tex]2Li^+(aq)+SO_4^{2-}(aq)+Ba^{2+}(aq)+2Cl^-(aq)\rightarrow BaSO_4(s)+2Li^+(aq)+2Cl^-(aq)[/tex]
In this equation, [tex]Li^+\text{ and }Cl^-[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]SO_4^{2-}(aq)+Ba^{2+}(aq)\rightarrow BaSO_4(s)[/tex]