The net ionic equation for the reaction between Pb(NO3)2 and NaOH is: Pb2+ + 2OH- → Pb(OH)2(s). This equation represents the formation of a precipitate of lead(II) hydroxide.
The net ionic equation for HF and NaOH is: HF (aq) + NaOH (aq) -> H2O (l) + NaF (aq).
The net ionic equation for the reaction between hydrofluoric acid (HF) and sodium hydroxide (NaOH) is: HF + OH- -> F- + H2O
The net ionic equation for NaOH (sodium hydroxide) in water is: Na⁺ + OH⁻ → NaOH. This represents the dissociation of sodium ion and hydroxide ion to form sodium hydroxide in solution.
To write the net ionic equation for NaOH + H2C4H4O6, first write the balanced molecular equation: NaOH + H2C4H4O6 -> NaC4H4O6 + H2O. Then, split the compounds into ions and remove spectator ions: Na+ + OH- + H2C4H4O6 -> Na+ + C4H4O6^2- + H2O. The net ionic equation is OH- + H2C4H4O6 -> C4H4O6^2- + H2O.
The chemical equation is:Mg+ + 2 OH- = Mg(OH)2(s)
The net ionic equation for HF and NaOH is: HF (aq) + NaOH (aq) -> H2O (l) + NaF (aq).
The net ionic equation for the reaction between hydrofluoric acid (HF) and sodium hydroxide (NaOH) is: HF + OH- -> F- + H2O
The net ionic equation for NaOH (sodium hydroxide) in water is: Na⁺ + OH⁻ → NaOH. This represents the dissociation of sodium ion and hydroxide ion to form sodium hydroxide in solution.
To write the net ionic equation for NaOH + H2C4H4O6, first write the balanced molecular equation: NaOH + H2C4H4O6 -> NaC4H4O6 + H2O. Then, split the compounds into ions and remove spectator ions: Na+ + OH- + H2C4H4O6 -> Na+ + C4H4O6^2- + H2O. The net ionic equation is OH- + H2C4H4O6 -> C4H4O6^2- + H2O.
The chemical equation is:Mg+ + 2 OH- = Mg(OH)2(s)
Dissolving is not a chemical reaction; any chemical equation.
The net ionic equation for the reaction between NaOH and HCl is: OH- + H+ --> H2O. This equation represents the formation of water from the combination of hydroxide ions and hydrogen ions. Sodium and chloride ions are spectators in this reaction and are not involved in the formation of the products.
2NaCl + Zn(OH)2
The net ionic equation for NaOH and Na2SO4 when they form a precipitate is simple. It will contain only the atoms that participate in the reaction. Both of these compounds are soluble.
The chemical reaction is:2 NaOH + CoCl2 = 2 NaCl + Co(OH)2
Ni2+(aq) + SO42-(aq) + 2 Na+(aq) + 2 OH-(aq) ==> Ni(OH)2(s) + 2 Na+(aq) + SO42-(aq), the ionic equation is ; Ni2+(aq) + 2 OH-(aq)==> Ni(OH)2(s)
To determine the net ionic equation, write out the balanced molecular equation first. Then, write the complete ionic equation with all ions separated. Finally, cancel out spectator ions (ions that appear on both sides of the equation) to arrive at the net ionic equation, which shows only the reacting ions.