Answer:
Here's what I get.
Explanation:
a. HCl + KOH
[tex]\rm Molecular: KOH(aq)+ HCl(aq) \longrightarrow \, KCl(aq) + H_{2}O(l)\\\rm Ionic: K^{+}(aq) + OH^{-}(aq)+ H^{+}(aq) + Cl^{-}(aq) \longrightarrow \, K^{+}(aq)+ Cl^{-}(aq) + H_{2}O(l)\\\rm Net ionic: OH^{-}(aq)+ H^{+}(aq) \longrightarrow \, H_{2}O(l)\\[/tex]
b. HCHO₂ + LiOH
Only the first H in HCHO₂ is acidic
[tex]\rm Molecular: HCHO_{2}(aq)+ LiOH(aq) \longrightarrow \, LiCHO_{2}(aq) + H_{2}O(l)\\\rm Ionic: HCHO_{2}(aq) + Li^{+}(aq) + OH^{-}(aq) \longrightarrow \, Li^{+}(aq)+ CHO_{2}^{-}(aq) + H_{2}O(l)\\\rm Net ionic: HCHO_{2}(aq) + OH^{-}(aq) \longrightarrow \, CHO_{2}^{-}(aq) + H_{2}O(l)\\[/tex]
c. N₂H₄ + HCl
Both N atoms in N₂H₄ are basic, so the compound reacts with 2 mol of HCl.
[tex]\rm Molecular: N_{2}H_{4}(aq)+ 2HCl(aq) \longrightarrow \, N_{2}H_{6}Cl_{2}(aq) \\\rm Ionic: \qquad N_{2}H_{4} (aq) + 2H^{+}(aq) + 2Cl^{-}(aq) \longrightarrow \, N_{2}H_{6}^{+}(aq) + 2Cl^{-}(aq)\\\rm Net ionic: \ N_{2}H_{4} (aq) + 2H^{+}(aq) \longrightarrow \, N_{2}H_{6}^{+}(aq)\\[/tex]
