In water solution potassium ion (K+) is a spectator ion, it does not react because both KOH and KBr are soluble salts (the first strongly basic, the last a neutral salt)
The molecular equation is 2KOH(aq) + CoBr2(aq) ==> 2KBr(aq) + Co(OH)2(s)The spectator ions are K^+ and Br^-
The ionic equation is 2K+(aq) + 2OH-(aq) + Co^2(aq) + 2Br^-(aq) ==> 2K+(aq) + 2Br-(aq) + Co(OH)2(s)net ionic equation is Co2+(aq) + 2OH-(aq) ==> Co(OH)2(s)So spectators are K+ and Br- ions.
There will be no reaction.
There is no reaction
KOH + HCl ---> KCl + H2O
The molecular equation is 2KOH(aq) + CoBr2(aq) ==> 2KBr(aq) + Co(OH)2(s)The spectator ions are K^+ and Br^-
The ionic equation is 2K+(aq) + 2OH-(aq) + Co^2(aq) + 2Br^-(aq) ==> 2K+(aq) + 2Br-(aq) + Co(OH)2(s)net ionic equation is Co2+(aq) + 2OH-(aq) ==> Co(OH)2(s)So spectators are K+ and Br- ions.
There will be no reaction.
There is no reaction
KOH + HCl ---> KCl + H2O
NaOH + KHT = KOH + NaHT
imidazole will form
When alc. KOH react with alkyle halide it for Alkene, KX (X Is stande for halide) and water. And this reaction also called Dehydrohalogenation...
If the KOH is in a moderately concentrated aqueous solution, the net reaction can be: 2 Al + 6 H2O => 2 Al(OH)3 + 3 H2. In this instance, the KOH does not undergo any net reaction; instead it catalyzes the reaction between aluminum and water by preventing the solid aluminum from maintaining a passivation layer on its surface. If the KOH is in a still more concentrated aqueous solution, the reaction can be: 2 Al + 4 KOH => K2Al2O4 + H2.
There will be no reaction. These two compounds do not react with each other at all.
K+ and NO3- because KOH + HNO3 (yields) H2O + KNO3
Potassium chloride and water result from this reaction: KOH + HCl = KCl + H2O