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SeO2 (s) + 2KOH(aq) > K2SeO3(aq) +H2O(l) SeO2(s)+2KOH(aq)→K2SeO3(aq)+H2O(l)
CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2CO3(aq) is the balanced chemical equation H2CO3 is not stable in this form, so it splits into CO2(g) + H2O(l) Therefore the true equation will be: CaCO2(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l) You NIE will start like this: CaCO2(s) + 2H+(aq) + 2Cl-(aq) -> Ca2+(aq) + 2Cl-(aq) + CO2(g) + H2O(l) and will end as: 2H+(aq) + CaCO2(s) -> Ca2+(aq) + CO2(g) + H2O(l)
The chemical equation is:Na + OH- + H+ + Cl- = Na+ + Cl- + H2O(l)
Kb = [CH3NH3 +] [OH-] / [CH3NH2]
Sulfurous Acid = H2SO3(aq) -> SO2(g) + H2O(l) Carbonic Acid = H2CO3(aq) -> CO2(g) + H2O(l) Ammonium Hydroxide = NH4OH(aq) -> NH3(g) + H2O(l) Ammounium Sulfide = (NH4)2S(aq) -> 2NH3(g) + H2S(g)
Kb=[(Ch3)3 NH+][OH-]/[(Ch3)3 N]
Kb=[(Ch3)3 NH+][OH-] __________ [(Ch3)3 N]
KHC8H4O4(aq) + NaOH(aq) → KNaC8H4O4(aq) + H2O(l) NaOH(aq) + KHP(aq) => Na+ + KP- + H2O(l)
HNO3 (aq) + NaOH (aq) --> H2O (l) + NaNO3 (aq)
Will sodium carbonate react with hydrochloric acid. 2HCI(aq) NA2CO3(aq)--- 2NaCI(aq) H2O(aq) CO2(aq)?
H^+(aq) + C2H3O2^-(aq) + Na^+(aq) + OH^-(aq) = Na^+(aq) + C2H3O2^-(aq) + H2O(L)Reducing (by crossing out repeated [spectator] ions) gives H^+(aq) + OH^-(aq) = H2O(L)
Balanced Molecular Equation:HCl(aq) + NaOH(aq) --> NaCl(aq) + H2O(l)Complete Ionic Equation:H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq) -->Na+ (aq) + Cl- (aq) + H2O (l)Cancel out the "spectator" ions, or ions that do not change in structure or charge (on both sides of the equation; "cancel out")Net Ionic Equation:H+ (aq) + OH- (aq) --> H2O(l)User forgot to place the yield arrows (separating products from reactants)
SeO2 (s) + 2KOH(aq) > K2SeO3(aq) +H2O(l) SeO2(s)+2KOH(aq)→K2SeO3(aq)+H2O(l)
Overall ionic equation: K+(aq) +OH-(aq)+ H30+(aq) + Cl-(aq)------>2H20 (l) + K+(aq) + Cl-(aq) Net ionic Equation OH-(aq)+ H30+(aq) ------>2H20 (l)
HClO4 (acid) (aq) + H2O (base) = CLO4- (conj. base) (aq) + H3O+ (conj. acid) (aq)
KOH(aq) + HNO3(aq) = KNO3(aq) + H2O(l)
Equilibrium:NO2- (aq) + H2O HNO2 (aq) + OH- (aq)