They are all dissolved in water.
KOH (aq) + HCl (aq) = KCl (aq) + H20 so potassium chloride and water - KCl solution
AgNo3 (aq) + HCl (Aq) -> HNO3 (aq) + AgCl (s) there's no gass given off, only solid and aquous substances.
Na+(aq), I-(aq)
1)Write the BALANCED equation for the reaction with CORRECT state symbols: Mg(s) +2HCl(aq)---------->MgCl2(aq) +H2(g) 2)Now rewrite the equation replacing aqueous substances as they would appear in solution( ie)as ions.Leave everything else as is in the standard state. Mg(s) + 2H+ (aq) + 2Cl-(aq)---->Mg2+(aq) +2Cl-(aq) +H2(g) 3)Now cancel ions common to both sides ,that is,spectator ions. Cancel the 2Cl-(aq) on each side. 4)Rewrite the equation without what you have canceled. Mg(s) + 2H+(aq)------>Mg2+(aq) + H2(g). Good Luck.
Carbonic Acid
These substances are in solution.
If you mean pure substances then:HCl(aq) and NaCl(aq) are not pure substances but solutions. The (aq) means the substance is dissolved in water.HCl(g) and NaCl(s) is are pure substances
(aq) is from aqueous, a substance in solution.
KOH (aq) + HCl (aq) = KCl (aq) + H20 so potassium chloride and water - KCl solution
AgNo3 (aq) + HCl (Aq) -> HNO3 (aq) + AgCl (s) there's no gass given off, only solid and aquous substances.
cleaners which is solvent
Na+(aq), I-(aq)
1)Write the BALANCED equation for the reaction with CORRECT state symbols: Mg(s) +2HCl(aq)---------->MgCl2(aq) +H2(g) 2)Now rewrite the equation replacing aqueous substances as they would appear in solution( ie)as ions.Leave everything else as is in the standard state. Mg(s) + 2H+ (aq) + 2Cl-(aq)---->Mg2+(aq) +2Cl-(aq) +H2(g) 3)Now cancel ions common to both sides ,that is,spectator ions. Cancel the 2Cl-(aq) on each side. 4)Rewrite the equation without what you have canceled. Mg(s) + 2H+(aq)------>Mg2+(aq) + H2(g). Good Luck.
Carbonic Acid
Ionisation in solution is not a chemical reaction. It can be written as one but still it's NOT: only physical change of state ( subscripted between brackets (s) or (aq) ) Na2S(s) --> 2Na+(aq) + S2-(aq)
NaNO3(aq) + KCl(aq) ---> NaCl(aq) + KNO3(aq) Na+(aq) + NO3-(aq) + K+(aq) + Cl-(aq) ----> Na+(aq) + Cl-(aq) + K+(aq) + NO3-(aq) NO NIE since they are all spectator ions it is soluble in water and no precipitate forms, there is no net ionic equation
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