NaOH(s) is solid and mostly pure sodium hydroxide. NaOH(aq) is sodium hydroxide dissolved in water.
Na+H2O → NaOH+H2 Then you balance it to get: 2Na + 2H2O → 2NaOH + H2
In the reaction between Li(s) and NaOH(aq), Li loses an electron and gains a positive charge, changing its oxidation state from 0 to +1. This occurs because Li donates its outer electron to Na, which reduces Na+ to Na(s).
There's two possible reactions Zink with solid NaOH gives double salt Zn + 2NaOH --t--> Na2ZnO2 + H2 And I Think you need the other one: Zink with NaOH(aq) gives the complex salt: Zn + 2NaOH + 2H2O --> Na2[Zn(OH)4] + H2
The reaction "Zn (s) + 2HCl (aq) -- H2 (g) + ZnCl2 (s)" is not a double-displacement reaction. It is a single displacement reaction where Zn displaces H from HCl to form ZnCl2 and H2 gas.
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)
The chemical equation is:3 NaOH + FeBr3 = 3 NaBr + Fe(OH)3
The dissociation reaction of NaOH in water is as follows: NaOH (s) → Na+ (aq) + OH- (aq)
When NaOH dissolves in water, it dissociates into Na+ and OH- ions. The equation representing this ionization reaction is: NaOH (s) → Na+ (aq) + OH- (aq). If you want a chemical equation showing the complete dissociation of NaOH in water, it would be: NaOH (s) → Na+ (aq) + OH- (aq).
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)
2NaOHaq + 2HClaq --> 2NaClaq + H2Ol is the perfect balanced equatiion,except the solubility (aq) of NaCl, so it is notNaCls but NaClaq
HCl (aq) + NaOH (s) → NaCl (aq) + H2O (l)
BeCl2(aq) + 2NaOH(aq) -> 2NaCl(aq) + Be(OH)2(s) Be(aq) + 2OH(aq) -> Be(OH)2(s) Hope this helps!
It goes from 0 to +1
The balanced chemical equation for the reaction between hydrobromic acid (HBr) and sodium hydroxide (NaOH) is: HBr + NaOH → NaBr + H2O This equation shows that one molecule of HBr reacts with one molecule of NaOH to form one molecule of NaBr and one molecule of water.
Na+H2O → NaOH+H2 Then you balance it to get: 2Na + 2H2O → 2NaOH + H2
Zn(s) + 2 OH-(aq) -> ZnO22-(aq) + H2 (g) Perhaps a bit more may be needed Zn(s) + H2O(l) +2OH-(aq) -> Zn(OH)42-(aq) + H2(g)
In the reaction between Li(s) and NaOH(aq), Li loses an electron and gains a positive charge, changing its oxidation state from 0 to +1. This occurs because Li donates its outer electron to Na, which reduces Na+ to Na(s).