Sodium (Na) plus sulfur (S) creates sodium sulfide (Na2S). The reason is that Na gives up it's valence electron and S takes it. It takes 2 sodiums because S need TWO electrons to complete the octet. This forms the ionic compound sodium sulfide.
Not sure what L(s) is supposed to be, but regardless, the oxidation state of Na changes from 1+ in NaOH to zero in Na(s).
No, it should be:2 Na(s) + 2HCl(g) --> H2(g) + 2 NaCl(s)By the way: sodium at normal conditions is a solid metal (s), not gaseous (g)
The net ionic equation is Cd2+(aq) + S2-(aq) -> CdS(s). This equation represents the actual chemical change happening in the reaction where cadmium ions and sulfide ions combine to form solid cadmium sulfide. The spectator ions, Na+ and Cl-, are present on both sides of the equation but do not participate in the reaction.
+1 for Na, +1 for H, -2 for each O, +4 for S
The net ionic equation for the given reaction is H+ (aq) + OH- (aq) → H2O (l)
seriously
Lithium atom become the cation Li+.
it goes from +1 to 0
Not sure what L(s) is supposed to be, but regardless, the oxidation state of Na changes from 1+ in NaOH to zero in Na(s).
If you mean a reaction of AgNO3 + Na(s) ==> NaNO3 + Ag(s), there would be ONE electron transferred.Ag^+ + 1e- ==> Ag(s) Reduction reaction Na(s) ==> Na^+ + 1e- Oxidation reaction
Na+ and Cl- are spectator ions.
The elements involved in this reaction are sodium (Na) and magnesium (Mg). The reaction occurs between solid sodium (Na) and solid magnesium fluoride (MgF2) to form solid sodium fluoride (NaF) and solid magnesium (Mg).
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).
Na+is bigger
No, it should be:2 Na(s) + 2HCl(g) --> H2(g) + 2 NaCl(s)By the way: sodium at normal conditions is a solid metal (s), not gaseous (g)
Na+-K+ ATPase
Na+ plus OH- plus H+ equals H2O plus Na+ plus Cl-