net ionic equation of NaOH(aq)+HCl(aq) -->NaCl(aq)+H2O(l)?
548546
OH- is an alkali (pH is more than 7) and H+ is an acid (pH is less than 7).\n\nAcid + Alkali -> Neutrality\n\nSince it's neutral, water (pH 7) is produced.\n\nHope my explanation helped c",)
764101
In all reactions one uses that 2 OH -> H2O + O, the split of oxygen goes to create the ion. Finally balance the equation:\n \nSO3 + 2 NaOH -> Na2SO4 + H2O\n\nCO2 + 2 KOH -> K2CO3 + H2O \n\nSO2 + Ba(OH)2 -> BaSO3 + H2O
262916
4 OH-(aq) --->  O2(g) + 2H2O(l) + 4e-
160963
K2SO4 ---> 2K+  +  SO4-\n\nHCl ---> H+  +  Cl-\n\nNaHCO3 ---> Na+  +  HCO3-\n\nNH4OH ---> NH4+  +  OH-\n\nBaCl2 ---> Ba-2  +  2CL-\n\n(NH4)3PO4 ---> 3NH4+  +  PO4-3
353745
Net ionic is\n\nAg+ + Cl- => AgCl\n\nThis is double replacement.
424592
H2SO4 + 2NaOH ----> 2H2O + Na2SO4\n\nYOU MUST SEPARATE AND CANCEL THE COMMON IONS (AQUEOUS ONES)
474531
CH3COOH + NAOH =.CH3COONA +H2O
164045
What do you want a step by step set of instructions or a balanced equation?\n\n2C6H5OH + NaOCL = 2C6H50 + NaCl + H2O
838918
Its a decomposition equation. In essence Sodium Chloride (salt) breaks down into the Sodium ion and the Chlorine Ion. Chances are it's talking about NaCl when it's an aqueous solution (a solution who's solvent is water)
706351
KCl(aq) + AgNO3(aq)  --> KNO3(aq) + AgCl(s)\n\nNet ionic: Cl-(aq) + Ag+(aq)  --> AgCl(s)\n\nNa2SO4(aq) + Ba(NO3)2  --> 2NaNO3(aq) + BaSO4(s)\n\nnet ionic: SO4(-2)(aq) + Ba(+2)(aq)  --> BaSO4(s)
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
742926
Cu(s) + 2HNO3(aq) = Cu(NO3)2(aq) + H2(g)\n\nCu + 2H+ + 2NO3- = Cu+2 + 2NO3- + H2 (leaving off the s,aq,g)\n\nCu + 2H+ = Cu+2 + H2 (net ionic equation, the NO3-'s just cancel out, they don't change in the reaction).
811861
KCl(s)  ---->  K+(aq)  +  Cl-(aq)
497024
H2S---> 2 H^+ + S^2-\nK(eq) = \n[2H^+]^2 [S^2-]\n ---------------\n     [H2S]\n\nNH4+----> NH3  + H^+\nK(eq) =\n [H^+][NH3]\n       ------------\n         [NH4+]\n\nC6H5COOH----> C6H5COO-  + H+\n\nK(eq) = \n[H^+][C6H5COO-]\n        ---------------\n         [C6H5COOH]
777003
CH3COOH + NaOH ---> NaCH3COO + H2O
781593
I always have to derive the equations from scratch:\n\nKa = [H+][A-]/[HA]\n\nlog(Ka) = log(H+) + log(A-) - log(HA), multiple both sides by -1\n\npKa = pH + log(HA) - log(A)\n\npH = pKa + log(A-/HA)\n\n4.54 = 4.74 + log(A-/HA)\n\n-0.20 = log(A-/HA)\n\n10^(-0.20) = A-/HA = 0.631\n\nplease check for errors
747348
Hi Kitty!  Dr. J to the rescue with a full answer.\n\n1) 4Li + O2 ---> 2Li2O\n\n2) 4HF + SiO2 ---> SiF4 + 2H2O\n\n3) It is simpler if we eliminate the K and the sodium and write the equation ionic because it is a redox reaction and we need to balance.\n\nThe overall redox equation will be:\n\nCr2O7(=) + SO3(=) + H(+) ---> Cr(+3) + SO4(=).\n\nNow you balance by the half reaction method.  The two reactions are:\n\n6e(-) + Cr2O7(=) + 14H(+) ---> 2Cr(+3) + 7H2O\n3SO3(=)  + 3H2O --->  3SO4(=) + 6H(+) + 6e(-)\n\nEliminating the electrons:\n\nCr2O7(=) + 3SO3(=) + 8H(+) ---> 2Cr(+3) + 3SO4(=) + 4H2O\n\nThis one was tough!!!  Thumbs up!\n\nIf you want you could put the K and the Na back.\n\nBest answer?