Write the net ionic equation for the reaction of each acid with water.?
814262
H2SO4 + H20 -> H2SO4:H2O (water decreased)\nHNO2 + H2O-->NH3 + O2   (you balance it)\nNaHSO3 + H2O-->H2SO3 + NaOH(you balance it)
866210
2NO+O2--->2NO2\n\nCaO+H2O--->Ca(OH)2\n\nCaCO3--->CaO+CO2
283215
[Cu(NH3)4]SO4*H2O (s) + 2HCl(aq) --->\n\nCu(NH3)4 +2 (aq) + H2SO4(aq) + H2O(l) + 2Cl-(aq)
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
861717
2 HNO3(aq) + Sr(OH)2(aq) --> Sr(NO3)2 (aq) + 2 H2O\n\n2 H3PO4(aq) + 3 Ba(OH)2(aq) --> Ba3(PO4)2 (cr)* + 6 H2O\n\n* Barium Phosphate will precipiate out of an aqueous solution in this reaction
129663
Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
1008973
HF+ KOH  ---> KF +H2O can also be written as \nH+F- + K+OH- . ---> K+F- + H2O\nWe can canel out K+ and F-  since they appear on both side of the equation.  Thus the net ionic equation is H + OH ---> H2O.  The answer is D.
250762
2 HCl + CaCO3 ----> CaCl2 + H2O + CO2
639844
[H3O+]=1 x 10^-4 => pH=-log[H+]=4 --- acidic\n[OH-]=1X10^-6 => pOH=-log[OH-]=6 => pH=14-6=8 --- basic\n\nHCl --> H+ + Cl-\n[H+]= 5.6 x 10^-4\npH = -log[H+] = -log[5.6 x 10^-4] = 3.25 --- acidic\npOH = 14 - 3.25 = 10.75\n-log[OH-] = 10.75 => [OH-]=1.78 x 10^-11\n\nSince HBr --> H+ + Br-\npH of 1 x 10^-4 of HBr = -log[1x10^-4] = 4 --- acidic\n\nSince Ca(OH)2 --> Ca2+ + 2OH-\npOH = -log[OH-] = -log(2 * 5 x 10^-4) = 3\npH = 14-3 = 11 --- basic
1029892
Na2S (aq) + 2H2O (l) = 2NaOH + H2S (g)
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
746933
You could use the Van't Hoff equations to calculate this.\nOP = i * c * RT\nwhere i is the number of ions per particle and the membrane will be retaining the ions.\n\nBecause RT will be constant for all of them, you only need to compare the product i *c for each solution.  So OP/RT for the solutions above will be.\n\nNa2S ===> 0.30 * 3 = 0.9\nMgCl2 ===> 0.2 * 3 = 0.6\nEtOH ===> 0.50 * 1 = 0.5\nKCl ===> 0.50 * 2 = 1.0\nso the answer is KCl, Na2S, MgCl2, EtOH.
904934
2HNO3 + Mg(OH)2 -->Mg(NO3)2 +2H20\n\n2H+(aq) + 2NO3-(aq) + Mg+(aq) + 2(OH)-(aq) -->Mg+(aq) + 2(NO3)-(aq) +2H20(l)\n\n2+(aq) + 2(OH)-(aq) --> 2H20(l)
806865
Na2SO4 + H2O\n\nYou always get a type of salt plus water in acid-base neutralization.
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
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
304632
Okay, so the first thing that happens it that the NaF completely dissociates:\nNaF ---> Na+ + F-\n\nCool.  Now, the Na+ is just floating around, so we can ignore it.  Let's deal with the F-.  The F- is in the water, so it's going to do this:\nF- + H2O <---> HF + OH-\nNote that the arrow is going in both directions; it's an equilibrium problem, so I need to set up an equilibrium equation.  \n\nBut first, because I'm dealing with OH- ions, I'll need a Kb value.  Since Kw, the product constant of water, equals Kb * Ka, and Kw= 1E-14:\n(1E-14)/(6.8E-4)= Kb = 1.47E-11\n\nNow,\nKb=[OH-][HF]/[F-]\nor\n1.47E-11 = x^2/(.00339-x)\nsolve for x (I just set it equal to zero, graph it, and find the x-intercpets on my calculator):\nx=2.23E-7 M \nSince x=[OH-], [OH-]=2.23E-7 M\n\nWe can calculate pOH by taking the negative log of [OH-]:\n-log 2.23E-7 M = 6.65\nSince pH + pOH = 14.00:\npH + 6.65 = 14.00\npH = 14.00-6.65\npH = 7.35, the answer within one one-hundredth, which is probably some minor rounding discrepancy.  Yay!