KOH + CH3COOH  -->  H2O + ______?
209995
KOH + CH3COOH -----> H+ , OH-, CH3COOK (Its a salt of weak acid and strong base.)\nH+, OH- will give you undissocialted water molecule H2O\n\nCLear?
395918
this is the electrolytic dissociation of acetic acid in water\nCH3COOH --> CH3COO- + H+
393267
Actually it is not a solubility problem, but a buffer.\n\nLet's use the Henderson-Hasselbalch equation for simplicity\n\npH=pKa+log[CH3COONa]/[CH3COOH]\n\nIf V is the volume of the solution in L then\n[CH3COOH]= mole/V= 0.010/V\n[CH3COONa]= mole/V =x/V\n\nso substitute in the equation and you get\n\n5.37=4.74+ log((x/V)/(0.01/V))=>\n0.63=log(x/0.01) =>\nx/0.01= 10^0.63 =>\nx =0.01*10^0.63= 0.0427 mole =42.7 mmole or since MW=82\nm= 0.0427*82= 3.50 g
247376
I don't know if this is right ... but check it and tell me\n\nV*C KOH = V*C HBr\n\nV KOH = 18.6ml C KOH = 0.752M\nV HBr = 22.1ml C HBr = ?\n\n18.6*0.752 = 22.1*C HBr\n\nC HBr = 0.6329M\nIts seems reasonable cause the difference between concentrations isn't significant
376020
H3PO4 + 3NaOH ----> 3H2O + Na3PO4
498999
C6H6O + 7O2 yields 3H2O + 6CO2
62742
C6H12O6 + 6O2->6CO2+6H2O\n\nOh you are right, I never did see you have a completely incorrect answer that magically changed to an answer with faint resemblence to mine! Man, I should wear glasses!
443139
Assuming KOH is a strong base,\npOH = - log [-OH] = - log (1.00 M) = 0\npH + pOH = 14\npH = 14 - pOH\npH = 14\npH = - log [H+]\n[H+] = antilog(-pH)\n[H+] = 1 x 10^-14 M
227602
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]
553114
H2SO4 will dehydrate and remove water.\n\n\n(CH3)2 CH - CH2-CH=CH-(CH3)2
302451
NaOH (Base) + HNO3 (Acid) -> NaNO3 (Salt) + H2O (Water)\n   H  H H\n   |  | |\nH-C-C-C-H\n   | | |\n   H H H\n\nReplace the last H by an -O-H\n\nDraw the carbon skeleton first.
478180
H2+O2-->H2O\nNaOH+HCl-->NaCl+H2O\nKOH-->(on heating,deomposition)K2O+H2O\ni guess this is what u want\nand i f u didn't get it i am sorry......
371445
CH4 + 2O2-> CO2 + 2H2O
338927
IAnd for your other request, I would request you to send a seperate mail to me through yahoo Answer so that I am able to send it part by part.These are long subjects and would take some time,The balanced equation would be                                           H3PO4+3KOH=K3PO4+3H2O
439197
If C is your initial concentration (C=0.114 M) then\n\n.. .. .. .. .. .. .. .. HC3H5O2 <=>  C3H5O2- + H+\nIntial .. .. .. .. .. .. .. C\nDissociate .. .. .. .. x\nProduce .. .. .. .. .. .. .. .. .. .. .. .. .. .. x .. .. .. .. x\nAt Equilbrium .. .. C-x .. .. .. .. .. .. .. x. .. .. .. .. x\n\nKa=[C3H5O2-][H+] / [HC3H5O2]= x^2/(C-x)\n\nYou find x either by solving the quadratic equation or by assuming that C >> x and then C-x=C so Ka=x^2/C. You calculate x and if it is truly x << C the assumption is valid (otherwise you have to solve the quadratic)\n\nOnce you have the value of x,\n[H+]=[C3H5O2-]=x\n[HC3H5O2]=C-x= 0.114-x\npH=-logx\n[OH-]= Kw/[H+]= 10^-14/x, or if you prefer pOH=14-pH and [OH=]=10^-pOH\n\npercent dissociation is (x/C)*100%
391420
CO2\nCarbon Dioxide\nHere is the reaction:\nHC2H3O2(aq) + NaHCO3(aq) ----> NaC2H3O2(aq) + H2O(l) + CO2(g)
494033
(NH4)2CO3+MgSO4->(NH4)2SO4+MgCO3