How do you convert H2O into HHO?
541170
New in Chemistry?????\n\nThere is no conversion of H2O in HHO.\nIt is simply way of writing a formula. HHO is wrong, it is HOH.\n\nActually Oxygen has two separate bonds with two different atoms of Hydrogen so it is written as H-O-H.\n\nWhen H-O-H is broken it usually converts to H+ and OH-
222974
Two hydrogen atoms, one oxygen atom.  H2O
880087
De = 2\nHydrogen = H\nmono = 1\nOxide = Oxygen\nOxygen = O\nH2O
265538
well, the [H+] (which is the same as [H3O+]) concentration and the [OH-] concentration always add up to 1.0x10^=14. so an equation for that would look like\n\n[H+] x [OH-] = 1.0x10^-14\n\nso just divide 1.0x10^-14 by your concentration of H30+. its as easy as taht!
745644
H3PO4 + 3NaOH ----> 3H2O + Na3PO4
168718
The eq constant is 5.286E-16, so this means that 5.287E-16 = [H][OH]. (The [H2O] doesn't matter as it is a pure liquid.) The H and OH concentrations are the same (because of mole ratios), so solve this equation:\n\nx^2 = 5.287E-16\n\nx is the eq. concentration of H, and also the eq. concentration of OH.
212429
dots..\nits jux 2 hydrogen and 1 oxygen for every mole of h2o
470703
reduced: 2H2O + 2e- ---> H2 + 2OH-\n\noxidized: 2I- ---> I2 +2e-
1111771
two parts hydrogen.......one part oxygen...hence.....H2O
630069
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.
952215
[H3O+][OH-] = 1.0 x 10^-14 M always at 25C\nso if [H3O+]= 5.0 x 10^-10M\n[OH-]= 2.0x10^-5
851681
Write the balanced equation - you need the mole ratios.\n\nChange 1g hexane to moles of hexane.\n\nSee how many moles of oxygen are USED and how many moles of  CO2 and H2O are formed.\n\nYour total number of moles in the reaction mixture would then be:\n\nmoles of Oxygen NOT used + moles of CO2 (we will assume the water formed consenses to a liquid at 20 degrees).\n\nUse the Ideal Gas Law to figure out the volume.\n\nPV = nRT\n\nwhere P = given pressure - water vapor pressure\nwhere V = your unknown\nwhere n = your total moles of gas\nwhere R = the appropriate Universal Gas Law constant\nwhere T = the temperature in KELVIN
545577
H2(g)+ O3(g)=> H2O(L) + O2(g)\n\nBecause H2O is the prime factor of 02G if only 2H2 equals 2H20. If you take the "Smithy" equation into consideration and change the X into L factor, you will find that D is the correct answer!
650541
pH = -log ([H3O+])\npOH = -log ([OH-])\n\npH + pOH = pKw = 14\npOH = 14 - pH\n\n[OH-] = 10^-(pOH)\n\nPlug in and solve.
835897
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
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
1185900
H2SO4 + 2KOH --> K2SO4 + 2H2O