what is the balanced formula for KOH dissolving in H20?  Is it KOH + H2O --> ???
468312
Potassium hydroxide is a strong base.\nKOH -> K+ + OH-\n\nWater doesn't participate directly in the reaction, so you can leave out the H2O on both sides. Because KOH is a strong base, it's pretty accurate to just have the arrow pointing one way in the reaction -- almost all of the KOH disassociates. For weak acids and bases, instead of an arrow pointing just in one direction, you need a bidirectional arrow to show that both sides of the equation are present in the solution. For example,\nH2CO3 <---> H+ + HCO3-
335806
Because KOH is virtually completely soluble, there is no real Kb value for it.  A Kb value is calculated by multiplying the concentrations of the ions that KOH forms in solution, then dividing that number by the concentration of actual KOH still bonded together in that solution, which is almost nothing.  Mathematicians will tell you that dividing by a number close to zero will give you a number very close to approaching infinity, so the Kb value for KOH that you find in any table or whatever will never give you a number; it will probably just say "very large."
551370
Well, the equation is..\nCH3COOH + NaHCO3 ----> CH3COONa (Sodium Acetate) + H2CO3 (Carbonic Acid) \n(or)\nCH3COOH + NaHCO3 ----> CH3COONa (Sodium Acetate) + H2O (Water) + CO2 (Carbon-dioxide)\n\nSo you might have noticed the equation is already balanced. It is a double-displacement reaction. So the sides are already balanced.\nHope this helps.\n\nOne small note...I don't think there is a compound such as NaCO3. The valency of CO3(^-2) ion is 2, so it has to combine with two Sodiums (two positive charges to neutralise its two negative charges) unless the compound one of the above users mentioned is NaCO3-.\n\nrule wid style n luv@vinny
459648
Themes > Science > Chemistry > Miscellenous > Help file Index > Precipitation Equilibrium > Ksp and Ion Concentration \n\nThe solubility product constant can be used to compute the equilibrium concentration of ions in solution in the same way that you use other equilibrium constant expressions to compute equilibrium concentrations. \n\nUse the Initial, Change, Equilibrium method to compute the concentrations: write the equilibrium expression, the initial amounts of each thing in the expression, how much they change by, then the equilibrium amounts. From that, you can compute the concentrations of all species. \n\nExample: BaF2 has a Ksp of 1.8*10-7. What is the concentration of barium in a saturated solution of BaF2? \n\nSolution:. First, write the balanced equation and the solubility constant expression: \n\nBaF2(s) < = > Ba+2(aq) + 2F-(aq) \nKsp = [Ba+2][F-]2\nWe start with no barium or fluoride ions in solution. If we set the amount of barium ions formed to x, then the number of flouride ions formed is just 2x \n\n [Ba+2] (M) [F-] (M) \nInitial 0 0 \nChange +x +2x \nEquilibrium x 2x \n\n\nNow plug in the equilibrium values shown above and solve for x \nKsp = [Ba+2][F-]2 \nKsp = (x)(2x)2 \nx = 3.6*10-3\nSince we defined the concentration of Ba+2 to be x, we've solved the problem \n[Ba+2] = 3.6*10-3 \n\nInformation provided by: http://learn.chem.vt.edu\n \n**try to visit the site below..** You can calculate the Ksp based on the above example..I'm sure the answer is in your books and notes..have fun learning..
288115
It would largely depend on the polarity of the bond.\n\nEver hear of the "like dissolves like" rule?\nWater, being a polar molecule, will dissolve other polar molecules as well as ionic compounds.  If the colvalently bonded molecule you are trying to dissolve is polar, you have a good chance at dissolving it in water (actual solubility values will varry).  However, if the molecule is non-polar, the molecule is not going to be soluble.\n\nExample,\nWater will easily dissolve "Salt" (NaCl) since it is an ioic compound.\nWater will dissolve alcohol since alcohol has an OH group, making it polar.\nWater will not dissolve gassoline for example since it is a non-polar hydrocarbon (that is why you cant mix oil and water).
887797
It looks like you got it. This is how it came out for me:\n\nThis is the reaction:\nC10H8(s) + 12 O2(g) ® 10 CO2(g) + 4 H2O(l)\n\nand Hc=(Summation of Hf of products)-(Summation of Hf of reactants)\nSo...\n\nHc (C10H8) = 10 Hf(CO2) + 4 Hf(H2O) – Hf(C10H8)\n-5156.8kJ/mol=(10)(-393.5 kJ/mol)+(4)(-285.9kJ/mol0-Hf (C10H8)\n-5156.8kJ/mol=(-3935 kJ/mol)+(-1143.6 kJ/mol)- Hf (C10H8)\n-5156.8kJ/mol=(-5078.6)- Hf(C10H8)\n-78.2kJ/mol=-Hf (C10H8)\n78.2kJ/mol=Hf (C10H8)\n\nI believe that's how it's done, but it's been a while since I've done that kind of problem, so it's probably not a terrible idea to double check with someone else.
470703
reduced: 2H2O + 2e- ---> H2 + 2OH-\n\noxidized: 2I- ---> I2 +2e-
930806
When you want to balance ANY chemical equation, always follow these steps:\n\n1. Balance metals\n2. Balance non-metals\n3. Balance hydrogen\n4. Balance oxygen\n\nLet's do it\n\nFirst let's balance K (metal):\n\n"2" KHCO3 + __H3PO4 yields __K2HPO4 + __H2O + __CO2\n\ndone. now let's balance P (non metal): if you check, it is balanced already.\n\nnow C (also non metal):\n\n2 KHCO3 + __H3PO4 yields __K2HPO4 + __H2O + "2" CO2\n\nnow hydrogen:\n\n2 KHCO3 + __H3PO4 yields __K2HPO4 + "2" H2O + 2 CO2\n\nfinally, oxygen (its already balanced!!)\n\nSo you end with\n\n2 KHCO3 + H3PO4 yields K2HPO4 + 2 H2O + 2 CO2
270274
ur equation is wrong....\nu will get H2O + C...\ncheck again\nif it's correct, u have to find out which one is the limiting reagent by calculating the number of moles of each reactant used...thus figuring out the number of moles of product and multiplying with the molecular mass of methanol...
777672
S2O3(2-)-->SO4(2-)          balance S first\nS2O3(2-)-->2SO4(2-)         balance O with water\n5H2O+S2O3(2-)-->2SO4(2-)        balance H with H+\n5H2O+S2O3(2-)-->10H++2SO4(2-)       balance charge with e-\n\n5H2O+S2O3(2-)-->10H++2SO4(2-)\n                                        +8e-\nthe 8e- is with the product side
224213
well, pH = -log[H+]. so if you just undo the log you will have the concentration of H+ (which is the same as H30+). then .. [H+} + [OH-] = 14 good luck!
946866
11/27:Please post the answer so I can find the error in my work or determine if the answer key is wrong.  Lets see if we can get to the bottom of this one!\n\n\nFirst find the masses of each acid:\n\nHCl:  48.8 ml x 1.19 g/ml x 0.38 = 22.1 g\n\nHNO3:  17.5ml x 1.42 g/ml x 0.70 = 17.4 g\n\nSecond, convert grams to moles:\n\nMass HCl = 35.45 + 1.01 = 36.46 = 36.5 g/mol\nMass HNO3 = 16.0 x3 + 14.0 + 1.0 = 63.0 g/mol\n\nmoles HCl = 22.1 g/36.5 g/mol = 0.605 mol HCl\nmoles HNO3=17.4g/63.0 g/mol = 0.276 mol HNO3\n\nSince you're diluting to one liter, these are also the molarities.\n\n[HCl] = 0.605 mol/L ; [HNO3] = 0.276 mol/L\n\nEach of these acids are strong acids and can be assumed to dissociate completely.  If they were weak acids we would have to account for the acid dissociation equilibria here.  \n\nEach of these acids contribute only 1 equivalent of hydrogen ion.  Thus [H+] is the sum of the two concentrations.\n\n[H+] = [HCl] + [HNO3] = 0.605 + 0.276 = 0.881 mol/L = 0.88 M\n\nReport two sig figs since your mass % vals only have two sig figs.\n\nThe equilibrium equation for water, H+ and OH- is:\n\nH2O <--->  H+  + OH-   K = 1.00 x 10^-14\n\nH2O is a pure liquid and by definition has an activity of unity, so the equilibrium expression is:\n\n[H+][OH-] = 1.00 x 10^-14\n\nNow we can solve for [OH-] :\n\n0.881 X [OH-] = 1.00 x 10^-14\n\n[OH-] = 1.14 x 10^-14 mol/l = 1.1 x 10^-14 M\n\nAgain, two sig figs in final answer.\n\nFinally, pH = - log [H+]\n\nso pH = - log 0.881 = 0.055
250273
Enough to cancel out the H's. For the reaction with 2H's, add 2OH's to both sides. For the reaction with 3H's, ad 3OH's to both sides.\n\nH + OH --> H2O\n\nEDIT: Oh, well in that case, the H's cancel out, so you have just one H on the side that previously had 3. Then, just add one OH to both sides.
275919
2H20 --> 2H2 + O2\n\ntheoreticalyy spk it's exactly half the vol of H2, just look over the stoechiometrical balance above; but due to gas losses, it may be less
700313
NaOH + K+ = Na+ + KOH
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
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]