How many moles of ions are present when .10 moles of the following compounds are dissolved in WATER?
612107
a. 2k+ = .20moles , SO42-=.10moles\n\nb.Fe3+=.10moles,3NO3-= .30moles\n\nc.2Al3+=.20moles,3SO42-=.30moles\n\nd. Ni2+=.10moles,SO42-=.10moles
770958
1/1000 x molar mass (grams/litre) and solute in 1 litre
884150
Well, in pure water, if you let x = [Mg2+] the [OH-] = 2x.  Per the equilibrium expression 5.61 x 10E-11 = 4xE3 so x=2.4 X 10E-4 M\n\nIn the NaOH solution, the [OH-] is 0.180 M (the contribution from the Mg(OH)2 is negligible).  So in this case\n                5.61 x 10E-11 = [Mg](.180)E2  \nSolving for [Mg], I get 1.7 X 10E-9 M\n\nCheck my arithmetic, I often get careless punching the calculator.
781203
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
904575
pH=-log(H+)\nthen H+ or H3O+ if done in water is:\n10^(-1*pH)\nFor plugging in 7.5 in above returns: 3.16 x 10 -8 M (molar concentration)\nsimilarly pH 9.6= 2.51 x 10 -10 M
1173010
If you add one molar **water** to the equation?  There would be no change, since the reaction is already occurring in water!\n\nWater, by the way, is by definition 55.56M -- there are 55.56 moles of water in every liter of water (1000 g/18 g/mole).
553835
Well to determine [H+] you use the equation that \npH=-log [H+] \nthus a solution with pH=3 will have a [H+] of 1x10^-3, \nand a solution with pH=8 will have a [H+] of 1x10^-8\nTo find [OH-] there are several equations you can use\n[OH-]=Kw/[H+], \nKw is the ionization constant of water which is equal to 1.0x10^-14\nThus you will have [OH-] of 1.0x10^-11 and 1.0x10^-6\nSo you answer will be that the solution with pH=3 will have a [H+]>[OH-]
954955
HCl_aq + NaOH_aq -> NaCl_aq + H2O_l\n(H+)_aq + (Cl-)_aq + (Na+)_aq + (OH-)_aq -> (Na+)_aq + (Cl-)_aq + H2O_l\n(H+)_aq + (OH-)_aq -> H2O_l\n\nFor all the other equations, the (Cl-), (K+), (NO3-), (Na+) are spectator ions, and the n.i.e is the same:\n(H+)_aq + (OH-)_aq -> H2O_l
379843
Using the formula for pH: pH = -log(H+)\nwe can find the molarity\n\nFirst, we plug in what we know\nThe pH is 2.65\nand the H+ is the concentration of hydrogen ions measured in moles/ liter. Hey, that's what we're looking for!\n\n2.65 = -log(moles/liter)\n\nusing algebra we can solve for moles/liter\n\n-2.65 = log(moles/liter)\n\nLet's raise 10 to the power of each side\n\n10^(-2.65) = 10 raised to the negative 2.65 power\n10^(log[moles/liter]) = 10 raised to the power of log(...)\n\nOk, so..\n\n10^(-2.65) =10^(log[moles/liter])\n\nNow 10 with log as a power cancel each other out so...\n\n10^(-2.65) = moles/liter\n\nmoles/liter is the molarity, let's call it "M"\n\nM = 10^(-2.65) \n\nM = approx. 3.16 x 10^(-3)
684276
Answer is A.\n\nNote that the other three are listed as (aq) = aqueous?  That means they are ions in solution.  But PbI2 is listed as (s) = solid = precipitate\n\nIn truth, even the PbI2 has SOME ions in solution, but the solubility product is 7.47x10^-9, so there isn't very much.
725707
number of moles of NaCl = Mass/Mr\nalso\n\nNumber moles NaCl = molarity X vol(cm3)/1000\n\ntherefore\n\nMass/Mr = molarity X vol (cm3)/1000\n\nSo, if you rearrange the above:\n\nMass NaCl you need to dissolve = Mr X molarity X vol(cm3)/1000\n\nIn the question molarity is 2M and the volume is 300cm3.  I'll let you figure out the Mr of NaCl (just add together the relative atomic masses of Na [23] and Cl[35.5] and arrive at your answer.\n\nHope it helps.
445882
the formula mass of Cu(NO3)2 is 187.55g/mol\n\nthe mass of anhydrous salt is 3.9g therefore the number of moles of Cu(NO3)2 is 0.02079mols.\n\nThe mass of water lost was 1.1g, as the formula mass of water is 18.02g/mol then 0.0610mols of water were released.\n\n0.0610/0.02079=2.9\n\nSo the answer is 3.
921604
theoretically yes 10m acid has a pH of -1, and 10 moles of base pH 15.\n\nBut , in fact, for very high concentrations, you must not take in account the concentration of ions H+ but the activity . The actvity takes in account the high concentration and is \n\nactivity = gamma * pH where gamma is lower than one for high concentrations
786992
b) small fraction of the maximum number of possible hydronium ions, H3O+.\n\nHNO2 + H2O <===> H3O+ + NO2-
853922
These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
411083
If the pH is 4, the pOH is 10 (pH + pOH = 14).  10^(-pOH) = [OH-].\n\nAnswer is 1E-10
692123
1. pH = -log [H+] which is the hydronium ion concentration.  [H+] = 10^-2.85 = 0.00141 M\n[OH-] = 1 x 10^-14/0.00141 = 7.09 x 10^-12\n\n2.  pH = -log 12 = -1.08\n\n3.  neutral water would be when [H+] = [OH-], so [H+] would be the square root of Kw, so [H+] = 1.58 x 10^-7.  pH = 6.80.