how do u calculate the theoretical % P2O5 in a KH2PO4 sample?
149084
I am not very sure but here it goes: Calculate the molecular weight(MW) of P2O5 and of KH2PO4. Then divide the MW of P2O5 with the MW of KH2PO4. Multiply the result by 100. The answer is the percentage of P2O5 present in KH2PO4! The way I just described is about how to find out the actual percentage. In order to calculate theoretical percentage you use a different formula: (Theoretical - Actual)/ Theoretical. Once again multiply the result by a 100! But in order to use the previous formula you must know both the values! Anyhow, without confusing you any further, I hope this helped!
315574
I think this is how you do it:\nPb + F --> PbF\n\nSo Ksp = [Pb][F]\nBecause they react in a 1 to 1 ratio, their concentrations will be the same.\nSo Ksp = X^2 or [Pb]=[F]=(3.2E-8)^2\n-Not 100% sure but I think thats it
494638
P3O10 does not exist; it's P4O10.\n\nBalanced equation is:\n\nP4O10 + 6H2O --->  4H3PO4
950743
theoretical amount of product=10g lead acetate/ mw lead acetate x mw lead iodide\n\ntake the result of this formula as the denominator and divide 5.00 grams product/theoretical x 100%
942373
From the reaction\n\nS8(g) ===> 4S2(g)\n\nKp = (pS2(g))4/pS8(g) where p is the partial pressure\n\nGiven the partial pressure of S8(g) at equilbrium 0.25 atm, clearly 0.75 atm of S8(g) has been consumed to produce (from the equation) 4 volumes of S2(g), Given the volume was fixed this mean dS2(g) = 3 atm\n\nSubstituting into Kp yields\n\nKp =3^4/0.25 = 324\n\nThus Kp at 1325 K is 324
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!
861843
Start with the Ca then balance the number of PO4 ions, then balance the Oxygen and the hydrogens should balance themselves out.  Its just trial and error. Monkey with it until it works.
904431
Just the take the [H+] ion concentration divided by the initial [HF]:\nfrom your info i gather it should be (10^-1.59)/1.0 \n\nConvert it into a percentage.
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.
444216
Write the Kc and the Kp\n\nChange PV=nRT to P/RT=(n/V) to P/RT=(concentration) note n/V is concentration.\n\n\nSubstitute P/RT for each concentration in the Kc.  You will see the Kp in the equation with extra R's and T's.  Finaih solving.
914017
I will suggest that first you write the equation in a compact mode, balance it and then add all the bells and whistles.\n\nCuSO4 +K3PO4 ---> Cu3(PO4)2 + K2SO4\n\n3CuSO4 +2K3PO4 ---> Cu3(PO4)2 + 3K2SO4\n\n3CuSO4(aq) +2K3PO4(s) ---> Cu3(PO4)2(s) + 3K2SO4(aq)\n\nNow that it is balanced it is easier to break down:\n\n3Cu(+2) + 3SO4(-2) + 2K3PO4(s) --->  Cu3(PO4)2(s) + 6K(+) + 3SO4(-2)
141442
KE per mole = (3/2) RT where R = 8.3145 J/mole*K and T is temperature in kalvin, so:\n\nKE = (3/2) RTn where n is the number of moles of gas\n\nYou can solve PV=nRT for n and substitute to get KE = (3/2) PV where P is pressure and V is volume.
235216
mass P = 283.9 * .4365 = 123.92g P\n\n  moles of P =  123.92/30.97 = 4\n\n so = P4\n\nunknown molecular mass = 283.9 - 123.92 = 160g\n\n P has a charge of 5+ times 4 = total charge of 20+\n\n    \n20 can divide into 1-,2-,4-,5-,10-,20- evenly \n   \n the trial and error method\n   \n   160g /10 =16\n\nthe unknown F.W. is 16g which is Oxygen that has an anion charge  Of 2-\n  \n so (5+ * 4 ) + ( 2- *10) = 0\n \n    oxygen moles = 160 / 16  = 10\n\n     O  =  O10\n\n the molecular formula is:\n\n\n P4O10
694411
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
715055
First off, you need the concentration of the NaOH solution.  From that you can determine the number of moles of NaOH that were used.  This is equal to volume (V) time the concentration, in mol/L (C).\nn = V * C = 0.02260 L * ? mol/L = ? mol\nn = 0.02200 L * ? mol/L = ? mol\n\nThe reaction of KHP (C8H5O4K) and NaOH is this:\nHOOC-C6H4-COOK + NaOH -> NaOOC-C6H4-COOK + H2O\nSo, 1 mol of NaOH will react with 1 mol of KHP.\nThat means that the number of moles of NaOH you calculated above is also equal to the number of moles of KHP.\nThe molecular weight of KHP is 12*8 + 1*5 + 16*4 + 39 = 204 g/mol\nMultiply that by the number of moles: 204 g/mol * ? mol = g actual KHP\n\nYou can divide that number by the 0.911 and 0.950 g and then multiply by 100% to find the purity.\n\nHere’s an example if your NaOH solution was 0.1 M (M means ‘molar,’ or mol/L):\nn = 0.02260 L * 0.1M = 0.002260 mol of NaOH = 0.002260 mol of KHP\n204 g/mol * 0.002260 mol of KHP = 0.461 g pure KHP\n0.461 g pure KHP / 0.911 g impure KHP * 100% = 50.6 % KHP\n\nHope this helps!
979796
2H3PO4+6NaOH-------->6H2O+2Na3PO4\n\nThere has to be two moles of phosphoric acid and six moles of sodium hydroxide for it to balance out
504715
You need the formula:\ndelta Tf=iKfXmolality\nyour delta Tf is 1.96\nyour i is 1 (because it does not dissociate)\nyour Kf is 5.12\nmolality is moles of solute divided by kilogram solvent, which is .0250 kg\nYour final equation should read:\n1.96=(1)(5.12)(x/.0250)\nThen use algebra to find x.