Max Mass of Trichlorophenol....?
926524
1 mole of phenol produces 1 mole of trichlorophenol.\nChange to grms.\n94 grms of phenol makes 197.5 grm off trichlorophenol.\nSo 10 grms makes a max of 10/94 x 197.5 = about 20 grms (no calculator available on this computer)\nIf you actually make 19,54 grms this is 19.54/20 x 100 percentage of what you could get in ideal circumstances. About 97% (again do this arithmetic)\nYou know this stuff is sometimes called TCP and is a good antiseptic. Best wishes from a British chemist.
903371
I have a serious feeling that you haven't given us all the info.\nIf you know the density of HCl you can get the first one.\nBut we don't have enough info to even begin to tell you the relevance of the NaOH.   More info, or concentrate on your own homework...
316922
Ok.... \nfirst: 1atm = 760 torr\n Now, as per the gaseous  STP conditions., 1 mole of oxygen will occupy 22.4 lit volume at STP conditions.\nSTP: T=273K\n        P= 760 torr\n       V =22.4 lit\n       Mass of O2 at STP = 32g\n\nNow. second set of conditons:\n P2= 1400 torr\nT2 =41 K\nMass= ?\n\nPV = nRT\nPV = (w/Mwt) XRT\n\nP1/w1T1 = (P2 /W2 T2)\n\nSubstitute these values;;\nW2= 392g of oxygen.\n\nIs the solution is clear enough?
847153
PV = nRT  .... n/R = P/RT\n\n800 torr x (1.3158×10^−3 atm/1 torr) = 1.053 atm\n\nn/R = (1.053 atm)/(8.206 x 10^-2 L atm/Kmol  x 298 K)\nn/R = 0.043 mol/L\n\nNH3 mass is (14 + 1x3)=17 g/mol\n\ndensity = n/R but we need it in g/mL\nn/R = 0.043 mol/L x (17 g/mol NH3) x (1L/1000mL)\nn/R = 7.31 x 10^-4 g/mL
9807
Well lots of chemistry questions lately....finals coming up? :)  Good Luck!!!  \n\nTo calculate this type of question first of all calculate the #of moles of HCl used.  So \nyou used 2100 mL = 2.1 L HCl \nThis HCl is 0.3 mol/L so you used:\n\n2.1 L HCl* (0.3 mol HCl/L) = 0.63 mol HCl\n\nnow use the ratio of Cl/HCl  (which is 1 to 1)\n\n0.63 mol HCL *(1mol Cl/1mol HCl)= 0.63 mol Cl ion\n\nSo far so good I hope.  \n\nNow do your solution diltion to calculate volume,\n2.1L + .2385L = 2.3385 L of mixed solution\n\nNow simply divide \n0.63 mol Cl/2.3385 L solution = 0.269 mol Cl ion in final solution.\n\nAgain these numbers are quick and dirty but the method of solving this type of problem is the same.
874480
2AgNO3 + CaCl2 --> 2AgCl + Ca(NO3)2\n\nMoles of AgNO3 given\n= (115/1000) dm3 x (0.23) mol/dm3 = 0.02645 mol\n\nMoles of CaCl2 given\n= (115/1000) dm3 x (0.14) mol/dm3 = 0.0161 mol\n\n2 moles of AgNO3 needs 1 mole of CaCl2 for complete precipitation. Hence for 0.02645 mol of AgNO3  provided, (0.02645/2 = 0.013225) mole of CaCl2 is needed.\n\n2 moles of AgNO3 produced 2 moles of AgCl. Hence 0.02645 mole of AgCl is produced.\nMass of 0.02645 mole of AgCl\n= (0.02645 x Mr of AgCl) = 0.02645 x 143.5 = 3.80g (to 3 s.f.)\n\n0.02645 mole of AgNO3 produces 0.02645 mole of Ag+ and 0.02645 mole of nitrate ions.\n0.0161 mole of CaCl2 produces 0.0161 mole of Ca2+ and (0.0161 x 2 = 0.0322) moles of Cl-.\n\nAfter precipitation, with AgNO3 as the limiting reagent,\nMoles of Ag+ left = 0 (all precipitated as AgCl)\nMoles of Cl- left = 0.0322 - 0.02645 = 0.00575\nMoles of nitrate ions left = 0.02645 (nitrate not precipitated)\nMoles of Ca2+ = 0.0161 (Ca2+ not precipitated either)\n\nTotal volume of solution is now (115 + 115 = 230 ml).\nFor concentration of each ion left, please divide the moles of each ion left by the volume (230 ml) respectively.\n\nHope this helps. :)
759463
Atomic masses\nCu - 63.546g/mol\nN - 14.007g/mol\nO - 15.999g/mol\n\nmolar mass of Cu(NO3)2 - 187.554g/mol\n\nno of moles in 2.35x10^3g = 12.53
836016
Here is a good guide\n\nhttp://dbhs.wvusd.k12.ca.us/webdocs/Stoichiometry/Limiting-Reagent.html
893997
maximum moles per liter or maximum grams per liter would do.\n\nBut since solubility changes with temperature, you need to specify the temperature too, typically 0, 4, or 20 degrees Celsius depending on the solvent.  0 C is the freezing point of water so that is not a good temperature to use for water as a solvent.\n\nThe second poster is correct in saying that density is grams per liter if he means that as a ratio.  But if you know the density and you have a fixed volume you know the amount by mass (grams) dissolved in that volume, and that is the solubility.  It is common practice to express solubility in terms of mass, but since the molar mass provides an easy switch to moles, it is a trivial point of whether it is mass or moles we are talking about.\n\nBut to be exact, according to the CRC, p 8-122, all solubility values are expressed as mass percent of solute, 100xw2, where w2 = m2/(m1 + m2).  m2 is the mass (grams) of the solute and m1 is the mass (grams) of the water, or whatever your solvent is.
91712
83.93 grams of GaCl3 are produced.\nHow?\nWell, note the following:\n35.453 g Cl/mol Cl\n70.91 g cl2/mol Cl2\n69.723 g Ga/mol Ga\n\n1 mole of GaCl3 = 3(35.453 g Cl) + 1(69.723 g Ga) = \n176.082 g GaCl3\n\n(50.7 g Cl2)*(1 mol Cl2 / 70.91 g Cl2)*(2 mol Cl/1 mol Cl2)*(1 mol GaCl3 / 3 mol Cl)*(176.082 g GaCl3 / 1 mol GaCl3) = \n83.93 g GaCl3\n\nThis means that for every 3 moles of Cl used, there will be 1 mole of GaCl3 produced.  It also means that for every mole of Cl2 used, 2 moles of Cl are going into the product.  Therefore for every 3/2 moles of Cl2 used, 1 mole of GaCl3 is produced.
859471
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 9.630 g x (12 / 44) = 2.626 g\nhydrogen: 4.928 g x (2 / 18) = 0.548 g\noxygen: 4.055 - 2.626 - .548 = 0.881 g\n\n2) convert to moles.\ncarbon: 2.626 / 12 = 0.219 mol\nhydrogen: 0.548 / 1 = 0.548 mol\noxygen: 0.881 / 16 = 0.0551 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.219 / 0.0551 = 4\nhydrogen: 0.548  / 0.0551 = 10\noxygen: 0.0551 / 0.0551 = 1\n\nThe empirical formula of the substance is (C4H10O)n\nThe atomic masses add up to 74, which is also the molar mass of the compound. Therefore n=1 and we find C4H10O
381243
Are you actually working on silazanes? They are hot stuff. Wanted to do them too but had problems with shipping of some reagents. Anyway, I was tryig to access this article but I was unsuccessful. Maybe you ca try it yourself: "Rates of methanolysis of several organosilazanes" by William L. Budde, J. Phys. Chem.; 1969; 73(11); 3718-3721. Hope this helps. Or click here: http://pubs.acs.org/cgi-bin/abstract.cgi/jpchax/1969/73/i11/f-pdf/f_j100845a028.pdf\n\nIf you were not able to access this, then contact me so i can send you the pdf file.
804090
ooo this will be fun!\n\nok..\nA)\n2.207g of H2O MW. 18.0\nmol = weight of sample (g) / molar weight (g/mol) (n=m/M)\nn = 2.207 / 18\nn = 0.1226 moles of H2O\n\nnow, since there is 2 hydrogens in H2O, then when the molecule is broken down, there is twice as many moles of hydrogen then the overall molecule,\ntherefore,\n\n2x 0.1226 = 0.2452 moles of H.\n\ndetermining number of moles of CO2 is also helpful,\n4.313g, n = 4.313 / (12.01 + 16 +16) = 0.0980moles\napprox, 5(moles H2O) :  4(moles CO2)\n\nB) \nworking the number of moles of this compound will be useful:\nn = 1.816 / 74.122 = 0.0245 moles, which is 1/10 of the moles of H produced. (or 1:5 of the moles produced of water)\n\nif you think about it, the reaction is something like this:\nCH2 + O2 --> CO2 + H2O\nclearly the above doesn't make sense, because 1) it isn't balanced, and 2) the reactant needs a higher molecular wt.. now thinking about it, the following can be deteremined:\nand from the ratio of moles found above:\n       1     : (excess):  4        :    5\nC4H10 + 13/2O2 --> 4CO2 + 5H2O\n\nnow, I amoung most allow fractions in equations, but if you don't, multiply through by 2.\n2C4H10 + 13O2 --> 8CO2 + 10H2O\n\nnow, you want the empirical forumla,\ntherefore,\ndivide by highest common factor,\nC2H5 \n\nthe empircal fomula is just a ratio and doesn't make huge sense, as the formula for an alkane is CnH2n+2\nhope that makes sense :)
1041359
Apparently, the EPA proposed a limit of 300pci/L 10 years ago (numerous web-based sources including the EPA). The site at this link talks more about the problem and has a lot of information as well as actual data of levels in many supplies.\nhttp://www.ewg.org/tapwater/contaminants/contaminant.php?contamcode=4004
768102
p,p-dichloro diphenyl trichloroethane
701172
1. molar mass of Ar: Mr = 40, so\nm = n x Mr = 8.44 x 40 = 337.6 g\n\n2. molar mass of ZnCl2: Mr = 136.4, so\nn = m/Mr = 288/136.4 = 2.11 mol (approx)\n\n3. The right formula of ethanol is CH3CH2OH and it's molar mass: Mr = 46, so\nm = n x Mr = 6.75 x 46 = 310.5 g\n\n4. Atomic mass of Na: Ar = 23, so we have:\nn = m/Ar = 5.4/23 = 0.235 mol of Na.\nAccording to the chem. equation:\n\n2Na + F2  --> 2NaF\n\n2 mol of Na react with 1 mol of F2, so\n0.235 mol of Na react with 0.235/2 = 0.117 mol of F2\n\nMolar mass of F2: Mr = 19\nm = n x Mr = 0.117 x 19 = 2.23 g of F2
844407
Easy, the molecular weight of HCl is 36.461 gr per mole. \n\nIf your trying to reach 4 molar solution then you need 4 x 36.461g HCL diluted to 1 liter. you need 145.844g HCl in the liter\n\nBut since your probably using 32-38 % HCl (actual amount might be on the bottle) you need to multiply you'll need to adjust for inactives by assuming your active molar mass is ~35% of 36.461 or 12.761g \n\nin order to make 1 liter of 4 molar solution you'll need 416.71g \n\n36.461/12.761 * 145.844 = 416.71\n\nassuming it is 35 % active. \n\nYou should always standardize your solution after mixing to determine final activity!!!!!