How many moles are there in .007 L H2 and .004 L O2?
548946
22.4 L = 1 mol\nConvert the liters to moles, and then moles to grams with\n 16g = 1 mol\nThen, if you need to find atoms, it's \n6.02 x 10^23 atoms = 1 mol
805169
alright...\nconsider the R.H.S.\nh2o has two o-h bonds which account for 2x467 and since thre are 2 mol. u get 2x2x467--------(1)\nco2 has two c=o bonds which account for 2x799 and since u hav 4 mol. u get\n4x2x799-------------(2)\naddin 1 and 2 u get 1868+6392=8260\nconsider the L.H.S.\nu hav in c2h2, two c-h bonds which account for 2x413--------(3)\nand one triple bond accounting for 1x839---------(4)\nnow u hav 2x(3+4)=2x(826+839)=3330------(5)\nand u hav 5x(o=o) which acc for 5x(495)=2475----(6)\naddin 5 and 6 u get 5805\ntherefore ur ans is 5805-8260=-2455\nand since u want for one mol of acetylene u get -1227.5 kj/mole
524657
molarity = moles/volume\n\n125 g C2H5OH / 46 g/mol = 2.717 moles\n\nmolarity = 2.717/0.250 = 10.87 M
384394
Well, let's see:  you're releasing 1000 kg/hr of liquid N2, which equilibrates to 1 atmosphere and 20C (STP).\n\nAt STP, one mole of a gas occupies 22.4 liters;\nyou're releasing 1000 kg/hr = (1000 kg*1000g/kg)/28g/mole = 35,714 moles of N2 per hour, and thus:\n\n22.4 liters/mole * 35,714 moles = 800,000 liters (at STP) per hour
502184
5H2O2(aq)+2KMnO4(aq)+\n3H2SO4(aq)-->\n2MnSO4(aq)+K2SO4(aq)+\n8H2O(l)+5O2(g)
847416
With only 2 gases, the partial pressure of the O2 is 0.440 atm - 0.160 atm = 0.280 atm O2\n\nThe temperature is 40°C, or (40°C + 273°K) = 313°K\n\nThe volume is 478 mL or 0.478 L\n\nYou will use the ideal gas law to calculate the number of moles (n) of O2:\n\nPV = nRT\n\nUse the partial pressure for P.\n\n0.280 atm * 0.478 L = n * R * 313°K\n\nR = 8.2057 E-2 L * atm / (mol * °K)\n(this will probably be in your textbook's appendix, but you can also search the internet for "gas constant")\n\n0.280 atm * 0.478 L = n * 8.2057 * 10^-2 L * atm / (mol * °K) * 313°K\n\nn = 0.280 * 0.478 / (.082057 (1/mol) * 313)\n\nn = 0.005211 mol\n\nThe molecular weight of O2 is (16*2) = 32 g/mol\n\nSo, the mass is 32 g/mol * 0.005211 mol = 0.1667 g or 166.7 mg\n\nYou might want to work it out yourself to be sure you can do it.
778656
1 mol of Mg + 2 mol of OH = 1 mol of Mg(OH)2 because there is only one Mg used and 2 OH molecules
781068
The balanced equation for the reaction is:\n\n2 H2O2 --> 2 H2O + O2\n\n1]  If one mole of Oxygen is produced, the 2 moles of Hydrogen peroxide would be needed.  One mole of Hydrogen peroxide contains 2*H + 2*O = 2(1) + 2(16) = 2 + 32 = 34 grams.\n\n2 moles of Hydrogem peroxide = 2*34 grams = 68 grams.\n\n2]  Is it 0.3 moles or 0.3 cubic decimeters of Hydrogen peroxide?\n\nAssuming 0.3 moles, then from the above, 0.15 moles of Oxygen would be produced.  The volume would be:\n\n0.15 moles * 24 cubic dm/mole  = 3.6 cubic dm of Oxygen at rtp.
948311
1) Calculate the number of moles of HCl:\n3.75/(1.008+35.45) = 0.103 mol\n\n2) Calculate the molarity, that is the number of moles per liter:\n0.103 mol / 20.2 x 10^-3 l = 5.1 mol/liter
754271
2H2 + O2 --> 2H2O\nPV = nRT\nP = 2 + 3 = 5\nT = 25 + 273 = 298\n5.00 atm x 55.0 L = n x .08206 x 298\nn = 11.25 mol H2 and O2 mixture\n11.25 mol x (2 mol H2O / 3 mol) = 7.50 mol H2O\n\nP x 55.0 = 7.50 x .08206 x 298\nP = 3.33 atm
468007
At STP, 1 mole of Ne will occupy 22.4 liters\nT1 =273\nP1=760 torr\nV1=22.4 L\nn1=1\n\nT2=300\nP2=750\nn2=2.3 moles\nV2=?\n\n(P1 V1)/ (n1 T1) = (P2 V2)/ (n2 T2)\n\nSubstitute u will get V2= 57.37 L (check this again)\nThis is the solution...\nIs it clear?
961959
0,220M/L*2,20L= 0,484 moles of HCl\n12,00M = 0,484 moles/?\n? = 0,0403 L or 40,3 mL of 12,00 M HCl solution + 2,16 L of pure water, so you can get you 2,20 L of 0,220 M HCl solution
137179
you need the balanced rxn which has a 2:1 ratio of Mg: O2\n 314 gms Mg = 12.9 moles divided by 2 = 6.46 moles O2. At STP that = 144.7 L volume.
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
543389
I will explain it by an example.\nA compound is 75.46% carbon, 4.43% hydrogen, and 20.10% oxygen by mass. It has a molecular weight of 318.31 g/mol. What is the molecular formula for this compound?\n\nGet the mass of each element by assuming a certain overall mass for the sample (100 g is a good mass to assume when working with percentages).\n\n(.7546) (100 g) = 75.46 g C \n(.0443) (100 g) = 4.43 g H\n\n(.2010) (100 g) = 20.10 g O\n\n\n\nConvert the mass of each element to moles.\n\n(75.46 g C) (1 mol/ 12.00 g C) = 6.289 mol C \n(4.43 g H) (1 mol/ 1.008 g H) = 4.39 mol H\n\n(20.10 g O) (1 mol/ 16.00 g O) = 1.256 mol O\n\n\n\nFind the ratio of the moles of each element.\n\n(1.256 mol O)/ (1.256) = 1 mol O \n(6.289 mol C)/ (1.256) = 5.007 mol C\n\n(4.39 mol H)/ (1.256) = 3.50 mol H\n\n\n\nUse the mole ratio to write the empirical fomula. \n\nMultiplying the mole ratios by two to get whole number, the empirical formula becomes: \nC10 H7 O2\n\n\n\nFind the mass of the empirical unit.\n\n10(12.00) + 7(1.008) + 2(16.00) = 159.06 g/mol\n\n\nFigure out how many empirical units are in a molecular unit.\n\n(318.31 g/mol) / (159.06 g/mol) = 2.001 empirical units per molecular unit\n\n\nWrite the molecular formula.\n\nSince there are two empirical units in a molecular unit, the molecular formula is: \nC20 H14 O4\n\nI think,now u got an idea about the compound's formula
759497
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 7.432 g x (12 / 44) = 2.027 g\nhydrogen: 4.057 g x (2 / 18) = 0.451 g\noxygen: 3.383 - 2.027 - .451 = 0.905 g\n\n2) convert to moles.\ncarbon: 2.027 / 12 = 0.169 mol\nhydrogen: 0.451 / 1 = 0.451 mol\noxygen: 0.905 / 16 = 0.0566 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.169 / 0.0566 = 3\nhydrogen: 0.451  / 0.0566 = 8\noxygen: 0.0566 / 0.0566 = 1\n\nThe empirical formula of the substance is (C3H8O)n\nThe atomic masses add up to 60, which is also the molar mass of the compound. Therefore n=1 and we find C3H8O
813197
0.0821 moleP1V1/T1=P2V2/T2\n375*4/293=760*V2/273.5\nV2=375*4*273.5/(293*760)\n22.4 litres contains 1mole\n375*4*273/293*760\n=375*4*273/293*760*22.4\n=0.0821 mole