Consider the flask apparatus below, which contains 2.00 L of H2 at a pressure of 372 torr and 1.00 L of N2 at?
957415
Use PV = n RT, RT = constant. Therefore n = PV * constant in the three cases.\nWe have \nH2 :  P = 372 torr V = 2 liter  n_H2=  744 * constant\nN2:  P = p torr V = 1 liter  n_N2 = p * constant\nCombined: P = 303 torr V=3 liter  n_combined = 909 * constant\n\nTherefore  744 + p = 909 => p = 165 torr
984937
The equation is: 2H2 + O2 ---> 2H2O\nSo you would form 2L of water VAPOR\nIf you want this in terms of LIQUID water, remember that one mol of a gas occupies 22.4L, so you have 2/22.4 mols of water (MW=18) formed, so...\n2/22.4 x 18 = 1.6 grams or 0.0016L of LIQUID water.\nThis assumes that all three gasses act as 'ideal gasses' (which is only close) and conditions are reduced to STP...so answer is not exactly correct, but 'close enough for government work'  :-))
1151439
Ideal Gas Law: PV = nRT\n\nYou are holding n and T constant, and R is a constant, so:\n\nP1V1 = P2V2\n\n(750 torr)(500ml) = (x torr)(1500 ml)\n\nx torr = (750 torr)*(500/1500) = 750 torr *1/3 = 250 torr
883737
Hydrogen gas molecules, H2
552193
Can be solved by simply plugging values into the Ideal Gas equation without using two states.  Pressure must be converted to atmospheres, however, and temperature to Kelvin.  P = 1.243 atm, T = 306 K\n\nn = PV/RT = 1.243 * 2.74 / (0.08206 * 306) = 0.136 mol\n\nmass = moles * MM = 0.136 * 28.01 = 3.81 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.
918910
See if this helps:\nPV = nRT = NkT, where P is pressure, V is volume, n is the number of moles, N is the total number of molecules, T is temperature, k = 1:38 £ 10¡23 J/K is the Boltzmann constant and R is a universal constant, R = 8:314 J/mol.K.
105912
1 atm. = 760.0 mM Hg = 760.0 Torr = 14.70 psi = 1.013 Bars = 1033 cM water = 101.325 kPa.\n\n745 mM Hg = 745/760 = 0.983 or so atmospheres.\n\nA great conversion program can be found at:
849763
Water vapor.\n2 H2 + O2 = 2 H2O
792353
P1V1/T1=P2V2/T2\n\nV2=P1V1T2/(T1P2)\n\nV2=25 Liters* 1 atm* 298 Kelvin/( (19.71/760)atm*373 kelvin)\n\nV2= 770 Liters\n\nuse temperature in kelvin and convert mm Hg to atm  760 mm Hg in 1 atm
887467
Im a little rusty on the thermodynamics, but lets see if this helps.\nV/T (initial)=V/T (final) since no gas is lost or added  Raising the temperature on a container that is free to expand does not change the pressure, so you can find the change in volume with temperature.  Since V=A*h, you can find h.\n\nNow, the problem is that the vessel is not entirely free to expand, and acts on a spring, with force equal to F=k*h.  This force is related to the pressure created by the expanding vessel as simply F/A.\n\nAll values are given, so I think you are set.
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
385149
density of H2 at STP = d= PM /RT\n\nd= density  =X\n\nP= pressure  =1atm\n\nM= molar mass = 2.016g of H2\n\nR= gas constant = .0821\n\nT=temperature  = 273K\n\n\nX= .0899g/L\n\n\n 1.6g H2 corrected to STP: \n\n\n1.6gH2(.0899g/L / 2.016gH2)= .0713g/L
828582
Remember, all ideal gas molecules exert the same pressure in the same situation, so the partial pressure of oxygen is the same fraction of the total pressure as oxygen molecules are of the total molecules.  Since 21% of the molecules are O2, 21% of the pressure comes from O2.  So the partial pressure is 0.21(751) or 157.71 torr.
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?
726095
3.19/18 moles of H2, so double that for moles of H atoms.
766934
Assume it is an ideal gas and use PV = nRT\n\nP is in atm 1 atm = 760 torr\n    225 torr / 760 torr = 0.296 atm\nV is in L so 5.7 L\nR is 0.0821 atm*L/mol*K\nT is in K so C + 273 = K\n     27 + 273 = 300 K\n\n(0.296)(5.7) = (0.0821)(300)n\nn = 0.0685 mols