Oxygen gas is contained in a cylinder with a volume of 0.750L at a temperature of 25C and a pressure  of?
397497
I would use the ideal gas law...more specifically...the combined ideal gas law...\n\nP1*V1/T1 = P2*V2/T2
732197
The density can be assumed to be 1.0 g/ml.  \n\nSo you have 25 ml x 1.0 g/ml = 25 g H2O\n\nThe molar mass of water is 18 g/mol, so you have:\n\n25 g H2O/18 g H2O/mol H2O = 1.4 mol H2O
173870
1.Since Volume is proportional to Temperature, it could also be written with a constant(k): V = k x T , V/T = k , k = 0.923L/ (25+273)K = 3.10x10^-3 L/K\n2.I'm not too sure about this one, but here is my best shot.\nUsing the ideal gas law PV=nRT, same Temp & Pressure, let Va be the volume of the first baloon and Vb be the volume of the second balloon. Va = 1.25Vb, Let na be the number of moles in baloon 'a' and nb be the number of moles in balloon 'b'.\n\nBalloon a: PVa = naRT\nBalloon b: PVb = nbRT\nsolve for P,T or R.\nThis case I solved for T:\nT = PVa/naR  & T = PVb/nbR, since same temp (or pressure or constant) which is PVa/naR = PVb/nbR. Divide by P & R, Va/na = Vb/nb\nVa x nb = Vb x na, Va = 1.25 Vb, @75%\n1.25(.75)Vb nb = 0.75Vb x na, divide by .75Vb\n1.25nb = na\nSo there is 25% more moles (or mass) in baloon A.
551286
oxygen is made by distilling liquid air. liquid air is regular air cooled to about -200 degrees.  it becomes a liquid.\n\nthe oxygen doesnt need cleaning,  the distillation process removes any impurities.\n\noxygen is put into the cylinders by a machine similar to a super heavyduty air compressor.  the oxygen in a full cylinder is at about 2000 psi.
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
840037
PV = NRT\n\nN - Mols of argon\n\nR = .0821\n T = 26+273.15\n\nV  = 75/1000 to convert to L\n\nThen you will find your P, pressure, in atmospheres.\n\nconvert that to torr
167746
At STP, 1 mole of any ideal gas will have a volume of 22.4 Liters.\nUsing this logic, we can deduce that at STP, and equal volumes of (what we will assume are ideal gasses) gasses will each contain the same number of moles.\nIf each flask contains the same number of moles, then they much each contain the same number of molecules.\n\nBy saying that the flasks are at STP, we are given all the information we need to solve this.  STP (Standard temperature and pressure) implies a temperature of 273 K and a pressure of 1 atm.
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
757981
Ideal gas law\n\nPV=nRT\n\nP is pressure\nV is volume\nn is moles of gas\nR is the ideal gas constant\nT is temperature\n\nGood luck!
791970
Use the Gay-Lussac gas law: V/T is constant. Note that T here is the temperature in Kelvin, not in Celsius. Luckily 27 C = 300 K
569971
This problem only has enough information if you assume it is an ideal gas.  In this case, PV=nRT is the equation of state.\n\nn and R are constants in this case, so PV/T must also be constant.  These values must also all be in absolute scales - the celcius must be converted to kelvin or any other scale based on absolute zero.  This means:\n\nP(i)V(i)/T(i)=P(f)V(f)/T(f)\nV(f)=T(f)P(i)V(i)/[P(f)T(i)]\n    =(250+273)K(740mmHg)(410mL)/[(680mmHg)(27+273)K]\n    =[523*740*410/(680*300)]mL\n    =777.83mL\n\nBy the significant digits you've presented, this can only be evaluated to any accuracy as 780mL.  (two significant digits)
336167
oxygen and carbon dioxide
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.
924197
PCO2 = 35-45\nPO2 = 75-100
774704
S + O2 --> SO2\n\nmole of S needed to be converted = 5.00 / 32 = 0.15 mole\nmole of O2 = 0.15 mole\nBy using ideal gas theory:\nP * V = n * R * T\n5.15 * V = 0.15 * 0.082 * 310\nV = 0.74 litre\n\npeace\nvixklen
570434
when all parameters r given use ideal gas formula i.e;pv=nrt when temp is kept constant use boyles law ie pv=constant or while solving problems use the form p1.v1=p2.v2 and when pressure is kept constant use charles law and when volume is kept const use pressure temp law morover when u hav rms velocity(=c) given use kinetic gas equation ie pv=mn(c.c)/3 [m=mass of gas molecule,n=total no of gas molecule]
939108
Given:\nPV = nRT, 862 torr x (1 bar/750torr) = 1.15 bar, 117 C +273 = 390 K, CH4 = 16 g/mol\n\nn/V = P/(RT)\n\nn/V = (1.15bar)/[(0.08315 L bar/K mol)x(390K)\nn/V = 0.035 mol/L x (16 g/mol) x (1L/1000mL)x(1mL/1cm^3)\ng/cm^3 = 5.67 x 10^-4 g/cm^3