Given the following conditions for an ideal gas, what is the number of moles of gas ?
921632
PV = NRT, solve for N, note that T is absolute temperature in kelvins
918968
(See answer to very similar question at the second source listed below)\n\nAssuming ideal gas behavior, we have that the molar density (moles per unit volume) of a gas is given by:\n\nmolar density = n/V = P/(R*T)\n\nwhere R is the gas constant = 62.363 (liter*torr)/(mol*K) -see source\nP is the pressure\nT is the absolute temperature\n\nTo convert from molar density to mass density (mass per unit volume) we simply multiply by the molecular mass of the material in question. In this case, the molecular mass of propane is 44.1. \n\nPutting all this together, and plugging in values:\n\nmass density = (44.1 gm/mol)*(874 torr)/((62.363 liter*torr/mol*K)* (273.15+95)K)\n\nmass density = 1.679 gm/liter = 1.679e-3 gm/cm^3
876756
Depends on what base was used.  Was it NaOH?\n\nOne mole of anything is 6.02 x 1023 (Avogadro's number) of that thing. One mole of an element is 6.02 x 1023 atoms of that element and is a mass of that element equal to its atomic weight in grams. One mole of a compound is 6.02 x 1023 molecules of that compound and equals its formula weight (the sum of the atomic weights of all the atoms in the formula) in grams. For example, 1 mole of C = 12 g C = 6.02 x 1023 atoms; 1 mole CO2 = 44 g CO2 = 6.02 x 1023 molecules of CO2. And (grams of compound) × (formula weight) = number of moles of compound (number of moles of compound) × (formula weight) = grams of compound.\n\n\nNa = 23\nO = 16\nH = 1\n1 mole of NaOH = 40 grams
434788
well, the ideal gas law is PV = mRT.  You need to write this expression for the butane that escaped the lighter, because we don't have enough information about the butane still in the lighter to solve the problem (we don't know the pressure).\n\nFirst you need to convert all your units into the units found in your gas constant, R. looking at the information given, you can see that you'll need the gas constant with units of PV/(mT).\n\nR for an individual gas is the universal gas constant, R_u, divided by the molar mass, which is what we're trying to find. \n\nR = R_u/M\nR_u = 8.314 kPa*m^3/(kmol*K)   [I chose this form because it most closely matched the form I know I need]\n\nThis means that really, the ideal gas law is:\n\nPV = m(R_u)T/M\n\nConversions:\nvolume:  1m^3* = 1000 L\n           477ml = 0.477L = 0.000477 m^3\n-note: you need to know the volume of butane that escaped the lighter.  To do this, find the density of butane (mass/volume) of the origional amount of butatane, and then divide the mass that escaped by the density. You're assuming that the density is constant, because we don't know the volume of butatane that escaped the lighter, or the pressure inside the lighter.  If you have to list sources of error, this would be one.\n\ndensity  = (0.01311 kg/0.000477m^3) = 27.28 kg/m^3\n\nV = 0.00072 kg/27.28 kg/m^3 = 2.6 x 10^-5 m^3\n              (see how the kg units cancel, leaving m^3)\n             \nTemperature:\n23 degrees Celsius  = 23 kelvin\n\nPressure:\n100.33 kPa  - this would be about atmospheric pressure\n\nMass:\n1kg = 1000g\n\n13.11g = 0.01311 kg\n12.39g = 0.01239 kg \n0.72g = 0.00072 kg\n\n\nPutting the numbers in pv  = mrT, with units :\n\n(100.33 kPa)(2.6x10^-5 m^3)=0.00072 kg*(8.314 kPa*m^3/(kmol*K))*(23 K)/M \n\nI would write all that down on a piece of paper, so you can see it more clearly.  The key to getting this kind of problem right is making sure that all your units cancel, leaving you with only the units you want- in this case, kg/kmol.\n\nSolving for M, and canceling all the units, I got:\n\n0.0189 kmol/kg = 1/M\n\nM = 52.9 kg/kmol\n\n\nAccording to my tables, the actual value for Butane is 58.124 kg/kmol, so this is pretty close!\n\n\nHope this helped.  Feel free to email me if you need more help.
53064
Assuming your gas is an ideal gas, you can use the ideal gas law.\n\nPV=nRT where \nP=Pressure in atm (1 atm= 760 mm Hg so 730 mm Hg= 0.96atm\nV=volume in L\nn= moles\nT=Temperature in K (C=K+273 so 30 C = 303 K)\nR= the universal gas constant= 0.08206 L-atm/mol-K\n\nSolve the ideal gas equation for n=PV/RT\n\nn=[(.96atm)(10L)]/[(.08206 L-atm/mol-K)(303K)]\n\nn=0.386 moles
1030041
Given the answer you are trying to work towards, the ideal gas law is sufficient. But beware, in reality, the ideal gas law is accurate only at low pressures (not far from 1 atm). That said, it is a simple matter of algebra at this point:\n\n        T=(P*V)/(n*R)    with    R=0.0821 L*atm*mol^-1*K^-1\nNote that your answer is in degrees Kelvin and the required value in celcius would be degrees Kelvin minus 273.15
260662
The ideal gas law shows the relationship between pressure and atoms / molecules.  PV=nRT.  P = pressure, V = volume, n = moles of the gas, R = gas constant, T = temperature.  Pressure goes up linearly as the number of atoms / molecules of a gas increase.  See the links for more info.
450501
Sorry this is a brief response: figure out the moles of gas in the reactants side of the equation. 5 moles of oxygen means there are 5 moles of gas. Then use Ideal gas law:\n\nPV = nRT\n\nSo, you know Volume (1500 mL), n (5 moles x molecules of O2 per mole), and T (25C + 273C = 298 K). Calculate P.\n\nFor n:\n5 moles (6.022x10^23atoms/moles) (2 atoms/molecule)\n\nHope that helps....
751375
Yeah, that's right.\n\nThere are 2 atoms of O in each molecule of O2, so 1 mole of O2 molecules contains 2 moles of O atoms.
256772
ideal gas law:\n\nPV = nRT\n\n\n solve for T\n\n  1.21*2 = .0625* .0821* T\n\n    T =  472 k\n\n    now slove for C\n\n C + 273 = k \n\n     \ndeg C =  199
685887
Use the ideal gas law    pv = nRT. You will be solving for moles so rearranged it is pv/RT = n\n\nP = pressure (1.0 ATM)  V = volume (3.38L) R = .0821 (Ideal gas constant) and T = Kelvin temperature (Celsius temp + 273)\n\nThis will give you moles of propane. \n\nMUltiply this number by the molar mass of propane (44 g/mol) and you will have it.
952068
a) 0.01 mol\n\n1 mole of gas, at standard tremperature and pressure occupies 22.4 l.  You have exactly 1/100 of that, therefore 1/100 of a mole.
453509
i'm not familiar with the unit barg, but just use the ideal gas equation PV=nRT solved for n so n=(PV)/(RT) with an apropriate value for the universal gas constant (R).  that will give you the number of moles of N2 which you can then determine the volume under standard conditions if you wish using the same equation.
851404
At room temperature and pressure one mole (6 x 10^23 particles/atoms/molecules) of a gas fills 24 dm^3 (Litres).\n\nThis means that in every cubic centimeter (one millilitre) of air there are 2.5 x 10^19 molecules.\n\nIf we forget about the rapid movement and think how many static molecules this means in each dimension we find that each 1-dimensional 1-cm long line contains 2,924,017 molecules (let's say 3 million to keep things simpler).\n\nOver 70% of air is Nitrogen and the length of an N2 molecule is about (keeping things simple) 100 pm (picometers; 10^-12 m) so 3 million molecules take up 300 million picometers or 300 micrometers (following conversion)\n\nThere are 10,000 micrometers in a centimeter so the molecules only 3% of the available distance.\n\nIf all 3 million molecules were equally spaced they would be approximately 3.3 nm (nanometers; 10^-9 m) apart.\n\nA division of the molecular length (100 pm) by the separation (3.3 nm) shows that they are about 33 molecules length apart.\n\nNow I've made plenty of assumptions here. I'm more than happy for people to point out where I've gone wrong (if I have). Please PM me.
851681
Write the balanced equation - you need the mole ratios.\n\nChange 1g hexane to moles of hexane.\n\nSee how many moles of oxygen are USED and how many moles of  CO2 and H2O are formed.\n\nYour total number of moles in the reaction mixture would then be:\n\nmoles of Oxygen NOT used + moles of CO2 (we will assume the water formed consenses to a liquid at 20 degrees).\n\nUse the Ideal Gas Law to figure out the volume.\n\nPV = nRT\n\nwhere P = given pressure - water vapor pressure\nwhere V = your unknown\nwhere n = your total moles of gas\nwhere R = the appropriate Universal Gas Law constant\nwhere T = the temperature in KELVIN
755048
The ideal gas equation is pV = nRT.\n\nSo if you double n on one side, you must double something on the other. In your case it is p.
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.