A 2.0 L. cannister of an ideal gas has a pressure of 11.2 atm. at a temperature of 273 degrees Celsius. How many moles of gas does it contain?
212585
PV=nRT if R= 8.31*10^-3 and the temp is 273 celcius = 476K\n\n2 L* 11.2 atm = 22.4 \n22.4= 476 K * 8.31 nK/Latm *n\nn=0.0056629149854887803496850003539322\nwith sig digits it would be .0057 moles
194190
Use the General Gas Equation:\n\nPV = nRT\n\nwhere\nP = pressure \nV = volume \nn = number of moles\nR = Universal Gas Constant (value depends on units of P,V, n, T eg R = 62.3638 when V is in litres, P is in mmHg, T in Kelvins and n in moles.)\n\nUse the above equation to solve for n. If you want the answer in grams, simply multiply the number of moles by the molar mass of H2.\n\nIf you require a more precise answer, then you should use the Van der Waals equation (a more sophisticated version of the General Gas Equation that takes into account interactions between particles and the volume of those particles).
520992
Convert Liters of CH4 into moles of CH4 with PV=nRT. n= (2 L x 1 atm)/(.082058 x 273.15 K) = .08923 moles CH4. Then, using the mole ratio of 2 moles of O2 per 1 mole of CH4, you multiply .08923 by two to get .17846 moles O2. Plug that into the PV=nRT. V= (.17846 x .082058 x 273.15 K)/1 atm, which equals 4.
306309
A. \n\nOne mole of an ideal gas at STP occupies 22.4 liters.
355010
mole of Argon = 100 / 40 = 2.5 mole\n\nBy using ideal gas theorem:\nP * V = n * R * T\n80 / 76 * 40 = 2.5 * 0.082 * T\nT = 205.39 K\nT = -67.75 C
415410
N2H4. The N/H mass ratio is .874/.126, so the volume ratio is (.874/14) / (.126/1), or 1/2. So we have NH2, N2H4 or any other such compound with the 1/2 ratio. The density at 760 torr and 0 C is 0.977*760/680*(273+84.1)/273 = 1.428 g/L. Then 22.4 L has a mass of 31.99 g, which fits N2H4 (MW=28+4).
22292
1 mole (also represented as mol)= 6.022x10^23.  This is just a number.  Molecules are represented in mol quantities because they're too small to work with otherwise.  So that when you are dealing with CaCl2 in these problems, you're actually using 6.022x10^23 molecules of CaCl2.  Additionally, when you see the molecular weight of a molecule, that is actually the weight of 1mol of that molecule.\n1M= 1mol/Litre.  You need 1.27M.  Divide 1.27 by 1,000 to find the number of mol CaCl2 per mL you need.  (1.27/1,000=0.00127mol CaCl2/mL)\nMultiply this by 147.5 to get the number of mol CaCl2 you will have.  (0.00127*147.5= 0.187325mol CaCl2)\nNow multiply this number by the molecular weight of CaCl2 to get your answer!  (0.187325*110.984= 20.79g CaCl2)
385221
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!
399732
C:H = 85.63:14.37 (mass ratio)\n= 85.63/12.01:14.37/1.008 (molar ratio)\n= 7.13:14.26\n= 1:2\nSo the empirical formula is CH2 and its empirical molecular weight ≈ 14.03\n\nP1V1/T1 = P2V2/T2\n0.980 * 1/(273.2 + 51.7) = 1.V2/273.2 (at STP of 1 atm and 0°C)\nSo V2 = 0.980 * 273.2/324.9\n= 0.824 L\nThus 3.09 g occupies 0.824 L\nSo 3.09 * 22.4/0.824 g occupies 22.4 L at STP\nie 1 g mole = 84\nThus MW of this gas is 84\nSo number of empirical units = 84/14.03 ≈ 6\nSo the molecular formula is C6H12
453043
2.81 L. convert the temperatures to Kelvins and the 180. torr to atm. then plug it in into the Combined Gas Law.
454646
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.
451748
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.
19624
first, u should find the no. of moles in 11 grams of CO2 gas\n\n1 mole = 44g\n\ntherefore, 11g = 11 divided by 44\nwhich is equal to 0.25 moles\n\n1 mole of co2 gas occupies 24000cm cubed.\n0.25 moles of co2 occupies x cm cubed\n\nx = o.25 times 24000\n\nwhich is equal to 6000 cm3\n\nthats the volume...\n\nby da way, r u kashyap from canada? my email id harshini317. contact me if u r kashyap sripada from canada
450619
This may not be the way your teacher taught it, but trust me it works.\n\nYou can use the ideal gas law for each bulb:\nPV = nRT\nWith a constant T, you can cancel out RT and just solve PV = k, where k is a function of n, the number of moles.\n\nP1 * V1 = k1 and P2 * V2 = k2\nk1 = 672 mm Hg * 1.00 L = 672. mm Hg * L\nk2 = 442 mm Hg * 0.500 L = 221 mm Hg * L\n\nYou can add these together: k1 + k2 = k3 = 893 mm Hg * L\n\nThe final volume is 1.5 L\n\nP3 = k3/V3 = 893 mm Hg * L / 1.5 L = 595.3 mm Hg\n\nconvert that to atmospheres, multiply by 0.001316 atm/mm Hg = 0.783 atm
513695
24 litres of air (at RTP) will contain 1 mole of gas. You are told that you have 12.9 cubic kilometres of water vapour. Convert this to litres, and then to moles The enthalpy of vaporisation of water is 40.65kJ/mol,  so you can now work out the total energy released.\nYou would also need to research the power consumption of a city of 1 million people!
333981
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.
316676
Here's how to solve it (I am convinced that you should do your own homework, but hints are OK as long as you learn something):\n\nStart solving by finding the molecular mass of the compound.\nA. Volume occupied by one mol of gas is 22.4 L at standard temperature and pressure (STP: 760 mmHg and 273 K (0 C))\nTo find the volume of gas that weighs 2.57 g at STP, use formula: P1V1/T1=P2V2/T2, where R is 0.08206, P1=319 mmHg, V1=2 L, T1=298 K; P2=760 mmHg, T2=273 K and V2 is unknown (it will be somewhat less than 1 L).\nAfter you get this volume, see how many mols do you have in it: divide 22.4 by V2 you just got. Then multiply this number by 2.57 and get the molecular mass of your compound (M(X)).\n\n2. Multipy M(X) separately by 0.6481, 0.136 and 0.2159 to get number of atomic weight units that are allocated to C, H, and O respectively. \nThen divide first number by 12, 2nd by 1 and 3rd by 16 to see how many atoms it amounts to. The resulting integer numbers are X, Y, Z.\n\n3. CxHyOz is the empirical formula of your compound. Google it to get the possible molecular ones. This is a trick question, so there will be more than one but less than 8. All of them are alcohols or ethers.