How many moles is 0.224 L of hydrogen gas at STP?
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.
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
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
852876
Assuming STP, one mol of oxygen (64 grams) would be 22.4 L.\n\nSo divide 1 by 22.4 and that's how many mols you have. Multiply by 64 grams and there's your mass of one Liter. It'd be useful to convert to kg now, just multiply by 10^(-3)\n\nWeight is mass multiplied by the acceleration of gravity. Assuming this is on earth, acceleration is 9.8.\n\nSo final answer should be 28*10^(-3)  kgm/s^2==>.028 Newtons\n\nIf you want pounds, multiply by 0.2248 to get 0.0062944\n\nShould be right, although I'm not sure. I'm mixing physics and chemistry...unless you wanted mass, not weight, then ignore everything after mass.\n\nEDIT: Just wanted to point out a mistake in the answer above...he forgot that oxygen gas is O2. Weight of one mol of O is 32 g, but molecular weight of one mol of O2 is 64 g.\n\n2nd EDIT: Just realized...I was wrong, Maussy right. Atomic weight of oxygen is 16, so molecular weight of O2 is 32 g. I didn't even look at a periodic table; I was thinking of sulfur. Sorry for any confusion.
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.
891313
The Cp of Hydrogen gas is 28.82 J/mol/k or 14.30 J/g/k\nTherefore, the Cv = Cp - R\n\nCv = 28.82- 8.3\n     => 20.52 J/mol/k\n\nTo covert this to B.T.U\nhttp://www.gordonengland.co.uk/conversion/specific_heat.htm\n\nRate mine as the best answer if you found this helpful.
242296
22.49 L = 2 moles of HNO3 will react with one  mole of K2HPO4\n\nyou only have .0014 mol  of K2HPO4 available\n\nSO, YOU WILL NEED 31.45 mls of HNO3 to react with .0014mol of K2HPO4
702103
1) Determine how many moles of C3H8 we have:\n147 g/ (3*12.01+8*1.008) =3.33 mol\n\n2) Write up to reaction to determine the amount of O2 in moles:\nC3H8 + 5 O2 -> 3 CO2 + 4 H2O \nWe need 5*3.33 = 16.67 mol of O2\n\n3) The volume of 1 mol gas is 22.4 liter at STP. Determine at the given conditions:\nV(1mol) = 22.4 * (303/273) = 24.86 liter \n\nTherefore 16.67 mol = 414 liter = 0.414 m^3
35280
The Ksp of the reaction\n\nHgS <--> Hg + S\n\nis 2*10^-53. Extremely insoluble.
192134
1.74g should = 17.4g approx.\n\nso,\n\n x  = H (Hydrogen) = 1.008g
367911
For example, hydrogen burns in oxygen to form water vapor. \n\n2H2(g) + O2(g) -> H2O(g) \nHow much hydrogen must be burned to form 1.00 L of water vapor at 1.00 atm pressure and 3000C? \nTo work this problem, we need to figure out how many moles of water vapor we need to form. Use the ideal gas equation to compute this: \n\nPV = nRT \n1.00 atm * 1.00 L = n*0.0821 (L*atm/mol*K)* 473 K \nn = 0.0258 moles \nThus, we need to form 0.0258 moles of water. Now it's just a basic stoichiometry problem: for each mole of water formed, you need 2 moles of hydrogen gas, so 0.0258 moles water * 2 moles H2/1 mole water = 0.0516 moles of hydrogen \n\n=====================================================\n\nthis will help you a lot\n\nhttp://www.chem.hawaii.edu/asato161/Ch10NotesP2.pdf#search='Gas%20Stoichiometry%20PV%3DnRT'\n\nhttp://members.aol.com/profchm/gasstoic.html
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
765956
A) 1\nB)2 moles of gas. ,1 mole of gas takes up 22.4 liters, so 44.8 L
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.
690917
1) By definition, one mole of any gas at STP occupies 22.4 liters.  So, if you have 100 liters of CO2 at STP, you have 100/22.4 = 4.464 moles of CO2\n\n2) 28 grams of N2 = one mole of N2\nuse the ideal gas law: PV = nRT, so V = nRT/P\nn = 1 mole, \nT = 180C + 273.16 (to convert to Kelvins) = 453.16\nP = 620 kPa = 620 kPa * (1 atm/101.325 kPa) = 6.11 atm\n\n3) P1*V1/T1 = P2*V2/T2\nor V2 = P1*V1*T2/(P2*T1)\nP1 = 1 atm, P2 = 3.8 atm; T1 = 293K, T2 = 346K, V1 = 80L\nsubstitute those values into the equation to get V2\n\n4) P1*V1 = P2*V2, or (P1*V1)/P2 = V2; T1 = T2\nP1 = 1 atm, V1 = 66L, P2 = 6atm, and substitute into the equation\n\n5) 19 moles of any gas at STP occupies 19moles*22.4L/mole = 425.6 liters
242747
Ans. 20) For this question, you should remember that \n\nPV = nRT or PV/T remains constant.\n\nP = pressure, V = volume, n = no. of moles of gas, R = gas constant and T = temperature in Kelvin.\n\nSo initially,\n\nP = 1, V = 30, T = 273. PV/T = 30/273.\n\nFinally V = 20, T = 303.\nTherefore PV/T = 20P/ 303\n\nSo 20P/303 = 30/273\n\nSo P = (303*30)/(20*273)\n= 1.66 atmospheres.\n\nFor 56) The volume the gas occupies is proportional to the no. of moles of the gas.\n\nSince the no. of moles of O2 in SO2 is the same, it will be the same == 3.5 liters\n\nFor 72) STP = standard temperature and pressure = 0 Celsius and 1 atmospheric pressure.
112430
193,000 Amperes per second to produces 1 mole of H2 or 2.016 grams of x.