How many moles of gas does it contain?
469279
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
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....
736571
there are .833 moles. divide the mass by 18 (2 hydrogens + 1 oxygen) oxygen= 16g/mole Hydrogen = 1 g/mole
859377
Use Avogadro's number. Avogadro stated that 6.02*10^23 molecules make up 1 mole. To find the number of moles of something, divide the number of molecules (or atoms, in your case) by Avogadro's number:\n\n(4.81*10^24) / (6.02*10^23) = 8 moles.\n\nI hope this helps!
882871
From the ideal gas law, we can get the number of moles (n):\nPV=nRT\nT = 50.8°C = 323.95 K\n.980 atm * 1.00 L = n * 8.2057 E-2 L*atm / (mol*K) * 323.95 K\nn = 0.03687 mol\nSo, the molecular weight, MW = m/n = 2.58 g / .03687 mol = 69.98 g/mol\n\nThe mass % of C is 85.62%, so the mass% of H has to be 14.38%\n\nIt you had 1 mole of the gas, it would weigh 69.98 g\n\n69.98 g * 85.62% = 59.92 g\n69.98 g * 14.38% = 10.06 g\n\nThe atomic weight for carbon is about 12:\n59.92g / 12 g/mol = 4.99 mol = 5 mol\n\nThe atomic weight of hydrogen is about 1:\n10.06g / 1 g/mol = 10.06 mol = 10 mol\n\nSo, your molecular fomula is C5H10\n\nThis can be several different forms:\n\nCH3-CH=CH-CH2-CH3, 2-pentene\nCH2=CH-CH2-CH2-CH3, 1-pentene\nC5H10, Cyclopentane\nCH3-CH(CH3)-CH=CH2, 2-methyl-1-butene\nThe list goes on...
54895
To be precise, the answer is no. Do u noe why? The mean mol wt of air is 29 whereas that of CO is 28.\n\nlet me show u how i get these answers:\nC=12,O=16,mol wt of O2=32g/gmol,mol wt of N2=28g/gmol\nmol wt of CO = 12+16\n             =28\nas for air, by common sense, we know that air is made up of O2,N2, and other gases,rite? As the mol composition of other gases are very negligible(<1%), we assume air is only made up of O2 and N2. Air consists of 21mol% O2 and 79mol% N2.\n\nlet's take a basis of 100gmol(or 100mol)\nmoles of O2=21gmol\nmoles of N2=79gmol\nmass of O2=21X 32g/gmol\n          =672g\nmass of N2=79X28g/gmol\n          =2212g\ntotal mass= 2212+672\n          = 2884g\naverage mol wt of air=total mass/total moles\n                     =2884g/100gmol\n                     =28.84g/gmol\n                     =29g/gmol(whole no.)\n\nsince the ave mol wt of air is greater than that of CO, so aCO is not heavier than air,instead,slightly lighter than air.\n\nDoes it answer ur question?
756159
n   = pV/RT = 1.03*1.02/284*0.0821 = 0.045moles\n\nmoles of Nitrogen = 0.045*0.78=0.0351\n\nmoles of oxygen = 0.022*0.045 = 0.0098mole\n\nweight of the gases = 0.0098*32 + 0.0351*28.02 =0.316+ 0.986=1.3g\n\ndensity 1.3 g/l
777427
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
205383
Divide each mass by their molar weight to get the number of moles of each gas. 0.856g He/(4.003 g He/mole) = 0.214 mole of He. Do that for all the masses and add them up. I got a total of 1.034 moles. \nFrom that you could find the total pressure using PV=nRT.\nNext, find the mole fraction (I think this is the same as molar concentration) which is the # of moles of each gas/total number of moles.\nThen multiply each mole fraction by the total pressure to get each partial pressure. \nHope this helped. Let me know if you want me to go through each step.
1112530
Moles=moles/Liter. Moles-solution + some substance,=amount of substance(which contain 12g of Carbon)/a Liter amount of aqueous solution(not always water).\n\nAvagadro's #- the amount of something(precisely 6.02*10^22) which contains exactly 12g of C in it.\n\nI don't think I ever have to know that stuff again, but it's in my head for now.
765956
A) 1\nB)2 moles of gas. ,1 mole of gas takes up 22.4 liters, so 44.8 L
920990
methane: CH4\n\nMr;\ncarbon 12\nhydrogen 1\n\nMr of methane= 1*carbon+4*hydogen\n=12+4\n=16\n\ntherefore one mole of methane weighs 16 grams\n\n4 g of methane= 4/16\n=0.25 moles\n------------------------------...\n\none mole of any gas occupies 24 dm3 at r.t.p\n\nmoles of oxygen in 7 dm3= 7/24\n=0.292 moles
776910
1 mole of an ideal gas occupies 22.4 liters, so multiply that by your number of moles and you would have the total volume occupied by this amount of gas at STP.
716563
The first thing you need to do is work out how many moles of oxygen you have. You do that by dividing 96 by the molecular mass of oxygen (which is 32). As it's 1:1 stoichiometry, you just take that nymber and multiply it by the molecular mass of methane. (that's 16)
782235
Avagardro's Number = one mole, and all of the masses on the periodic table are in grams/mole.One mole of P equals 30.97g, so d. (c is the only other logical answer, but O2 has a molar mass of 32g/mole beacuse 1 oxygen has a molar mass of 16)\n\nGet used to moles, you'll see them all throughout chemistry.
760656
mole of O2 = 1 mole\nmole of CH4 needed = 0.5 mole\n\nBy using ideal gas theorem:\nP * V = n * R * T\n1.5 * V = 0.5 * 0.082 * 300\nV = 8.2 litre\n\npeace\nvixklen
664728
One mole of *anything* = 6.023x10^23 units of that item.\n\nSo if you have 6.10x10^24 molecules of ethane, that's:\n\n6.10x10^24/6.023x10^23 = 10.13 moles.\n\nMultiply 10.13 moles by the molecular weight of ethane to get your answer.