What volume of   O2 (m3)   is needed?
963278
1) Determine how many moles of CH4 we have:\n175 g/ (12.01+4*1.008) =10.9 mol\n\n2) Write up to reaction to determine the amount of O2 in moles:\nCH4 + 2 O2 ->  CO2 + 2 H2O\nWe need 2*10.9 = 21.8 mol of O2\n\n3) The volume of 1 mol gas is 22.4 liter at STP. At the given conditions:\nV(1mol) = 22.4 * (290/273) = 23.79 liter\n\nTherefore 21.8 mol = 519 liter = 0.519 m^3
781068
The balanced equation for the reaction is:\n\n2 H2O2 --> 2 H2O + O2\n\n1]  If one mole of Oxygen is produced, the 2 moles of Hydrogen peroxide would be needed.  One mole of Hydrogen peroxide contains 2*H + 2*O = 2(1) + 2(16) = 2 + 32 = 34 grams.\n\n2 moles of Hydrogem peroxide = 2*34 grams = 68 grams.\n\n2]  Is it 0.3 moles or 0.3 cubic decimeters of Hydrogen peroxide?\n\nAssuming 0.3 moles, then from the above, 0.15 moles of Oxygen would be produced.  The volume would be:\n\n0.15 moles * 24 cubic dm/mole  = 3.6 cubic dm of Oxygen at rtp.
976256
BALANCED . . . \n\n2Sr + O2 ---> 2SrO\n\n. . . and solved . . .\n\nGet an "A".
734011
to get the volume, you would need to use the formul V=dwl (d-depth, w-width, l-length).\n\nbefore you can plug in the numbers, you have to make sure your numbers' units are the same. KHDBdcm.\n\n6km=6000m, 2km=2000m, and .4km=400m\n\nnow, 1 cubic meter equals 1000 L, so you want to do this in m.\n\nV=(6000)(2000)(400)\nV=4,800,000,000 m^3\n  =(4.8)10^12L\n\n--\n\nnow youve got 6,000,000 people who use 420 L of water each day.\n\ndivide how much water youve got by how much water each person uses. \n\n=1.142857 x 10^10 \n\nthats how many people you can serve.\n\n\nthen divide that by 6million, and you find you can serve them for 1904.8 days.
761497
1) The first one is wrong on the last person who answered, O2 weighs 32g/mol because there are 2 O's in it, 16 each.  So in 1.33x10^4 (13,300mol) you have 4.256x10^5g of O2 (425.6kg).\n\n2) Recognize that hydrogen, oxygen, and chlorine are diatomic gasses, so they always occur in pairs (H2, O2, Cl2).  Others like this are nitrogen (N2), fluorine (F2), bromine (liquid @ room temp. Br2) and Iodine (solid at room temp, I2).  So your equations would be:\n\n2H2+O2-->2H2O\nH2+Cl2-->2HCl\n\n3) PV=nRT and solve for n, number of moles.  Put everything in correct units, liters, atmospheres, moles, and kelvins, and R is a constant R=0.0821 L*atm/mol K.  Kelvin temp=Celcius temp +273\n\nSolving for n:\nn=PV/RT=(1.09atm)(0.942L)/(0.0821L*atm/mol*K)(287K)\nn=0.0436 mol\n\nSo, when we usually want to know a number of moles given a mass and having a molar mass we can look up, we say,\nn=mass/molar mass\n\nin this case, we need to find molar mass, so:\nmass/n=molar mass\n3.17g/0.0436mol=72.71g/mol
202815
100g O2 x (1 mole O2)/(32g O2) x (22.4 L O2 @ STP)/(1 mole O2 @ STP) x (1/1.5) x (298/273)\n\n= 50.9 L\n\nThe 22.4 L = 1 mole is only true @ STP.  The higher pressure reduces that volume and the higher temperature increases it.
275919
2H20 --> 2H2 + O2\n\ntheoreticalyy spk it's exactly half the vol of H2, just look over the stoechiometrical balance above; but due to gas losses, it may be less
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
477933
Check out the link below:\n\nhttp://www.respiratorygroup.com/products/high_pressure/med_o2_spec.aspx
354284
The volumne of a gas increases to fill its container and both H and O are gases at room temperature.  Is volume really what you want to know?
190428
22.4 l=>1 mole N2O\n0.1 l=> 0.0045 mole N20\nsince 1 mole NH4NO3 gives 1 mole N20,\n0.0045 mole of NH4NO3 is required.\nsince 1 mole NH4N03=>80 grams,\nmass of NH4N03 required is 80*0.0045=0.3571 grams\n\nsince 4 moles of Fe reacts with 3 moles of 02\nand 52 g of Fe=1 mole Fe,\n0.75 mole of 02 required.\nsince 1 mole 02=>22.4 l,\nvolume of 02 required is 16.8 litres\n\nhope that helps
239849
.7M x 20mL = .25M x (Volume of new IA solution)\nVol = (.7 x 20) / .25\nVol= 56mL
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
837288
Use the formula M1V1=M2V2\n\nThat is the first Molarity times the first volume are equal to the second Molarity times the second Voume.\n\n.2098M x70mL = .662M x V2\n\nSolve for V2.\n\n(.2098 X 70)/.622 = V2\n\nV2 = 23.61mL NaOH to titrate the solution to neutrality.
1149287
pV = n R T => if pressure is constant => p = n R T / V = const. => p / n R = T/ V = const.\n=> T / V = const.\nT1 = 333 K\nT2 = 533 K\nV1 = 0.2 m3\n\nT1 / V1 = T2 / V2 =>\nV2 = V1 x T2 / V2 = 0.2 x 533 / 333 = 0.32 m^3\nanswer D is correct
1140664
You're missing a unit. Do you mean 2.5 kg solution?
658673
As someone said, this is a mass not a volume but that doesn't mean you can't do it : this is all about conversions.\n\n1) Gas : The only thing you need to know is that a gram mole of any gas at standard conditions occupies 22.41 liters. You can work out the equivalent volumen for a kilogram mole for yourself (hint, 1000 time as much) and don't forget to change the molar volume (use the ideal gas law) for the conditions if they are not standard. This will be to a first approximation the same volume for a kmol of any gas (providing its not at not too extreme conditions)\n\n2) liquid : you need to work out the molar volume from the density and the molecular weigh. If you don't have that data, it can't be done. Unlike the gas, this cam be very differnent for different liquids, and also chages with conditions but in a manner that cannot be calculated with a simple formula \n\nWhe you come to do real work with any of these things, you usually find that volumes are given in standard and actual conditions. Be careful to use the conversion factors at the appropraite conditions or you will get it wrong\n\nEnjoy.