Calculate the mass, in grams, of 2.74 L of CO gas (28.01 g/mol) measured at 33°C and 945 mmHg?
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
609831
pv=nrt\nmeasure pressure, volume, and temperature to find the number of moles, n\n\nfind the mass of the gas\n\nmass of gas/ moles of gas= molecular mass (g/mol)
878226
The method is very simple.  I can do the first and you can try the others.\n\n1.  First write the reaction and balance it.\n\n2HgO ---> 2Hg + O2.\n\n2. Convert grams to moles of whatever mass you have.  In this case you have O2 as 96 grams.  You convert to moles by dividing the mass by the molecular weight.  In this case, for O2 will be 32 g/mol\n\n96 grams / 32 g/mol = 3 mol.\n\n3. Now you look back to the equation.  In the equation for producing 1 mol O2 you need 2 moles of HgO.  If now have 3 moles O2, then you will need 2 x 3 moles of HgO, 6 moles.\n\n4. Now you need to convert the moles of HgO to grams.  To do this calculate the molecular weight of HgO.  It is 216.6 g/mol\n\n6 moles * 216.6 g/mol = 1299.6 g of HgO.
900861
As usual, we first check for a balanced equation.  It isn't, so we fix it: 2 LiOH + CO2 -> Li2CO3 + H2O.  Also as usual, get the molecular weights: LiOH = 24, CO2 = 44, and the others don't matter.  There are 4 moles of CO2, so we need 8 moles of LiOH, and that is 192 grams.  Extra credit: find the error in the next response.
709574
I get 1.69522 g/L, using constants with the most significant figures I can find:\n\nN/V = P/(R*T)\nP = 101.325e3 Pa\nV = 1 m^3\nR = 8.314472\nT = 273.15 K\n\nN/m^3 = 44.6149 mol/m^3 = 0.0446149 mol/L\n1 mol of F2 = 37.9968 g\n0.0446149*37.9968 = 1.69522 g/L\n\nIf I use 100 KPa (1 bar) as standard pressure, I get 1.67305 g/L\n\nThe difference in the values given in the answer and the "Handbook of Chem. & Physics" is apparently due to the shift from 1 atm to 1 bar as the definition of "standard pressure."\nI think the slight difference in my calculated value and the published one may be due to the fact that when you start getting out to that many decimal places even "ideal gases" aren't totally ideal....maybe I should try the Redlich-Kwong EOS...\n\nOooooops!  I didn't want to do your homework for you...maybe I should erase all this....:)
803561
Since pressure is a colligative property, its additive. Since you have .125 atm of methane, you can subtract that from the total pressure. That will give you .36 - .125 = .235 atm of Choride gas. \n\nThen using the ideal gas law (PV = nRT) to solve for moles. \nP = .235 atm (from Cl2) V = .268 L (given) T = 293 K (given and adjusted to kelvin) and R = .08206 Latm/mol K (adjusted gas constant to cancel units.\n\n.235 atm(.268 L) = n .08206 Latm/mol K (293K)\nn = .0026 mols. \n\nThen use moles to solve for grams. \nAMU of Cl2 is 70.9 AMU\nmoles = grams / AMU\n.0026 = x grams / 70.9 AMU\n.0026*70.9 = 0.186 grams\n\n0.186 grams of Cl2 present in the mixture.
759383
First find the number of moles by dividing by the avogadro number, say n.\n\nSecond find the molecular weight of CO2, say M. This will give you the weight of one mole\n\nso\n1 mol CO2  gives you M g\nn mol CO2 gives you x g\n\nx = n * M.
795154
2CO(g) + O2(g) --> 2CO2(g)
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
262778
73.93 grams\nFirst you use the Ideal Gas Law: PV=nRT >P is pressure in atm., V is volume, n is moles, R is the constant: .0821, T is the temperature in Kelvins> (1atm.)(3.16L)=(n)(.0821)(305K) > .126195563 moles. Now you use a proportion. 9.33grams/.126195563moles = x grams/ 1 mole > 73.93 grams
895405
use dalton's law to get the pressure of CH4(total pressure - H2 pressure)\nthen use ideal gas law to get the moles of CH4(PV=nRT)\n*watch your units*\nchange moles of CH4 to grams of CH4(moles x molar mass)
187568
You can come up with a molecular weight of air - use the percentages of the different molecules comprising air and the molecular weight of the components.  The answer is around 29 g/mol (I think).
116688
from the equation: 2CO + O2 -----> 2CO2,\n\n0.69 g CO2 x (1 mol CO2/ 44 g CO2) x (2 mol CO/ 2 mol CO2) x (28 g CO/ 1 mol CO)\n\n\nthe answer is 0.4391 g of CO needs to react with O2 to produce 0.69 g of CO2
1053372
3.425 BTU/lb F in gas form
860730
Multiply moles by the molecular weight (grams per mol) to get mass (grams).
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
92170
The ideal gas law is another tool that we can use to determine the molar mass of a gaseous substance. To determine a molar mass, we need the mass of a given sample and also the number of moles in that sample. You recognize immediately that given the volume, temperature, and pressure of a gaseous sabstance, you can calculate the number of moles.\n\nfirst:\n \n V = Volume = 987.56g / 1g/cm^3 = 987.56cm^3 (1 liter/ 1000cm^3) = .987 liter\n\n T = temperature = 23 + 273 =  296K\n\n P = pressure = 97.7 kPa (  1 atm / 101.3 kPa ) = .964 atm\n \n r = .0821 L'atm/mol.K\n \n n = mole = ?\n     \nThe ideal gas law equation:\n \n   PV  =  nRT\n\nsolve for  n\n\n   n = PV /RT = .964*.98756 / .0821*296 = .039 mol \n\n The molar is simply the mass divided by the number of moles:\n\n  mass of gas in flask =  187.942g - 183.257g = 4.685g\n\nso, the answer is:\n\n            4.685g / .039 mol = 120.1 g/mol