What is the mass of 3.35 L of methane gas at STP?\nShow your work.
120317
First of all, it's spelled stoichiometry.\nSTP=22.4L per mole\nmethane = CH4\nmolar mass of methane = 4*1.0079(mass of 4H) + 12.011(mass of C)=16.0426gCH4\n\nSo it would be like this\n3.35L / 22.4L * 16.042g = 2.39923g methane\n\nWith significant figures, it would be 2.40g methane
262008
dont use PV=nRT... you dont need it for these problems... just requires more work because you have to figure out more than one variable\n\nfor everything here it seems you need different laws\n\nBoyles Law is P1V1=P2V2\nCharles Law is V1/T1=V2/T2\nGuy Lussac's Law is P1/T1=P2/T2\n\nall of those laws are just proportions.. just substitute in the numbers you need.  for temperature problems you need to convert them to kelvin... just add 273 to the celcius temperature and you have the temperature in kelvin.\n\nfor example, look at number one...just plug in all the numbers\n\nsince you're dealing with volume and pressure, you'd use Boyle's Law.\n\nP1V1=P2V2\n\n(3atm) * (15mL) = (2.5atm) * (x)\n\nsolve for x\n\n3 times 15 is 45...  2.5 times x is 2.5x\n\n45=2.5x\n\ndivide both sides by 2.5\n\n45/2.5 = 18\n\nthe volume will increase because there is less pressure.  the volume would increase by 3 mL.\n\nlook for what you are given, find out which law you need, and just plug in the numbers\n\nfor number 4.... since you're dealing with pressure and volume again you'd use Boyle's Law.  Given pressure is STP\n\n(15L) * (1atm) = (x) * (3atm)\n\nsolve for x...\n\n15 times 1 is 15... x times 3 is 3x\n\n15=3x\n\ndivide by 3 on both sides...\n\nx=5L of methane gas.\n\nfor number 5 you're dealing with volume and temperature.  with that we would have to use Charle's law.\n\ncharle's law = V1/T1=V2/T2\n\nnow plug in the numbers...\n\n(450L)/(30+273K)= (x)/(90+273K)\n\n(450L)/(303K) = (x)/(363K)\n\n1.485 = x/363\n\nmultiply 363 on both sides\n\n539.01 = x\n\nthe volume of ethane would be 539.01 L\n\ngood luck
137554
molecular weight's 305.4188 g/mol
369935
The mass percentage of carbon in CH2O is 4%.  You can solve this by first calculating the molar mass of CH2O (which is 30.0262 - you obtain this by adding the atomic mass of each element, 12.011 for C since there is only one, 2.0158 for H2 since there are two Hydrogen, and 15.9994 for oxygen since there is only one oxygen).  After you got the molar mass, you divide the amount of Carbon that is in the compound (in this case, it's 12.011 since there is only 1 carbon), and you get approximately 0.4000.  Multiple that by 100 and you get approximately 4%.
227604
Lol, trying to get people to do your chemistry homework?\nUse PV=NRT \nP = Pressure in ATM (1 ATM in STP)\nV = Volume in liters (convert 65cc to liters)\nN = number of moles (will be constant)\nR = gas constant (0.0821 L.ATM/ Mol.K)\nT = temp in Kelvin (273 + degrees C) (STP = 273 K or 0 C)\n\nwith moles being held constant, and the gas constant being constant, your equation should look like;\n\nPV/T (initial) = PV/T (final). solve for P (final)\n\nIf you can't solve this in an hour, i'll post the answer.\n\n....... i was hoping you'd try to solve this and post your answer.....\n\ni got .9383 ATM as the new pressure. mind, that answer is not in sig fig.
416534
The question is if you start with gas at 20.0 atm and 20.3 C and heat it to 125 C what will be the pressure?  It is a changing condition problem making use of Charles' Law.  The answer is 27.1 atm.
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.
531660
51kg x 25 m = energy used\n\ndivide the energy used by 0.30 to get the second answer
506461
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
474593
Assuming 1 mole (28.01348 g) N2 = 22.4 liters at STP, I get (28.01348*.00220102/22.4) =  0.0027526 g if collected at STP. Correcting by 273/293 * (1.5*760-17.5)/760 (for temp., press. and water) = 0.003788 g. \nSo I basically agree with your results. What's the ans. supposed to be?
458132
Use the gas law\nPV=nRT
123069
2 CH3OH + 3 O2 → 2 CO2 + 4 H2O + 726 kJ/mol
366192
Start with 74.4L N2 gas. End with g NaN3. Multiply the 74.4L N2 by 1mol N2/22.4L N2. This is the molar volume of an ideal gas at STP. The L N2 cancel to give moles of N2 gas. Multiply by 2mol NaN3/3mol N2. This comes from the balanced equation. The mol N2 cancel to give the moles NaN3. Multiply by 65g NaN3/1mol NaN3. The mol NaN3 cancel, giving g NaN3, the mass. Atomic weights: Na=23, N=14.
487292
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)
276162
Ack... Munir got here before I did agian.  A agree with Munir!  Volume of any gas at STP is 22.4L.
350108
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....:)
108404
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