A mixture of 6.80g of N2 and 10.8g of Xe exerts a pressure of 0.475 atm.?
817814
moles of N2  6.8/14= 0.486\nmoles of Xe  10.8/131= 0.0824\n\ntotal mole= 0.5684\n\npartial pressure = mole fraction * total pressure\n                          = 0.486/.568 * 0.475\n                          = 0.406atm
766934
Assume it is an ideal gas and use PV = nRT\n\nP is in atm 1 atm = 760 torr\n    225 torr / 760 torr = 0.296 atm\nV is in L so 5.7 L\nR is 0.0821 atm*L/mol*K\nT is in K so C + 273 = K\n     27 + 273 = 300 K\n\n(0.296)(5.7) = (0.0821)(300)n\nn = 0.0685 mols
1097483
P/T=P/T\n1/(20+273)=P/(600+273)\nP=30 atm\n\nThe new pressure would be 30 atms, but this is considering that the volume remains constant.
46596
Use mole concept. Heres how it's done:\n1)Neon's atomic number's 10 and it has acomplete orbit. therefore it's molecular formula is Ne.\n2)1 mole of Ne weighs 20.18g&i mole is 22.4 l\n3) use the following equation and then cross multiply:\n\n    22.4   litres    weigh          20.18 a.m.u\n     2     litres    weigh          x a.m.u.\n\nNow ur equation is 'x'= (2 X 20.18)22.4\nwhich is equal to 1.801 gms or a.m.u
343770
Because gas doesn't have a definable 'volume'.  You can compress it and change it as long as you are willing to deal with changing the pressure.  To compensate, you have to PV = nRT
803930
I assume you want to find the partial pressures.\n\nmoles of O2 = 6.20 / 32.00 = 0.194 moles\nmoles of N2 = 11.6 / 28.02 = 0.414 moles\n\nTotal moles = 0.608\n\nmole fraction of O2 = 0.194 / 0.608 = 0.319\nmole fraction of N2 = 0.414 / 0.608 = 0.681\n\nPartial pressure of O2 = 0.570 * 0.319 = 0.182 atm\nPartial pressure of N2 = 0.570 * 0.681 = 0.388 atm
1187434
I think that's dalton's law.  Whoever came up with it, the pressure is additive.  So convert so everything is in the same units, multiply by the percent, add, and then compare to the answers.  I bet it's more fun than you predict.  Whooo!  Conversion factors!  I don't want to take that kind of enjoyment away, so get cracking!
634933
Density and phase 0.626 g/cm3, liquid
558084
Start with the ideal gas law: PV=nRT.  Since no gas enters or leaves the system, n is constant, and R is the universal gas constant, the product nR has the same value at the beginning and as at the end.  So, collect it to one side, and move the other factors to the other side.\n\nP1 * V1 / T1 = nR = P2 * V2 / T2\n\nSolve for T2\n\nT2 = T1 * (P2 * V2) / (P1 * V1)\n= 300e4 K (.0003e15 atm * .046e-15 cc)/(50e3 atm * .0004e-7 cc)\n=2.07 * 10^7 K
595268
Use the formula: \n\nPV(1)=PV(2) where P is pressure and v is volume.\n\n400(1)=(2000)x(P)\n\nSolving, P =.2atm\n\nPS: If you ever forget the formula, use common sense. When you place any gas under pressure, what happens to its volume? It goes down, because the pressure is forcing the gas into a smaller volume. So when you increase your gas volume by 5 (400x5=2000) you must lower that pressure by the same amount. So the pressure goes down by 5 (1/5=.2).
278691
Use the combined law of gases formula P1*v1/T1=P2*v2/T2.  Where P=pressure, v=volume, and T=temperature.  If the problem doens't have either the temperatures or the pressures assume them to be 1.  I.E. The answer to problem number 7 is:  8 liters.
49337
At room temperature, 0.968 g/cm³
787380
PV=nRT\nP=pressure\nV=volume\nn=moles\nR=constant(.08026atm*L/mol*K)\nT=temperature\n\nThis is the ideal gas law.
438665
Its not pure...there is not a noble gas that would give a m/z at 10 or 11.\na molecule with bond angles of 90 would be SF6\n\nIf you have collected this mass spec it could also be a result from background interference.  Subtract you background and see what happens.
384394
Well, let's see:  you're releasing 1000 kg/hr of liquid N2, which equilibrates to 1 atmosphere and 20C (STP).\n\nAt STP, one mole of a gas occupies 22.4 liters;\nyou're releasing 1000 kg/hr = (1000 kg*1000g/kg)/28g/mole = 35,714 moles of N2 per hour, and thus:\n\n22.4 liters/mole * 35,714 moles = 800,000 liters (at STP) per hour
198092
Use:\n\npressure = density x height x g\nwhere g = 980 cm/s^2\n\nfor each liquid.  the answer (in dynes) will be the sum of the two pressures.
259405
Total volume =( 1 + 2 ) dm3 = 3 dm3.\nUse Boyles  pV = constant on both gasses.\nThe partial pressure of helium is 2/3 kPa.\nThe partial pressure of neon is also 2/3 kPa.\nDalton's law:\nthe pressure of the mix is 2/3 kPa + 2/3 kPa = 1.3 kPa.