the given massof a gas occupies 572cm3 at130c&725mm pressure.what will be it's volume at 240c&729mm pressure
492280
The equation for this is (PV)/(nT).  Pressure must be in kPa, volume must be in L, and temperature must be in Kelvin.\n\nTherefore, 572 cm3 = 0.572 L, 725 mm = 96.635 kPa, and 130 C = 403 K.\n\nIn the first equation, n = 0.137 moles\n\nIf you input it into the second equation, where:\n240 C = 513 K\n720 mm = 97.168 kPa\n\nYou get V = 0.724 L, or 724 cm3.\n\nOf course, these are all approximations due to rounding.  If you want, you can try it with a calculator, but make sure the units are correct.  Also, 101.3 kPa = 760 mm.\n\nThis is used in basic chemistry, which I just took.
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)
821966
Not sure where the 600 came from, but it sounds like the question wants the total pressure of all the gases combined.  So you would just add together all the partial pressures, and the total pressure is 330 mm Hg.\n\nDividing the total pressure by the number of gases would give you the average partial pressure, but I don't think that's very useful.
923282
a) 98.5 L\nb) 0.000763 cg\nc) 57.0 m\nd) 9500 sec OR 9.50 * 10^3 sec\ne) 12.2 degrees C\nf) 0.00750 x 10^4 mm OR 7.50 * 10mm OR 75.0 mm\ng) 1760 mL or 1.76 * 10^3 mL\nh) 8.86 km
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
755048
The ideal gas equation is pV = nRT.\n\nSo if you double n on one side, you must double something on the other. In your case it is p.
412124
Do you know the Ideal Gas Law:\n\nPV = nRT?\n\nP = pressure in kPa\nV = volume the gas occupies\nn = the number of MOLES of gass\nR = 8.381\nT = Temperature in Kelvins\n\nYou might need to check for the units for volume and pressure, but that's pretty much it.  Solve for n, and use the density given to find the mass (assuming you know the MW for methane [16.04 g/mol])\n\nI'm not gonna do it for ya :-p, but it should be self explanatory
850176
Change grams to moles:  0.285 g/122.5g/mole = 0.00233 moles\n\nChange to moles O2 (ratio from equation is 3/2)\n                        0.00233 x 3/2 = 0.00349 moles O2\nCalculate vol. from relationship 22.4L/mole for any gas at STP\n                        22.4L/mole x 00349 mole = 0.0782 L at STP\n\nUse Combined Gas Law to change to vol. at 24.0 degrees and 729.9mm. Reduce pressure by the water vapor pressure.\n\nP1 = 760.mm          V1 = 0.0782    T1 = 0 + 273\nP2 = 729.8-22.4      V2 =   x           T2 = 24 + 273\n\nx = (0.0782)(760)(297)/(273)(707.4) = 0.0914 L or 91.4 mL
74771
hecka!!  damn.  i miss california.\n\nwhat you have to do is realize that this question is asking about the volume of a rectangular shape that is 1 meter by 1 meter by X meters, where X is the height of the column.\n\nso, you know the density.. density is defined as mass divided by volume.  and you know the weight.. which is mass times acceleration.  here the acceleration is the acceleration of gravity "g" = 9.8 m/s^2.  \n\nso you can do a little algebra to get the equation volume = mass / density \nvolume = weight / (acceleration * density)\n\nyou know the weight (100,000 N), the acceleration (9.8 m/s^2) and the density (1.2 kg/m^3).  plug those numbers in and you get the VOLUME of the atmosphere, and since you know 2 of the dimensions of the box (it is 1 square meter across the bottom face, remember) you can figure out the 3rd pretty easily.\n\nsince this is your homework i wont just tell you the answer, but you should be pretty much able to crank it out from here.
792853
The pressure of a fluid colum is P = (rho) * g * h\n\nYou know that for P = 1 atm and (rho)*g corresponding to Hg, h = 785 mm.\n\nNow change to H20. (rho) is 1/14 of the previous value, g is the same, P is the same, so h must be 14x greater.\n\n--> h = 14 * 785 = 10990 mm.
193517
You don't provide enough data to determine this.  The gas law PV=nRT has variables you need to define.
741784
(21 g x 52.7 C) + (54.9 g x 31.5 C) / (21 g + 54.9 g)\n\n= 37.4 C
789416
PV = nRT\nsince n = g/MM\nMM = gRT/PV\n\nso...\n\nMM = 1.097g * (62.3637 L·Torr/Kmol) * 372.15K / 743 torr * 0.256L\nMM = 133.85 g/mol
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
256772
ideal gas law:\n\nPV = nRT\n\n\n solve for T\n\n  1.21*2 = .0625* .0821* T\n\n    T =  472 k\n\n    now slove for C\n\n C + 273 = k \n\n     \ndeg C =  199
878016
V1T2 = V2T1\n147mL * (57+273K) = V2 * (28 + 273K)\nV2 = (147mL * 330K)/ 301K\nV2 = 161mL\n\nTherefore, volume at 57C is 161mL. Its logical since when temperature rises, volume increases.
1080767
It depends on the pressure and temperature. 50 lbs. of O2 represent a fixed number of moles of CO2. From the ideal gas law PV=nRT. Cannot solve for volume V without knowing P and T.