"At 27 C and 750 torr pressure, what wil be the volume of 2.3 mole  of Ne?
468007
At STP, 1 mole of Ne will occupy 22.4 liters\nT1 =273\nP1=760 torr\nV1=22.4 L\nn1=1\n\nT2=300\nP2=750\nn2=2.3 moles\nV2=?\n\n(P1 V1)/ (n1 T1) = (P2 V2)/ (n2 T2)\n\nSubstitute u will get V2= 57.37 L (check this again)\nThis is the solution...\nIs it clear?
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.
189614
Gas density d is formulated as d=MP/RT where M is molar mass, P is pressure, R is universal gas constant (R=0.0821) and T is Kelvin temperature. \n\nSTP is standart pressure and temperature and after 1997 IUPAC (International Union of Pure and Applied Chemistry) started to use 0°C (273K) and 1 bar (0.987 atm).\n\nSo for Argon:\n\nd = [(39.95 g/mol) x (0.987 atm)] / [(0.0821 atm-L/mol-K) x (273K)] \n\nd = 1.759 g/L
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.
793636
shawty's NOT good at stoichiometry...has made a fundamental error in the final line so end result is rubbish...\nhow can you end up with MORE Cu than you started with when it's 2 moles : 2moles? \n\n(8/159)x 127 = 6.4 g
755542
[.634 moles Ne/(.884 total moles of gas)]* 0.420 atm = .301 atm exerted by the Ne.
350090
Use the formula PV=rnT\n\np= pressure in atm\nV= volume in L or dm^3\nr=number of moles\nn= constant: 0.0821 atm x L/ moles x K (This constant is used to balance out units)\nt= temperature in K\n\nSo for your problem:\n2.3 atm x 120 L = r x (0.0821 atm x L/ moles x K) x 340 K\n\nthe constant should allow you to be left with moles as your unit\nusing algebra, r is left, which also stands for the number of moles (meaning the problem coordinates with your question)\n\nusing this setup, i got 9.89 moles
665350
As the volume is constant,\n(P/T)initial = (P/T) final\n1/(25+273.15) = P/(50+273.15)\nP = 1.084 atm\n\n(Taking the pressure of the initial gas was 1 atm.
497026
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.
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....:)
856026
Use the gas law\nPV=nRT
1081631
T = 8239\nE = 1/2\nA = 1\nM = 1/3
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.
785530
Answer D.\n\nThis is the Nernst equation at standard conditions.
977807
M = moles/L\nso, 0.220M = 0.220 moles per liter and there are 2 liters so there are...\n0.440 moles of Mg(NO3)2\n\nBUT... notice that there are 2 ions of NO3 for each molecule, so double it again and get 0.880 moles of NO3
141442
KE per mole = (3/2) RT where R = 8.3145 J/mole*K and T is temperature in kalvin, so:\n\nKE = (3/2) RTn where n is the number of moles of gas\n\nYou can solve PV=nRT for n and substitute to get KE = (3/2) PV where P is pressure and V is volume.
469279
PV=nRT if R= 8.31*10^-3 and the temp is 273 celcius = 476K\n\n2 L* 11.2 atm = 22.4 \n22.4= 476 K * 8.31 nK/Latm *n\nn=0.0056629149854887803496850003539322\nwith sig digits it would be .0057 moles