a sample of gas is collected over water at 25.0C and a total pressure of 641 torr. the volume of the?
827978
PV = nRT\nwhere P must be in atmospheres, V must be in L, n is moles, R is a constant, and T must be in Kelvins.\nyou have to compensate for the water vapor.\n(see source, 25 degrees is 23.76 mmHg which is the same as torr.)\nP = 1 atm @ STP\nV = unknown\nto find n: (641-23.76/ 760) atm x .5 L = n x (R is constant: .08206) x (25 + 273 to be in Kelvins)\nn =  .0166 mol\nR = .08206\nT = 273 K @ STP\n\nsolve for V\n1 x V = .0166 x .08206 x 273\nV = .372 L = 372 mL
856026
Use the gas law\nPV=nRT
984937
The equation is: 2H2 + O2 ---> 2H2O\nSo you would form 2L of water VAPOR\nIf you want this in terms of LIQUID water, remember that one mol of a gas occupies 22.4L, so you have 2/22.4 mols of water (MW=18) formed, so...\n2/22.4 x 18 = 1.6 grams or 0.0016L of LIQUID water.\nThis assumes that all three gasses act as 'ideal gasses' (which is only close) and conditions are reduced to STP...so answer is not exactly correct, but 'close enough for government work'  :-))
569971
This problem only has enough information if you assume it is an ideal gas.  In this case, PV=nRT is the equation of state.\n\nn and R are constants in this case, so PV/T must also be constant.  These values must also all be in absolute scales - the celcius must be converted to kelvin or any other scale based on absolute zero.  This means:\n\nP(i)V(i)/T(i)=P(f)V(f)/T(f)\nV(f)=T(f)P(i)V(i)/[P(f)T(i)]\n    =(250+273)K(740mmHg)(410mL)/[(680mmHg)(27+273)K]\n    =[523*740*410/(680*300)]mL\n    =777.83mL\n\nBy the significant digits you've presented, this can only be evaluated to any accuracy as 780mL.  (two significant digits)
985910
You have 2 38.0 mL volumes, so the total volume is 76.0 mL.\nThe reaction is H2SO4 + 2 KOH --> 2 H2O + K2SO4\nThe rise in temperature (delta T) is 30.17°C - 23.50°C = 6.67°C\n\nThe heat capacity of water is 1.0 cal/(g*°C)\nThe density of water is 1.0 g/mL\n\nThe H of the reaction is 1.0 cal/(g*°C) * 1.0 g/mL * 76.0 mL * 6.67°C = 506.9 cal or (506.9 cal * 4.186 J/cal * 0.001 kJ/J) = 2.122 kJ\n\nThere were 0.82 mol/L of KOH, and 38.0 mL, so there were .82*.038 mol = 0.03116 mol of KOH\n\nDeltaH per mole of KOH is 2.122 kJ / 0.03116 mol = 68.1 kJ/mol
905377
According to Boyles Law, P1V1 = P2V2\nlet p = rou, the density of pure water (assuming tt the lake is pure)[1000 kg/m^3]\nTake g to be 10N/kg, and atmospheric pressure [P(atm)] to be 1.013x10^5 Pa \n\n[P(pgh) + P(atm)] x 1.55 = P(atm) x New Volume\n(1000x10x120 + 1.013x10^5) x 1.55 = 1.013x10^5 x new volume.\ntherefore, \nnew volume = 1210130 x 1.55 / 10130\n                     = 185.16cm^3\n\nHope this helps =)\n\n**The Boyles Law which i stated above states that the volume of gas in an enclosed area is inversely proportionate to its pressure. Initially under the water, the pressure on the bubble is taken to be atmospheric pressure + the weight of the column of water above it. As it rises, the pressure gradually decreases to only atmospheric pressure when it is at the surface. As such its volume increases proportionately, and tt's how i found out the ans.
782970
The expansion ratio is 9.8/8 = 1.225, so that is also the pressure ratio compared to 1 atmosphere (101325 Pa)---> P = 101325*1.225 = 124123 Pa\n\nAssuming atmospheric pressure at the surface of the water, the depth to cause the additional .225 atm = .225*10.33m/atm = 2.324 m\n\nSummary:\n1) 2.324 m\n2) 124123 Pa
776910
1 mole of an ideal gas occupies 22.4 liters, so multiply that by your number of moles and you would have the total volume occupied by this amount of gas at STP.
960502
Since the oxygen is collected over water, you must subtract out the water vapor pressure to determine the pressure due to just the oxygen.\n\n722.2 mmHg - 21.1 mmHg = 701.1 mmHg = 0.922 atm due to O2\n\n0.276g KClO3 = 0.00175 moles KClO3 (molar mass = 158 g/mol)\nSince KClO3 decomposes to form O2 in a 2:3 ratio, the number of moles of O2 produced = 0.00262 moles O2\n\nPV=nRT\n(0.922 atm) V  = 0.00262 mol (0.08206)(296.2K)\n\nSolving for the volume due to the O2,  V = 691 mL
433864
Use the General Gas Equation:\n\nPV = nRT\n\nwhere\nP = pressure \nV = volume \nn = number of moles\nR = Universal Gas Constant (value depends on units of P,V, n, T eg R = 62.3638 when V is in litres, P is in mmHg, T in Kelvins and n in moles.)\n\nUse the above equation to solve for n. If you want the answer in grams, simply multiply the number of moles by the molar mass of H2.\n\nIf you require a more precise answer, then you should use the Van der Waals equation (a more sophisticated version of the General Gas Equation that takes into account interactions between particles and the volume of those particles).
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.
961596
You first balance this equation:\n\nCH4 + 2O2 ---> CO2 + 2H2O\n\nTotal number of mole of 64.16 g CH4:\n= 64.16 / (12 + 4)\n= 4.01 moles\n\nTotal number of moles of H2O produced:\n= 4.01 * 2 = 8.02 moles\n\nTotal mass of H2O:\n= 8.02 * (2 + 16)\n= 144.36 g
700759
Let \nQ = volume flow rate in Litre/sec, \nD = diameter of the nozzle in cm and \nV = velocity of water in metre/sec. \n\nThen,  V = (40*Q)/ [pi*(D^2)]\n\nwhere pi = 3.14
780627
WATER BOIL AT:  27*C at 26.739 (pressure(torr)  thru 30*C at 31.824\n pressure(torr)
447026
this is 250 liters, that is 250 cubic meters.\n\nVolume of the cylinder is (area of the base)*height.\nAs the base is a circle, its area is (squared radius)*Pi.\n\nIf you assume Pi almost equal to 22/7, you can take its inverse, 7/22, and multiply by 10: you get 70/22. Take this value in meters as the height and 5 meters as the radius of the base.
354060
0.2 gal. or 25.6 fluid oz.
898154
PV=nRT\n(.65)(.5 L)=(vol/22.4)(.082)(-33+273)\n.325=(vol/22.4)(19.68)\n.0165=vol/22.4\nvol= .369 L\nvol= 369 mL