A mixture of   7.20 g   of   H2   and   10.3 g   of   Kr   exerts a pressure of   0.605 atm.?
963461
Assuming both gasses are ideal, and the mixture is ideal, we know from Dalton's Law P1+P2=Ptotal.  We can find the individual pressures by P1V1=n1RT1 and P2V2=n2RT2.  Since the two gases are in the same volume at the same temperature we can combine the above equations  to get P2/P1=n2/n1 this P1=Ptotal*n1/(n1+n2)\n\nMoles of H2 = 7.20g * 1mole/2.016g= 3.57 moles H2\nMoles of Kr = 10.3g * 1mole/83.80g= 0.123 moles Kr\n\nSo the partial pressure of H2 is 0.585 atm.  Sorry for skipping some steps above, its just ratios but I don't have much time.
918910
See if this helps:\nPV = nRT = NkT, where P is pressure, V is volume, n is the number of moles, N is the total number of molecules, T is temperature, k = 1:38 £ 10¡23 J/K is the Boltzmann constant and R is a universal constant, R = 8:314 J/mol.K.
521266
This sounds like homework, so I'll give you the ideas but not the data.\n\nThe enthalpy of the mixture at standard conditions is the sum of standard heats of formation of the components of the mix. It will be given in kJ/mol (Ar is 0 kJ/mol being an element, the other two you can look up in tables or on internet). \n\nThen you need to adjust for the sensible energy difference compared to standard conditions : you will also need to look up the specific heat of each component (usually expressed in kJ / mol ºK) and multiply the specific heat of each component by the amount and by the difference in temperature. Make sure you have everything in the same units.\n\nRemember if you are working in moles its easier because to a first approximation, one mole of any gas occupies the same volume (if you undrestand how to allow for deviations from the ideal gas laws, you wouldn't be asking this question) and I assume that the percentages are volume percent, not by weight\n\nEnjoy
843756
USE PRINCIPLE OF CONSERVATION OF linear MOMENTUM.\n\nearth gas lower KE because it has huge mass and huge inertia
783604
volume=pir^2h=1000\nsince there are two variables we need a minimum of two equations
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.
45446
The gas has enough pressure to support 165 mm of mercury, therefor the pressure is 165 mmHg.  To get the pressure in different units, simply look up the unit conversion factors.  The National Institute of Science and Technology is the standards body in the US, so you should refer to their conversion factors as being "official".  For units of pressure, go to:\nhttp://physics.nist.gov/Pubs/SP811/appenB9.html#PRESSURE\nAs is obvious from this table, to convert mmHg to Pascal you multiply 165 times 133.3224 to get 22000 Pa, which is 22.0 kPa.  To convert from Pascal to torr you divide by 133.3224, which gets you back to 165 torr.
885936
we have PV = m/M * R * T\nT = t(celsius) + 273\nR = 0.082 = const\nM is the molecular mass\nP is the pressure (1atm = 760 mmHg)\n\nso M = (m*R*T)/(P*V) = 3.6 * 0.082 * (23.4+273)/(4042.4/760 * 0.183 )= 90 g/mol\n\nwith the percentages we can find the compound \n\nthe formula is CxHyOz (x,y,z is the number of atoms)\n\nx : y : z = %C/M(C) : %H/M(H) : %O/M(O) = 40/12 : 6.71/1 : 53.28/16 = 3.33 : 6.71 : 3.33 = x : 2x : x\n\nwe can use this table to find the answer\nx=1 ; M = 30 : CH2O\nx=2 ; M = 60 : C2H4O2\nx=3 ; M = 90 : C3H6O3\nso the formula is C3H6O3
569322
Use the ideal gas law \n\nPV/T = constant for the final and initial states.
56363
Dalton's law states that the total pressure is equal to the sum of the individual partial pressures of the gases in the mixture: therefore if the vapor pressure of water @ 21.5 degrees C is 2.6kPa and the total pressure in the flask is 99.4kPa then your answer should show 99.4-2.6=96.8kPa\n\nHope that helps...
926935
PV = nRT\n\n685 mmHg / 760 = .901 atmospheres\n\n18.6 mmHg / 760 = .0245 atm\n\n.901-.0245 = .8765 atm\n\n21 degrees c = 294 K\n\n.8765 atm x 7.80 L = n x .08206 x 294 K\n\nsolve for n then use the molar mass of H2 (2.016) to convert to grams.\n\nmy calculator's out in the car or else i'd do it for you.
656021
The first answerer had the correct answers.  If the temperature of the box after compression is the same as when you started out, then you can use the ideal gas law (as the second answerer did) to show that if you halve the volume, the pressure will double.\n\nIn the case where you do not let the heat that's produced by compressing the gas escape from the box, then you need to use the equation of state for an adiabatic process:\n\nP*V^g = constant,\n\nwhere g is the ratio of the molar specific heat at constant pressure to the molar specific heat at constant volume.  For an ideal diatomic gas (a good approximation for air at these pressures and temperatures), this ratio is equal to 7/5.  In which case, \n\nP_final = P_init*(V_init/V_final)^(7/5).  \n\nIn the case at hand, \n\n V_init/V_final=2, 2^(7/5) = 2.639, \n\nso the final pressure is 2.639 times that of the initial pressure.\n\nThe temperature-volume relationship for an adiabatic process obeys a similar law:\n\nT*V^(g-1) = constant \n\nFor an ideal gas:\n\nT_final = T_init*(V_init/V_final)^(2/5)\n\nIf you compress air to half its original volume in a perfectly insulated container (no heat is allowed to escape) then the temperature will increase by a factor of 2^(2/5) = 1.32.  (Note that the temperature here refers to the absolute temperature expressed in Kelvins.)
321421
0.215 kL = 215L\n(1kL = 1000L)
58327
Pressure goes up as temperature goes up.
889941
Use the state equation of the ideal gases and solve for volume V:\n\np*V = n*R*T, V = n*R*T/p\n\nNow p = 657 Torr = 657/760 atm = 0.86 atm and R = 0.082 L*atm/mol*K, T = 283.8 oC = (283.8 + 273) K = 556.8 K so:\n\nV = 2.82*0.082*556.8/0.86 = 148.940 L
303776
Gasoline + Oxygen = CO2 + H2O + NOx + SO4 + energy
504628
you'll have a mixture of hydrogen and oxygen