A mixture of 6.40 g of N2 and 12.7 g of Ne exerts a pressure of 0.405 atm."?
913556
did you forget the rest of the question?
638067
Nitrogen\nAcetone chilled with dry ice (CO2)
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!
635000
The easiest way to work this out is from the Ideal Gas Law, PV=nRT. Assuming that they start at the same temperature, this is even easier because both R and T will be constants.\n\nLets make the first gas 1, the second gas 2 and the final mix f so that :\n\nPressure Gas 1 =P1, Volume Gas 1 = V1, Temperature Gas 1 = T1\nPressure Gas 2 =P2, Volume Gas 2 = V2, Temperature Gas 2 = T2\nPressure final mix =Pf, Volume final mix = Vf, Temperature final mix = Tf\n\nthen :  P1V1/RT1+P2V2/RT2=PfVf/RTf\n\ntake out the constant R\n\nP1V1/T1+P2V2/T2=PfVf/Tf\n\nIf T is different for each mix\n\nPf=(Tf*(P1V1/T1+P2V2/T2))/Vf\n\nif T is constant (which I assume from your question)\n\nP1V1+P2V2=PfVf\n\nPf=(P1V1+P2V2)/Vf\n\nOkay?
377149
Nitrogen (N2) 78.084 % by volume\nOxygen(O2)20.9476 % by volume\nArgon(Ar)0.934 % by volume\nCarbon Dioxide(CO2)0.0314 % by volume
700450
Letters A, B, D.\nIn not sure about C, its wording is tricky.\nIdeal gas law : PV = nRT\nReal gas law : (P + (a(n / V)^2)) * (v - nb) = nRT\nThe corrected pressure times the corrected volume.
787380
PV=nRT\nP=pressure\nV=volume\nn=moles\nR=constant(.08026atm*L/mol*K)\nT=temperature\n\nThis is the ideal gas law.
856026
Use the gas law\nPV=nRT
797227
You have Pressure (being asked) You have the volume (5.43L). You have the number of moles (1.82)You have temperature (69.5C or 342.5K). I assume you know R.\nSo yes it is PV=nRT
852232
Pv = nRT\nn = m/M (M Molecular weight)\nPV = (m/M)RT\nP = (m/V)RT/M\nPM/RT = m/v = Density.\nPlug the numbers, watch the units and get the density.\n\nOh a numerical answer, yes, ...42.
53064
Assuming your gas is an ideal gas, you can use the ideal gas law.\n\nPV=nRT where \nP=Pressure in atm (1 atm= 760 mm Hg so 730 mm Hg= 0.96atm\nV=volume in L\nn= moles\nT=Temperature in K (C=K+273 so 30 C = 303 K)\nR= the universal gas constant= 0.08206 L-atm/mol-K\n\nSolve the ideal gas equation for n=PV/RT\n\nn=[(.96atm)(10L)]/[(.08206 L-atm/mol-K)(303K)]\n\nn=0.386 moles
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).
107278
the pressure rises.  this is because the molecules move faster.  if you want a formula use PV=nRT.  solve for P to get  P = (nRT)/V .\n\nsince the can is sealed, Volume stays the same.  R is a constant value anyway, n is the number of moles, which stays the same.  if you increase T, then P has to increase as well
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.
186855
each one exerts 623N so overall it exerts 3742N.\n\ni dont care if its wrong (im too lazy to actually try to figure it out properly and check it n stuff)
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.
940335
In your crucible, you have a mixture of mostly MgO and some Mg3N2.  That combined mass is 0.470 g.\n\nWhen you add water, all you are really adding is oxygen, since the hydrogen leaves as part of the ammonia gas.\n\nBasically, you are adding oxygen (atomic weight 16) and removing nitrogen (atomic weight 14).\n\nBecause of the formula Mg3N2(s) + H2O(l) -> MgO(s) + NH3(g),\nyou are 'taking out' 2 nitrogen atoms for every 3 oxygens you add (because there are 2 nitrogens to every 3 magnesium atoms, and each magnesium atom uses 1 oxygen to make MgO)\n\nSo you are gaining (0.486 - 0.470) = 0.016 g from the reaction of Mg3N2 to MgO\n\nThe mass from adding the oxygen is 16 g/mol * (3) * X\nThe mass from taking out the nitrogen is -14 g/mol * (2) * X\n\nWhere X is the number of moles of Mg3N2\n\nSo, 16 g * 3 * X - 14 g * 2 * X = 0.016 g\n\nSolving for X, X = 0.016 / (48 - 28) = 0.0008 mol\n\nThe molecular weight of the Mg3N2 is 100.9 g/mol, so the mass of Mg3N2 is 100.9 g/mol * 0.0008 mol = 0.0807 g\n\nSo, the mass percent is .0807 / 0.470 = 17.2 % Mg3N2\n\n.............\n\nAn another note, one reason you might be confused is your second reaction:\n\nThere are actually 2 reactions going on:\nMg + 1/2 O2 -> MgO\n3Mg + N2 ->Mg3N2\n\nThe first reaction reacts much faster than the second, meaning that more MgO will be produced than Mg3N2.\n\nHope this helps.