What is the formula for the total kinetic energy for a sample of gas?
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.
753535
Just use the formula\n\nE = m*Cp*(Delta T)\n\nE=energy\nm=mass\nCp= Specific Heat\nDelta T = change in temperature\n\nCp= 0.51998 J/(C*g)\n\nGood Luck with the test!
573522
kinetic energy is 1/2 mv² so it depends on mass and velocity
895405
use dalton's law to get the pressure of CH4(total pressure - H2 pressure)\nthen use ideal gas law to get the moles of CH4(PV=nRT)\n*watch your units*\nchange moles of CH4 to grams of CH4(moles x molar mass)
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
843034
1. Modify the ideal gas equation such that molecular mass (MW)and weight of gas appear.\nIdeal gas equation: PV = nRT\nsince n (moles) = weight / MW = g/MW\nso, your equation will become PV = gRT/MW\nGiven your values, you will end up:\nMW = gRT/ PV\n\n2. Plug in the values:\n\nMW = 3.06g * 0.08206 L-atm/mol-K * (51.4 + 273 K) / 0.970atm * 1.00L\nMW = 83.98 g/mol\n- this is the molecular mass of your gas.\n\n3. Determine the molecular formula of the gas.\n-since you have 14.37% H, your C is 85.63%\n-assume 100g gas so that you'll have 14.37g H and 85.63g C\n-get the number of moles:\nH: 14.37g / 1g/mol = 14.37 mol H\nC: 85.63g / 12g/mol = 7.13 mol C\n\n- get the mole ratio: that is, 1 mol C is to 2 mol H.\n- determine your multiplier such that MW result will be approximately 83.98g/mol. In this case, its 6. \n\nSo, your molecular formula is C6H12.\n\nTo check,\n\nC: 6 * 12g/mol = 72g/mol\nH: 12 * 1g/mol = 12g/mol\ntotal: 84g/mol ~ 83.98g/mol
856297
V=55.0L\nP[H2 = 2.00 atm\nP[O2 = 3.00 atm\nT[1 = 25°C = 298 K\nT[2 = 125°C = 398 K\nR = 0.082057 L · atm / (K · mol)\n\nIdeal gas law: PV = nRT\n\nAt 25°C:\nsolving n for H2:\nn = PV/RT = 2.00 * 55 / (0.082057 * 298) mol = 4.498 mol\n\nsolving n for O2:\nn = PV/RT = 3.00 * 55 / (0.082057 * 298) mol = 6.748 mol\n\nAt 125°C:\nsolving n for H2:\nn = PV/RT = 2.00 * 55 / (0.082057 * 398) mol = 3.368 mol\n\nsolving n for O2:\nn = PV/RT = 3.00 * 55 / (0.082057 * 398) mol = 5.052 mol\n\nThe Reaction that occurs is:\nH2 + 1/2 O2 = H2O\n\nSo, hydrogen is the limiting reactant, leaving water and oxygen as the resulting composition.\n\nAt 25°C, 4.498 mol of H2 and 2.249 mol of O2 form 4.498 mol of H2O, leaving 4.499 mol of O2.  The total number of moles (n) is 8.997 mol\n\nAt 125°C, 3.368 mol of H2 and 1.644 mol of O2 form 3.368 mol of H2O, leaving 3.408 mol of O2.  The total number of moles (n) is 6.776 mol\n\nUsing the ideal gas law again:\nPV=nRT\n\nP = nRT/V\n\nAt 25°C:\nP = 8.997 mol * 0.082057 L · atm / (K · mol) * 298 K / 55 L = 4.00 atm\n\nAt 125°C:\nP = 6.776 mol * 0.082057 L · atm / (K · mol) * 398 K / 55L = 3.00 atm
802951
Kinetic energy is energy due to motion and potential energy is energy due to position in a force field.  The formula for translational kinetic energy is KE = 0.5*m*v^2, where m is mass and v is velocity.  The most common type of potential energy is gravitational potential energy, and near the surface of the Earth (or some other planetary body) the formula is PE = mgh, where m is mass, g is acceleration due to gravity, and h is the height above the surface.  Another common one is elastic potential energy in a compressed spring.  In that case the formula is PE = 0.5*k*x^2, where k is the spring constant and x is the compression.  An object can also have electrical potential energy, if it has an electrical charge and is positioned some distance from another charge.
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
994673
I am not sure of the final answer you are looking for.  If you are looking for the number of molecules transferred after a certain time t, the answer is simply n=-DAt(dc/dz).  This of course assumes that dc/dz remains relatively constant.  If not you will need a dc/dz as a function of t.
933724
We will assume CH4 is an ideal gas.\n\nPV=nRT\nP=nRT/V= 0.123mol * (0.0821 atm*L/(mol*K)) * 760 torr/1atm * (30+273.15)K / 20.0L = 116 torr\n\nUsing Dalton's Law of partial pressures, Ptotal = P1 + P2 + P3 + .... + Pn, the total vapor pressure is just the sum of the two partial pressures.\n\nP= 31.82 torr + 116 torr = 148 torr
945784
PV = nRT\nn = PV/RT\n= ([2.5 * 101.325] * 30 ) / (8.31 * [273 + 327])\n= 1.5 mol of CH4 (2 Sig Fig)\n\n1 CH4 = 1 CO2 Therefore 1.5 mol of CO2\nm(CO2) = 1.5 * (12 + 32)\n= 66 g
816920
Call p1 = p(Cl2), p2 = p(O2)\n\np1 = 0.15 atm, p = p1 + p2 = 0.365 atm\n\n-> p2 = 0.365 - 0.15 = 0.215 atm\n\nV = 0.538 l\n\nT = 273 + 32 = 305 K\n\nR = 0.082 L*atm / K\n\nFor a mixture, all gases share the same V and T:\np2.V = n2.RT -> n2 = p2.V/(R.T) = 0.004625 mol\n\nTherefore, the mass of O2 is:\n0.004625 * 32g = 0.148 g
363124
detlaH=H(products) - H(Reactants)\n\n200KJ/mol
606680
If by force you are refering to the kinetic energy of a moving object, then the equation is as follows: kinetic energy = one half momentum squared/mass.  Momentum of course can be calculated by mass times velocity.
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?
752459
This is a complete combustion.\nReaction:            aX + bO2 --> dCO2 + eH2O\nd = 0.811818182, derived from mass of CO2 and its Mr.\n\nX has 0.811818182 mol of C.\n\nOriginal sample has 0.811818182/ 1.626 x 60.095\n                                  = 30.00382143mol of C