I need chem help!! (Ideal Gas Law)?
450501
Sorry this is a brief response: figure out the moles of gas in the reactants side of the equation. 5 moles of oxygen means there are 5 moles of gas. Then use Ideal gas law:\n\nPV = nRT\n\nSo, you know Volume (1500 mL), n (5 moles x molecules of O2 per mole), and T (25C + 273C = 298 K). Calculate P.\n\nFor n:\n5 moles (6.022x10^23atoms/moles) (2 atoms/molecule)\n\nHope that helps....
173870
1.Since Volume is proportional to Temperature, it could also be written with a constant(k): V = k x T , V/T = k , k = 0.923L/ (25+273)K = 3.10x10^-3 L/K\n2.I'm not too sure about this one, but here is my best shot.\nUsing the ideal gas law PV=nRT, same Temp & Pressure, let Va be the volume of the first baloon and Vb be the volume of the second balloon. Va = 1.25Vb, Let na be the number of moles in baloon 'a' and nb be the number of moles in balloon 'b'.\n\nBalloon a: PVa = naRT\nBalloon b: PVb = nbRT\nsolve for P,T or R.\nThis case I solved for T:\nT = PVa/naR  & T = PVb/nbR, since same temp (or pressure or constant) which is PVa/naR = PVb/nbR. Divide by P & R, Va/na = Vb/nb\nVa x nb = Vb x na, Va = 1.25 Vb, @75%\n1.25(.75)Vb nb = 0.75Vb x na, divide by .75Vb\n1.25nb = na\nSo there is 25% more moles (or mass) in baloon A.
193517
You don't provide enough data to determine this.  The gas law PV=nRT has variables you need to define.
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?
863456
I assume the reaction in question is:\n\nA(g) <-> 2B(g)\n\nThe equilibrium constant, Keq, for this reaction is given by:\n\nKeq = ([B]^2)/[A]\n\nwhere the quantities in square brackets mean "the concentration of" the relevant species at equilibrium (or, in the case of gases, one commonly uses equilibrium partial pressures instead of concentrations).\n\nPlugging in the values we are given for the equilibrium concentrations, we have that:\n\nKeq = (0.154^2)/0.381\n\nKeq = 0.0622
695888
I assume the temperature is high enough that the O2 & Ne behave as ideal gases.\n\nThe partial pressure of each gas in a mixture of gases is proportional to its mole fraction.\n\nMoles of Ne = 12.6/20.18 = 0.6244\nMoles of O2 = 7.10/32.00 = 0.2219\nTotal moles = 0.6244 + 0.2219 = 0.8463\nMole fraction of Ne = 0.6244/0.8463 = 0.7378\nPartial pressure of Ne = 0.7378*0.435 = 0.321 atm (answer)\nMole fraction of O2 = 0.2219/0.8463 = 0.2622\nPartial pressure of O2 = 0.2622*0.435 = 0.114 atm\nTo check to make sure there are no errors:  0.114 + 0.321 = 0.435
434788
well, the ideal gas law is PV = mRT.  You need to write this expression for the butane that escaped the lighter, because we don't have enough information about the butane still in the lighter to solve the problem (we don't know the pressure).\n\nFirst you need to convert all your units into the units found in your gas constant, R. looking at the information given, you can see that you'll need the gas constant with units of PV/(mT).\n\nR for an individual gas is the universal gas constant, R_u, divided by the molar mass, which is what we're trying to find. \n\nR = R_u/M\nR_u = 8.314 kPa*m^3/(kmol*K)   [I chose this form because it most closely matched the form I know I need]\n\nThis means that really, the ideal gas law is:\n\nPV = m(R_u)T/M\n\nConversions:\nvolume:  1m^3* = 1000 L\n           477ml = 0.477L = 0.000477 m^3\n-note: you need to know the volume of butane that escaped the lighter.  To do this, find the density of butane (mass/volume) of the origional amount of butatane, and then divide the mass that escaped by the density. You're assuming that the density is constant, because we don't know the volume of butatane that escaped the lighter, or the pressure inside the lighter.  If you have to list sources of error, this would be one.\n\ndensity  = (0.01311 kg/0.000477m^3) = 27.28 kg/m^3\n\nV = 0.00072 kg/27.28 kg/m^3 = 2.6 x 10^-5 m^3\n              (see how the kg units cancel, leaving m^3)\n             \nTemperature:\n23 degrees Celsius  = 23 kelvin\n\nPressure:\n100.33 kPa  - this would be about atmospheric pressure\n\nMass:\n1kg = 1000g\n\n13.11g = 0.01311 kg\n12.39g = 0.01239 kg \n0.72g = 0.00072 kg\n\n\nPutting the numbers in pv  = mrT, with units :\n\n(100.33 kPa)(2.6x10^-5 m^3)=0.00072 kg*(8.314 kPa*m^3/(kmol*K))*(23 K)/M \n\nI would write all that down on a piece of paper, so you can see it more clearly.  The key to getting this kind of problem right is making sure that all your units cancel, leaving you with only the units you want- in this case, kg/kmol.\n\nSolving for M, and canceling all the units, I got:\n\n0.0189 kmol/kg = 1/M\n\nM = 52.9 kg/kmol\n\n\nAccording to my tables, the actual value for Butane is 58.124 kg/kmol, so this is pretty close!\n\n\nHope this helped.  Feel free to email me if you need more help.
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
415862
R is the gas constant (= 8.31 J/mol*K).  The value should be with the list of constants in your book.\n\nOk Nic, you seem to be a bit confused on this issue so lets go back and start from scratch.\n\nFirst you obtained you values for T and K (experimentally or otherwise).  MAKE SURE YOU CONVERT YOUR TEMP VALUES TO KELVIN.  It's important for the units of the constant (R) are in Kelvin.  So then you calculate 1/T and lnK and plot the values for each set to generate your graph.\n\nYou calculate slope by taking two of your points (x1,y1) and (x2,y2).  From the data you added, let's say we'll use your first (0.002976, -3.079) and last points (0.003165, -4.225).\n\nslope= [-4.225- (-3.079)]/[0.003165-0.002976]= -6063.49 kelvin\n\nSince slope = -delta H/R\nThen -delta H = slope * R\nSo -delta H = -6063.49kelvin * 8.31 J/molkelvin\nAnd -delta H= -50387.62 J/mol\nRemoving the negatives from both sides\nDelta H= 50387.62 J/mol\n\nFor the second part of the question, use one (x,y) experimental point to find the y-intercept.  They don't want you to determine it from your plot, they want you to calculate the exact value.\n\nThe equation of a line is y=mx+b where "m" is the slope and "b" is the intercept.  So take another experimental point from readings (ie 0.003012, -3.751) and calculate the intercept by using the slope value calculated above.\n\nIf y=mx+b, then b= y-mx\nPlugging in the values\nb= -3.751 - (-6063.49kelvin * 0.003012/kelvin)\nb= -3.751 - (-18.26)\nb= 14.51\n\nTo calculate delta S for that same point (for that's what is requested), first you have to calculate delta G for that point.  Use the  following equation;\n\ndelta G = -RTlnK\n\nYou know the value for R (8.31 J/molkelvin), and you have the temp value and lnK value for that point for it was expermientally determined\n\n1/T= 0.003012/kelvin so T=332.01 kelvin\nlnK= -3.751\n\nSo delta G= -8.31 J/molkelvin * 332.01 kelvin * -3.751\nand delta G = 10349.02 J/mol\n\nYou can then calculate delta S for\nintercept = delta S/ delta G (given in your original data)\n\nTherefore delta S = intercept * delta G\nSo delta S = 14.51 * 10349.02 J/mol = 150164.28 J/mol\n\nHope this clarifies things.
655398
Here's how to solve it (I am convinced that you should do your own homework, but hints are OK as long as you learn something):\n\nStart solving by finding the molecular mass of the compound.\nA. Volume occupied by one mol of gas is 22.4 L at standard temperature and pressure (STP: 760 mmHg and 273 K (0 C))\nTo find the volume of gas that weighs 2.57 g at STP, use formula: P1V1/T1=P2V2/T2, where R is 0.08206, P1=319 mmHg, V1=2 L, T1=298 K; P2=760 mmHg, T2=273 K and V2 is unknown (it will be somewhat less than 1 L).\nAfter you get this volume, see how many mols do you have in it: divide 22.4 by V2 you just got. Then multiply this number by 2.57 and get the molecular mass of your compound (M(X)).\n\n2. Multipy M(X) separately by 0.6481, 0.136 and 0.2159 to get number of atomic weight units that are allocated to C, H, and O respectively. \nThen divide first number by 12, 2nd by 1 and 3rd by 16 to see how many atoms it amounts to. The resulting integer numbers are X, Y, Z.\n\n3. CxHyOz is the empirical formula of your compound. Google it to get the possible molecular ones. This is a trick question, so there will be more than one but less than 8. All of them are alcohols or ethers.
716268
PV = nRT\nn = PV/RT\n\nP = standard pressure(don't know which one you're picking)\nV = 2 L\nT = standard temp(again don't know which one you're picking)\nR = constant you can decide which one you want to use at\nhttp://en.wikipedia.org/wiki/Ideal_gas_constant
105211
Pressure * volume = number of mol * temperature (in ^oK) * 8.314\n\nYou have all the information you need right there. \nFirst solve for n for the initial state. \nSince you haven't added any more material you can then use the same value of n for the second state and solve for P.\n\nThis is sixth grade algebra. Anyone being asked to solve gas equations had better be able to do it standing on their head or they are in real trouble.
685887
Use the ideal gas law    pv = nRT. You will be solving for moles so rearranged it is pv/RT = n\n\nP = pressure (1.0 ATM)  V = volume (3.38L) R = .0821 (Ideal gas constant) and T = Kelvin temperature (Celsius temp + 273)\n\nThis will give you moles of propane. \n\nMUltiply this number by the molar mass of propane (44 g/mol) and you will have it.
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
805169
alright...\nconsider the R.H.S.\nh2o has two o-h bonds which account for 2x467 and since thre are 2 mol. u get 2x2x467--------(1)\nco2 has two c=o bonds which account for 2x799 and since u hav 4 mol. u get\n4x2x799-------------(2)\naddin 1 and 2 u get 1868+6392=8260\nconsider the L.H.S.\nu hav in c2h2, two c-h bonds which account for 2x413--------(3)\nand one triple bond accounting for 1x839---------(4)\nnow u hav 2x(3+4)=2x(826+839)=3330------(5)\nand u hav 5x(o=o) which acc for 5x(495)=2475----(6)\naddin 5 and 6 u get 5805\ntherefore ur ans is 5805-8260=-2455\nand since u want for one mol of acetylene u get -1227.5 kj/mole
851681
Write the balanced equation - you need the mole ratios.\n\nChange 1g hexane to moles of hexane.\n\nSee how many moles of oxygen are USED and how many moles of  CO2 and H2O are formed.\n\nYour total number of moles in the reaction mixture would then be:\n\nmoles of Oxygen NOT used + moles of CO2 (we will assume the water formed consenses to a liquid at 20 degrees).\n\nUse the Ideal Gas Law to figure out the volume.\n\nPV = nRT\n\nwhere P = given pressure - water vapor pressure\nwhere V = your unknown\nwhere n = your total moles of gas\nwhere R = the appropriate Universal Gas Law constant\nwhere T = the temperature in KELVIN
728397
You need to first balance the equation.\n\n2CO + O2 --> 2CO2\n\nUsing this you find the answer to be D.