What is the mass of a 199.4 mL sample of carbon dioxide gas at 23.1oC and 115.56 kPa
123253
PV = nRT (equation)\n\nP=115.56Pa,  V=199.4mL=0.1994L, n=?, R=8.314472 LkPa/Kmol, T=23.1ºC=296.2K  (stuff to plug in)\n\nn = PV/RT = 0.009356 mol CO2  (solve for moles)\n\n\nMM = 44.00g/mol\n0.009356 mol x (44.00g/mol) = 0.412g  (switch moles to grams)\n\nHope that helps!
423168
The first material is not what you give but N2O4\n\nAssume you start with 1 mole of N2O4. After dissociation you have 1 - .1605 = .8395 mole\nWhen 1 mole of N2O4 dissociates it gives 2 moles of product. Hence the .1605 moles gives .321 moles NO2.\nTotal moles .8395 + .321 = 1.1705\nPartial pressure for each gas is moles of gas/total moles multiplied by the total pressure.\neg for N2O4 the partial pressure is .8395/1.1705 x 1 X 10^5 pa. \n\nTake it from here.
371445
CH4 + 2O2-> CO2 + 2H2O
436322
Let's assume that the flask is upside down in a water bath.\nAt 100 C, it contains some water (volume = x), and a volume of hot gas (VH).  When cooled to 0 C, the volume of the gas decreases, and some water (214 mL) is pulled into the flask. The volume of the cool gas is now VC. \n\nSome relationships to get you started:\n\nVH - x = 300 mL\nVH - VC = 214 mL\n\nCharles' Law\n\nVC / TC  =   VH / TH\n\nor\n\nTH /TC    =    VH / VC
208982
E = Fd = (2E-9 N) (200E-12 m) = x J\nx J (1 kJ / 1000 J) = xE-3 kJ\n\nI think this is already per mol...
66125
The question says that 5.97 grams of Glucose reacts to form 1.44 Liters of CO2 (at a given temp./pressure).\nConvert both of these to moles.\n5.97 grams Glucose --(use molar mass)--> x moles of Glucose.\n1.44 L of CO2 --(Use ideal gas law)--> y moles CO2\n\nFrom the chemical reaction, find the ideal molar ratio between the moles of Glucose reacted and moles of CO2 produced.  You earlier found the actual moles (y) of CO2 produced by x moles of Glucose reacting, how does this compare?\n\n% yeild = (actual / theoretical) * 100
341689
1 mol the gas occupies 24dm3\n\n1 mol of the gas is 44 grams\n   (C=12, O= 16, 12+16+16=44)\n\n1 kg = 1000 g\n\n1000 g * 1mol/44g = 22.727 mol of CO2\n\n22.727 mol CO2 * 24dm3/1mol = 545.448 dm3 of CO2
435544
Change grams to moles: 0.254 g/122.5g/mole = 0.00207 moles\n\nChange to moles O2 (ratio from equation is 3/2)\n0.00207 x 3/2 = 0.00311 moles O2\nCalculate vol. from relationship 22.4L/mole for any gas at STP\n22.4L/mole x 00311mole = 0.0697 L at STP\n\nUse Combined Gas Law to change to vol. at 24.0 degrees and 729.1mm. Reduce pressure by the water vapor pressure.\n\nP1 = 760.mm V1 = 0.0697 T1 = 0 + 273\nP2 = 729.1-22.4 V2 = x T2 = 24 + 273\n\nx = (0.0697)(760)(297)/(273)(706.7)... = 0.0815 L or 81.5 mL
104791
Well...for CO2 I have:\n\na = 3.658 bar L^2/mol\n\nb = 0.04286 L/mol\n\n\nIf you are doing research at this level...have you forgottn about the trusty CRC handbook for Chem/Phys?\n\nThat's where I found my data. Mine is 77th edition page 6-48.\n\nHope that helps.
70902
Although I know I am not getting this $20.  199 kJ of energy.\n\nIn response to a later answer that claimed I was wrong he was right and wrong\n\n He wrote "first find the amount of mol of nitrogen\nmol = 1/14" But infact the amount was 1 kg and that divided by 14g/mol would be 71.43 moles and multiplying that times his enthalpy of vaporization which is 2.7928kJ/mol would give 199 kJ of energy. So I was right, but wrong with my significant figures because the asker said it was 1.0 kg which would gie me 2 sig figs and the answer would be 2.0*10^2 kJ.
419069
Since pressure is a colligative property, its additive. Since you have .125 atm of methane, you can subtract that from the total pressure. That will give you .36 - .125 = .235 atm of Choride gas. \n\nThen using the ideal gas law (PV = nRT) to solve for moles. \nP = .235 atm (from Cl2) V = .268 L (given) T = 293 K (given and adjusted to kelvin) and R = .08206 Latm/mol K (adjusted gas constant to cancel units.\n\n.235 atm(.268 L) = n .08206 Latm/mol K (293K)\nn = .0026 mols. \n\nThen use moles to solve for grams. \nAMU of Cl2 is 70.9 AMU\nmoles = grams / AMU\n.0026 = x grams / 70.9 AMU\n.0026*70.9 = 0.186 grams\n\n0.186 grams of Cl2 present in the mixture.
468054
C57H110O6(l)+109O2(g)--heat>\n57CO2(g)+55H2O(l)\n\n\nabout  15.05kg of  CO2
359737
I am assuming that the values of H° and G° are the standard enthalpy and Gibbs free energy changes for the formation of phsogene from CO(g) and Cl2(g).\n\nFor part (a), you should already know that:\n\ndelta-G = delta-H - T*delta-S\n\nSo, for the values you are given for 298 K,\n\n-206 kJ/mol = -220 kJ/mol - 298 K * delta-S°\n-46.98 J/(mol*K) = delta-S°\n\nFor part (b), we now have (assuming delta-H and delta-S are independent of temperature) that \n\ndelta-G_601K = -220kJ/mol - 601 K * (-4.698*10^-02 kJ/mol*K)\ndelta-G_601K = -191.8 kJ/mol
569482
Propane  = C3H8. The molar mass is\n(3 x 12 + 8 x 1) gram = 44 gram.\nYou give 44.1 g. That a little bit better.\n\nThe volume of the cylinder is\nAh = pi r2 h = 3.14 x (0.050 m)^2 x 1.30 = 0.212 m3.\n\nBefore: pV = nRT\n1.4×10^6 Pa x 0.212 m3 = n x 8.31 J/(mol x K) x (273 + 23)K\nSo n(begin) = 119 mol.\n\nAfter: pV = nRT\n2.4×10^5 Pa x 0.212 m3 = n x 8.31 J/(mol x K) x (273 + 23)K\nSo n(end) = 20 mol.\n\nUsed is (119 - 20) mol propane, being 99 x 44.1 gram = 4.4 kg
453648
What is the weight of 15L of Carbon dioxide at STP?\n\nAssuming you want to use the Ideal Gas law.... PV = nRT\n\nFrom this, you know:\nP (1 atm at STP)\nV (15 Liters...given)\nR (0.08206 ...constant)\nT (25 degrees, or  298 kelvin)\nPlug in everything into the equation, adn find n, which is # of moles.\n\nNow use the formula \nmass = molar mass * moles (m = mm * n)\nyou know the molar mass (just mass of C + mass of O*2)\nand find mass!\n\n-------------------------------------------------------------------------------------------------\nFor the second question, use the m=mm*n formula to find the moles. Divide by Avagadro's number to get number of atoms!
429096
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
494642
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 65.45% x 4.13 g = 2.703 g\nhydrogen: 5.49% x 4.13 g = 0.227 g\noxygen: 29.06% x 4.13 g = 1.200 g\n\n2) convert to moles.\ncarbon: 2.703 / 12 = 0.225 mol\nhydrogen: 0.27 / 1 = 0.227 mol\noxygen: 1.200 / 16 = 0.0750 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.225 / 0.075 = 3\nhydrogen: 0.227 / 0.075 = 3\noxygen: 0.075 / 0.075 = 1\n\nThe empirical formula of the substance is (C3H3O)k\nThe atomic masses add up to 55. To find k we need the molar mass of the compound. Use PV=nRT and plug in the data given *in the correct SI units*. Solve for n the number of moles in the sample. You problably find 5.51/55 or 5.51/110 so that k=1 or 2