Why is the density of a gas much lower than that of

Name: Justin RothCollege ID: 0360782Thomas Edison State CollegeGeneral Chemistry I (CHE-111)Section no.: 6Semester and year: Jan 2016Written Assignment 6: GasesAnswer all assigned questions and problems and show all work.1. A gas occupying a volume of 725 mL at a pressure of 0.970 atm is allowed to expandat constant temperature until its pressure reaches 0.541 atm. What is its finalvolume?(5 points)V2=725x.970/.541=1299.91(Reference: Chang 5.19)2. The volume of a gas is 5.80 L measured at 1.00 atm. What is the pressure of the gasin mmHg if the volume is changed to 9.65 L? (The temperature remains constant.) (5points)1*5.8=P2*9.655.8/9.65=P2P2=.601.601*760=456.788mmHg(Reference: Chang 5.21)3. Why is the density of a gas much lower than that of a liquid or solid underatmospheric conditions? What units are normally used to express the density ofgases? (5 points)In solids and liquids the particles are more packed together causing not as muchmovement. Where in a gas the particles are very loose and everywhere.Units-L mL cm^3 dm^3 m^3(Reference: Chang 5.30)4. A sample of nitrogen gas kept in a container of volume 2.3 L and at a temperature of32°C exerts a pressure 4.7 atm. Calculate the number of moles of gas present. (8points)PV=nRT1 4.7*2.3=n*.0821*305K10.81=n*25.04n=.43moL(Reference: Chang 5.31)5. What volume will 5.6 moles of sulfur hexafluoride (SF6) gas occupy if thetemperature and pressure of the gas are 128°C and 9.4 atm? (8 points)9.4*V=5.6*.082*(273 128) V=19.6L(Reference: Chang 5.33)6. A gas-filled balloon having a volume of 2.50 L at 1.2 atm and 25°C is allowed to riseto the stratosphere (about 30 km above the surface of Earth) where the temperatureand pressure are –23°C and 3.00 × 10–3 atm respectively. Calculate the final volumeof the balloon. (8 points)((1.2atm*2.50L)(298K)=((3.00×10^-3atm*V2/250K).0007*250K=2.5175/3.00*10^-3atm=8.39×10^-4 = 839L(Reference: Chang 5.35)7. A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0°C. (a) Calculate thedensity of the gas in grams per liter. (b) What is the molar mass of the gas? (10points)PV/RT=1.00*2.10/.08206*300=.08534.56g/.0853mol=54.5g/mol54.5*1.00/.08206*300=2.21g/L(Reference: Chang 5.47)8. A certain anesthetic contains 64.9 percent carbon 13.5 percent hydrogen and 21.6percent oxygen by mass. At 120°C and 750 mmHg 1.00 L of the gaseous compoundweighs 6.90 g. What is the molecular formula of the compound? (10 points)PV=nRTn=PV/RT= (0.987 atm x 1.00 L)/(0.082 x 392.3 oK)= 0.0307 molesmoles= mass/molec. wt0.0307 moles= 6.90/molec. wt.molec. wt.= 224.8assume 100 g of anesthetic: 64.9 g C 13.5 H 21.6 g Omoles of each elementmoles C: 64.9 g C/12 g/mole C=5.40moles H: 13.5 g H/1g/mole H= 13.5moles O: 21.6 g O/16 g/mole O= 1.352 moles ratio of elements: C 5.40/1.35=4; H 13.5/1.35=10; O 1.35/1.35= 1C4H10O1 and its molec. wt= 74224.8/74=~3 so the molecular formula is C12H30O39. What is the mass of the solid NH4Cl formed when 73.0 g of NH3(g) are mixed withan equal mass of gaseous HCl? What is the volume and identity of the gasremaining measured at 14.0°C and 752 mmHg? (8 points)NH3(g) HCl(g) ? NH4Cl73.0g HCI x 1mol HCI/36.5g HCI=2.00mol73.0g NH3x 1molNH3/17.0g HCI=4.29 mol2.00mol NH4CI x53.5g NH4CI/mol NH4CI=107.0g NH4CIV=NRT/PV=2.29 mol x 62.4 L mmHg/752mmHg=54.4L10. A mixture of gases contains 0.31 mol CH4 0.25 mol C2H6 and 0.29 mol C3H8. Thetotal pressure is 1.50 atm. Calculate the partial pressures of the gases. (8 points)(Reference: Chang 5.67)total moles of gas = 0.31 0.25 0.29 = 0.85P/ P of CH4 = 0.31/0.85 * 1.5 atm = 0.547059 atmP/P of C2H6 = 0.25/0.85 * 1.5 atm = 0.441176 atmP/P of C3H8 = 0.29/0.85 * 1.5 atm = 0.511765 atm11. Propane (C3H8) burns in oxygen to produce carbon dioxide gas and water vapor. (a)Write a balanced equation for this reaction. (b) Calculate the number of liters of carbondioxide measured at STP that could be produced from 7.45 g of propane. (10 points)a.C3H8 5O2 —- 3CO2 4H20b.7.45g x 1mol/44g/mol=.169 moles propane.169×3=.508 moles of carbon dioxide22.44 mol x .508mol=11.4L(Reference: Chang 5.109)12. A 10.0 g piece of pure aluminum is placed in 75.0 mL of 0.54 M hydrochloric acid atSTP condition. They react as follows:2Al 6HCl ? 3H2(g) 2AlCl3Calculate the following:a. Volume in liters of hydrogen gas. (5 points)(0.0405 mol HCl) x (3 mol H2 / 6 mol HCl) x (22.414 L/mol) = 0.45 L H23 b. Molarity of Al 3. (Assume 75.0 mL solution.) (5 points)(0.0405 mol HCl) x (2 mol Al / 6 mol HCl) / (0.0750 L) = 0.18 mol/Lc. Molarity of Cl–. (Assume 75.0 mL solution.) (5 points)0.54 M CI4