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Name: Justin Roth College ID: 0360782 Thomas Edison State College General Chemistry I (CHE-111) Section no.: 6 Semester and year: Jan 2016 Written

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Name: Justin Roth College ID: 0360782 Thomas Edison State College General Chemistry I (CHE-111) Section no.: 6 Semester and year: Jan 2016 Written Assignment 6: Gases Answer 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 expand at constant temperature until its pressure reaches 0.541 atm. What is its final volume? (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 gas in mmHg if the volume is changed to 9.65 L? (The temperature remains constant.) (5 points) 1*5.8=P2*9.65 5.8/9.65=P2 P2=.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 under atmospheric conditions? What units are normally used to express the density of gases? (5 points) In solids and liquids the particles are more packed together causing not as much movement. 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 of 32°C exerts a pressure 4.7 atm. Calculate the number of moles of gas present. (8 points) PV=nRT 1
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4.7*2.3=n*.0821*305K 10.81=n*25.04 n=.43moL (Reference: Chang 5.31) 5. What volume will 5.6 moles of sulfur hexafluoride (SF 6 ) gas occupy if the temperature 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 rise to the stratosphere (about 30 km above the surface of Earth), where the temperature and pressure are –23°C and 3.00 × 10 –3 atm, respectively. Calculate the final volume of the balloon. (8 points) ((1.2atm*2.50L)(298K)=((3.00x10^-3atm*V2/250K) .0007*250K=2.5175/3.00*10^-3atm=8.39x10^-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 the density of the gas in grams per liter. (b) What is the molar mass of the gas? (10 points) PV/RT=1.00*2.10/.08206*300=.0853 4.56g/.0853mol=54.5g/mol 54.5*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 21.6 percent oxygen by mass. At 120°C and 750 mmHg, 1.00 L of the gaseous compound weighs 6.90 g. What is the molecular formula of the compound? (10 points) PV=nRT n=PV/RT= (0.987 atm x 1.00 L)/(0.082 x 392.3 oK)= 0.0307 moles moles= mass/molec. wt 0.0307 moles= 6.90/molec. wt. molec. wt.= 224.8 assume 100 g of anesthetic: 64.9 g C + 13.5 H + 21.6 g O moles of each element- moles C: 64.9 g C/12 g/mole C=5.40 moles H: 13.5 g H/1g/mole H= 13.5 moles O: 21.6 g O/16 g/mole O= 1.35 2
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