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Cuyamaca CHEM 141 - 141-Exam-2-Fa16-Key

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Part 1: Multiple Choice (2 points each)Directions: Please circle the best answer for each of the following questions.13. Part 2: Short Answer14. Directions: Answer each of the following questions. Be sure to use complete sentences where appropriate. For full credit be sure to show all of your work.Grossmont College Chemistry 141 Exam 2 Fall 2016Name: ___________________________________Date: ________Instructor: Diana VanceExam 2Part 1: Multiple Choice (2 points each) Directions: Please circle the best answer for each of the following questions. 1. The pressure exerted on a sample of a fixed amount of gas is doubled at constant temperature, and then thetemperature of the gas in kelvins in double at constant pressure. What is the final volume of the gas? a. The final volume of the gas is four times the initial volume. b. The final volume of the gas is one-fourth the initial volume. c. The final volume of the gas is one-half the initial volume. d. The final volume is twice the initial volume. e. The final volume of the gas is the same as the initial volume. 2. _____________ states that the volume of a gas varies inversely to the pressure on the gas, when temperatureand number of moles are kept constant (PV = constant). a. Avogadro’s lawb. Boyle’s lawc. Charles’ lawd. Gay-Lussac’s lawe. Dalton’s law 3.4. Assuming ideal behavior, which of these gas samples has the greatest volume at STP?a. 1 g H2b. 1 g O2c. 1 g Ard. 1 g CH4e. all of the above have the same volume5. What is the measurementshown by the calipers?a. 24 mmb. 27.4 cmc. 28.4 cmd. 29 cme. 30.4 cm 6. For which should thestandard heat of formation, ∆Hf°, be zero at 25 °C? a. oxygen gas, O2 (g)b. ozone gas, O3 (g)c. atomic oxygen gas, O (g) d. all of the abovee. none of the above7. For a process at constant volume: a. q = 0, w = 0, ∆E = 0Page 1 of 7Grossmont College Chemistry 141 Exam 2 Fall 2016Name: ___________________________________Date: ________Instructor: Diana Vanceb. w = 0 and ∆E = qc. w = 0 and ∆H = qd. w = 0 and ∆E = ∆He. none of the above8. For the reaction I2 (g)  I2 (s), ∆H° = -62.4 kJ at 25 °C then a. ∆H°vap = -62.4 kJ/mol b. ∆H°vap = 62.4 kJ/mol c. ∆H°sub = -62.4 kJ/mol d. ∆H°sub = 62.4 kJ/mol e. b and c9. A quantized variable a. can be continuously varied.b. can only assume certain values.c. consists of photons.d. is extremely small.e. is a state function.10. Which subshell letter corresponds to an 8 lobed pattern?a. sb. pc. dd. fe. more than one answer11. How many electrons does an f orbital contain? a. 2 b. 6c. 7d. 10 e. 1412.Page 2 of 7Grossmont College Chemistry 141 Exam 2 Fall 2016Name: ___________________________________Date: ________Instructor: Diana Vance13. Part 2: Short Answer 14. Directions: Answer each of the following questions. Be sure to use complete sentences where appropriate.For full credit be sure to show all of your work. 15.1. State whether each of the following samples of matter is a gas. If there is not enough information for you to decide, write “insufficient information.” (5 points) a. A material is in a steel tank at 100 atm pressure. When the tank is opened to the atmosphere, the material suddenly expands, increases its volume by 1%. 16. Not a gas. A gas would expand to an infinite volume. b. A 1.0 mL sample of material weighs 8.2 g. 17. Not a gas. A density of 8.2 g/mL is typical of a solid. c. The material is transparent and pale green in color.18. Insufficient informationd. Nitrogen, oxygen, and helium at room temperature and pressure. 19. Gase. One cubic meter of material contains as many molecules as 1.0 m3 of air at the same temperature and pressure.20. Gas 21.2. At an underwater depth of 250. ft, the pressure is 8.38 atm. What should the mole percent ofoxygen in the diving gas be for the partial pressure of oxygen in the gas to be 0.21 atm, thesame as it is in air at 1.0 atm (6 points)? 22.23.PO2=PtotalχO2⟹ χO2=PO2Ptotal=0.21 atm8.38 atm=0.02524.%O2=0.025× 100 %=2.5 % O225.26.27.28.29.30.31.32.33.34.35.36.37.38.39.Page 3 of 7Grossmont College Chemistry 141 Exam 2 Fall 2016Name: ___________________________________Date: ________Instructor: Diana Vance3. You have a gas, one of the three known phosphorus-fluorine compounds (PF3, PF5, and P2F4).To find out which, you decide to measure its molar mass (12 points). a. First you determine that the density of the gas is 5.60 g/L at a pressure of 0.971 atm and atemperature of 18.2 °C. Calculate the molar mass and identify the compound. 40.MM=DRTP41.MM=(5.60gL)(0.08206L atmmol K)(18.2+273.15)K0.971 atm42.MM=DRTP=(5.60gL)(0.08206L atmmol K)(291. 3 5 K)0.971 atm=137.884463gmol≈ 138gmol43. The molar mass is closest to that of P2F4, which has a molar mass of137.94 g/mol. 44.b. To check the results from part (a), you decide to measure the molar mass based on the relative rates of effusion of the unknown gas and CO2. You find that CO2 effuses at a rate of 0.050 mol/min, whereas the unknown phosphorus fluoride effuses at a rate of 0.028 mol/min. Calculate the molar mass of the unknown gas based on these results. 45.rate1rate2=√MM 2MM 1⟹ MM 2=(rate1rate 2)2MM 1=(0.050molmin0.028molmin)2(44.009gmol)=140.3348214gmol≈ 140gmol46. This is consistent with the results from part a. 47.4. A 13.8 g piece of zinc is heated to 98.8 °C in boiling water and then dropped into a beaker containing 45.0 g of water at 25.0 °C. When the water and metal come to thermal equilibrium, the temperature is 27.1 °C (15 points). a. What is the specific heat capacity of zinc? 48.q¿=−qout49.qwater=−qzinc50.mwatercwater∆ Twater=−mzincczinc∆ Tzinc⇒ czinc=mwatercwater∆ Twater−mzinc∆ Tzinc51.czinc=(45.0 g)(4.184Jg ℃)(27.1℃ −25.0 ℃ )−(13.8 g)(27.1℃ −98.8℃ )=(45.0 g)(4.184Jg ℃)(2.1 ℃ )−(13.8 g)(−71.7 ℃ )=0.399599782Jg ℃≈ 0.40Jg ℃b. If the molar heat capacity of zinc at 300 K is 25.45 J/mol K, what is the percent error? 52.25.45Jmol K×1 mol Zn65.382 g Zn×1 K1℃=0.389250864Jg ℃≈ 0.3893Jg ℃Page 4 of 7Grossmont College Chemistry 141 Exam 2 Fall 2016Name: ___________________________________Date: ________Instructor: Diana Vance53.%error=observed value−accepted valueaccepted value× 10054.%error=0.40Jg ℃−0.3893Jg ℃0.3893Jg ℃× 100=0. 01 07Jg ℃0.3893Jg ℃× 100=2.7 %55.5. If the same amount of heat is supplied to samples of 10.0 g each of aluminum (specific heat0.897 J/g °C), iron (specific heat 0.450 J/g °C) and copper


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Cuyamaca CHEM 141 - 141-Exam-2-Fa16-Key

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