Exam 2 answer key

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FA23 CHM 113 Exam 2 Student Name: 1 INSTRUCTIONS • Answer any 10 questions. If you answer all 12 questions, I will count your best 10 scores. • Constants and conversion factors are printed on the back of the periodic table. • You may NOT look up any other resources using a computer, cell phone, or any other means. • Report all your answers in appropriate sig figs. • You must show your work to receive full credit. • If you answer on scratch paper, indicate that explicitly on the exam answer sheet. 1. Determine the volume occupied by 2.34 g of CO 2 gas at STP. V = nRT/P V = [(2.34 g / 44.0 g mol¯ 1 ) (0.08206 L atm mol¯ 1 K¯ 1 ) (273.15 K)] / 1.00 atm = 1.19 L 2. A sample of argon (Ar) gas at STP occupies 56.2 L. Determine the number of moles of argon and the mass of argon in the sample. n = PV / RT n = [(1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol¯ 1 K¯ 1 ) (273.15 K)] = 2.50728 mol = 2.51 mol 2.50728 mol X 39.948 g/mol = 100. g (to 3 sig figs). 3. 96.0 g of a gas occupies 48.0 L at 700.0 mm Hg and 20.0 °C. What is its molecular weight? n = PV / RT n = [(700.0 mmHg / 760.0 mmHg atm¯ 1 ) (48.0 L)] / [(0.08206 L atm mol¯ 1 K¯ 1 ) (293.15 K)] = 1.837 mol 96.0 g / 1.837 mol = 52.2 g/mol = MW 4. 20.83 g of a gas occupies 4.167 L at 79.97 kPa at 30°C. What is the molecular weight of this gas? n = PV / RT n = [(79.97 kPa / 101.325 kPa atm¯ 1 ) (4.167 L)] / [(0.08206 L atm mol¯ 1 K¯ 1 ) (303.15 K)] = 0.13220 mol 20.83 g / 0.13220 mol = 157.6 g/mol 5. A gaseous mixture of O 2 and N 2 contains 32.8% of nitrogen by mass. What is the partial pressure of oxygen in the mixture, in mm Hg, if the total pressure is 785.0 mmHg? Hint: assume you have 100.0 g of the mixture. If 32.8% is nitrogen, then 67.2% is oxygen. Assuming 100.0 g is the mass of the mixture, 32.8 g N 2 / 28.014 g/mol = 1.17 mol 67.2 g O 2 / 31.998 g/mol = 2.10 mol n (total) = 1.17 + 2.10 = 3.27 mol P (oxygen) = P (total) X (n (oxygen) /n (total) ) = (785.0 mmHg) (2.10 mol / 3.27 mol) = 504 mmHg 6. The mass percent of a gas mixture is 22.70% O 2 , 21.00% C 2 H 2 F 4 and 56.30% C 6 H 6 . Calculate the partial pressure (atm) of C 2 H 2 F 4 if the total pressure of the sample is 1444 torr. Hint: assume you have 100.0 g of the mixture. 22.70 g O 2 / 31.998 g/mol = 0.7094 mol 21.00 g / 102.03 g/mol = 0.205822 mol 56.30 g C 6 H 6 / 78.1134 g/mol = 0.720747 mol Mole fraction (C2H2F4) = 0.205822 mol / (0.7094 mol + 0.205822 mol + 0.720747 mol) = 0.1258

FA23 CHM 113 Exam 2 Student Name: 2 P (C2H2F4 ) = 1444 torr X 0.1258 = 181.7 torr X 1 atm/760 torr= 0.2391 atm 7. 8.00 g of methane (CH 4 ), 18.0 g of ethane (C 2 H 6 ), and an unknown amount of propane (C 3 H 8 ) were added to the same 10.0 L container. At 27.0 °C, the total pressure in the container was measured to be 4.43 atm. Calculate the partial pressure of each gas in the container. 8.00 g of CH 4 / 16.0 g/mol = 0.500 mol 18.0 g of C 2 H 6 / 30.0 g/mol = 0.600 mol N (total) = PV / RT = (4.43 atm) (10.0 L) / (0.08206 L atm / mol K) (300.15 K) = 1.7986 mol P (methane) = P (total) X (n (methane) /n (total) ) = (4.43 atm) X (0.500 mol / 1.7986 mol) = 1.231515 atm P (ethane) = (4.43 atm) (0.600 mol / 1.80 mol) = 1.47782 atm P (propane) = P (propane) – (P (methane) + P (ethane) ) = 4.43 − (1.23 + 1.48) = 1.72 atm 8. A 1.50 L bulb containing helium (He) at 155 torr is connected by a valve to a 2.00 L bulb containing methane (CH 4 ) at 245 torr. The valve between the two bulbs is opened and the two gases mix. a. What is the partial pressure, in torr, of He and CH 4 ? b. What is the mole fraction of helium? The combined volume of the bulbs after connecting them = 3.50 L P 1 V 1 = P 2 V 2 , or P 2 = P 1 V 1 / V 1 P (He) = (155 torr) X (1.50 L) / (3.50 L) = 66.4 torr P (CH4) = (245 torr) X (2.00 L) / (3.50 L) = 140. torr P (total) = 66.4 torr + 140. torr = 206.4 torr Mole fraction of He = 66.4/206.4 = 0.322 9. A sample of 1.43 g of helium and an unknown quantity of O 2 are mixed in a flask at room temperature. The partial pressure of helium in the flask is 42.5 torr, and the partial pressure of oxygen is 158 torr. What mass of O 2 is in the sample? P (total) = 42.5 torr + 158 torr = 200.5 torr P (oxygen) = mole fraction (oxygen) = 158 / 200.5 = 0.788 Mole fraction (He) = 1.000 – 0.788 = 0.212 1.43 g He / 4.0026 g/mol = 0.35727 mol 0.35727 mol He = 0.212 X n (total), therefore n (total) = 0.35727 / 0.212 = 1.685 mol n (oxygen) = 1.685 X 0.788 = 1.3279 mol of O 2 X 31.9988 g/mol = 42.5 g O 2 10. A 7.00 L bottle is filled with 1.00 mol of ammonia gas (NH 3 ) at 350.0 K. Calculate the pressure in the bottle assuming, a. Ammonia is an ideal gas. b. Ammonia is a real gas, with the following parameters: “ a ” = 4.170 atm L 2 mol –2 , and “ b ” = 0.03707 L mol –1 . Using the ideal gas equation, P = nRT / V = [(1.00 mol) X (0.08206 L atm / mol K) (350 K)] / (7.00 L) = 4.103 atm Using Van der Waals real gas equation, !࠵? + !" ! # ! $ (࠵? − ࠵?࠵?) = ࠵?࠵?࠵? or ࠵? = "$% #&"’ − !" ! # ! 1 ࠵?࠵?࠵? × 0.08206 ࠵? ࠵?࠵?࠵? ࠵?࠵?࠵? &( ࠵? &( × 350 ࠵? 7.00 ࠵? − (1.00 ࠵?࠵?࠵? × 0.03707 ࠵? ࠵?࠵?࠵? &( ) − 1.00 ࠵?࠵?࠵? ) × 4.170 ࠵?࠵?࠵? ࠵? ) ࠵?࠵?࠵? &) 49.0 ࠵?࠵?࠵? ) = 4.04 ࠵?࠵?࠵?

FA23 CHM 113 Exam 2 Student Name: 3 11. A 10.0 L cylinder contains 20.0 mol of helium under 120.0 atm of pressure. Calculate the temperature of the cylinder assuming, a. Helium is an ideal gas. b. Helium is a real gas with the following parameters “ a ” = 0.0341 atm L 2 mol –2 , and “ b ” = 0.0237 L mol -1 . Using the ideal gas equation, T = PV / nR = (120.0 atm) X (10.0 L) / (20.0 mol) X (0.08206 L atm / mol K) = 731 K Using Van der Waals real gas equation, !࠵? + !" ! # ! $ (࠵? − ࠵?࠵?) = ࠵?࠵?࠵? or ࠵? = *+, "# ! $ ! -(#&"’) "$ ࠵? = (120.0 ࠵?࠵?࠵? + 0.0341 ࠵?࠵?࠵? ࠵? ) ࠵?࠵?࠵? &) × 400. ࠵?࠵?࠵? ) 100. ࠵? ) ) × (10.0 ࠵? − (20.0 ࠵?࠵?࠵? × 0.0237 ࠵? ࠵?࠵?࠵? &( )) 20.0 ࠵?࠵?࠵? × 0.08206 ࠵? ࠵?࠵?࠵? ࠵?࠵?࠵? &( ࠵? &( ࠵? = 697 ࠵? 12. Calculate the energy and wavelength of a photon of light with a frequency of 6.165 X 10 14 Hz. ࠵? = ℎ࠵? = 6.62607015 × 10 &01 J Hz -1 × 6.165 × 10 (1 Hz = 4.087 × 10 &(4 J ࠵? = ࠵? ࠵? = 299,792,458 m Hz 6.165 × 10 (1 Hz = 4.863 × 10 &5 ࠵? 13. Calculate the energy and wavelength of a photon of light with a frequency of 2.68 X 10 6 Hz. ࠵? = ℎ࠵? = 6.62607015 × 10 &01 J Hz -1 × 2.68 × 10 6 Hz = 1.78 × 10 &)5 J ࠵? = ࠵? ࠵? = 299,792,458 m Hz 2.68 × 10 6 Hz = 1.12 × 10 ) ࠵? 14. Concerning electron quantum numbers, a. List the names and abbreviations of the four quantum numbers. b. What does each quantum number represent or define? Principal quantum number ( n ): defines the shell in which an electron exists. Secondary quantum number ( ℓ ): defines the different orbitals (or subshells) within each shell. Magnetic quantum number ( ࠵? ℓ ): defines suborbital within each orbital. Spin quantum number ( ࠵? 8 ): defines the spin state of the two electrons within a suborbital 15. Write the complete AND shorthand electron configurations of the two elements listed below. Hint: the complete electron configurate of Li is 1s 2 2s 2 . a. Chlorine (Cl). Complete: 1s 2 2s 2 2p 6 3s 2 3p 5 . Shorthand: [Ne]3s 2 3p 5 b. Arsenic (As). Complete: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 3 . Shorthand: [Ar]4s 2 3d 10 4p 3 16. Write the complete AND shorthand electron configurations of the two elements listed below. Hint: the complete electron configurate of Li is 1s 2 2s 2 . a. Silicon (Si). Complete: 1s 2 2s 2 2p 6 3s 2 3p 2 . Shorthand: [Ne]3s 2 3p 2 b. Selenium (Se). Complete: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 4 . Shorthand: [Ar]4s 2 3d 10 4p 4

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FA23 CHM 113 Exam 2 Student Name: 4 17. Since you have already determined the electron configurations of Cl and As in the previous question, write down the four quantum numbers of the highest energy atomic orbital of each element, i.e., the quantum numbers of the electrons that is added last. Hint: the quantum numbers of the highest energy atomic orbital of Li are n = 2, ℓ = 0, ࠵? ℓ = 0, ࠵? 8 = + ( ) a. Chlorine: n = 3, ℓ = 1, ࠵? ℓ = 0, ࠵? 8 = − ( ) b. Arsenic: n = 4, ℓ = 1, ࠵? ℓ = 1, ࠵? 8 = + ( ) 18. Since you have already determined the electron configurations of Si and Se in the previous question, write down the four quantum numbers of the highest energy atomic orbital of each element, i.e., the quantum numbers of the electrons that is added last. Hint: the quantum numbers of the highest energy atomic orbital of Li are n = 2, ℓ = 0, ࠵? ℓ = 0, ࠵? 8 = + ( ) a. Silicon: n = 3, ℓ = 1, ࠵? ℓ = 0, ࠵? 8 = + ( ) b. Selenium: n = 4, ℓ = 1, ࠵? ℓ = -1, ࠵? 8 = − ( ) 19. Concerning periodic table trends, a. Which element has a larger atomic radius: Na or Rb? Rb b. Which element has a larger atomic radius: K or V? K c. Which requires more energy: removal of the first valence electron of Mg to form Mg + , or removal of the second electron from Mg + to form Mg 2+ ? 2 nd ionization to form Mg 2+ d. Which element has higher ionization energy: Na or Rb? Na e. Which element has higher electron affinity: N or F? F f. Which element has higher ionization energy: F or Br? F 20. Concerning periodic table trends, a. Which element has a larger atomic radius: Ca or Ba? Ba b. Which element has a larger atomic radius: Ca or Ti? Ca c. Which requires more energy: removal of the first valence electron of Ca to form Ca + , or removal of the second electron from Ca + to form Ca 2+ ? 2 nd ionization to form Ca 2+ d. Which element has higher ionization energy: C or Ba? Ca e. Which element has higher electron affinity: Cl or Br? Cl f. Which element has higher electron affinity: Cl or P? Cl 21. Draw the Lewis dot structure of the following compounds. Hint: the central atom is underlined. a. H 2 S. b. (NO 3 ) – . 22. Draw the Lewis dot structure of the following compounds. Hint: the central atom is underlined. a. CH 2 Br 2 . b. (CO 3 ) 2– . 23. Using your knowledge of VSEPR, and the Lewis dot structures provided below, identify the molecular shape of the following compounds.

FA23 CHM 113 Exam 2 Student Name: 5 Bent Tetrahedral Trigonal pyramidal Linear 24. Using your knowledge of VSEPR, and the Lewis dot structures provided below, identify the molecular shape of the following compounds. Bent Trigonal Pyramidal Tetrahedral Trigonal Planar

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