Name Student ID Number TA Name Lab Section Winter 016 Enderle CHEMISTRY A Exam II Instructions: CLOSED BOOK EXAM! No books, notes, or additional scrap paper are permitted. All information required is contained on the exam. Place all work in the space provided. If you require additional space, use the back of the exam. (1) Read each question carefully. Circle Part I answers on front page. () There is no partial credit for the problems in Part I. None of the answers are E. You will lose 10 points if you do not circle your multiple choice answers on the front page or if you do not write your TA s name or section in the space above. (3) The last two pages contain a periodic table and some useful information. You may remove these pages for easy access. (4) Graded exams will be returned in lab sections next week. (5) If you finish early, RECHECK YOUR ANSWERS! Possible Points 1 14. 4 points Points 15 16. 16 points 17 18. 19 points 19 0. 18 points U.C. Davis is an Honor Institution Points Multiple Choice (circle one) 1. A B C D E. A B C D E 3. A B C D E 4. A B C D E 5. A B C D E 6. A B C D E 7. A B C D E 8. A B C D E 9. A B C D E 10. A B C D E 11. A B C D E 1. A B C D E 13. A B C D E 14. A B C D E 1. 10 points Total Score (105)
Exam II Page of 10 Part I: Multiple Choice, Concepts & Short Calculations Circle the correct answer and enter your response on the cover No partial credit (3 pts each) 1. Given the forward reaction MnO 4 - (aq) + I - (aq) I (s) + Mn + (aq), identify the oxidizing agent. A. MnO 4 - (aq) B. I - (aq) C. I (s) D. Mn + (aq). Which element would experience the least shielding? A. Se B. O C. Ga D. Be E. Ba 3. A solution is 0.0 M Ba(OH). How many moles of H + must be added to neutralize 100.0 ml of the solution? A. 0.00 mol B. 0.080 mol C. 0.040 mol D. 0.40 mol E. 0.80 mol 4. Which of the following words/phrases are used to characterize behavior within quantum mechanics? A. Discrete B. Discontinuous C. Allowed & forbidden regions D. All of the above E. None of the above 5. If the pressure of a fixed amount of gas is quadruples and the volume remains constant: A. The temperature (K) decreases B. The molecules move slower C. The density of the gas is smaller D. The final temperature (K) is four times the temperature (K) E. The final temperature (K) is ¼ the temperature (K) 6. A 10.0 g sample of a gas has a volume of 5.5 L at 5 C and 76 mmhg. If.5 g of the same gas is added to this constant 5.5 L volume and the temperature is raised to 6 C, what is the new gas pressure (in mmhg)? A. 1.41 B. 43 C. 867 D. 1070 E. 36!" 7. Simplify the units of the following units: (!!!!!)(m) A. Pa B. N C. kg D. J E. unitless 8. In the reaction Na + H O NaOH + H, how many liters of hydrogen at STP are produced from 50.0 grams of sodium (Na = 3.0, O = 16.0, & H = 1.0 g/mol)? A. (55.0/18.0)(.7) L B. (50.0/3.0)(.7/) L C. (50.0/3.0)()(.7) L D. (50.0/3.0)(.7) L E. (55.0/3.0)(.4) L 9. A gaseous hydrocarbon weighing 0.31 g occupies a volume of 10 ml at 3 C and 749 mmhg. What is the molecular formula of this compound? A. C 3 H 8 B. C 4 H 8 C. C 4 H 10 D. C 5 H 8 E. C 5 H 10
Exam II Page 3 of 10 10. The volume correction term in the van der Waals equation is present because: A. barometers are inaccurate B. molecules are diatomic C. molecules attract each other D. molecules occupy volume E. molecules repel each other 11. How many spin up (+½) electrons can have the following set of quantum numbers: n = 4 and l = 1? A. 1 B. C. 3 D. 4 E. 6 1. O gas is collected above dichloromethane in a burette at 11.9 C. A total of 4.6 ml of gas is observed in the burette. If the vapor pressure of dichloromethane at 11.9 C is 47.5 mmhg and the barometric pressure is 695 mmhg, how many molecules of O gas are collected? A. 5.79 x 10 0 B. 7.85 x 10 0 C..06 x 10 0 D. 3.73 x 10 0 E. 6.33 x 10 0 13. How many angular nodes and radial nodes respectively are present in a 3p y orbital? A., 1 B. 1, 1 C. 1, D. 3, 1 E. 1, 14. A sample of nitrogen gas effuses through a tiny hole in 38 s. What must be the molar mass of a gas that requires 64 s to effuse under identical conditions? A. 9.8 B. 8.0 C. 47. D. 79.4 E. 39.7
Exam II Page 4 of 10 Part II: Short Answer 15. (6 points) Fill in the blanks with what is missing. Fill in the correct ground state electron configuration (noble gas configuration in spdf condensed form). Ce 4+ [Xe] Br [Ar] 4s 3d 10 4p 5 16. (10 points) Write all acceptable values for the missing quantum numbers (n, l, m l, m s ) in each row of the table. Note that each row corresponds to a unique set of quantum numbers, and each box may have more than one acceptable answer. You must determine all possible values for full credit. n l m l m s 1 0 0 +½ 3 ½ or,3,4 1-1, 0, 1 +½ 0 0 ±½
Exam II Page 5 of 10 17. Consider an electron with the following four quantum numbers to answer parts A through C: n = 3, l =, m l =, m s = ½ A. ( points) What type of orbital does this electron belong to? (i.e. 1s, p, etc.) 3d B. ( points) How many radial nodes does the orbital contain? 0 C. ( points) How many angular nodes does the orbital contain? D. (3 points) Draw a 4d!!!!! orbital. Include labels for all axes and nodes. 18. (10 points) Fill in the blank: in the column marked Answer, fill in the answer that best describes the statement. Each answer is only two or four-words long. Answer Periodic Law Effective Nuclear Charge or Zeff Statement 1. This law states that when the elements are arranged in order of increasing atomic mass, certain sets of properties recur periodically.. This charge is the true nuclear charge minus the charge that is screened out by the electrons. Covalent Radius or Covalent Law of Combining Volumes or Combining Volumes Heisenberg Uncertainty Principle or Uncertainty Principle 3. This radius is one-half the distance between the nuclei of two identical atoms joined by a single covalent bond. 4. This law states that when gases react together to form other gases (assuming the same T & P for each gas), the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers. 5. This principle states that uncertainty is in fact a characteristic feature of the system rather than a bug.
Exam II Page 6 of 10 Part III: Long Answer Please show all work Partial credit 19. (8 points) Calculate the wavelength, in nanometers, of the spectral line produced when an electron in a hydrogen atom undergoes the transition from the energy level n = 4 to the level n =. What color is the spectral line? ΔE = (.179e-18 J)(1/ni -1/nf ) ΔE = (.179e-18 J)(1/4-1/ ) = -4.086e-19 (released) ΔE = hc/λ or λ = hc/δe λ = (6.66e-34 Js)(.9979e8 m/s)/4.086e-19 = 4.86 x10-7 m = 486.00 nm λ = 486.0 nm Color = Blue (violet or bluegreen accepted) 0. (10 points) Given argon gas at STP, fill in all your answers in the table below. Use three significant figures, else lose 1 point on the problem. For calculations, show your work next to the table for partial credit. If there is 1.0 mol of argon gas, how many liters of argon gas are present? Which one has the highest root mean square velocity: Ar, N, or O? What is the root mean square velocity of argon gas in m/s? What is the kinetic energy per mole (J/mol) of argon gas? Which of the following gases would have the largest kinetic energy per mole at STP: Ar, N, or O?.7 L N or nitrogen gas 413 3.40e3 Same u rms = 3RT M = 3(8. 3145)(73) 0. 03995 e K = 3 RT = 3 (8. 3145)(73)
Exam II Page 7 of 10 1. (10 points) 9.00 L of He (g) at 4.00 atm and 3.00 L of Ne (g) at.60 atm are mixed at room temperature while the total volume (1.00 L) is kept constant. If you don t get part A, use 1.8 total moles. If you don t get part B, use 3.60 atm. A. (4 points) Calculate the mole fraction of He (g) in the gas mixture. n He = PV/RT = (4.00)(9.00)/(0.0806)(98) = 1.47 mol He n Ne = (.60)(3.00)/(0.0806)(98) = 0.319 mol Ne n tot = n He + n Ne = 1.789 mol x He = n He /n tot = 0.8 x He = 0.8 B. (3 points) Calculate the total pressure (in atm) of the gas mixture. P = nrt/v = (1.789 mol)(0.0806)(98)/(1.00) = 3.65 atm P tot = 3.65 atm C. (3 points) Calculate the partial pressure of He (g) (in atm) of the gas mixture. P He = x He P tot = (0.8)(3.65) = 3.00 atm P He = 3.00 atm
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Exam II Page 9 of 10 Solubility Rules: Compounds that are soluble or mostly soluble Group 1, NH 4 +, chlorates, acetates, nitrates Halides (except Pb +, Ag +, and Hg + ) Sulfates (except Ca +, Sr +, Ba +, Pb +, and Hg + ) Compounds that are insoluble Hydroxides, sulfides (except above rule, and group sulfides) Carbonates, phosphates, chromates (except above rules) Conversions: 1 atm = 14.7 psi = 101,35 Pa = 760 mmhg = 1.0135 bar = 760 Torr; 1in =.54 cm; 1 in = 1 ft Constants: R = 8.3145 J / mol K = 0.0806 L atm / mol K Avogadro s number = 6.0 x 10 3 / mol c =.9979 x 10 8 m / s h = 6.66 x 10-34 J s R H =.179 x 10-18 J m (electron) = 9.109 x 10-31 kg m (proton) = 1.673 x 10-7 kg m (neutron) = 1.675 x 10-7 kg d (H O) = 1.0 g / cm 3 g = 9.81 m / s Equations and Various Tables: ax + bx + c = 0; x = b ± b 4 ac a 15 1 1 1 Z R ν = 3.881 10 s En = n n Δ xδp h 4π effusion ratea = effusion rateb M M B A p = mu H n h E k = 8mL h λ = P = d g h E = h ν c = λ ν mu 1 1 E = R H e K = ½mu = 3/RT ni n f Δ n a P + = V ( V nb) nrt nπx Ψ( x) = sin, n = 1,,3,... L L 3RT u rms = u = M PV = nrt P total = P 1 + P + x A = n A / n tot = P A / P tot = V A / V tot
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