CHEMISTRY 2A (D) Final Exam Winter 2009

Transcription

1 Last Name: First Name: Lab Sec. # TA: Lab day/time: Dr. Enderle CHEMISTRY 2A (D) Final Exam Winter 2009 Instructions: CLOSED BOOK EXAM! DO NOT OPEN the exam until instructed to do so. 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. A scientific calculator may be used (if it is a programmable calculator, its memory must be cleared before the exam). Sharing of calculators is not allowed. (1) Read each question carefully (27 problems, 15 pages). (2) ANSWER ALL THE QUESTIONS YOU ARE SURE OF FIRST, AND THEN USE THE REMAINING TIME FOR THE REST. (3) Please show all relevant work where it is indicated for full credit. (4) The last 2 pages contain a periodic table and some useful information. You may remove them for easy access and for scratch. (5) If you finish early, RECHECK YOUR ANSWERS! UC Davis is an Honor Institution Multiple Choice Circle one 1. a b c d e 2. 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 Possible Points Points 11. a b c d e # 113 (3 points each) / 39 # 1419 (5 points each) / 30 # 20 (6 points) / 06 # 21 (42 points) / 42 # 22 (10 points) / 10 # 23 (10 points) / 10 # 24 (06 points) / 06 # 25 (30 points) / 30 # 26 (06 points) / 06 # 27 (16 points) / a b c d e 13. a b c d e total points: 14. a b c d e 15. a b c d e 16. a b c d e 17. a b c d e 18. a b c d e 19. a b c d e 1419 total points: Total Score (195) / 195

2 WINTER 2009 FINAL CHEM 2A (Page 2 of 15) Concepts: Multiple Choice Questions 113: 3 points each (no partial credit) 1. Write the symbol of the species that has 28 protons, 26 electrons, and 32 neutrons. a) b) c) d) e) 58 Fe Ni 2 60 Ni Fe Ni In the reaction, H 2 O 2 H 2 O + 1/2O 2 oxygen is: a. Reduced b. Oxidized c. Both a. and b. d. Neither a. nor b. e. Annihilated 3. How many σ bonds does the molecule of guanine have? a) 17 b) 2 c) 3 d) 4 e) 0 HC N N H C C O C N NH C NH 2 4. Choose the INCORRECT formulaname combination. a. Na 2 CO 3 sodium carbonate b. P 2 O 5 diphosphorus pentoxide c. FeBr 2 iron dibromide d. N 2 O 3 dinitrogen trioxide e. K 3 N potassium nitride 5. In which of the following compounds does chlorine have the highest oxidation number? (a) ClO 2 (b) Cl 2 (c) FeCl 3 (d) KClO 3 (e) Mg(ClO) 2

3 WINTER 2009 FINAL CHEM 2A (Page 3 of 15) 6. Which of the following reactions is NOT a redox reaction? (a) (b) (c) (d) (e) 5 I (aq) + IO 3 (aq) + 6 H + (aq) 3 I 2 (s) + 3 H 2 O (l) Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) 2 H 2 O 2 (aq) 2 H 2 O (l) + O 2 (g) BaCl 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + 2NaCl (aq) CH 3 COOH (aq) + 2 O 2 (g) 2 CO 2 (g) + 2 H 2 O (l) 7. Which statement is FALSE about the 5d xy orbital? (a) It has a nodal plane in the xz plane. (b) It has a nodal plane in the yz plane. (c) For an electron in the 5d xy orbital there is a higher probability of finding the electron farther from the nucleus than if it is in the 5s orbital. (d) It can hold a total of 10 electrons. (e) None of the above. 8. Choose the species from which one electron could most easily be removed. (a) Cl (b) Na + (c) Na (d) K (e) Ar 9. Which of the following has the largest radius? (a) As 3 (b) Br (c) Sr 2+ (d) Cl (e) They all have approximately the same size 10. Using the VSEPR theory, which one of the following chemical species would you expect to have a dipole moment (polar)? a. I 3 b. O 3 c. CO 2 d. CH 4 e. None of the above

4 WINTER 2009 FINAL CHEM 2A (Page 4 of 15) 11. What is the bond order of the nitrogenoxygen bonds in the nitrate ion? (a) 2, 1, and 1 (b) 1, 2, and 2 (c) 3 (d) 1 (e) Choose the INCORRECT statement about the molecule of HCN. a. There are a total of two σ bonds in the molecule. b. There are a total of two π bonds in the molecule. c. There is a lone pair on nitrogen. d. The molecule is bent. e. The carbon atom has sp hybridization. 13. Which of the following is NOT TRUE about a mixture of gases? a. The sum of the partial pressures of each gas is equal to the total pressure. b. The sum of the moles of each gas is equal to the total number of moles. c. The sum of the temperature of each gas is equal to the total temperature. d. The sum of the partial volumes of each gas is equal to the total volume. e. All of the above statements are true.

5 WINTER 2009 FINAL CHEM 2A (Page 5 of 15) Calculations: Multiple Choice Questions 1419: 5 points each (no partial credit) 14. A hypothetical element, E, has two stable isotopes: E38 = u 75.68% E46 = u 24.32% The element s atomic mass would be closest to which element? (a) K A.M. = (b) Ar A.M. = (c) Ca A.M. = (d) Sc A.M. = (e) Sr A.M. = (38.012)(0.7568) + (45.981)(0.2432) = u 15. Find the empirical formula for a compound with the following mass composition: K 42.4%, Fe 15.2%, C 19.5% and N 22.8%. a. KFeCN b. K 4 FeC 6 N 6 c. KFeC 6 N 6 d. K 4 FeCN 6 e. K 4 FeC 6 N K 42.4 Fe 15.2 C 19.5 N 22.8 K 42.4/39.1 Fe 15.2/55.8 C 19.5/12.0 N 22.8/14.00 K 1.08 Fe C 1.63 N 1.63 K 1.08/0.272 Fe 0.272/0.272 C 1.63/0.272 N 1.63/0.272 K 4 FeC 6 N 6 (K 4 FeC 6 N 6 )

6 WINTER 2009 FINAL CHEM 2A (Page 6 of 15) 16. You have a M stock aqueous solution of Sr(OH) 2. What volume of the stock solution (in ml) must be diluted with water to prepare ml of a solution that has a hydroxide concentration [OH ] = M? a ml b ml c ml d ml e ml [OH ]= 2 [Sr(OH) 2 ] [Sr(OH) 2 ] after dilution = [OH ] after dilution /2 = M/2 = M M 1 of Sr(OH) 2 = M V 1 =? ml + H 2 O M 2 of Sr(OH) 2 = M V 2 = ml M 1 V 1 = M 2 V 2 V 1 = (M 2 V 2 )/M 1 V 1 = ml 17. When chlorine is added to acetylene, 1,1,2,2tetrachloroethane is formed: 2 Cl 2(g) + C 2 H 2(g) C 2 H 2 Cl 4(l) How many liters of chlorine gas at STP will be needed to make 75.0 grams of C 2 H 2 Cl 4? a L b L c L d L e L Molar mass of C 2 H 2 Cl 4 = g/mol (75.0 g C 2 H 2 Cl 4 ) (mol C 2 H 2 Cl 4 /167.8 g C 2 H 2 Cl 4 ) = mol C 2 H 2 Cl 4 Moles of Cl 2 gas needed = 2 x = PV = nrt V = (nrt)/p = [(0.894 mol Cl 2 ) x ( L atm mol 1 K 1 ) x (273 K)] /1 atm = = 20.0 L

7 WINTER 2009 FINAL CHEM 2A (Page 7 of 15) 18. Consider an electron moving at 1 x 10 6 m/s. Find its wavelength in angstroms. a. 1 Å b. 3 Å c. 5 Å d. 7 Å e. 9 Å λ = h mυ 19. The energy of a photon is used to completely remove the electron of a hydrogen atom. What is the wavelength of the photon in nanometers if the electron is removed from its second orbit? a) 365 nm b) 410 nm c) 434 nm d) 486 nm e) 656 nm E n = R H (Z 2 /n 2 ) R H = x J Z = 1 and n =2 E 2 = (2.179 x J) (1 2 /2 2 ) E 2 = x J E = hν = hc/λ λ = hc/e λ = (6.626x10 34 Js)( 3.00 x 10 8 m s 1 )/(5.448 x J) = 3.65 x 10 7 m = 365 nm

8 WINTER 2009 FINAL CHEM 2A (Page 8 of 15) 20. (6 points) Fill in the blanks with the correct ground state electron configuration (noble gas configuration) for the given atom or the atom for the given ground state electron configuration of the neutral atom. Atom Electron Configuration Sn [Kr] 5s 2 4d 10 5p 2 Cu [Ar] 3d 10 4s (42 points) Fill in the following table (see the ClF 3 example). For the determination of formal charge(s) consider the Lewis formula with the smallest formal charges (lowest energy). Molecule ClF 3 ElectronGroup Geometry (central atom) Hybridization Trigonal Bipyramidal Molecular Geometry Ideal Bond Angles Polar or Nonpolar Molecule (central atom) Formal Charge(s) of the underlined element Oxidation Number of the underlined element T shape 90º, 1 80º polar sp 3 d 0 1 CCl 2 Br 2 Tetrahedral Tetrahedral 109.5º Polar sp PCl 3 Tetrahedral Trigonal Pyramidal 109.5º Polar sp CO 2 Linear Linear 180º Nonpolar sp 0 +4 H 2 O Tetrahedral Bent 109.5º Polar sp XeF 4 Octahedral Squareplanar 90º, 180º Nonpolar sp 3 d SF 4 Trigonal Bipyramidal Seesaw 90º, 180º 120º Polar sp 3 d 0 +4

9 WINTER 2009 FINAL CHEM 2A (Page 9 of 15) Partial Credit. Show ALL Your Work. Explain. 22. (10 points) If 10.0 g of sodium hydrogen carbonate is added to 450 ml of a M aqueous solution of copper(ii) nitrate and left to react according to the reaction: 2NaHCO 3 (aq) + Cu(NO 3 ) 2 (aq) CuCO 3 (s) + 2NaNO 3 (aq) + H 2 O(l) + CO 2 (g) How many liters of carbon dioxide are produced at STP? Moles of Cu(NO 3 ) 2 = (0.450 L) x (0.155 mol/l) = mol Moles of NaHCO 3 = (10.0 g)/(84.01 g/mol) = mol Moles of NaHCO 3 < 2 times moles of Cu(NO 3 ) 2 so NaHCO 3 is the limiting reactant Moles of CO 2 = ½ moles of NaHCO 3 = mol /2 = mol V = nrt/p = ( mol x L atm mol 1 K 1 x 273 K)/1 atm = 3 L V = 3 L

10 WINTER 2009 FINAL CHEM 2A (Page 10 of 15) 23. (10 points) A civil engineer wants to reduce odors at a wastewater treatment plant by adding hydrogen peroxide to the sewage. The hydrogen peroxide is delivered as 50% by mass solution, but for maintenance and safety issues, the H 2 O 2 is diluted to a 3% by mass solution. If the engineer needs 20.0 gallons of the 3% by mass aqueous solution of H 2 O 2, how many gallons of water does the engineer need to add to the 50% solution? (Density of 50% solution = g/ml; density of 3% solution = g/ml) 50% by mass = 50 g H 2 O 2 per 100 g of aqueous solution Change both mass percents to molarity (mol/l) using density and molar mass. (50.0 g pure H 2 O 2 /100.0 g aq. solution) x (1.197 g solution/1 ml solution) x (1000 ml/l)x (1 mol H 2 O 2 / g H 2 O 2 ) = mol H 2 O 2 / L solution = 17.6 M (3.0 g pure H 2 O 2 / g aq. solution) x (1.015 g solution / ml solution) x (1000 ml / L) x (1 mol H 2 O 2 / g H 2 O 2 ) = mol / L = 0.90 M Using the dilution formula: M i V i = M f V f (17.6 M) V i = (0.90 M)(20.0 gal) V i = 1.0 gal Volume of water to add = V f V i = 20.0 gal 1.0 gal = 19 gal of water. Volume of water = 19 gal 24. (6 points) Consider the molecule of ozone, O 3. A) Draw its two resonance Lewis electron dot diagrams. B) What is the electron group geometry of its central atom? C) What is its molecular geometry? D) What is the hybridization of its central atom? E) What is the ideal bond angle? F) Is it polar or nonpolar molecule? A) B) Trigonal planar C) Bent D) sp 2 E) 120 o F) Polar

11 WINTER 2009 FINAL CHEM 2A (Page 11 of 15) 25. (30 points) Complete the MO energy level diagram for the cation B + 2. a. Fill in the electrons using arrows ( and/or ) for the atomic and molecular orbitals. b. Designate all the energy levels (i.e., σ 2s *, π 2p, σ 2s, etc.) c. Calculate the bond order for B + 2. d. Is B + 2 a paramagnetic or diamagnetic chemical species? a. and b. σ 2p * π 2p * E 2p σ 2s * π 2p σ 2p 2p B 2s σ 2s 2s B 2 + B + c. Bond order for B 2 + = 0.5 d. Paramagnetic or Diamagnetic? Answer: Paramagnetic

12 WINTER 2009 FINAL CHEM 2A (Page 12 of 15) 26. (6 points) Write the possible four quantum numbers n, l, m l, and m s for each of the following electrons. 2s 4d 5f n = 2 n = 4 n = 5 l = 0 l = 2 l = 3 m l = 0 m l = 2, 1, 0, +1, +2 m l = 3, 2, 1, 0, +1, +2, +3 m s = 1/2 m s = 1/2 m s = +1/2

13 WINTER 2009 FINAL CHEM 2A (Page 13 of 15) 27. (16 points) Balance the following chemical equation by the halfreaction method and show your work. Next, answer the following questions. CH 4 O + O 2 CO 2 + H 2 O Note! The above chemical equation is very easy to balance just by inspection. You are NOT asked to balance it by the inspection method. Balancing it by the inspection method will result in zero points for credit. CH 4 O + H 2 O CO H e 4 H e + O 2 2 H 2 O Or 2 CH 4 O + 2 H 2 O 2 CO H e 12 H e + 3 O 2 6 H 2 O 2 CH 4 O + 3 O 2 2 CO H 2 O a. The oxidizing chemical species is: O 2 b. The reducing chemical species is: CH 4 O c. The name of the oxidized element is: carbon d. The name of the reduced element is: oxygen e. The oxidant s molecular formula is: O 2 f. The reductant s molecular formula is: CH 4 O g. The name of the element that gains electrons is: oxygen h. The name of the element that loses electrons is: carbon i. The chemical species that causes oxidation is: O 2 j. The chemical species that causes reduction is: CH 4 O k. The oxidation number of _C (give the chemical symbol of the element) increases from _2 to _+4_ l. The oxidation number of _O (give the chemical symbol of the element) decreases from _0 to _2

14 WINTER 2009 FINAL CHEM 2A (Page 14 of 15) Solubility rules: Compounds which are soluble or mostly soluble: Group 1, NH 4 +, chlorates, acetates, nitrates Halides (except Pb 2+, Ag +, and Hg 2 2+ ) Sulfates (except Sr 2+, Ba 2+, Pb 2+, and Hg 2 2+ ) Compounds which are insoluble: Hydroxides, sulfides (except above rule, and sulfides of group 2) Carbonates, phosphates, chromates (except above rules) Some useful equations and data: PLEASE NOTE: Important values and equations required for calculations are given with the respective problem. The following may or may not be of any use. λ = h mυ d = PM RT Moles Molarity = Liter n (moles) = mass (g) molar mass (g/mol) PV = nrt N A = x h = 6.626x10 34 Js 1 J = 1 kg m 2 s 2 = 1 N m ml M = mmol P total = P 1 + P 2 + T K = T o C R = L atm mol 1 K 1 1 nm = 10 9 m c = 3.00 x 10 8 m s 1 c = νλ x A = n A / n tot = P A / P tot E E photon = hν Rate 2 /Rate 1 =(M 1 /M 2 ) 1/2 1 atm = 760 mmhg E = R H (1/n i 2 1/n f 2 ) n h x p = 1 g = 6.022x10 23 amu 1 Å = m 4π m Actual Yield d = m e = x10 28 g p = m υ % Yield = x100 V Theoretical Yield R H Z n 2 = R H = x J 2

15 WINTER 2009 FINAL CHEM 2A (Page 15 of 15) Potentially Useful Information (You may remove this page for ease of access) Key 1 H Atomic Number Symbol Atomic Mass Electronegativity 2 He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc (98) Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po (209) At (210) Rn (222) 87 Fr (223) Ra (226) Lr (260) 104 Unq 105 Unp 106 Unh 107 Uns 109 Une 57 La Ce Pr Nd Pm (145) Sm Eu Gd Tb Dy Ho Er Tm Yb Ac (227) Th Pa (231) U Np (237) 6 94 Pu (244) Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (251) 99 Es (252) 100 Fm (257) 101 Md (258) 102 No (259)