CHEMISTRY 2A (A) Final Exam Fall 2009

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1 Last Name: First Name: Lab Sec. # TA: Lab day/time: Dr. Enderle CHEMISTRY 2A (A) Final Exam Fall 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. (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. Possible Points UC Davis is an Honor Institution Points # 120 (3 points each) / 60 # 2122 (16 points) / 16 # 23 (54 points) / 54 # 2425 (21 points) / 21 # 2627 (11 points) / 11 # 28 (15 points) / 15 # 29 (14 points) / 14 # 30 (11 points) / 11 # 31 (11 points) / 11 Total Score (213) / 213 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 11. a b c d e 12. a b c d e 13. a b c d e 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 20. a b c d e

2 Fall 2009 FINAL CHEM 2A (Page 2 of 13) Concepts: Multiple Choice Questions 113: 3 points each (no partial credit) 1. How many hydrogen atoms are represented in the formula Fe(C 5 H 7 O 2 ) 3? (A) 3 (B) 7 (C) 14 (D) 21 (E) The formula of potassium sulfate is K 2 SO 4. The formula of iron(iii) chloride is FeCl 3. What is the formula of iron(iii) sulfate? (A) Fe 2 (SO 4 ) 3 (B) Fe(SO 4 ) 2 (C) Fe 2 SO 4 (D) Fe 3 (SO 4 ) 2 (E) Fe 3 SO 4 3. Which substance is not a gas at room temperature and atmospheric pressure? (A) Ar (B) CH 4 (C) N 2 (D) CO 2 (E) K 4. A 1.0 g sample of oxygen gas occupies a fixedvolume container at a pressure of 0.50 atm. If the temperature is increased, then (A) the density of the oxygen gas will decrease. (B) the density of the oxygen gas will increase. (C) the speed of the oxygen molecules will decrease. (D) the pressure of the oxygen gas will increase. (E) the pressure of the oxygen gas will decrease. 5. The average kinetic energy of the molecules of an ideal gas is proportional to its (A) temperature. (B) volume. (C) density. (D) pressure. (E) amount in moles. 6. When the chemical equation NH 3 (g) + O 2 (g) NO 2 (g) + H 2 O(g) is balanced, what is the smallest integer for the coefficient of O 2 (g)? (A) 3 (B) 9 (C) 5 (D) 7 (E) 1 7. In a chemical reaction, which of the following cannot change between reactants and products? (A) total mass (B) total moles (C) volume (D) pressure (E) temperature 8. Which species has the Lewis symbol Z? (A) Si (B) N (C) Cl (D) O (E) Ca Which bond is most polar? (A) H O (B) N O (C) F F (D) H C (E) C N 10. Which of the following represents a neutralization reaction of an acid with a base? (A) 2 H 2 (g) + O 2 (g) 2 H 2 O(l) (B) NH 4 OH(aq) + HCl(aq) NH 4 Cl(aq) + H 2 O(l) (C) 2 NaOH(aq) + CuCl 2 (aq) 2 NaCl(aq) + Cu(OH) 2 (s) (D) HBr(aq) + AgNO 3 (aq) HNO 3 (aq) + AgBr(s) (E) 2 H 2 O 2 (aq) 2 H 2 O(l) + O 2 (g)

3 Fall 2009 FINAL CHEM 2A (Page 3 of 13) 11. The balanced chemical equation 2 HCl(aq) + CO 3 2 (aq) 2 Cl (aq)+ H 2 CO 3 (aq) (A) involves the oxidation of C. (B) involves the reduction of C. (C) involves an acid and a base. (D) involves a change in oxidation numbers. (E) is an example of a combustion reaction. 12. What is the number of neutrons, protons, and electrons in the species 37 Cl? (A) 20 neutrons, 17 protons, and 18 electrons (B) 18 neutrons, 17 protons, and 20 electrons (C) 37 neutrons, 17 protons, and 17 electrons (D) 17 neutrons, 20 protons, and 18 electrons (E) 17 neutrons, 17 protons, and 19 electrons 13. The electronic configuration of P is (A) 1s 2 2s 2 2p 3 (B) 1s 2 2s 2 2p 11 (C) 1s 2 2s 2 2p 6 3s 2 3p 3 (D) 1s 2 2s 2 2p 6 3p 5 (E) 1s 2 2s 2 2p 6 2d The elements of Group 18 or VIII are best described as (A) monoatomic gases. (B) diatomic gases. (C) metals. (D) halogens. (E) metalloids. 15. Which substance is the strongest base in aqueous solution? (A) NH 3 (B) HNO 3 (C) CO 2 (D) HCN (E) NaOH 16. Which element is a metal? (A) H (B) Ne (C) Cl (D) P (E) Pb 17. What is the approximate percent by mass of hydrogen in C 2 H 6 (molar mass = 30 g/mol)? (A) 6 (B) 20 (C) 24 (D) 60 (E) How many moles of carbon are in 36 g of carbon? (A) 2.0 (B) 3.0 (C) 0.33 (D) 432 (E) Given the balanced chemical equation 2 CuSO KI 2 CuI + I K 2 SO 4, how many moles of I 2 can you make from 0.40 mol of CuSO 4 and excess KI? (A) 0.80 (B) 0.40 (C) 1.0 (D) 2.0 (E) What is the molar concentration of Cl when 0.20 mol of ZnCl 2 are dissolved in enough water to prepare 3.0 L of solution? (A) 0.67 (B) 1.5 (C) 3.0 (D) 0.13 (E) 0.60

4 Fall 2009 FINAL CHEM 2A (Page 4 of 13) Short Answer (no partial credit). 21. (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 I [Kr] 5s 2 4d 10 5p 5 Cu [Ar] 3d 10 4s (10 points) Circle the correct answer. Question: Circle the correct answer. Which element has the smaller atomic radius? Cl S Which element has the lower first ionization energy? Li F Which orbital is better at shielding (screening)? s f Which species will most readily lose an electron? Na + Ne Which element can form an acidic oxide? K C Which element is a better reducing agent? W I Which element will more readily lose an electron? Li F In the atom, what electrons are most shielded (screened) from the nucleus? Core Valence Which species has the larger radius? S 2 S Which species has the smaller radius? F + F

5 Fall 2009 FINAL CHEM 2A (Page 5 of 13) 23. (54 points) Fill in the following table (ClF 3 is done for you as an example). For the determination of formal charge(s) consider the Lewis formula/structure with the smallest formal charges (lowest energy). In determining the total sigma and pi bonds, use the Lewis formula/structure with the smallest formal charges. Molecule ClF 3 Electronic Geometry Trigonal Bipyramidal Molecular Geometry Ideal Bond Angles Polar or Nonpolar Molecule Hybridization (central atom) Formal Charge(s) of the underlined element Oxidation State of the underlined element Total σ Bonds in Molecule Total π Bonds in Molecule Tshape 90º,180º polar sp 3 d HCN Linear Linear 180º polar sp NO 2 Trigonal Planar Bent 120º polar sp PCl 5 Trigonal Trigonal Bipyramidal Bipyramidal 90,120º 180º nonpolar sp 3 d XeF 4 Octahedral Square Planar 90º 180º nonpolar sp 3 d SCN Linear Linear 180º polar sp H 2 SO 4 Tetrahedral Tetrahedral 109.5º polar sp

6 Fall 2009 FINAL CHEM 2A (Page 6 of 13) Partial Credit. Show ALL Your Work. 24. (12 points) A molecule that contains only carbon and oxygen is found to be 52.94% carbon by mass. a) What is the empirical formula for the compound? mol C = g C 1 mol C = mol C g C 1molO molo= 47.06gO = molO gO C O = C 1.5 O = C 3 O Empirical Formula: b) Draw the Lewis structure/formula of the molecule (in this case the empirical formula is also the molecular formula). O = C = C = C = O 25. (9 points) Vision is initiated by a photoisomerism of the molecule retinal. The molecular structure of all trans retinal is shown below. Considering this molecule: (A) (2 points) What is the total number of πbonds? 6 (B) (2 points) What is the hybridization of the Catom marked A? sp 2 (C) (2 points) What is the hybridization of the Catom marked B? sp 3

7 Fall 2009 FINAL CHEM 2A (Page 7 of 13) 26. (5 points) You are mixing X ml of wine, which is M in sugar, with 20.0 ml of wine, which is M in sugar. The resulting wine is mixed with 30.0 ml of wine, which is M in sugar. The resulting wine is M in sugar. What is the value of X? Assume that the volumes are additive. X(0.0030) = ( X) X = 8.7 ml Volume of wine = 8.7 ml 27. (6 points) A balloon is filled with nitrogen dioxide gas to an initial volume of 1.20 L. Two thirds of the nitrogen dioxide molecules react to form the dimer, dinitrogen tetroxide. The temperature, and the external pressure (and hence the total pressure of the balloon) are kept constant. What is the final volume of the balloon after the reaction is completed? Assume the balloon is indefinitely expandable. 2 NO 2 N 2 O 4 Assume the initial moles is z V 1 = 1.20 n 1 = z V 2 =? n 2 = moles of NO 2 + moles of N 2 O 4 = (1/3z) + (1/3z) = 2/3z V 2 = 0.80 L Final Volume =

8 Fall 2009 FINAL CHEM 2A (Page 8 of 13) 28. (15 points) A valve connects two gas reaction tanks. The first tank is 1.23 L and contains oxygen gas at 724 mmhg. The second tank is 2.05 L and contains nitrogen gas at 698 Torr. When the valve is opened the gases react to form N 2 O. Assuming the reaction takes place at 25.0 o C: (A) What is the limiting reagent and how many moles of N 2 O are produced? Determination of mols: N 2 : P = 698/760 = atm, V = 2.05 L, T = 298 K, n=pv/rt= 7.70x10 2 O 2 : P = 724/760 = atm, V = 1.23 L, T = 298 K, n = PV/RT = 4.79 x10 2 Balanced Reaction: 2 N 2 (g) + O 2 (g) 2 N 2 O (g) Moles of N 2 O Produced: N 2 : 7.70x10 2 mols N2 * (2 mols N 2 O / 2 mols N 2 ) = 7.70 x10 2 produced O 2 : 4.79 x10 2 mols O2 * ( 2 mols N 2 O / 1 mol O 2 ) = 9.58 x10 2 produced Limiting Reagent: Nitrogen Moles N 2 O Produced: 7.70x10 2 mols (B) How many moles of the excess reagent remain after the reaction? n excess = n initial n final = 4.79 x10 2 [7.70 x10 2 n excess = 9.4 x10 3 mol O 2 * (1 mol O 2 / 2 mols N 2 O)] Mols excess reagent: 9.4x10 3 mol O 2 (C) What is the pressure of N 2 O formed? P N 2 O : n = 7.70x10 2, T = 298 K, V = 1.23L L = 3.28 L P = nrt/v = atm Pressure of N 2 O: m (D) What is the total pressure in the tanks following the reaction? P O 2 = n O 2 = 9.4 x10 3 mol, V = 3.28, T = 298, P = nrt/v = 7.01 x10 2 atm P Total = P N 2 O + P O 2 = atm x10 2 atm = atm Total Pressure: atm

9 Fall 2009 FINAL CHEM 2A (Page 9 of 13) 29. (14 points) B 2 is a paramagnetic molecule. Below are shown two possible molecular orbital energylevel schemes for diatomic molecules of the second period elements. (i) For both diagrams, A and B, label the molecular orbitals (π 2p, σ 2s *, etc.) (ii) In both diagrams, A and B, fill in the electrons using arrows ( and ) both in the atomic and molecular orbitals. Which of the two diagrams is consistent with B 2 being paramagnetic (circle the correct diagram)? A

10 Fall 2009 FINAL CHEM 2A (Page 10 of 13) 30. (11 points) Write each of the four quantum numbers n, l, m l, and m s for all the electrons of neutral magnesium (electron configuration written below). Use the table below to fill in your answers. When filling in the table, start with the first electron (arrow) on the left and progressively move to the right until all electrons are accounted for (electrons are numbered for clarity). Enter only one quantum number per box. The first electron is done for you. You will receive one point per electron. Electron Number: Electron Number n l m l m s ½ ½ ½ ½ ½ ½ ½ ½ ½ ½ ½ ½

11 Fall 2009 FINAL CHEM 2A (Page 11 of 13) 31. (11 points) Balance the following chemical equation by the halfreaction method in acidic conditions. Show your work. P 4 + NO 3 H 2 PO 4 + NO (A) What is the oxidized element? P What is the reduced element? N (B) Write balanced half reactions for the oxidation reaction: 16 H 2 O (l) + P 4 (s) 4 H 2 PO 4 (aq) + 24 H + (aq) + 20e (C) Write balanced half reactions for the reduction reaction: NO 3 (aq) + 4 H + (aq) + 3e NO (g) + 2 H 2 O (l) (D) Write the overall balanced reaction Oxidation reaction x 3 + reduction reaction x 20 3 P 4 (s) + 20 NO 3 (aq) +8 H 2 O (l) + 8 H + (aq) 12 H 2 PO 4 (aq) + 20 NO (g)

12 Fall 2009 FINAL CHEM 2A (Page 12 of 13) 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 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 ) 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 E n 2 Z = R R H 2 H = x J n Potentially Useful Information

13 Fall 2009 FINAL CHEM 2A (Page 13 of 13) (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)