* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Chemistry 2008 Multiple Choice
Electrical resistivity and conductivity wikipedia , lookup
Photoelectric effect wikipedia , lookup
Liquid–liquid extraction wikipedia , lookup
Chemical reaction wikipedia , lookup
Atomic orbital wikipedia , lookup
Chemical equilibrium wikipedia , lookup
Computational chemistry wikipedia , lookup
X-ray photoelectron spectroscopy wikipedia , lookup
Gas chromatography–mass spectrometry wikipedia , lookup
Self-assembled monolayer wikipedia , lookup
Atomic nucleus wikipedia , lookup
Acid–base reaction wikipedia , lookup
Rutherford backscattering spectrometry wikipedia , lookup
Electrolysis of water wikipedia , lookup
Oxidative phosphorylation wikipedia , lookup
Photoredox catalysis wikipedia , lookup
Biochemistry wikipedia , lookup
Molecular orbital diagram wikipedia , lookup
Marcus theory wikipedia , lookup
Extended periodic table wikipedia , lookup
Transition state theory wikipedia , lookup
History of molecular theory wikipedia , lookup
Implicit solvation wikipedia , lookup
Physical organic chemistry wikipedia , lookup
Thermometric titration wikipedia , lookup
Metallic bonding wikipedia , lookup
Resonance (chemistry) wikipedia , lookup
Light-dependent reactions wikipedia , lookup
Electron configuration wikipedia , lookup
Hypervalent molecule wikipedia , lookup
Stoichiometry wikipedia , lookup
Atomic theory wikipedia , lookup
Chemical bond wikipedia , lookup
Geometrical frustration wikipedia , lookup
Multiple Choice 1. A Noble gases are the least able to lose electrons (oxidation) in their period. 2. D High atomic mass atoms (# > 82) tend to be radioactive. 3. C Halogens pick up electrons and become anions. 6. B 8. D 9. D 10. C 11. A ln(k1/k2) = (Ea/R)(1/T2 – 1/T1): Ea is determined using the reaction rates ( k) and T. 12. C 13. A 14. C 15. B 16. D Cations act as Lewis acids when combine with ligands (Lewis base) to form complex ion. 17. A Metal loses electrons to nonmetal (redox) producing a single compound (synthesis). 18. C 19. C Ions of elements # 33-39 are isoelectric with Kr. 21. B 1 ½ ¼ 1/8 3 t½: 120 s/3 = 40 s 2 H2 + O2 2 H2O 0.30 + 0.15 0.30 0.40 – 0.30 = 0.10 mol H2 (0.2 g) Ca3(PO4)2 + 4 H3PO4 3 Ca(H2PO4)2 35. Shift to the right add reactant (I), increase T (II), pressure has no effect. 36. 2.8 g C(2 mol C/28 g)(6 x 1023 atom/1 mol) = 1.2 x 1023 37. A SiO2 is covalent network. 38. D mol dilute = mol concentrate 0.5 L(2 mol/L) = (X L)(10 mol/L) X = .1 L 39. From first exp. SnSO4 is soluble ZnF2 is ppt. 40. 41. 42. D Excited state produces gap in electron order. d = m/V = 5 sig fig/4 sig fig 4 sig fig 34. C London forces between non-polar molecules. S: 30/32 1, F: 70/19 3.5 SF4 33. A 20. D C D 1 L O2(1 mol O2/22.4 L)(2 mol KClO3/3 mol O2) m = mol/kg = (¼ mol)/0.1 kg = 2.50 m 32. A Ions combine to form solid product. C2H6 fits CnH2n+2 saturated (all single bonds). 31. D Exothermic (H < 0), more gas (S > 0). Mass: 238 + 1 = 239, Charge: 92 + 0 = 92 D. 30. D Break bonds (H > 0), more disorder (S > 0). Compared to exp. 1, the rate for exp. 2 is ½R because [½X]1, but 4R because [2Y]2 (4/2)R. 29. A Lower energy (H < 0), greater disorder (S > 0). P = nRT/V = (1.85)(0.0821)(308)/(3.00) 28. B Entropy (S) is zero when motion stops (0 K). p = e = 55; n = 134 – 55 = 79 27. C Large molecules (HBr) have slow effusion rates (escape a small opening). 55Cs: 26. B HBr is a strong acid = strong electrolyte. 134 25. C NH3 has a trigonal pyramid shape. Most transition metals form 2+ ions (lose s e-). 24. D Buffer is formed with equal concentration of acid and conjugate, which is along flat section. 7. A B D Adding more base raises pH closer to pure base. F2 and CO2 are nonpolar, but only CO2 has polar bonds (bonds between unlike atoms). 23. A Equal moles of acid and base occur at equivalence, which is midpoint along vertical. 5. D C B 4. C 22. All are dispersion forces largest (Br2) has highest melting point. N2 + 3 H2 2 NH3 1 N2 + 3 H2 2 NH3 with 3 – 1 = 2 N2 left over Ne is 5 x heavier than He equal masses have 5 He/1 Ne PHe = (5 mol He/6 mol total)Ptot = 5 atm 43. C 64. Boiling occurs when Pvapor = Patmosphere. 44. B Cl2 takes electron form stronger ox. agent C D 67. t4 is at the beginning of freezing max. liquid. B 47. B "Do what you aughta, add acid to wata"—mixing is exothermic, need water to absorb heat. 68. NH2- + H+ NH3 69. 48. B According to the solubility rules, column 1 metals, ammonia and nitrates are soluble Phosphate. 70. 10 Br- 5 Br2 + 10 e- 71. 50. D 51. C B A 49. A B D Gases are most ideal at high T and low P. Volume changes with temperature (thermal expansion) only molarity (mol/L) is affected. 53. A Smallest molecule has the weakest London forces. [OH-] = ½(0.002 M) = 0.001 M pOH = -log(1 x 10-3) = 3 pH = 14 – 3 = 11 At the same temperature both gases have the same kinetic energy (K = 3/2RT). Amino acids: NH2–C(R)H–COOH (I hope you remember your biology). CO32- + 2 H+ CO2(g) + H2O Zn + 2 H+ H2(g) + Zn2+ Ba2+ + SO42- BaSO4(s) B O–O=O O=O–O There is an additional pair of electrons around the center oxygen, which makes the molecule bend (120o). C Least polar bond would be between atoms that are closest to each other in the periodic table. 74. Molarity is mol/L. The number of moles is given, so volume is the additional information. 55. D 3(NO2- + H2O NO3- + 2 H+ + 2 e-) Cr2O7 + 14 H+ + 6 e- 2 Cr3+ + 7 H2O 8 H+ + 3 NO2- + Cr2O72- 2 Cr3+ + 3 NO3- + 4 H2O 2- 73. 54. D Collision theory: reactants collide frequently, with sufficient force and correct orientation. 72. 52. C To equalize the pressure inside the tube to equal atmospheric pressure, the water levels must be equal. 66. mol CH3OH = (16 g/32 g) = 0.5 mol H2O = (90 g/18 g) = 5 mol 0.5/5.5 0.1 46. D pH > 7 is basic. Bases are anions other than from strong acids bonded to non-H+ cation. 65. I- 45. A D D Vapor pressure only depends on temperature and type of liquid, which are the same for both containers. 75. Pvapor = (mol H2O/mol total)Po Pvapor = (9/10)25.0 = 22.5 mm Hg B K = [Z]3/[X]2[Y] = (4)3/(2)2(.5) = 32 1. 1 mol Ni x 2 mol e- x 96,500 C x 1 s = 193,000 s 1 mol Ni2+ 1 mol e- 1 C 56. E Free Response Kp = PCO2/PCO2 57. A 2 bonds and no unshared electron pairs sp. 58. D mol C9H8O4 = 0.360 g/180 g = 0.002 volume solution = 0.2 L M = 0.002/0.2 = 0.01 59. B K = [N2O4]/[NO2]2 = 3/62 < 1 (but never neg.) 60. B NH3 is a weak base, so the initial precipitate is Ni(OH)2, but with excess NH3, the complex Ni(NH3)42+ is formed. 61. D –C–O–C=O– is an ester group. || O 62. B 25 mL pipet has accurate graduation marks and a capacity closest to the volume to be measured. 63. B a. F- has one more proton compared to O2-, which attracts the electrons closer to the nucleus. b. PV = nRT (5.00 atm)(2.00 L) = n(0.0821L•atm/mol•K)(1,160 K) n = 0.105 mol CO2 c. (1) PCO = Ptotal – PCO2 = 8.37 atm – 1.63 atm = 6.74 atm (2) Kp = PCO2/PCO2 = (6.74 atm)2/1.63 atm = 27.9 d. Equal to because the catalyst would increase the forward and reverse reaction rates, which would decrease the time to reach equilibrium, but the not the equilibrium position. e. Decrease because Qp = PCO2/PCO2 = (2.00 atm)2/2.00 atm = 2.00 < Kp the reaction will proceed to the right, which will increase the partial pressure of CO and decrease the partial pressure of CO2. 2. a. No additional mass was lost during the third heating, indicating that all the water of hydration has been driven off. b. (1) g H2O = 25.825 g – 22.977 g = 1.848 g H2O 1.848 g H2O x 1 mol/18.02 g H2O = 0.1026 mol H2O (2) g MgCl2 = 23.977 g – 22.347 g = 1.630 g MgCl2 1.630 g MgCl2 x 1 mol/95.20 g MgCl2 = 0.01712 mol MgCl2 0.1026 mol H2O/0.01712 mol MgCl2 = 6 MgCl2•6 H2O c. The calculated mass of water lost by the hydrate will be too large because the mass remaining in the container will be less the mass of the spattered solid. d. Add excess AgNO3. Mass the filter paper. Filter the mixture. Wash and dry the AgCl(s). Mass of AgCl + filter paper. Subtract mass from mass of filter paper. e. (1) 5.48 g AgCl x 1 mol AgCl x 1 mol MgCl2 = 0.0191 mol MgCl2 143.32 g AgCl 2 mol AgCl (2) 5. F(g) F+(g) + 1 eb. In both cases the electron removed is from the same energy level (2p), but fluorine has a greater effective nuclear charge due to one more proton in its nucleus (the electron is held more tightly and thus takes more energy to remove). c. Less because the electron removed from fluorine is from a 2p orbital, whereas the electron removed from xenon is from a 5p orbital, which is at a higher energy level and farther from the nucleus than 2p, hence it takes less energy to remove the electron from xenon. d. F \ .. O–Xe–O F–Xe–F | .. \ O F e. (1) 0.0191 g MgCl2/2.94 g mixture x 100 = 61.9% MgCl2 3. a. Ereaction = ECu2+ + ENO3- ENO3- = Ereaction – ECu2+ ENO3- = 0.62 V – (-0.34 V) = 0.96 V b. Go = -nFEo = (6)(96,600 C•mol-1)(0.96 J/C) Go = -360,000 J•mol-1 = -360 kJ•mol-1 c. So > 0 because of the formation of two moles of NO(g) from non-gaseous reactants. d. (1) Second order because the reaction rate for exp. 3 is 9 times the reaction rate for exp. 1 (0.767/0.0852) when [NO]o is tripled and [O2]o remains constant (9 = 32). (2) First order because the reaction rate for exp. 2 is 3 times the reaction rate for exp. 1 (0.256/0.0852) when [O2]o is tripled and [NO]o remains constant (3 = 31). e. rate = k[NO]2[O2] 4. f. k = rate/[NO]2[O2] k = 0.0852 mol L-1 s-1/(0.0200 mol L-1)2(0.0300 mol L-1) k = 7.10 x 103 mol-2 L2 s-1 a. (1) OH- + Al(OH)3 Al(OH)4(2) Decrease because the H+ would react with the OH-, reducing the [OH-] and shifting the equilibrium to the left, thus reducing the concentration of Al(OH)4-. b. (1) HCl + O2 HClO2 (2) Neither reactant would be present because they react in a 1 to 1 ratio, thus three moles of HCl would require 3 moles of O2, which is what is available. c. (1) K2O + H2O 2 K+ + 2OH- (2) The phenolphthalein would turn pink because of the presence of OH- in solution making it basic. a. reduction potential Eo for the reduction of NO3- in acidic solution. Trigonal pyramid (2) sp3d2 6. f. XeO3 is polar because the pair on non-bonding electrons push the bonding oxygens toward each other resulting in an asymmetrical structure, where the bond dipoles do not cancel each other out. a. Pyridine can form H-bonds with water around the :N, but benzene can not because it has no polar region. As a result, pyridine will dissolve in water whereas benzene will not. b. Liquid ethanol and dimethyl ether form London dispersion and dipole forces, but liquid ethanol will also form H-bonds with other ethanol molecules because of the presence of the OH-. Since H-bonds are stronger than London dispersion and dipole forces, the energy (temperature) needed to break the bond is higher for ethanol than it is for dimethyl ether. c. In the solid state SO2 forms London dispersion and dipole forces between distinct molecules whereas, SiO2, a covalent network solid, forms covalent bonds throughout. The much stronger covalent bonds, which are broken during melting of SiO2, require much more energy (higher temperature) to break compared to the relatively weak London dispersion and dipole forces, which are broken during the melting of SO2. d. Liquid Cl2 is held together by London dispersion forces, which although weak increase in strength as the number of electrons increases. Liquid HCl is held together by dipole forces in addition to London dispersion forces, but the addition of dipole forces between HCl molecules must not make up for the fewer electrons around the HCl molecule compared to Cl2.