2015 US National Chemistry Olympiad Local Section Exam Answers and Explanations (Ben Ku 3/23/2015)

Transcription

1 015 US National Chemistry Olympiad Local Section Exam Answers and Explanations (Ben Ku 3/3/015) Shaded questions are those that current AP Chem students have not yet covered. Topics not on the AP Test are noted. Q Ans Topic Explanation 1. B Moles and Stoichiometry (Balancing Equations). A Moles and Stoichiometry (Concentration) 3. D Moles and Stoichiometry (Concentration) 4. C Elements and Compounds (Ionic Compounds) 5. B Moles and Stoichiometry (Limiting Reactants, Solubility) 6. A Solutions (Colligative Properties) The balanced equation is: C 6H O 1 CO + 6 H O 1.0 mol C 6H 6 15 mol O mol C = 7.5 mol O 6H g BaSO 4 1 mol BaSO g BaSO 4 1 mol Ba + 1 mol BaSO = mol Ba + = mol barium salt 4 Molar Mass = mass moles = g mol = 08.3 g/mol BaCl 3.0 g Mg(NO 3) 0.45 L 1 mol Mg(NO 3) g Mg(NO 3) mol NO 3 1 mol Mg(NO = 1.0 M 3) TbPO 4 Tb PO 4 Since phosphate carries a 3 charge, Tb has a +3 charge here. Tb 3+ + SO 4 Tb (SO 4) 3 The balanced equation is AgNO 3 + CaCl AgCl + Ca(NO 3) Since AgCl is insoluble, silver chloride is the precipitate. Find the moles of each ion: Moles of Ag + : (0.05 L) 0.15 mol AgNO 3 1 L 1 mol Ag + 1 mol AgNO = mol Ag + 3 Moles of Cl : 3.58 g CaCl 1 mol CaCl g CaCl mol Cl 1 mol CaCl = mol Cl Because there are fewer moles of available Ag +, the silver nitrate is the limiting reactant. Freezing point depression is given by: T = i m K f, where i is the number of dissolved species, and m is the concentration in molality. To find the greatest freezing point depression, we look at the greatest product (i)(m) (A) 1.0 m KBr ( ions)(1.0 m) = (B) 0.75 m C 6H 1O 6 (1 molecule)(0.75 m) = 0.75 (C) 0.5 m MgCl (3 ions)(0.5 m) = 1.5 (D) 0.5 m Ga (SO 4) 3 (5 ions)(0.5 m) = 1.5 * COLLIGATIVE PROPERTIES NOT TESTED ON AP CHEM TEST.

2 7. C Solutions (Acid/base, solubility) 8. C Periodic Trends (reactivity of metals) 9. D Solutions (colors of salts) 10. A Solutions (lab equipment, preparing solutions) 11. B Solutions (colors and solubility of salts) 1. B States of Matter (Intermolecular Forces) 13. A States of Matter (Vapor Pressure) 14. B States of Matter (Laboratory, state changes) 15 A States of Matter (State changes) 16. C States of Matter (Gas laws) 17. D States of Matter (Types of solids) A calcium salt more soluble in HCl than in water will have an ion that will react with HCl. (A) CaCO 3 + HCl CaCl + H O + CO (B) Ca(OH) + HCl CaCl + H O (C) CaSO 4 + HCl (no reaction) (D) Ca 5(OH)(PO 4) HCl 5 CaCl + H O + H 3PO 4 Metals that can most easily lose electrons (lowest IE) will react most vigorously. Metals on the lower left of the Periodic Table are most reactive. Solutions of zinc salts are colorless. * COLOR OF SALT SOLUTIONS NOT TESTED ON AP CHEM TEST. To precisely measure out 37 ml, you need a 50 ml graduated cylinder. Beakers and flasks are not graduated precisely, and volumetric flasks measure out only the volume specified. Both AgCl and PbCl are white precipitates. PbCl dissolves more rapidly at high temperatures. * COLOR OF PRECIPITATES NOT TESTED ON AP CHEM TEST. The compound with the weakest intermolecular forces will have the lowest boiling point. HF has hydrogen bonding, so it has the strongest IMFs and the highest BP. HCl, HBr, and HI have relatively similar IMFs, but because HCl has the fewest electrons, it has the weakest London Dispersion Forces among them, and will have the lowest BP. According to the Clausius Clapeyron Equation, increasing the temperature increases the vapor pressure (more molecules will vaporize at higher temperature). According to Raoult s Law, adding a nonvolatile solute will lower the vapor pressure. * VAPOR PRESURE IS NOT TESTED ON THE AP CHEM TEST. In the condenser, positions C and D would have the cooled condensed CHCl 3. The CHCl 3/CHCl CHCl mixture in A is warmer than the boiling point of CHCl 3. Position B has only the vaporized CHCl 3, so it would probably be at 61 C. It takes more energy to vaporize a sample than it does to melt a sample. At constant volume, temperature and pressure are directly related (Gay-Lussac s Law). P 1 T = P 1.0 atm 1 T 98 K = P 398 K P = 1.3 atm Diamond is a network covalent solid.

3 18. B Thermodynamics (Laws of Thermodynamics) 19. B States of Matter (Crystal structure) The first law of thermodynamics is the conservation of energy: in a system, energy is neither created nor destroyed. If there is an energy change in the system, it is either due to heat transfer with the surroundings, or work done with the surroundings. * THIS IS NOT TESTED ON THE AP CHEM TEST. The empirical formula of this substance is the number of each type of element within the unit cube. There is only one atom of A in the center of the cube. There are eight atoms of B, but only 1 8 of each atom is inside the cube, so there is a total of one atom of B in the compound. There are 1 atoms of C, but only 1 4 of each atom is inside the cube, so there is a total of three atoms of C 0. B Thermodynamics (Enthalpy of formation) 1. A Thermodynamics (calorimetry). D Thermodynamics (entropy) 3. B Thermodynamics (Hess Law) 4. A Thermodynamics (Vant Hoff Equation, Equilibrium) in the compound. The formula for the compound is ABC 3. * UNIT CELLS ARE NOT TESTED ON THE AP CHEM TEST. The enthalpy of formation is the energy for form 1 mole of the substance from their elements. For NaOH (s): Na (s) + ½ O (g) + ½ H (g) NaOH (s) The energy lost by the cooling of the gold is equal to the energy gained by the water. m goldc gold T gold = + m waterc water T water (37.5 g)(0.19 J g 1 K 1 )(T f 83.0 C) = +(100. g)(4.184 J g 1 K 1 )(T f.0 C) T f =.7 C Entropy increases in a process that produces more gaseous substances. We can use Hess Law to arrange the reactions given to form the overall reaction: A + D E + F +0 kj/mol C + E F 15 kj/mol A + B C 35 kj/mol A + B + D F 30 kj/mol Using the Van t Hoff Equation: ln K a K = H 0 a1 R 1 T 1 T 1 ln K a = ( 1400 J/mol) (8.314 J/mol-K) 1 (333 K) 1 98 K K a = * THE VANT HOFF EQUATION IS NOT TESTED ON THE AP CHEM TEST.

4 5. C Kinetics (Reaction rates) 6. B Kinetics (Rate Law) 7. C Kinetics (Integrated rate Law) 8. C Kinetics (Integrated rate law) 9. B Kinetics (Color of solutions) 30. D Kinetics, Equilibrium (Collision Theory, Le Chatelier s Principle) 31. A Equilibrium (ph of salts) [H O] dt = 6 4 [NH 3] dt = 3 (0.50 M/s) = 0.75 M/s Rate = k[h O ] 1 rate k = [H O ] 1 = ( M/s) (0.150 M) 1 = s 1 According to the integrated rate law, a second order reaction is linear when 1/[A] is plotted vs. time. If 80% decays, then 0% remains. Also, k = ln h 3. D Equilibrium (solubility) K sp = [Mg + ][F ] = (s)(s) = D Equilibrium (Kc vs. Kp) for a first order reaction. Using the integrated rate law for a first order reaction: ln n n = kt ln(0.0) = ln d (t) t = 19 days None of the species in this reaction has acidic properties, since Br and Cl are both conjugates of strong acids, so there should be negligible changes in ph. Bromine water (Br (aq)) is red-orange, while dissolved Cl is colorless, so as the reaction proceeds, the mixture will turn more red-orange, so a spectrophotometer would be appropriate to monitor the reaction s rate. * THE COLORS OF SUBSTANCES IS NOT TESTED ON THE AP CHEM TEST. As temperature increases, the reaction rate increases because k f increases. This can also be shown using the Arrhenius equation: ln k k = E a 1 R 1 T 1 T 1 A reversible reaction shifts toward the left as temperature increases in an exothermic reaction, so Q > K eq because K eq decreases. This can also be shown using the Vant Hoff Equation ln K a K = H a1 R 1 T 1 T 1 since H < 0. + NH 4 and C H 3O ions have approximately the same value for their respective K a and K b, so NH 4C H 3O has a ph of approximately 7. Na + and Ba + ions do not contribute to the ph. The C H 3O is basic (since it s the conjugate base of weak acid HC H 3O ). Since Ba(C H 3O ) contains more C H 3O ions, it is more basic than NaC H 3O. s = M K p = K c (RT) n so Kc 1 1 K = p (RT) n = [(0.081)(996 K)] 3 4 = 81.8

5 34. C Equilibrium (acid/base equilibrium) 35. C Equilibrium (K f, K sp, combining equations) 36. C Equilibrium (Acid/base, polyprotic acids, titrations) 37. B Redox (Oxidation numbers) 38. B Redox (oxidation/reduction half reactions) 39. D Redox (Galvanic cell) 40. A Redox (electrolysis) We can substitute the values into the K b expression: g K b = [NH 4+ ][OH ] [NH = L 1 mol NH g [OH ] 3] (3.00 M NH = ) [OH ] = [H ] = = M The K sp of Cu(OH) : Cu(OH) (s) Cu + + OH K sp = [Cu + ][OH ] = The K f of Cu(NH 3) + 4 : Cu NH 3 (aq) Cu(NH 3) + 4 Cu(OH) (s) + 4 NH 3 (aq) Cu(NH 3) + 4 (aq) + OH K eq = [Cu(NH3)4+ ][OH ] [NH 3] 4 = K sp K f = ( )(K f) = K f = * WHILE K f IS NOT TESTED ON AP CHEM TEST, COMBINING EQUATIONS CAN BE. At the point indicated, all of the H SeO 3 has been titrated into HSeO 3, which is the dominant species. HSeO 3 is minimally dissociated, so [SeO 3 ] would be negligible. The oxidation number of H is +1, and O is. If we let x be the oxidation number of C, (x) + (+1) + 1( ) = 0, the oxidation number of C = 0 The half reactions of the reaction are: Reduction: Ce 4+ + e Ce 3+ Oxidation: Cr Cr e Reduction takes place at the cathode, so here Ce 4+ is reduced at the cathode. The largest voltage would use the reduction half reaction with the most positive reduction potential (Ag + + e Ag) and the oxidation half reaction with the most negative reduction potential (Al e Al). To find the greatest mass of metal: 3000 s 1.50 C 1 s 1 mol e C 1 mol cation n mol e MM of cation 1 mol cation, so the greatest mass of metal has the greatest ratio. (A) Tl + (C) Zn + = = 04.4 (B) Pb + = = 3.70 (D) In 3+ = 07. = = = 38.3

6 41. A Redox (Nernst equation, ph) According to the Nernst Equation, E = RT Q nf ln Q. Q for this reaction is 1 by one unit, [H + ] is dropped 1/10 th, and Q Q 1 = (8.314 J/mol-K)(98 K) E = (4 mol)(96500 C/mol) ln (104 ) = V * CALCULATIONS USING THE NERNST EQUATION IS NOT TESTED ON AP CHEM TEST. 4. C Redox First, we need to combine the reactions: (Combining equations, Nernst Ag CrO 4 (s) + e Ag (s) + CrO 4 E = V equation) Ag (s) Ag + + e E = V Ag CrO 4 (s) Ag + + CrO 4 E = V According to the Nernst Equation: E = E RT nf ln Q, and when Q = K, E = A Atomic Theory (quantum numbers) 44. D Atomic Theory (element properties) 45. D Atomic Theory (metallic properties) 46. B Atomic Theory (Electron configuration) 47. B Atomic Theory (Periodic Trends) 48. A Atomic Theory (Electron emission) E = RT nf ln K, so K = ee nf/rt = e ( V)( e )(96500 C/e)/(8.314 J/mol-K)(98 K) = * CALCULATIONS USING THE NERNST EQUATION IS NOT TESTED ON AP CHEM TEST. The l quantum number for p orbitals is 1. * QUANTUM NUMBERS ARE NOT TESTED ON AP CHEM TEST. 1 [H + ] 4, and when the ph increases Both N and As are in the same family as P, but since As has d orbitals, it is closer to P in its properties than N. A metal has a lower melting points when it has greater metallic character, or when it holds onto its electrons looser. Among the alkali metals, metallic character increases going down the family since outer electrons are less attracted to the nucleus, so Rb would have the lowest MP. Be s electrons fill its s orbital. Electron Affinity is the energy released when electrons are added to an atom. In general, elements decrease in electron affinity going down a family because of increased shielding. However, because F s outer electrons are in the p orbital, which is much smaller than the 3p orbital, adding an electron is less stable (more electron repulsions) than adding it to Cl. * EXCEPTIONS TO GENERAL PERIODIC TRENDS ARE NOT TESTED ON AP CHEM TEST. Only transitions between n = 3, 4, 5, and 6 to n = (Balmer series) release energy in visible light. * TRIVIAL KNOWLEDGE OF EMISSION SPECTRUM IS NOT TESTED ON AP CHEM TEST.

7 49. C Bonding (Molecular shapes) NO + has a Lewis Structure: Steric number is linear. 50. B Bonding (Lewis Structures) 51. A Bonding (sigma and pi bonds) 5. B Bonding (Molecular shapes) I 3 has a Lewis Structure: Steric number 5 with 3 lone pairs is linear. Given the name peroxymonosulfate, HSO 5 looks like: [H O O SO 3] so there are four S O bonds, and one O O bond. (B) is incorrect because in general, pi bonds are weaker than sigma bonds since they are around atoms. (C) is incorrect because a double bond has one sigma bond and one pi bond. (D) is incorrect because pi bonds result from the overlap of UNhybridized orbitals. The Lewis Structure of chlorate is shown below. Cl has a steric number 3 with one lone pair, so it has a trigonal pyramidal shape. 53. D Bonding (Bond order, MO Theory) 54. D Bonding (Formal charges) 55. A Organic Chemistry (hydrogenation reaction) 56. D Organic Chemistry (Isomers) 57. C Organic Chemistry (Functional Groups) There are seven electrons in the molecular orbitals that result from the p orbitals of N and O. The electron configuration looks like: (σ p) (π p) 4 (σ* p) 1 With 6 bonding electrons and 1 bonding electron, the bond order is 6 1 * MO THEORY IS NOT TESTED ON AP CHEM TEST. The Lewis Structure for ozone is: The formal charge of the central oxygen of ozone is 6 5 = +1. Hydrogenation reactions are additional reactions of H across the double bond of an alkene, which results in an alkane. * ORGANIC REACTIONS ARE NOT TESTED ON AP CHEM TEST. The two compounds are structural isomers because they have different arrangements of the carbon skeleton. * ORGANIC COMPOUNDS ARE NOT TESTED ON AP CHEM TEST. Amines contain N, not O. =.5

8 58. D Organic Chemistry (Properties of hydrocarbons) 59. C Organic Chemistry (Degree of saturation) 60. B Organic Chemistry (Polymers) Hydrocarbons are nonpolar, and would not be soluble in water, since water is polar. Also, hydrocarbons are held by IMFs (vs. ionic compounds are held by stronger ionic bonds), so they would have much lower melting points. An alkyne has a general formula C nh n. * ORGANIC COMPOUNDS ARE NOT TESTED ON AP CHEM TEST. Glucose is a monomer, not a polymer.