U.S. National Chemistry Olympiad
1995 National Test Answers

Return to 1995 National Test


# Ans.   # Ans.   # Ans.
1. A   21. B   41. B
2. D   22. A   42. A
3. C   23. D   43. A
4. A   24. B   44. D
5. B   25. C   45. B
6. B   26. C   46. A
7. A   27. D   47. A
8. B   28. A   48. C
9. C   29. B   49. C
10. C   30. C   50. D
11. B   31. D   51. C
12. C   32. A   52. A
13. D   33. D   53. A
14. D   34. B   50. B
15. A   35. D   55. D
16. D   36. B   56. A
17. A   37. B   57. D
18. A   38. D   58. B
19. A   39. B   59. C
20. B   40. C   60. B

Part II - Answers

1. a) 2NaN3 --> 2Na + 3N2 b) PV = nRT n = PV/RT (5.2 X 10 L X 1.5 atm)/(298 K X 0.0821 L atm/mol K) = 3.19 mole c) 3.19 mol of N2 X (2 mol NaN3 /3 mol N2) X 65.0 g/mol = 138.3 g NaN3 d) 52L/0.04s = 1300 L/s 2. a) 0.0500L X 0.120 M = 6.00 X 10 -3 mol HOBr 0.0250L X 0.100 M = 2.50 X 10 -3 mol KOH 3.50 X 10 -3 mol HOBr(l) 2.5 X 10 -3 mol OBr- pH = 8.45; [H+]~3.5 X 10 -9 Ka = [H+][OBr-]/[HOBr] =(3.55X10 -9)(2.5X10 -3 mol OBr-/0.075L)/(3.50X10 -3 mol HOBr9l)/0.075L) =2.5 X10 -9 b) 2.5X10 -9 = [H+]2/0.120 M HOBr [H+]2 = 3.04X10 -10 [H+] = 1.74X10 -5 pH = 4.76 c) At equivalence point, 6.00X10 -3 mol OBr- in 0.050 + 0.060 = 0.110L [OBr-] =0.0545 OBr- + H2O <==> HOBr + OH- [OH-] = sq. root((Kw/Ka)[OBr-]) = sq. root((1X10 -14/2.5X10 -9)/0.0545)) [OH-] = 4.67X10 -4 d) i. pH will increase since OBr- will shift equilibrium to right, giving more OH- ii. Ka will not change. It depends only on T(temperature) and not on concentration. 3. a)i. SnO2 + 2H2 --> Sn + 2H2O delta Ho = 2mol(-241.8kJ/mol) - 1mol(-580.7kJ/mol) = 97.1 kJ delta So = 0.0516+2(0.1887)-[0.0523+2(0.1310)]=0.290-(0.3143) =0.1147 kJ/K ii. Sn)2 + C --> Sn + CO2 delta Ho = -393.5-(-580.7)=187.2 kJ delta So = 0.1516+0.2136-[0.0523+0.0057]=0.2652-0.0580=0.2072 kJ/K b) delta Go = O T=delta H/ delta S H2 T=97.1kJ/0.1147kJ/K=847K C T=187.2kJ/0.2072kJ/K=903K c) H2 is a more expensive reducing agent that C. Since the required reduction temperatures do not differ significantly, the most economical choice would probably be C rather than H2. 4. a)BrO3- order=1 as[BrO3-]doubled, rate doubles Br- order=1 as[Br-]doubled, rate doubles H+ order=2 as[H+]doubled, rate increases by four b)Rate=k[BrO3-][Br-][H+]2 k=1.04X10 -2/[(0.2)(0.2)(0.2)2]=6.5 L3 mol -3 s -1 c)The orders are not the same as the coefficients because the reaction does not occur in a single step. The orders are a reflection of the involvement of the reactants before the rate determining step. 5. a) H3PO4 + NaBr --> NaH2PO4 + HBr b) Ba 2+ + 2OH- + Cu 2+ + SO4 2- --> BaSO4 + Cu(OH)2 c) BCl3 + H2O --> B(HO)3 + H+ + 3Cl- d) PbS + O2 --> PbO + SO2 e) Cr2O7 2- + Sn 2+ + H+ --> Cr 3+ + Sn 4+ + H2O f) Co 2+ + NH3 --> Co(NH3)6 2+ g) Al(NO3)3 + H2O --> Al(OH) 2+ + H+ + 3NO3 - 6. a) b)C=O(CO2) < C-O (HCO3 -) < C--O(CO3 2-) < H-O-C (HCO3 -) shortest delocalization delocalization longest double bond (resonance) (resonance) single bond~11/2 bond~11/3 c)CO2 sp O-C-O angle = 180 degrees CO3 2- sp2 O-C-O angle = 120 degrees HOCO2 - sp2 O-C-O angle = 120 degrees 7. a)Melting and boiling points increase with as increase in mlar mass of the compounds. As the molar mass increases, the greater number of electrons makes the intermolecular forces (van der Waals or London) stronger. b)NO2 does not follow the general trend. Its melting point is greater than those of both N2O3 and N2O5, and its boiling point is higher than N2O3. In addition the melting point and boiling point differ markedly from that of N2O even though their molar masses are very similar. c)Ideal gas is one in which the "molecules" have no volume and no attractive forces. Such gases should be compressible (to V=O) without liquifying. d)The nitrogen oxides that are gases at O degrees C are NO and N2O (boiling points and <0 degrees C). N2O is least likely to behave ideally. (It has the higher boiling point indicating higher T is needed to overcome attractive forces.) 8.a)CaCl2 is colorless since all existing electronsare held tightly in ions and only =2 ion is formed since Ca loses its two 4s electrons. Cr forms CrCl2 and CrCl3 since is can lose 4s electron and defferent numbers of d electrons. Both are colored since electrons in partially filled d orbitals can be excited, absorbing visible light. b)Oxygen can form 2 single bonds by using its unpaired p electrons. Sulfur cannot only use p electronsbut also can promote an electron to a d orbital to form as additionsl 2 bonds. c)Sodium and magnesium halides consist of ions. As the halide ions increase in size the lattice energy decrease (so melting point decreases). The Si, P, and S halides are covalent and molecrles are attracted by dispersion forces which increase with the number of electrons.