1. June 2013 Part 2 questions

3. 51. 3Mg + N 2 → Mg 3 N 2 52. Mg atom is 2-8-2. Mg +2 cation has a 2-8 electron configuration. Neon has a 2-8. Mg +2 is isoelectric to neon. 53. The Mg atom has 3 orbitals, the Mg +2 cation has only 2 orbitals, because when Mg loses it’s these 2e -1, it also loses the whole valence orbital.

5. 54. In the O 2 there is a double nonpolar covalent bond (they share 2 pairs of electrons, they share 4 total electrons) 55. Oxygen atom Lewis dot diagram looks like this: 56. In order to break the double bond to separate these 2 paired oxygen atoms, this requires energy. Energy must be absorbed to make this occur. O

7. 57.From A to B, this sample is being warmed from -25 to 0 centigrade. Ice is warming. To change temperature, you must increase kinetic energy, but potential energy is steady. 58.CD is warming of water, the amount of heat is on the bottom axis of the graph, from 7→14 kJoules, so a total fo 7 kJ/mole 59. BC is melting, the heat of fusion formula. The H F for water is 334 J/gram. DE is vaporizing, the heat of vaporization formula is used. The H V for water is 2260 J/gram. It takes much more energy to vaporize 1 gram of water than to melt 1 gram of ice.

9. 60.Equal volumes of different gases at the same temperature and pressure have the same number of particles and same number of moles. (Avogradro’s Hypothesis). They have the same number of molecules in each container. 61.Real gases act most ideal when hotter and at lower pressure, they will have less collisions that way. So lower pressure, increase the temp to be more ideal. 62.Use the Combined gas law, (1.2 atm)(1.25 L) (1.0 atm)(V 2 ) 293 K 273 K solve for V 2 =

11. 63. The answer is on Table Q, since these are all single bonds between carbons, they are all alkanes. 64. C 6 H 14 formula C 3 H 7 (dumb) empirical formula 65. Higher BP means stronger intermolecular attractions, caused by the polarity of the molecule (no radial symmetry)

13. 66. Mendeleev used how atoms bonded to oxygen to put his atoms into groups. 67. In group 3, the atoms all lost 3 electrons 68. Table J puts Ag LEAST ACTIVE on this list of metals. (only gold is less active, but that metal is not on this table at all. 69. Mendeleev only included atoms that could react. The noble gases do not make compounds, they are UNREACTIVE (mostly) and were likely not even recognized then as elements.

15. 70. H 2 ⁰ → 2H +1 so, 0 to +1 71. Pb +2 + 2e - → Pb ⁰ 72. Because it’s an OPEN SYSTEM, hydrogen and water gas escapes, it must be closed to reach equilibrium. 73. Faster by using more heat, or finer powder PbO, or higher pressure (more) hydrogen gas.

17. 74. Only the mass is different, these are the same substance, with identical properties. 75. aluminum oxide 76. Aluminum collects there, the melting point from table S is 933 K (hot!) 77. The density of pure aluminum must be less than the density of the mixture if the aluminum ends up on the bottom.

19. 78. Here, when 2 moles of SO 3 form, 394 kJ of energy are released. So when just 1 mole SO 3 forms, 187 kJ are released. 79. V 2 O 5 80. This is exothermic, so: this is fine:

21. 81. Freon 12 is CCl 2 F 2, so the diagram looks like EITHER of these two: 82. Both Freon 12 and Freon 14 are halocarbons, or halides (table R)

23. 83. If sugar dissolves in water (it does) and water is polar, and LIKE DISSOLVES LIKE, sugar must be polar too (it is). 84. IN STEP 1 (the trick here) is to see the total mass is just the sugar and the water. No corn syrup in this problem. Write the formula, solve: % comp = X 100% = 70.1% 414 g 591 g 85. As water evaporates, the amount of sugar remains constant. With less water (solute) the solution becomes more concentrated. The more concentrated a solution is with solute, the greater the BP is elevated (or the FP depressed).