1. (4) Write the names of the elements below next to their atomic symbols:

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ity ollege, hemistry Department hemistry 10301, sections T, T2, Τ3. Prof. T. Lazaridis Final exam, Dec 20, 2007 Name (last name first): I.D. Number last 4: Note: There are 18 questions in this exam. Fill in your answer in the blank space provided immediately following each question. 1/2 point will be subtracted every time you report a numerical result with an incorrect number of significant figures. A Data Sheet with useful information is at the end. 1. (4) Write the names of the elements below next to their atomic symbols: Ar Al S u Argon Aluminum Sulfur opper 2. (4) Write the molecular formula next to the names of the following compounds: Sodium Nitrate NaN 3 Barium arbonate Ba 3 Magnesium Sulfate MgS 4 Potassium ydroxide K 3. (8) Balance the following chemical equations: 2 + I 2 2 I 2 As + 3 2 2 As 3 2 Kl 3 2 Kl + 3 2 2 K + 2 2 2 K + 2 1

4. (7) The equation for preparing chlorine gas from l is 4 l (g) + 2 (g) 2 l 2 (g) + 2 2 (g) ow many kilograms of l are required to produce 1750 kilograms of l 2? Molar masses: l 2 : 70.90 g/mol l: 36.46 g/mol 1750 Kg X 1000g/Kg X (1/70.90 g/mol) X (4 mol l/ 2 mol l 2 ) X 36.46 g/mol X (1 Kg/1000 g) = 1800 Kg l 5. (7) hlorobenzene is prepared from benzene by the following reaction: 6 6 (l) + l 2 (g) 6 5 l (l) + l (g) A 10.0-Kg sample of benzene treated with excess chlorine gas yields 10.4 Kg of chlorobenzene. alculate the percent yield of chlorobenzene. Molar masses: 6 6 : 78.11 g/mol 6 5 l: 112.55 g/mol 10000 g/78.11 (g/mol) = 128.0 mol benzene theoretical yield = 128.0 mol X 112.55 (g/mol) = 14406 g chlorobenzene percent yield: 10.4/14.4 X 100% = 72.2 % 6. (7) ow many L of oxygen, measured at 0 o and 1.00 atm, are required for the complete combustion of 5.00 g of ethane ( 2 6 )? 2 6 + 7/2 2 ----> 2 2 + 3 2 Molar mass: 2 6 : 30.07 g/mol 5.00/ 30.07 (g/mol) = 0.166 mol, this requires 0.166X 7/2 = 0.581 mol 2 IG equation: V= nrt/p = 0.581 X 0.082 X 273.15 K / 1 atm = 13.0 L 2

7. (7) xidation of ammonia gives nitric oxide and water: 4 N 3 (g) + 5 2 (g) 4 N (g) + 6 2 (g) Δ o = -905.4 kj Use this equation and data from your data sheet to calculate the enthalpy of formation of N. Δ o = Δ o f (products) Δ o f (reactants) => -905.4 = {6X(-241.8) + 4 Δ o f(n)} {4X(-46.11)} => Δ o f(n) = +90.2 kj 8. (7) 30.0 g of P 4 10 is mixed with 75.0 g of water to form phosphoric acid: P 4 10 (s) + 6 2 (l) ---> 4 3 P 4 (aq) a) (3) Which one is the limiting reactant? b) (3) ow many grams of phosphoric acid will form? a) Molar masses: P 4 10 : 283.88 g/mol 2 : 18.02 g/mol 3 P 4 : 97.99 g/mol 30.0/283.88 = 0.106 mol P 4 10 75.0/18.02 = 4.16 mol 2 0.106 mol P 4 10 requires 6X0.106 mol = 0.636 mol water. We have 4.16. So water is in excess and limiting reactant is P 4 10. b) 0.106 X 4 X 97.99 g/mol = 41.55 g 3 P 4 3

9. (4) alculate the molarity of each of the following solutions: a) (2) 45.0 g of Nal in 250 ml of solution Molar mass : 58.44 45.0/58.44 = 0.770 mol 0.707 mol / 0.25 L = 3.08 M b) (2) 40.0 g of 2 S 4 in 2.00 L of solution Molar mass : 98.09 40.0/98.09 = 0.408 mol 0.408 mol / 2 L = 0.204 M 10. (4) ow many orbitals are there in a) (2) the n=3 shell? 9 (1 3s, 3 3p, and 5 3d) b) (2) the 3d subshell? 5 11. (4) Write an abbreviated (noble gas core) electron configuration for the following: Si Ne]3s 2 3p 2 l - Ar] d Kr]5s 2 4d 10 Fe 3+ Ar]3d 5 12. (2) Place the atoms B, K, Ga in order of increasing ionization energy. K < Ga < B 4

13. (5) Draw the orbital diagram for the partially filled subshells of r. 4s 3d (r is an exception to Aufbau rules) 14. (10) Draw Lewis structures that satisfy the octet rule for, 2, and 3-2 and predict the order of the... bond lengths (which molecule will have the shortest, the longest, and the intermediate bond lengths). ] ] ] The... bond is shortest in (triple), intermediate in 2 (double), and longest in 3-2 (single with partial double bond character) 15. (5) Use Lewis structures and the VSEPR method to predict the molecular geometry and the bond angles of l 3 - ] Electron group geometry: tetrahedral (bond angles slightly smaller than 109.5), molecular geometry: trigonal pyramidal 16. (5) ompare the shapes and bond angles of 2 and 3 +. - l 5

E.G. geometry tetrahedral, mol. geometry bent (angular), bond angle somewhat smaller than 109.5 due to lone pair repulsion ] + E.G. geometry tetrahedral, mol. geometry trigonal pyramidal, bond angle somewhat smaller than 109.5 due to lone pair repulsion (but probably not as small as in water) 17. (5) Write the structural formula for propene ( 3 6 ), determine the hybridization of each central atom, and estimate all bond angles. sp2 sp2 sp3 The angles around the left and center are 120 o and around the on the right 109.5 o 18. (5) Give the oxidation number of each element in the following compounds: a) XeF 2 F : -1, : -2, Xe: +4 b) Sn() 4 c) P 4 + d) ul 4 : -2, : +1, Sn: +2 : +1, P: -3 l : -1, u: +2 e) N - N : -3, : +2 6

DATA SEET Speed of light : 2.9979 X 10 8 m/s Planck's constant : 6.626 X 10-34 Js E n = -2.179 X 10-18 J/n 2 Mass of a proton: 1.67262 X 10-24 g Mass of an electron: 9.10939 X 10-28 g R= 0.082058 L atm / (mol K) = 62.364 L torr / (mol K) = 8.3145 J / (mol K) Substance Δ fo, kj/mol N 3 (g) -46.11 2 (g) -241.8 N 4 l (s) -314.4 l (g) -92.31 2 4 (g) 52.26 2 6 (g) -84.68 7