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Chem 152 Exam 3 Fall 2003 Name: KEY R = 0.0821 L atm / mol K 1 atm = 760 mm Hg STP = standard temperature and pressure: 273 K and 1 atm All questions are worth 4 points. Multiple Choice, circle the correct answer: 1. When the equation NaOH + H2SO4 is completed and balanced, one of the terms in the balanced equation is: a) NaSO4 b) 2Na2SO4 c) H2OH d) 2H2O 2. The correct Lewis structure for N2 includes: a) one single bond b) one double bond c) one triple bond e) none of these d) none of these 3. The bonding in MgO is: a) ionic b) covalent 4. The bonding in HCl is: a) ionic b) covalent 5. Which compound has a trigonal pyramidal molecular structure? a) NaCl b) CO2 c) NH3 d) H2O e) CH4 6. When the reaction NaHCO3 + HNO3 is completed and balanced, one of the terms in the balanced equation is: a) CO2 b) H2 c) NaHNO3 d) H2CO3 e) none of these 7. Which of these is a strong electrolyte: a) HC2H3O2 b) CCl4 c) NaOH d) NH3 8. Which of these is a strong acid: a) HC2H3O2 b) NH3 c) HNO3 d) HClO True or False 9. The O2 molecule is polar. ______F_______ 10. When the temperature of a gas is decreased, the pressure increases. ______F_______ 11. Water is a polar molecule. ______T_______ Calculations: You must show your setup for credit on all problems (even if you can do the math in your head). Please circle your final answers. 12. How many mL of 6.0 M HCl do you need to prepare 500.0 mL of 0.10 M HCl? Dilution: M1V1 = M2V2 (6.0 M) V1 = (0.10 M)( 500.0 mL) V1 = 8.3 mL 13. How many grams of NaNO3 are need to make 750.0 mL of a 0.10 M solution? Describe in words how you would make the solution. M x L = moles (0.10 mole/L) (0.7500 L) = 0.075 mole NaNO3 0.075 mole NaNO3 x 85.00 g = 6.4 g NaNO3 1 mole Weigh 6.4 g NaNO3 , dissolve it in some water in a 750 mL volumetric flask, then bring the volume up to the line with water. 14. When 20.0 g FeS2 are reacted with 10.0 g O2, how many grams of Fe2O3 are produced? 4FeS2 + 11O2 20.0 g FeS2 x 1 mole FeS2 1 mole O2 32.00 g O2 2Fe2O3 + x 2 mole Fe2O3 x 159.7 g Fe2O3 119.99 g FeS2 10.0 g O2 x 4 mole FeS2 = 13.3 g Fe2O3 1 mole Fe2O3 x 2 mole Fe2O3 x 159.7 g Fe2O3 11 mole O2 8SO2 = 9.07 g Fe2O3 1 mole Fe2O3 ANSWER: 9.07 g Fe2O3 15. In problem #14, if you run the reaction and produce 5.3 g Fe2O3, what is your percent yield? % Y = AY/TY x 100 = 5.3 g/ 9.07 g x 100 = 58 % 16. A gas occupies a volume of 205 mL at 27C and 0.974 atm pressure. Calculate the volume the gas would occupy at STP. P1V1/T1 = P2V2/T2 (0.974 atm)(205 mL) / 300 K = (1 atm) V2 / 273 K V2 = 182 mL or 0.182 L 17. How many grams of O2 gas are in a sample occupying 0.500 L at 170.0 mmHg and 25C? PV = nRT P = 170.0 mm Hg x 1 atm / 760 mm Hg = 0.2237 atm PV/RT = n (0.2237 atm)( 0.500 L) = 0.00457 mol O2 (0.0821 L atm / mol K)(298 K) 0.00457 mol O2 x 32.00 g / 1 mole = 0.146 g O2 18. What is the density of He gas at STP? Assume you have 1 mole of He, then mass = molar mass = 4.003 g. Since we are at STP, 1 mole has a volume of 22.4 L. d = mass/vol = 4.003 g / 22.4 L = 0.179 g/L 19. For the following reaction, label each reactant as an acid or a base (according to the Bronsted / Lowry definition): HCO3base + HF H2CO3 + F- acid (reasoning: acids are H+ donors, bases are H+ acceptors) 20. A HCl solution is used to titrate a Ca(OH)2 solution. What is the Molarity of the HCl if 50.0 mL of it is required to titrate 50.0 mL of 0.500 M Ca(OH)2 ? 2 HCl(aq) + Ca(OH)2 (aq) CaCl2 (aq) + 2 H2O(l) (0.0500 L)(0.0500 mol/L) = 0.0250 mole Ca(OH)2 MHCl = moles HCl / L HCl = x 2 mole HCl 1 mole Ca(OH)2 0.0500 mole HCl / 0.0500L = 1.00 M HCl = 0.0500 mole HCl Draw Lewis structures for the following molecules or ions: 21) NH4+ 22) SO2 Note: these are too hard to type so I will describe the # of bonds and lone pairs. You should draw them. 4 single bonds, no lone pairs 1 single bond, 1 double bond. 3 lone pairs on O with single, 2 lone pairs on O with double, one lone pair on S. 23) NO3- 24) HNO2 one double bond, two single bonds 3 lone pairs on O's with single, 2 lone pairs on O with double, none on N. O=N–O–H with 2 lone pairs on each O, one on N and none on H. 25) H2O 2 single bonds, 2 lone pairs on O, none on H's Group Period 1 IA 1A 1 H 1 1.008 2 3 4 5 7 2 IIA 2A 13 14 IIIA IVA 3A 4A 15 VA 5A 16 17 VIA VIIA 6A 7A 2 He 4.003 3 4 5 6 7 8 9 10 Li Be B C N O F Ne 6.941 9.012 10.81 12.01 14.01 16.00 19.00 20.18 11 12 17 18 Na Mg 22.99 19 24.31 3 IIIB 3B 4 IVB 4B 5 VB 5B 20 21 22 23 6 7 VIB VIIB 6B 7B 24 25 8 9 10 ------- VIII ------------- 8 ------- 11 IB 1B 12 IIB 2B 13 14 15 16 Al Si P S 26.98 28.09 30.97 32.07 35.45 39.95 26 29 30 31 32 33 34 35 36 27 28 Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.47 58.69 63.55 65.39 40 41 42 43 44 45 37 38 39 Rb Sr Y 85.47 87.62 56 55 6 18 VIIIA 8A 46 47 48 49 Zr Nb Mo Tc Ru Rh Pd Ag Cd In 88.91 91.22 92.91 106.4 107.9 112.4 114.8 57 72 73 80 81 95.94 (98) 101.1 74 75 76 Cs Ba La* Hf Ta 132.9 137.3 138.9 178.5 180.9 183.9 186.2 87 88 89 104 105 106 107 Fr (223) 102.9 Lanthanide Series* Actinide Series~ 72.59 74.92 78.96 79.90 50 51 52 53 54 I Xe 126.9 131.3 85 86 Sn Sb Te 118.7 127.6 83 84 77 78 79 Pt Au Hg Tl Pb 190.2 190.2 195.1 197.0 200.5 207.2 108 109 110 111 112 114 116 118 --- --- --- --- --- --- () () () () () () (227) (257) (260) (263) (262) (265) (266) 58 59 60 61 62 63 64 66 Dy Ho Er Tm Yb Lu 140.1 140.9 144.2 (147) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0 90 91 92 93 94 95 96 97 98 99 100 101 102 103 U (238) 70 (210) 65 (231) 69 209.0 Tb Th Pa 68 Bi Po Ce Pr Nd Pm Sm Eu Gd 232.0 67 204.4 82 121.8 71 Np Pu Am Cm Bk Cf Es Fm Md No Lr (237) (249) (254) (242) 83.80 Ir W Re Os Ra Ac~ Rf Db Sg Bh Hs Mt (226) 69.72 (243) (247) (247) (253) (256) (254) (257) At Rn (210) (222)