CH150 15 Dec 08 Name: ________________________ Final Exam (100 points) You will only need a pen or a pencil and a calculator to finish this exam YOU MUST SHOW YOUR WORK TO RECEIVE CREDIT FOR PROBLEMS REQUIRING CACULATIONS You are encouraged to look over the entire exam before beginning to write your answers Problem Score Section I A First Look at Chemistry (60 points) Section II Chemical Reactions (70 points) Section III Real and Ideal Gases and Thermochemistry (70 points) Total: (200 points) Section I A First Look at Chemistry Periodic Table, Trends and Unit Conversions Name 3 of the halogens Cl, Br, I Name 3 diatomic elements O2, N2, F2 Arrange in order of increasing atomic radii: As, F, N, Na, C smallest F, N, C ,Na, As largest Arrange in order of increasing electron affinity: C, F, Si, Br, Rb smallest Rb, Si C Br F largest Arrange in order of increasing ionization energy: As, F, N, O, Br smallest As,N,OBrF Name the following compounds FeSO4 Iron(II) sulfate NH4CN Ammonium Cyanoate IF5 Iodine pentaflouride LiNO2 Lithium nitrite Give formulas for the following compounds. vanadium(III) iodide VI3 calcium perchlorate hexahydrate Ca(ClO4)2 . 6H2O dichlorine heptoxide Cl2O7 perchloric acid HCLO4 largest What is the mass percent of silver in AgBr? 107 +79 = 187 g/mol AgBr percent Ag = 107/187*100 = 58% Santa visits about 95,000 chimneys a minute on a certain night of the year (ask Einstein how he does it). How many sleigh loads per hour is this? [USE: 12 presents = 1 chimney, and 27,000 presents = 1 sleigh load] 95,000 Chim/Min * (12 Presents/Chim)* (1 load/27,000) * (60 min/1 hour) = 2533 sleigh loads Section II Chemical Reactions An aqueous solution was prepared by dissolving 20.93 g of KBr in enough water to make 340. mL of solution. What is the molarity of KBr in the solution? (molar mass of KBr: 119.00 g/mol) 20.93/119 = 0.1758 mol .1758/.340 = 0.517 M Suppose you need to prepare 1000. mL of 0.300 M HCl(aq), and all you have on hand is 0.500 M HCl(aq). What volume of the 0.500 M solution should be diluted to 1000. mL to give the desired 0.300 M HCl(aq) solution? M1V1= M2V2 (0.5)(X) = 0.300(1000) x=300/0.5 = 600mL 25.00 mL of a solution of oxalic acid are titrated with 0.301 M NaOH(aq). The stoichiometric end point is reached when 40.42 mL of the solution of base is added. What is the molarity of the oxalic acid solution? Oxalic acid reacts with sodium hydroxide as shown below: H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(l) 0.301*0.04042= 0.0121 mol NaOH * (1 mol acid/2 mol NaOH) = 0.00601 mol Acid / 0.025 = 0.243 M Give the oxidation number of each element in the following chemical reaction: 3HNO3(aq) H: 1 Al(s)  Al3+ + N: 5 O: -2 Al: 0 + Al 3 3NO2(g) N: 4 + 3OH- O: -2 O: -2 H: 1 Which element got oxidized (give a reason for your answer)? Which element got reduced (give a reason for your answer)? Aluminum chloride, an important reagent used in many industrial chemical processes, is made by treating scrap aluminum with chlorine according to the following unbalanced equation. 2Al(s) + 3Cl2(g) (molar masses, in g/mol: Al 26.98, → 2AlCl3(s) Cl2 70.91, AlCl3 133.34) If 98.0 g of Al and 345 g of Cl2 are mixed, which is the limiting reactant? 98/26.98 = 3.63 mol Al = (2 Al/2 AlCl3) = 3.63 mol 345/70.91 = 4.86 mol Cl2 * (2 AlCl3/ 3 Cl2) = 3.24 mol LR cause makes less What is the theoretical yield (in grams) of AlCl3 that can be produced when the quantities in the question above are mixed? 3.24 * 133.34 = 432.5g If 412 g of AlCl3 are actually obtained from the reaction, what is the percent yield? 412/432 *100 = 95% Section III Real and Ideal Gases and Thermochemistry Define both the a and b parameters of the real gas equation: a parameter – Molecular attraction term b parameter – Excluded volume term 1. Calculate the pressure exerted by 0.500 mol of N2 in a 10.000L flask at 25.0 oC using both the idea gas law and the real gas equation. Compare the two results. P =nRT/V = (0.5)(0.0821)(298)/(10.0L) = 1.20 ATM P = (0.5)(0.0821)(298)/(10.000L -‐ 0.5 – b) -‐ [(0.5)2a / (10.0)2 ] 2. A piece of solid carbon dioxide weighing 22.0g is placed in an otherwise empty 4.00 L flask at 27oC. a) What is the pressure of the flask after all of the carbon dioxide vaporizes? PV=nRT moles of CO2 = 22.0g / 44 g/mol = 0.5 moles of CO2 P = (0.5)(0.0821)(300K)/4.00L b) If 22.0g of carbon dioxide were placed in a similar flask already containing air at a pressure of 740 torr, what would be the partial pressure of carbon dioxide and the total pressure in the flask after the carbon dioxide has vaporized. 760 torr = 1 atm Pair = 0.95 atm in flask Total pressure in flask = 0.95 + answer in a. Partial pressure of CO2 is same as pressure in a. 3. Consider separate 1.0 L flasks of He, CH4, Ne and Ar all at 298K 1 atm a) Rank the flasks in order of increasing average energy. All have the same energy, it only depends on T not the gas b) Rank the flasks in order of increasing average speed. Rank them in order of molecular weight, lowest is fastest. 5. A 46.2 g of copper is heated to 95.4 oC and placed in 75.0 g of water at 19.6 oC. Given the final temperature is 21.8 oC, what is the specific heat (s) of copper? The specific heat of water is 4.184 J/(g o C) -‐ qlost = qgained -‐ 46.2g(X)(21.8 – 95.4) = 75.0g(4.184)(21.8-‐19.6) Solve for X 6. Given the following data: 2O3 (g)  3O2 (g) ΔH = -‐427 kJ O2 (g)  2O (g) ΔH = 495 kJ NO (g) + O3 (g)  NO2 (g) + O2 (g) ΔH = -‐199 kJ Calculate the ΔH for the following reaction: NO(g) + O (g)  NO2 (g) Reverse 1st equation 3O2 (g)  2O3 (g) ΔH = 427 kJ Reverse 2nd equation 2O (g))  O2 (g ΔH =-‐ 495 kJ Multiply last equation by 2 2NO (g) + 2O3 (g)  2NO2 (g) +2 O2 (g) ΔH = -‐398 kJ 2NO(g) + 2O (g)  2NO2 (g) ΔH = -‐466 kJ Useful Formulas and Constants Physical Constants: Speed of light (vacuum) c 3.00 X108 m/s Plank’s constant h 6.62 x 10-‐34 J·∙s Mass of a proton p 1.0073 amu Mass of a neutron n 1.0087 amu Mass of an electron e-‐ 0.005486 amu Gas Constant R 0.0821 L atm / (mol K) 8.314 J/mol k 1 L atm = 101 Joules Equations: M1V1 = M2V2 PV = nRT ΔE = q + w w = -‐ PΔV q = nsΔT qp = nspΔT qv = nsvΔT ΔH = ΔE + PΔV ΔHp = qp ΔHv = qv