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Chem 105 Exam #1 Chapters 7-8 Winter, 2006 Show all work, including any unit conversions for full credit. Use correct number of significant figures. Soluble K , Li , Na+, NH4+, NO3-, CH3COOHalides (Halogens) Sulfates + Exceptions: none Ag, Pb, Hg Ba, Sr, Pb, Hg + Insoluble Exceptions: 2- Carbonates (CO3 ) Group 1A and NH4+ Phosphates (PO43-) Group 1A and NH4+ For colligative properties, t = nKM, for water, Kfreezing = 1.86 C/M, Kboiling = 0.52C/M Osmotic pressure = nMRT, R= 0.0821 L atm/(K moles) Answers and explanations are in red The # points for each problem are found in blue Notes where similar problems can be found are in green. The letter in front refers to Homework (H), Worksheet (W), or Practice Test (P). The number before the decimal is the chapter number, after the decimal is the problem number. For instance H5.30 is homework problem #30 in chapter five. This is not a comprehensive list, but indicates where to go for some similar problems with perhaps lengthier explanations. Name_______________________ Chem 105 Exam #1 Chapters 7-8 Winter, 2006 1. While working on an experiment, you find that you need some 10.48 M sulfuric acid. After checking the stock, there is only 18.00 M sulfuric acid. Tell me how you would mix up 500.0 mL of 10.48 M sulfuric acid from the 18.00 M stock solution. (12 pts) H7.26, H7.48, W7ii.10, W7ii.11, P7.3, P7-10.5 C1V1 = C2V2 V2 = C1V1/C2 Keep in mind that the 500mL goes with the 10.48M soln. It doesn’t matter which you call C1 and which C2 as long as you keep the correct volume with the concentration. In this case, we will call the 18 M stuff C2 and solve for how much C2 we need (this will be V2). If you reverse which V goes with which C, you will get a Volume of 18 M stuff greater than the 500 mL of 10.48 M soln. This would be an answer that does not make sense. You’ve made a mistake; go back and start again. V2 = C1V1/C2 = 10.48 M x 0.5000 L/ 18.00 M = 0.2911 L Take 291.1 mL of 18.00 M sulfuric acid and add distilled water until the total volume = 500.0 mL 2. What, if anything, will be the precipitate when the following solutions are mixed together? For those that produce no reaction, write NR. Chose one that gave a precipitate and write balanced molecular, total ionic, and net ionic equations for it. Include physical states (s, l, g, aq) for all species. (16 pts) H5.30, W7.1, W7.2, P7-10.1 a. SrCl2 (aq) + MgSO4 (aq) SrSO4 (s) + MgCl2 (aq) SrSO4 (s) is the precipitate Sr2+ (aq) + 2 Cl- (aq) + Mg2+ (aq) SO42- (aq) SrSO4 (s) + Mg2+ (aq) + 2 Cl- (aq) Sr2+ (aq) + SO42- (aq) SrSO4 (s) b. AgCH3COO (aq) + NaNO3 (aq) NR (Neither acetate (CH3COO-) nor nitrate(NO3-) forms a precipitate with any counter ion. c. mercury (I) nitrate (aq) + potasium carbonate (aq) potassium has a plus one charge (group I), as does Hg (roman numeral I). Nitrate has a minus one charge, and carbonate has a minus 2 charge (see potentially useful information). 2 HgNO3 (aq) + K2CO3 (aq) Hg2CO3 (s) + 2 KNO3 (aq) so Hg2CO3 (s) is the precipitate 2 Hg+(aq) + 2 NO3-(aq) + 2 K+ (aq) + CO32- (aq) Hg2CO3 (s) + 2 K+ (aq) + 2 NO3- (aq) 2 Hg+(aq) + CO32- (aq) Hg2CO3 (s) 3. What are the boiling and freezing points of the solutions below? (12 pts) H7.66, P7.4 a. 3.00 M NaBr Calculate T, then add to the boiling pt and subtract from the solvent’s freezing pt (a solute increases the temperature range on both ends at which the solvent remains a liquid) Tf = nKfM = 2 x 1.86 C/M x 3.00 M = 11.2 C Tf = 0.00 - 11.2 C = -11.2 C Tb = nKbM = 2 x 0.52 C/M x 3.00 M = 3.1 C Tb = 100.00 + 3.1 C = 103.1 C b. 1.49 M Fe(NO3)3 Tf = nKfM = 4 x 1.86 C/M x 1.49 M = 11.1 C Tf = 0.00 - 11.2 C = -11.1 C Tb = nKbM = 4 x 0.52 C/M x 1.49 M = 3.1 C Tb = 100.00 + 3.1 C = 103.1 C c. 0.27 M glucose (C6H12O6) Tf = nKfM = 1 x 1.86 C/M x 0.27 M = 0.50 C Tf = 0.00 – 0.50 C = -0.50 C Tb = nKbM = 1 x 0.52 C/M x 0.27 M = 0.14 C Tb = 100.00 + 0.14 C = 100.14 C 4. Draw an energy diagram for a very exothermic reaction (large energy of reaction) with a small activation energy. Next to it, draw one for a slightly endothermic reaction with a large activation energy. Indicate which is which. On both diagrams, label both axes, reactants, products, activation energy (Ea), and energy of reaction (Erxn, also called heat of reaction). (12 pts) H8.26, W8.1, W8.2, P8.2 I have no program that will allow me to draw this. Maybe I can draw by hand and scan this and add it. I will look into that. If not done by Weds, than I can do this in class. BTW, if, after you read these answers and explanations, you still have questions, feel free to ask. It is the only way you will learn. 5. 23.48 mL of a 6.00 M solution of HCl is used to react with 50.00 mL of a solution of sodium carbonate of unknown concentration. What is the concentration of sodium carbonate (Na2CO3)? (12 pts) H7.56, W7ii.15 2 HCl (aq) + 1 Na2CO3 (aq) 1 H2O (l) + 2 NaCl (aq) + 1 CO2 (g) We have molarity and volume of HCl so we start there, and end up with sodium carbonate, which is what the question asked for L HCl moles HCl moles Na2CO3 molarity Na2CO3 0.2348 L HCl x 6.00 moles HCl/L HCl x 1 Na2CO3/ 2 HCl x 1/ 0.05000L = 1.41 M Na2CO3 6. For each of the situations below, the two solutions are initially at the same level on opposite sides of a U-shaped tube. Then a semipermeable membrane is placed between the solutions. Which solution will rise? Don’t forget to show your work. (12 pts) W7ii.14 a. 1.6 M glucose (C6H12O6) and 0.6 M Li2O Osmotic pressure (o.p.) = nMRT. Since both liquids are in the same container, T is the same. Since R is a constant, R is the same for both solutions. To see which solution has higher osmotic pressure, just compare the osmolarity (nM). Water will flow from low concentration to high concentration (from low o.p. to high o.p.). I am ignoring sig figs in my molarity calculations because I am not required to report any numbers. No numbers, hence no s.f.’s Glucose o.p. = nM = 1 x 1.6 M = 1.6 M Li2O o.p. = nM = 3 x 0.6 = 1.8 M The lithium oxide side will rise b. 0.10 M NaCl and nM = 2 x 0.1 M = 0.2M 0.19 M NH4Br nM = 2 x 0.19 M = 0.38 M the ammonium bromide side will rise c. 0.26 M FeI3 nM = 4 x 0.26 = 1.04 M the iron (III) iodide side will rise and 0.30 M Ca(NO3)2 nM = 3 x 0.30 M = 0.90 M 7. What is the osmotic pressure of 2.00 L of a 4.78 M solution of Fe2(SO4)3 solution at room temperature (25.0C)? How many moles of the iron (III) sulfate are present in this solution? (12 pts) H7.70 OP = nMRT = 5 x 4.78 M x 0.0821 L atm/mole K x 298 K = 585 atm Molarity = moles/L; so moles = Molarity x L = 4.78 moles/L x 2.00 L = 9.56 moles 8. The reaction below is exothermic and has reached equilibrium. What happens when the following changes occur? (12 pts) H8.52, H8.56, W8.5, P8.3 H2 (g) + Br2 (g) ↔ 2 HBr (g) a. the temperature is decreased Le Chatelier’s principle says that if you add to one side of an equilibrium, you tilt to the other. If you remove from one side, you draw the equilibrium towards the side that has stuff removed. Since the rxn is exothermic, heat can be considered a product. As we are removing product, we force the equilibrium to the right, towards product. b. some Bromine gas is added add to reactant side and we force the equilibrium towards the product side (or right) c. some Hydrogen gas is removed remove from reactant side and we force the equilibrium towards the reactant side (or left) d. a catalyst is added this has no effect on the equilibrium. Yes, the activation energy is lowered, but the activation energy for BOTH the forward and reverse direction is lowered. This has no net effect on equilibrium Extra Credit: Invariably, the student studies something that is not asked on the exam. Here is a chance to make good on that. Write a question on some aspect of chemistry covered in chapters 7 and/or 8 but was not on the exam today. Then, solve the problem. Extra credit points will be based on clarity, complexity and originality of the problem. (up to 10 pts)