Name: Final Grading key Page 2/9 Chemistry 125/126, Exam 1 Tuesday, March 13, 2012 Welcome to the first hourly exam for Chemistry 125/126. This exam consists of 6 questions worth a total of 75 points plus a bonus question worth 3 points for a possible total of 78 points. It is 9 pages long; 7 pages of questions plus one page with periodic tables for your reference (page 9). A few notes about the exam. To receive credit, your answers must be placed in the indicated spaces or boxes. If calculations are required to obtain an answer, show your calculations; you will not receive credit for numerical answers alone. Do not write in any box marked "For use by grader". The exam should take about l hour, however, you may use up to 1.5 hours to complete the exam. All exams must be turned in at 7:45 pm. Save any questions you have during the exam for Prof. Kerner. The GSIs have been told not to answer questions during the exam. Graded exams will be available starting tomorrow. If, after checking the exam key, you feel there was a grading mistake, turn your exam in with a statement describing the mistake to Prof. Kerner s mailbox in 1500o chem. All re-grades must be turned in by noon, Monday, March 26, 2012. Course Information Section GSI Section GSI Section GSI 127 T AM Hyung Ki Yoon 195 T PM2 Eli Fahrenkrug 237 Th PM2 Anisha Shakya 129 T AM Jake Lapping 199 T PM2 Yue Xie 239 Th PM2 Yue Xie 135 T PM1 Laura Kiefer 225 ThPM1 Wendi Hale 241 Th PM2 Chuan Leng 137 T PM1 Charity Haynes 227 ThPM1 Charity Haynes 251 F PM1 Meng Zhang 139 T PM1 Hyung Ki Yoon 231 Th PM1 Jake Lapping 253 F PM1 Anisha Shakya 141 T PM1 Wendi Hale 233 Th PM1 Laura Kiefer 119 F PM2 Zheng Zheng 143 191 T PM2 T PM2 Chuan Leng Zheng Zheng 235 Th PM2 Eli Fahrenkrug 123 F PM2 Meng Zhang For use by grader Page Points Score Name: 2 16 3 09 GSI: 4 07 5 10 Section: 6 09 7 14 e-mail: 8 10 (Bonus) (+03) Total 75 (+03)
Name: Final Grading key Page 3/9 Question one (16 points) asks you to make predictions based on your knowledge of periodic trends and the relationship of structure to properties and reactivity. Periodic tables are provided at the back of this exam for your reference. A. An unidentified white solid is either NaNO 3 or Ba(NO 3 ) 2 or Ni(NO 3 ) 2. The white solid dissolves in water. Upon addition of K 2 CO 3 a white precipitate (carbonate) is formed. Circle any soluble white compound forming a carbonate precipitate. NaNO 3 Ba(NO3)2 Ni(NO 3 ) 2 B. Circle any species that is a weaker oxidizing agent than Ca 2+. Cl 2 K+ Mg 2+ C. Predict the comparative solubility of the salts RbCl, SrCl2, and BaCl2 Most soluble least soluble RbCl > SrCl2 > BaCl2 D. Circle any combination of species likely to participate in a oxidation-reduction reaction. Cl - and I - F2 and Br- Na + and Cl - E. Predict if aqueous solutions of nitrate salts with cations of the indicated electron configurations will be colored or : Cation electron configuration colored or? Au + [Xe] 6s 0 5d 10 colored Cr 3+ [Ar] 4s 0 3d 3 2 pts or zero each = 4 points
Name: Final Grading key Page 4/9 Question 2 (12 points) deals with the precipitation reaction that occurs when a solution of silver fluoride, 0.1M AgF, is mixed with a solution of barium acetate, 0.1M Ba(C 2 H 3 O 2 ) 2. AgF + Ba(C 2 H 3 O 2 ) 2 precipitate A. What could the precipitate be? Give the formulas for all chemically reasonable possibilities; you may assume that this is not a redox reaction). Formulas of possible precipitates: BaF2 AgC2H3O2 Stoichiometry of each formula has to be correct for credit (2 pts each); extra formulas receive 1 point. (4 points) B. Based on your knowledge of cation color, indicate if the precipitate/s (recorded above under A) will be white or a color other than white. Note: fluoride and acetate ions are. Indicate if precipitate/s recorded under A will be white or a color other than white. BaF 2 = white AgC 2 H 3 O 2 = white 1 point each for white ; any other response (e.g. ) receives no credit. (2 points) C. In any attempt to undertand the AgF + Ba(C 2 H 3 O 2 ) 2 reaction, one team substituted a solution of 0.1M NaF, for the 0.1M AgF. This time, no precipitate is observed: NaF + Ba(C 2 H 3 O 2 ) 2 no precipitate What does the test result tell you about the reaction under investigation? Information about AgF + Ba(C 2 H 3 O 2 ) 2 reaction. Ag + ion is (2 points) (Ag + = 2 points or if charge on Ag ion matches stoichiometry cited for part A answer = 2 points; Ag = 0 points) critical to precipitate formation (1 pt) (OR identity of ppt is AgC 2 H 3 O 2 = 1 pt OR is a reactant = 0.5 pt) (3 points)
Name: Final Grading key Page 5/9 D. You filter the precipitated product mixture. To half of the filtrate you add 0.1 M NaNO 3. To the other half, you add 0.1 M NaCl: filtrate + NaNO 3 no precipitate filtrate + NaCl precipitate (identified as silver chloride) What do the above observations tell you about the comparative solubilities of silver acetate, silver chloride and silver nitrate? Most soluble least soluble silver nitrate > silver acetate > silver chloride 3. (15 points) This question deals with your determination of the identity and concentration of a salt solution by spectrophotometric analysis. A. The graph below shows the absorption spectrum of a sport s drink: Information: λ400-450 = violet; 450-500 = blue; 500-550 = green; λ550-600 = yellow; 600-625 = orange, 625-700 = red Indicate (circle one choice) the color of the sport s drink. Violet Blue Green-Yellow Red-Orange Colorless 4 points or zero B. Use the spectrophometric data below to: 1. Determine the concentration of a salt solution 2. Detrmine the formula weight of the salt.
Name: Final Grading key Page 6/9 1. A) A diluted sample has an absorbance = 0.45 What is the molarity of the diluted sample? For complete credit indicate how you use the graph. Calibration Curve for Salt Solution Absorption 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 [M + ] (Molar) Absorbanceλ = 0.45 = Elc 0.45 = 1.75 c (1 pt.for correct equation and 1 pt for correct correct slope calculation (two data points) **(Award only 1 pt if eyeballing the graph or or other less appropriate method)** 0.26M = c (1 point for correct M value) NO WORK, NO POINTS! Molarity of diluted sample*: 0.26 M (range of 0.25 0.27 depending on slope) 3 points B). The diluted sample (B.1, above) What is the molarity of the undiluted sample? MiVi = MfVf Mi x 3.00mL = (above calculated M) x 10.00mL Mi = above calculated M X 10.00mL 3.00mL ( 2 pts for correct set-up; no double penalty for incorrect Molarity of diluted sample) = 0.86M * (1 pt for correctly calculated value; no triple penalty for incorrect M e.g. OR correct answer = Molarity of diluted sample indicated under 1 x 3.33 Molarity of undiluted sample: 0.86M 3 points 2. The undiluted sample contains 290.40 g What is the formula weight (FW) of the salt? 290.40 (g) = (above calculatedm) mol per liter formula wt(g) 1 mol per liter OR 290.40 = formula wt. above calculated M. (3 points for CORRECT set-up; NO double penalty for incorrect undiluted Molarity) Improper setup = zero points) 290.40 = 337.67 = formula wt* (*1 point for correct formula weight) 0.86 Formula wt = 337.67 g/mole 4 points
Name: Final Grading key Page 7/9 Question 4 (9 points) You are to prepare 200 ml of 0.10 M cadmium nitrate. The reagent available for sample preparation is Cd(NO 3 ) 2 4H 2 O. A. What is the formula weight (FW) of Cd(NO 3 ) 2 4H 2 O? No Calculations = no points. 112.41 + 2(14.01 + 3 x 15.99) + 4 (2 x 1.01 + 15.99) 112.41 + 28.02 + 95.94 + 72.04 = 280.8 (2 pts or zero) formula wt. Cd(NO 3 ) 2 4H 2 O = 308.41 (acceptable range 307.4 309.5) 2 points B. How many grams of Cd(NO 3 ) 2 4H 2 O do you weigh out? moles needed (1 pt): 0.200L x 0.10M = 0.02 moles* * moles needed must be correct or no credit for problem. Calculation for grams needed (1 pt): FW* (above calculated value) x 0.02 mol = g 1 mol * no double penalty for incorrect formula wt under A. 308.4 g x 0.02 mol = 6.17 g 1 mol Correct calculated value (1 pt) g Cd(NO 3 ) 2 4H 2 O = 6.17 3 points C. Instead of using a volumetric flask, your lab teammate used a buret to add 200 ml of water to a beaker containing the solid. The resulting concentration was subsequently found to be in error. Was the resulting solution too concentrated or too dilute? Why? Circle the concentration error too concentrated too dilute 2 points or zero Explain your reasoning total volume of solid + 200mL water = > 200mL. (or reasonable clarification of why the volume >200mL when using a buret) OR There is less than 6.17g (or 0.02 moles) per volume of 200mL 2 points or zero
Name: Final Grading key Page 8/9 Question 5 (14 points) deals with a study of redox reactions: 1. CuBr 2 (aq) + SnCl 2 (aq) loss of blue color and formation of a white precipitate. blue 2. CuBr 2 (aq) + NaCl (aq) no reaction. blue 3. CuBr 2 (aq) + Sn(NO 3 ) 2 (aq) loss of blue color. A clear, solution forms. blue 4. Cu (s) + CdCl 2 (aq) formation of blue color of Cu 2+ and loss of Cu(s). A new solid is formed. A. What does reaction 2, by itself, tell you about reaction 1? What can you conclude about the CuBr 2 (aq)+ SnCl 2 (aq) reaction from reaction 2? Sn2+ ion (2 pts) is (or Sn 2+ acceptable; Sn not acceptable = 0 points total) is critical to reaction(1 pt) (or critical to loss of blue color and formation of a white ppt ) - 1 pt if another conclusion stated (3 points) B. What does reaction 3, by itself, tell you about reaction 1 What can you conclude about the CuBr 2 (aq)+ SnCl 2 (aq) reaction from reaction 3? Cl- ion (2 pts) is (or Cl - acceptable; Cl 2 not acceptable = 0 point total) critical for formation of the white precipitate and NOT responsible for the loss of blue color (1 pt) ; or Critical for formation of the white ppt = 1 pt; NOT responsible for the loss of blue color = 0.5 pt ) - 1 pt if another conclusion stated (3 points) C. Analyze test reaction 1 to complete the table below. Do not record spectators for product species. Reaction Oxidizing Agent Reducing Agent Net Reaction Product species CuBr2 SnCl2 1 Sn4+ Cu+ (or Cu2+) (or Sn2+) (or CuCl) *1 pt for net product only or 1 point 1 point zero each (2 points) D. From reactions 1 and 4 together, you can determine the relative strengths of Cu 2+ vs. Sn 4+ and Cd 2+ vs. Cu 2+ as oxidizing agents. Complete the tables below. Oxidizing agent strength of Cu 2+ vs. Sn 4+ : Strongest Weakest Oxidizing Agent Cu 2+ > Sn4+ 2 points ; -1pt if charge missing. Oxidizing agent strength of Cd 2+ vs. Cu 2+ : Strongest Weakest Oxidizing Agent Cd 2+ > Cu 2+ 2 points; - 1 pt. if charge missing.
Name: Final Grading key Page 9/9 Question 6 (10 points). This questions deals with an investigation of redox reactions. Information: Cl 2, Br 2, Fe 3+, I 2, Sn 4+, and K + are oxidizing agents (giving Cl -, Br -, Fe 2+, I -, Sn 2+, and K). Best oxidizing agent Cl 2 > Br 2 > Fe 3+ > I 2 > Sn 4+ > K + Worst oxidizing agent A. You add a few crystals of SnCl 2 to a test tube (A) with water. The crystals dissolve. Complete a balanced equation for the dissociation of SnCl 2(s) in water. Balanced equation for the dissociation of SnCl 2 into ions when placed in water: SnCl 2(s) Sn2+(aq) + 2Cl-(aq) Water +3 points for correct product species including charges or zero; -1 point if (aq) not indicated; -1 point if equation not balanced. 3 points B. You now mix the contents of test tube A with bromine water and add hexane to the resulting mixture. Based on the oxidizing agent strengths (above), determine if a redox reaction occurs. If reaction occurs assume that complete reaction occurs. Identify (circle) any species present in the hexane phase or indicate (circle) NONE. NONE Cl 2 Cl - Br 2 Br - Sn 4+ Sn 2+ C. Based solely on the relative oxidizing agent strengths indicated above, which reactions (below), if any, should occur? Put an X in the appropriate box. Will Occur Br 2 + 2Fe 2+ 2 Br - + 2Fe 3+ Won't occur Sn 4+ + 2I - Sn 2+ + I 2 BONUS QUESTION (3 points) 2 points or zero each = 4 points This is a continuation of question 5. One additional observation is recorded: CuBr 2 (aq) + SnCl 4 (aq) no reaction. blue Identify the white precipitate formed in reaction one (CuBr 2 (aq) + SnCl 2 (aq)). The white precipitate formed in the reaction between CuBr 2 (aq) + SnCl 2 (aq) is: CuCl.