Chem 255 Exam 2 May 9, 2023 PDF
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2023
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This is a past Chemistry exam for Chem 255, given on May 9, 2023. The exam focuses on electrochemistry, and atomic emission spectroscopy. The document contains multiple choice questions and calculations.
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Chem 255 Name: __________________________________ Exam 2 May 9, 2023 Show all work for which you want to receive credit. Report all your answers to calculations using the correct number of significant figures and the proper units. Use of a scientific...
Chem 255 Name: __________________________________ Exam 2 May 9, 2023 Show all work for which you want to receive credit. Report all your answers to calculations using the correct number of significant figures and the proper units. Use of a scientific calculator is permitted. Useful Information: DG = -nFE = -RT ln Keq E = E + - E- A = ebC t r = t r - tm ' A [ A ]2 t -t V o E=E - 0.05916V logQ N = g e -DE kT * * K= 2 » k = r m =K S A 1 [ A ]1 tm VM n N0 g0 æ n V1 ö 𝑡𝑜𝑡𝑎𝑙 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 2 q= q =ç n V1 ÷ 𝐷= 𝑡𝑜𝑡𝑎𝑙 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 1 𝐴𝐴 𝐴𝐵 KV2 + V1 è KV2 + V1 ø =𝐹 𝐶𝐴 𝐶𝐵 16t 2 5.55t 2 L 𝐾 t 'r2 k2 N = 2r = 2 r H= 𝐷= 𝛼0 𝑜𝑟 𝐷 = 𝐾𝛼0 a= = w w1/2 N org to aqu or aqu to org t ' k1 r1 N æ a - 1ö æç k'2 ö÷ R = Dt r H» A+ B + Cux k=1.38x10-23 J/K R= R=8.314 J/Kmole 4 è a ø è1 + k av ø wav ux ' F=96485 C/mole e- Al1 e yl1b e xl 1b Al1 A e yl 2 b e b Al 2 [ X] = l 2 [Y] = xl 2 e xl1b e yl 1b e xl1b e yl 1b e xl 2 b e yl 2 b e xl 2 b e y l2b Reduction Reaction E0(V) Au+ + e- Au(s) 1.69 Ni2+ + 2 e- Ni(s) -0.25 Cl2 + 2e- 2 Cl- 1.36 Cd2+ + 2 e- Cd(s) -0.40 Ag+ + e- Ag(s) 0.80 Fe2+ + 2 e- Fe(s) -0.44 I2(aq) + 2e- 2 I- 0.62 Cr3+ + 3 e- Cr(s) -0.74 Cu2+ + 2 e- Cu(s) 0.34 Zn2+ + 2 e- Zn(s) -0.76 Sn4+ + 2 e- Sn2+ 0.14 Mn2+ + 2 e- Mn(s) -1.18 2H+ + 2 e- H2(g) 0.00 Al3+ + 3 e- Al(s) -1.68 Pb2+ + 2 e- Pb(s) -0.13 Mg2+ + 2 e- Mg(s) -2.36 Multiple Choice (4 points each; 32 total) Circle your choices. There is only one correct answer for each question. 1. Which of the following statements are true about Atomic Emission Spectroscopy? a. Signal results from electrons moving from higher to lower energy levels, so temperature of the flame must be stable. b. Signal results from electrons moving from higher to lower energy levels, so temperature of the flame need not be stable. c. Signal results from electrons moving from lower to higher energy levels, so temperature of the flame must be stable. d. Signal results from electrons moving from lower to higher energy levels, so temperature of the flame need not be stable. 2. The following data were obtained when a Ca2+ ion-selective electrode was immersed in standard solutions containing Ca2+. The least squares regression information is provided on the graph. A 25.00 mL unknown Ca2+ solution was diluted to Calcium Ion Selective Electrode Calibration Curve 50.00 mL. The resulting solution was 60.0 analyzed using the calibrated Ca ion- 2+ 40.0 selective electrode and the reading was 20.0 E = -28.14±0.09(-log[Ca ]) + 51.1±0.2 2+ -22.5 mV. What is the [Ca ] in the 2+ 0.0 E (mV) original unknown sample? -20.0 -40.0 a. 2.42 x 10-3 M -60.0 b. 4.85 x 10-3 M -80.0 c. 2.62 M -100.0 0.00 1.00 2.00 3.00 4.00 5.00 6.00 d. 5.23 M -log[Ca2+] 3. In which order will hexane, benzene, and dichlorobenzene elute from a normal phase liquid chromatography column? a. dichlorobenzene, benzene, hexane b. benzene, hexane and dichlorobenzene c. hexane, benzene, dichlorobenzene d. dichlorobenzene, hexane, benzene 4. A certain battery (spontaneous) is represented in line notation as: Pb(s) | Pb2+(aq) || H+(aq) | H2(g) | Pt Which of the following best describes the reactions taking place at the anode and cathode? anode reaction cathode reaction a. Pb(s) → Pb2+(aq) + 2 e− H2(g) → 2H+(aq) + 2 e− b. Pb(s) → Pb2+(aq) + 2 e− 2H+(aq) + 2 e− → H2(g) c. 2H+(aq) + 2 e− → H2(g) Pb(s) → Pb2+(aq) + 2 e− d. H2(g) → 2H+(aq) + 2 e− Pb(s) → Pb2+(aq) + 2 e− 5. This week in Chem 126 lab students are studying the kinetics of the redox reaction between iodide (I −) and peroxodisulfate (S2 O2− 8 ). The products of this reaction are iodine (I2) and sulfate (SO2− 4 ). If you titrated 15 mL of 0.15 M S2 O2− 8 with 0.15 M I , − where do you expect the equivalence point to occur? a. 7.5 mL b. 15.0 mL c. 30.0 mL d. 37.5 mL 6. The silver–silver chloride, sat’d KCl electrode is filled with a solution saturated in both AgCl and KCl. Its potential is +0.197 V with respect to the standard hydrogen electrode (SHE). If an cell potential is 0.344 V with respect to SHE, what is the potential with respect to a silver–silver chloride electrode? (a) -0.541 V (b) -0.147 V (c) 0.147 V (d) 0.541 V 7. The Merck Index gives the following data for the solubility for 1 g of aspirin or salicylic acid. The numbers in this table are the amount of solvent needed to dissolve 1 g of the compound. You have a mixture of equal weights of aspirin and salicylic acid dissolved in water and wish to extract both compounds into an organic solvent. Which organic solvent (ethyl ether or chloroform) would give the most efficient extraction from water and why? Solvent Aspirin (acetylsalicylic acid) Salicylic acid Water, H2O 300 mL 460 mL Chloroform, CCl4 17 mL 42 mL Ethyl Ether, (CH3CH2)2O 12 mL 3 mL (a) Chloroform, because the partition coefficients are larger for both aspirin and salicylic acid than ethyl ether. (b) Ethyl ether, because the partition coefficients are larger for both aspirin and salicylic acid than chloroform. (c) Chloroform, because the difference in the partition coefficients between CCl4/H2O for aspirin and salicylic acid is larger than the partition coefficients for (CH3CH2)2O/H2O. (d) Ethyl ether, because the difference in the partition coefficient between (CH3CH2)2O/H2O for aspirin and salicylic acid is larger than the partition coefficients for CCl4/H2O. 8. One of the mechanisms for band broadening in chromatography is mass transfer. Which of the following best describes this band broadening phenomenon? (a) results from solute molecules taking different paths through the stationary phase (b) results from solute molecules diffusing from regions of higher concentration to regions of lower concentrations (c) results from some solute molecules spending time in the stationary phase while other solute molecules continue to move in the mobile phase (d) is inversely proportional to flow rate 9. (8 pts) Match each chromatography concept in the table with the types of chromatography listed at the top. In some cases more than one type of chromatography could be listed. It is only necessary to include one of the correct terms. Adsorption Chromatography Affinity Chromatography Partition Chromatography High Performance Liquid Chromatography (HPLC) Ion Exchange Chromatography Gas Chromatography (GC) Size Exclusion Chromatography Also known as gel permeation chromatography C-18 columns Separates larger molecules from smaller molecules Solute is first bound to the stationary phase and later eluted by changing the mobile phase Stationary phase is a cross-linked polymer resin with covalently attached ionic functional groups. Used to isolate a single compound from a complex mixture Utilizes isocratic or gradient elution strategies Utilizes open tubular columns 10. (12 pts) Analysis of bronze 37a for copper content was done by electrolysis. In this experiment a 0.1248 gram sample of bronze 37a was digested in aqua regia and quantitatively transferred to a 100.00 mL volumetric flask. A 25.00 mL aliquot of this bronze solution was placed into an electrochemical cell that was used to reduce the copper from Cu2+ to Cu(s). This cell was allowed to run until the copper was all reduced. During this electrolysis the same amount of charge flowed as would flow at a current of 12.5 mA for 1.5 hours. Calculate the percentage of Cu in the bronze sample. The atomic mass of Cu is 63.546 g/mole. 11. (12 pts) You are measuring the concentration of methyl benzoate in a manufacturing plant using gas chromatography. You prepare a sample of butyl benzoate to use as an internal standard. A preliminary run with known concentrations for methyl benzoate (peak A) and butyl benzoate (peak B) is shown at the right. The concentrations and areas for this run are given in the table. methyl benzoate (A) butyl benzoate (B) Concentration (mg/mL) Area Concentration (mg/mL) Area 1.71 346 1.88 402 A sample was prepared by mixing 1.00 mL of the manufacturing plant waste stream with 1.00 mL of a 2.19 mg/mL butyl benzoate solution and diluting to 10.00 mL. Chromatographic analysis gave a peak area of 471 for peak A and 397 for peak B. What is the concentration of methyl benzoate in the plant stream? 12. (10 pts) The graph of charge vs. +q distance from the electrode surface for a particular metal-metal ion solution interface is shown at the right. Charge (C) (a) (5 pts) Sketch a microscopic 0 Distance picture of the charge distribution for this interface. Assume the cations and anions in solution are both -q monovalent. Clearly label your drawing. (b) (5 pts) Indicate on your drawing in (a) the location of the electrical double layer. Describe what the electrical double layer is and why it forms when a metal is placed in solution with a soluble salt. 13. (12 pts) Ion selective electrodes may respond to solutes other than the analyte, if those solutes also interact with the membrane surface. One such ion selective electrode is the potassium electrode, which also responds to H+ concentration. The selectivity coefficient, 𝐾K+,H+, is a number that determines the magnitude of the effect of the presence of the interfering ion. For a K+ ion-selective electrode 𝐾K+,H+, = 0.00432. The equation for the measured potential is modified to account for this interference as follows: 𝐸 = 𝑐𝑜𝑛𝑠𝑡 + 0.05916 𝑉 ∗ 𝑙𝑜𝑔([𝐾 + ] + 𝐾𝐾+ ,𝐻 + [𝐻+ ]) When the electrode is placed into a 6.53×10−5 M K+ solution at pH 6.61, the potential is −0.168 V vs. SCE. If the pH of the solution is lowered to 1.22 and the ionic strength is kept constant, what is the potential of the electrode? 14. (14 pts) The chromatogram below was obtained using a reverse-phase HPLC column with a mobile phase of 50% methanol:50% water. The sample contains seven nonpolar compounds (#2 and #3 co-elute). 1.6 7 5 1.4 1.2 unretained solvent 1 6 2&3 absorbance 0.8 1 0.6 0.4 0.2 4 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 -0.2 Time (minutes) (a) (5 pts) Using this chromatogram, determine the retention factors k1 and k5, for analytes 1 & 5. Estimate any quantities needed from the chromatogram provided. (b) (5 pts) Determine the resolution of analytes 6 & 7. Estimate any quantities needed from the chromatogram provided. (c) (4 pts) What are some specific things you can do to improve the separation of the peaks in this chromatogram? Your Pledge I pledge my honor that during this examination I have neither given nor received assistance not explicitly approved by the professor, and that I have seen no dishonest work. Signed: ________________________________________ I have purposefully NOT signed the pledge (check only if appropriate) 1-8 _____ (32) Exams (200): 9 _____ (8) Achieve (121): 10 _____ (12) DAIT (90): 11 _____ (12) 12 _____ (10) 13 _____ (12) 14 _____ (14) Exam Total: