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Course: CHEM 122, Winter 2007
School: Ohio State
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pts) 1. (6 In the following reaction 2 MnO4-(aq) + 5 H2O2(R) + 6H+(aq) 6 2 Mn2+(aq) + 8 H2O(R) + 5 O2(g) the concentration of MnO4- dropped from 0.5684 M to 0.5522 M in 2.23 minutes. What was the average rate of formation of O2, in M/min, during this time interval? a) 2.14 x 10-3 b) 1.82 x 10-2 c) 2.09 x 10-2 d) 6.37 x 10-2 e) 8.50 x 10-1 2. (5 pts) Concentration and time data is obtained for a reaction. Which...

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pts) 1. (6 In the following reaction 2 MnO4-(aq) + 5 H2O2(R) + 6H+(aq) 6 2 Mn2+(aq) + 8 H2O(R) + 5 O2(g) the concentration of MnO4- dropped from 0.5684 M to 0.5522 M in 2.23 minutes. What was the average rate of formation of O2, in M/min, during this time interval? a) 2.14 x 10-3 b) 1.82 x 10-2 c) 2.09 x 10-2 d) 6.37 x 10-2 e) 8.50 x 10-1 2. (5 pts) Concentration and time data is obtained for a reaction. Which of the following statements is incorrect about the reaction? a) If Rn[A]t is plotted against time and the graph gives a straight line, the reaction was first order with respect to A. b) The half-life of a first order reaction is a constant. c) When 1/[A]t is graphed against time, the slope of the line is equal to the rate constant for a second order reaction. d) The half-life of a second order reaction decreases as the concentration of the reactants decrease. e) The rate of reaction at a specific time can be determined from the slope of the line tangent to the curve at that time on a graph of concentration versus time. 3. (5 pts) The half-life of of U-235. a) 4.7 x 108 yr-1 b) 8.0 x 104 yr-1 c) 19 yr-1 d) 6.5 x 10-5 yr-1 e) 9.8 x 10-10 yr-1 , a naturally radioactive isotope is 7.1 x 108 years. Radioactive 235 92 U decay is a first-order kinetic reaction. Calculate the rate constant for the radioactive decay (1) The initial rate of a hypothetical reaction 3A + B 6 2C was measured at different concentrations of A and B. The results are summarized below: Experiment Number 1 2 3 4 Initial Concentration [A] 1.20 M 0.40 M 1.20 M 0.72 M Initial Concentration [B] 0.50 M 0.50 M 1.00 M 0.48 M Initial Rate of Reaction M/s 1.08 x 10-1 1.2 x 10-2 1.08 x 10-1 ? Questions 4 - 6 refer to this reaction. 4. (6 pts) The rate law for the reaction is a) rate = k[A] b) rate = k[A][B]2 c) rate = k[A][B] d) rate = k [A]2[B] e) rate = k[A]2 5. (5 pts) The numerical value of the rate constant for this reaction is a) 0.0750 b) 0.0905 c) 0.150 d) 0.384 e) 0.628 6. (5 pts) The initial rate of the reaction for experiment 4 is a) 7.4 x 10-3 M/s b) 1.7 x 10-2 M/s c) 3.9 x 10-2 M/s d) 4.3 x 10-1 M/s e) 8.5 x 10-1 M/s (2) 7. (6 pts) A reaction is second order in C. The rate constant is determined to be 2.35 x 10-3 M-1s-1. The reaction flask initially contains 0.530 M of C. How much C will remain after 12.0 minutes? a) 0.360 M b) 0.279 M c) 0.182 M d) 0.149 M e) 0.114 M 8. (5 pts) The rate constant of a reaction which is first order in A is 7.8 x 10-2 s-1. What is the initial rate of the reaction when the initial concentration of A is 0.0030 M? a) 2.3 x 10-4 M/s b) 7.9 x 10-4 M/s c) 9.1 x 10-4 M/s d) 2.6 x 101 M/s e) 8.5 x 101 M/s 9. (6 pts) A reaction is first order with respect to A. The concentration of A was initially 3.50 M and after 764 seconds [A] = 0.826 M. Calculate the concentration of [A] after 1080 seconds. a) 0.810 M b) 0.703 M c) 0.614 M d) 0.455 M e) 0.381 M 10. (5 pts) The reaction of A is a second order reaction. 10.0 g of A is placed in container I while 50.0 g of A is placed in container II. Both containers are kept at the same temperature and pressure. The rate constant for the reaction is 7.6 x 10-2 M-1s-1. Compare the time required for half of A to decompose in each container. I II a) Container I requires the same amount of time as container II. b) Container II requires five times as much time as container I. c) Container I requires five times as much time as container II. d) Select this answer if the correct answer is not given above. 11. (3) (6 pts) A reaction obeys the rate law expression: Rate = k[B][C]2 at 47.3C. The rate constant was determined to be 2.8 x 10-3 M-2s-1 at that temperature. The activation energy for the reaction is 95.0 kJ/mole. Calculate the rate constant at 61.9C. a) 8.7 x 10-3 M-2s-1 b) 7.2 x 10-2 M-2s-1 c) 1.3 x 10-2 M-2s-1 d) 9.9 x 10-2 M-2s-1 e) 1.8 x 10-1 M-2s-1 12. (6 pts) Examine the following potential energy diagrams of two reactions. Which of the following statements is incorrect? a) The forward reaction of Reaction 1 has a smaller rate constant than the forward reaction of Reaction 2. b) The activation energy of Reaction 1 in the forward direction is greater than the activation energy of the reverse direction of Reaction 2. c) The rate of the forward reaction of Reaction 1 is greater than the rate of the reverse reaction of Reaction 1. d) The activated complex for the forward reaction of Reaction 2 is the same as the activated complex for the reverse reaction of Reaction 2. e) Ea for the reverse reaction of Reaction 1 is 93 kJ/mole. (4) The following mechanism is proposed for a reaction. Step 1: Step 2: Step 3: Br2(g) Br(g) + 2 Br(g) kf kr fast, equilibrium CHBr3(g) k2 HBr(g) + CBr3(g) slow fast Br(g) + CBr3(g) CBr4(g) k3 Questions 13 and 14 refer to this reaction and mechanism. 13. (5 pts) Which of the following statements is incorrect about the reaction or mechanism above? a) The rate determining step is a bimolecular elementary step. b) The balanced equation for the reaction is Br2(g) + CHBr3(g) 6 HBr(g) + CBr4(g). c) Br(g) and CBr3(g) are intermediates. The d) forward reaction of step 1 is a unimolecular reaction. e) k2 is greater than k3. 14. (6 pts) The rate law predicted by the mechanism is a) rate = k[Br2]1/2[CHBr3] b) rate = k[Br2][CHBr3]2 c) rate = k[Br]2[CHBr3] d) rate = k[HBr][CBr4]/[Br]2 e) rate = k[2 Br]2/[CBr4] 15. (5 pts) In the following mechanism B + AY2 BY + AY BY + AY 6 BY2 + A A + Y2 6 AY + Y AY + Y 6 AY2 fast slow fast fast ____________ is an intermediate and ___________ is a catalyst. a) AY, B b) Y2, BY c) BY, A d) Y, AY2 e) BY2, AY 16. (6 pts) Which of the following mechanisms is an acceptable mechanism for the reaction O3(g) + NO(g) 6 NO2(g) + O2(g) (5) which was determined to be first order with respect to both O3 and to NO? a) O3 6 O2 + O O + NO 6 NO2 b) NO 6 N + O O + O3 6 2 O2 O2 + N 6 NO2 c) 2 O2 + NO 6 NO2 + O3 2 O3 6 3 O2 d) NO + O3 6 NO3 + O NO3 + O 6 NO2 + O2 e) NO N + O O + O3 6 2 O2 O2 + N 6 NO2 slow fast slow fast fast slow fast slow fast fast slow fast 17. (6 pts) Hydrogen and nitrogen are placed in an empty 5.00 L vessel at 500C. When equilibrium is established, 3.01 mole of N2, 2.10 mole of H2 and 0.565 mole of NH3 is present. Calculate Kc for the reaction at 500C. N2(g) + 3 H2(g) 2 NH3(g) a) 1.13 x 10-2 b) 0.286 c) 0.509 d) 17.2 e) 63.8 (6) 18. (5 pts) Which of the following reactions is not correctly paired with its equilibrium constant expression? a) CO2(g) + H2(g) CO(g) + H2O(g) b) (AlF6)3-(aq) Al3+(aq) + 6 F-(aq) c) 2 BrF5(g) Br2(g) + 5 F2(g) Kc = [CO][H2O] [CO2][H2] Kc = [Al3+][F-]6 [AlF63-] Kp = (PBr2)(PF2)5 (P BrF5)2 Kc = [Fe2S3] [Fe3+]2[S2-]3 Kp = (PCO2)2 (PCO)2(PO2) d) 2 Fe3+(aq) + 3 S2-(aq) Fe2S3(s) e) 2 CO(g) + O2(g) 2 CO2(g) 19. (6 pts) A mixture of 0.100 moles of N2 and 0.150 mole of H2 is placed in a 1.00 L vessel. At equilibrium the concentration of [N2] = 0.070 M. Calculate Kc for the reaction at this temperature. N2(g) + 3 H2(g) 2 NH3(g) a) 4.80 x 10-6 b) 2.67 x 10-3 c) 2.38 x 102 d) 5.64 x 103 e) 1.99 x 104 20. (6 pts) The value of Kc = 1.55 x 103 for the reaction at 17C 2 H2S(g) + 3 O2(g) 2 H2O(g) + 2 SO2(g) What is the value of Kp? a) 5.34 b) 65.1 c) 4.70 x 102 d) 8.12 x 103 e) 3.69 x 104 (7) 21. (6 pts) At 1133 K, Kp = 4.32 x 10-3 for the reaction CO(g) + 3 H2(g) CH4(g) + H2O(g) What can you conclude about a reaction flask containing PCO = 1.58 atm, PH2 = 1.70 atm, PCH4 = 0.736 atm, and PH2O = 0.142 atm? a) The system will react to form more CH4 and H2O. b) The system will react to form more CO and H2. c) The system is at equilibrium. d) It is impossible to predict which direction the reaction will move. 22. (6 pts) A 10.0 L flask is filled with 5.00 moles each of H2 and I2 at 448C. Kc is equal to 50.5 at 448C for the reaction H2(g) + I2(g) 2 HI(g) What is the concentration of HI at equilibrium? a) 0.144 M b) 0.279 M c) 0.503 M d) 0.625 M e) 0.780 M 23. (6 pts) Calculate the value of Kc for the decomposition of NH4Cl(s) at some temperature. Solid NH4Cl was placed in a 2.00 L flask and decomposed. At equilibrium the flask contained 0.175 moles of NH3. NH4Cl(s) NH3(g) + HCl(g) a) 1.61 x 10-2 b) 7.66 x 10-3 c) 5.07 x 10-4 d) 2.53 x 10-5 e) 8.41 x 10-6 (8) 24. (6 pts) Which of the following stresses will shift the equilibrium to form more products? K2PtCl4(aq) + 2 NH3(g) Pt(NH3)2Cl2(s) + 2 KCl(aq) a) remove some NH3 b) add KCl solution )H<0 c) remove some Pt(NH3)2Cl2 d) cool the reaction e) decrease the pressure 25. (6 pts) Which of the following systems favor reactant formation when both the temperature is lowered and the pressure is raised? A. B. C. D. 2 Cu(NO3)2(s) 2 CuO(s) + 4 NO2(g) + O2(g) Na2CO3(s) + SO2(g) + O2(g) Na2SO4(s) + CO2(g) 3 CuO(s) + 2 NH3(g) 3 Cu(s) + 3 H2O(g) + N2(g) N2O4(g) 2 NO2(g) )H = +439 kJ )H = -350.2 kJ )H = -1099 kJ )H = +60.2 kJ a) both A and D b) both B and C c) C only d) B only e) A only 26. (5 pts) The strongest acid in the following system is ___________. H2PO4- + H2SO3 HSO3- + H3PO4 a) b) c) d) HSO3H3PO4 H2SO3 H2PO4statements is incorrect? 27. (5 pts) Which of the following a) The conjugate acid of SO32- is HSO3-. b) The stronger the base, the weaker its conjugate acid. c) In an acid-base system, the equilibrium lies to the side with the weaker acid and base. d) The conjugate base of tartaric acid, C10H18O6, is C10H17O6-1. e) The conjugate acid always has one less proton than its base. 28. (9) (6 pts) Calculate the pH of a solution made by dissolving 0.1155 g Ca(OH)2 (f. wt. = 74.09 amu) in enough water to make 500.0 mL of solution. a) 11.7949 b) 11.1762 c) 9.9088 d) 9.3464 e) 7.9665 29. (6 pts) The pH of a banana is 4.6. Calculate the hydroxide ion concentration. a) 8 x 10-6 M b) 5 x 10-7 M c) 2 x 10-8 M d) 7 x 10-9 M e) 4 x 10-10 M 30. (6 pts) Calculate the pH of a 0.136 M nicotinic acid, HC6H4O2N (mol. wt. = 123.111 amu). Ka for nicotinic acid is 1.4 x 10-5. a) 1.67 b) 2.86 c) 4.52 d) 6.26 e) 7.06 31. (6 pts) Calculate Ka for uric acid, HC5H3O3N4 (mol. wt. = 168.112 amu). A 0.500 M solution of the acid has a pH of 3.06 at 25C. a) 1.5 x 10-6 b) 8.2 x 10-5 c) 2.0 x 10-5 d) 7.4 x 10-4 e) 1.3 x 10-4 (10)
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Ohio State - CHEM - 122
THE MEAN WAS 115.00 (65.71%) THE STANDARD DEVIATION WAS 31.21, 453 students scored THE POSSIBLE RANGE WAS FROM 0.00 TO 175.00 THE ACTUAL RANGE WAS FROM 24.00 TO 175.00 (13.7% TO 100.0%)High: 175Low: 24DISTRIBUTION OF SCORES 0.00 - 9.99 0 10.00 -
Ohio State - CHEM - 122
1.The volume of one cylinder in an automobile is 0.44 L. The cylinder is filled with a gasoline and air mixture to a total pressure of 1.0 atm and then is compressed to 0.073 L prior to ignition. What pressure must be applied to produce this compre
Ohio State - CHEM - 122
Ohio State - CHEM - 122
Pop Quiz Number 0Chemistry 122, Dr. MathewsFirst Week, WI, 2005Name_ Lab Instructor:_ 1. The average speed of a nitrogen molecule in air at 25 oC is 515 m/s. Convert this speed to miles per hour. Recall that 1 hr = 60 min, 1 mi = 1.6093 km, 1 k
Ohio State - CHEM - 122
MQ-1 122, WI 07:THE MEAN WASTHE MODE WAS 95.00 WITH 10 STUDENTS. THE MEDIAN WAS 112.50 THE STANDARD DEVIATION WAS 29.35112.00 (64.00%) High: 175 Low 38N = 472THE POSSIBLE RANGE WAS FROM 0.00 TO 175.00 THE ACTUAL RANGE WAS FROM 38.00 TO 175.0
Ohio State - CHEM - 122
Ohio State - CHEM - 122
Some Useful informationgas constant R = 0.08206 L-atm/mol-K = 8.314 J/mol-K for water at 25 oC Kw = 1.00 x 10-14 Arrhenius equation,ln k = ln A -Ea RTThe quadratic equation has the solutionsax 2 + bx + c = 0- b b 2 - 4ac x= 2aFrom the XR
Ohio State - CHEM - 122
Some Useful Information Avogadro's number = 6.022 x 1023 gas constant, R = 8.314 J/mol-K R = 0.08206 L-atm/mol-K van der Waals Equation: n 2a P + 2 (V - nb) = nRT V Clausius Clapeyron Equationln P = -H vap 1 +C R TorH vap 1 P 1
Ohio State - CHEM - 122
Dr. C. Weldon Mathews Chem 122 Office: 0042 Evans Lab Telephone: 292-1574 email: mathews.6@osu.edu course web site: http:/www.chemistry.ohio-state.edu/~mathews/chem122wi07/ Office hours: TR 12:30 - 2:00 pm TR 4:00 - 5:00 pm or by appointmentChapter
Ohio State - CHEM - 122
Week 2 lectures-tentative10.7 10.8 Kinetic-Molecular Theory 420 Application to the Gas Laws Molecular Effusion and Diffusion 423 Graham's Law of Effusion Diffusion and Mean Free Path Real Gases: Deviations from Ideal Behavior The van der Waals Equat
Ohio State - CHEM - 122
Week 2 lectures-tentative10.7 10.8 Kinetic-Molecular Theory 420 Application to the Gas Laws Molecular Effusion and Diffusion 423 Graham's Law of Effusion Diffusion and Mean Free Path Real Gases: Deviations from Ideal Behavior The van der Waals Equat
Ohio State - CHEM - 122
The quiz next week will be given during your recitation. It will cover material starting with section 10.7.First Midquarter Exam Monday, Jan 29 6:30-7:48 pm Rooms TBA Chaps 10, 11, 13Dr. Zellmer's web site has very useful information on BP data an
Ohio State - CHEM - 122
First Mid-quarter Examination MQ-1 on Monday, Jan. 29 at 6:30 pm Covering Chapters 10, 11, and 13 room TBAFor students with a scheduled-class conflict with the first Mid-Quarter Exam:(1) Send me an email today with a copy of your class schedule, a
Ohio State - CHEM - 122
First Mid-quarter ExaminationMQ-1 on Monday, Jan. 29at 6:30 pm Covering Chapters 10, 11, and 13 room TBAFor students with a scheduled-class conflict with the first Mid-Quarter Exam:(1) Send me an email today with a copy of your class schedule,
Ohio State - CHEM - 125
Ohio State - CHEM - 122
Demo today ?1Colloids Hydrophilic and Hydrophobic Colloids Focus on colloids in water. &quot;Water loving&quot; colloids: hydrophilic. &quot;Water hating&quot; colloids: hydrophobic. Molecules arrange themselves so that hydrophobic portions are oriented towards
Ohio State - CHEM - 125
Ohio State - CHEM - 122
Demo today ?Colloids Hydrophilic and Hydrophobic Colloids Focus on colloids in water. &quot;Water loving&quot; colloids: hydrophilic. &quot;Water hating&quot; colloids: hydrophobic. Molecules arrange themselves so that hydrophobic portions are oriented towards
Ohio State - CHEM - 122
Quiz #4 Next Week (Feb 12-16)Covers sections 14.5-14.7During lectures next week, we will complete ALL OF Chapter 15 and parts of Chapter 16! The Second Midquarter Exam will cover Chapters 14, 15, and 16.1-16.3Week six, a continuation ofChapter
Ohio State - CHEM - 122
Quiz #4 Next Week (Feb 12-16)Covers sections 14.5-14.7During lectures next week, we will complete ALL OF Chapter 15 and parts of Chapter 16! The Second Midquarter Exam will cover Chapters 14, 15, and 16.1-16.3Week six, a continuation ofChapter
Ohio State - CHEM - 122
LAB INSTRUCTORLOCATION170 - 209 W.18TH AVE. 170 - 209 W.18TH AVE. 250 KNOWLTON HALL 250 KNOWLTON HALL 250 KNOWLTON HALL 250 KNOWLTON HALL 103 KOTTMAN HALL 103 KOTTMAN HALL 103 KOTTMAN HALL 103 KOTTMAN HALL 1180 POSTLE HALL 1180 POSTLE HALL 1180 PO
Ohio State - CHEM - 125
Ohio State - CHEM - 122
16.1 16.2Acids and Bases: A Brief Review Bronsted-Lowry Acids and BasesH+ in water, Proton-transfer, Conjugate acids and bases, Relative strengthsQuiz number 5 will be given in recitation next week, Feb 26-Mar 2 on the first part of Chapter 16,
Ohio State - CHEM - 122
Quiz number 5 will be given in recitation next week, Feb 26-Mar 2 on the first part of Chapter 16, to be covered in lectures this week.16.1 16.216.3 16.4Acids and Bases: A Brief Review Bronsted-Lowry Acids and BasesH+ in water, Proton-transfer,
Ohio State - CHEM - 122
Makeup Midquarter Exams Wed., Mar 7 5:30-7:20 pm 1000 McPherson You MUST Sign up in 100 CE Please do so as soon as possible, but no later than Friday, Mar 2. Final Exams for Chem 122 - Mathews 2:30 Class: Thursday, Mar. 15 at 1:30 in (TBA) 6:30 Class
Ohio State - CHEM - 122
Makeup Midquarter Exams Wed., Mar 7 5:30-7:30 pm 1000 McPherson You MUST have Signed up in 100 CEChapter 17 17.1 17.2Additional Aspects of Aqueous Equilibria17.3Final Exams for Chem 122 - Mathews 2:30 Class: Thursday, Mar. 15 at 1:30 (TBA) 6:
Ohio State - CHEM - 211
Analytical Chemistry Chemistry 211 Quantitative AnalysisLecturers: Dr. Allen (first-half) (firstDr. Gustafson (second-half) (secondDr. Allen's Office: Room 3105 in Newman &amp; Wolfrom (100 W. 18th Ave) My website will be used for distribution of some m
Ohio State - CHEM - 211
Practice Midterm I, Problem #7: one way to do this problem is. AgsS=2Ag+ + S2S2- + H2O= HSHS- + H2O= H2S + OHH2O= H+ + OHSolve/Write out all Ks: Ksp= 6 x 10-50 Kw=10-14 Kb1=8.3 Kb2=1.8 x 10-7 MB: [Ag+] = [S2-] + [HS-] + [H2S] CB: [Ag+] + [H+] = 2 [S
Ohio State - CHEM - 211
QUANTITATIVE CHEMICAL ANALYSIS FOR BIOLOGICAL AND MEDICINAL SCIENCES Chemistry 211 Summer Quarter 2005 Lecture Syllabus Lecturers: Heather C. Allen, 3105 NW, allen.697@osu.edu (June 20 to July 22) Terry L. Gustafson, 0103 NW, gustafson.5@osu.edu (Jul
Ohio State - CHEM - 211
KEY (tables were provided; Harvey tables) Partial Credit was given on all problems if deemed appropriate. Sig figs were checked, typically worth 1-2 points depending on the problem. As stated in class many times, if one did not state the assumptions
Ohio State - CHEM - 211
Name_Practice Midterm IChemistry 211 1. (30 points) An adrenal cell has about 2.5 x 104 vesicles that contain epinephrine. a. An entire cell has ~ 150 fmol (f = femto = 10-15) of epinephrine. How many amol (a = atto = 10-18) of epinephrine are in
Ohio State - CHEM - 211
Chemistry 211: Quiz worth 20 points total KEY - Quiz 1 (Take Home); June 29th, 2005 Due July 1st, 2005, before 8:30am (BEFORE class)A few notes: You must list the steps that you went through to solve each problem. (be sure to conduct all necessary s
Ohio State - CHEM - 211
Name_ Coworkers_ Chemistry 211 Quiz 2 (Take Home); July 13th, 2005 Due July 15th, 2005, before 8:30am (BEFORE class) 1. Calculate the pH of a 0.011030 M NaCH3COO (sodium acetate) solution. The pKa of acetic acid (CH3COOH) is 4.7570. Report the pH wit
Ohio State - CHEM - 211
Copyright 2000 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Copyright 2000 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Example of a Gaussian distributionCopyright 2000 Th
Ohio State - CHEM - 211
PolyproticsAcid speciation as a function of pHsome terms defined&quot;TOTA&quot; the total amount of the acid ATOTAHAA&quot; i&quot; the fraction of CHA (TOTA) that has lost i protonsHA0&quot;0&quot;: zero protons lostTOTA A TOTA1&quot;1&quot;: one proton los
Ohio State - CHEM - 211
Chemical EquilibriumTo Solve Multiple Equilibria Problems: 1. Write down the relevant chemical reactions 2. Write the &quot;charge balance&quot; equation a. Magnitude of charge = # in front of the species 3. Write the &quot;mass balance&quot; equation (1 or more) a. If
Ohio State - CHEM - 125
Ohio State - CHEM - 125
Ohio State - CHEM - 125
Ohio State - CHEM - 125
Ohio State - CHEM - 125
Ohio State - CHEM - 125
Ohio State - CHEM - 125
Ohio State - CHEM - 125
Ohio State - CHEM - 125
Ohio State - CHEM - 125
Ohio State - CHEM - 125
Ohio State - CHEM - 125
Ohio State - CHEM - 125
I will make the following suggestions for how you can do well in this course. 1) will be covering that week before coming to lecture. it to get an idea of what we will be covering. try to understand it at that point but it will help you follow what I
Ohio State - CHEM - 221
Constants R = 8.314 J-K-1 -mol-1 N0 = 6.022 1023 mol-1 Related Information Quadratic Equation: if ax2 + bx + c = 0, then x= -b kB = 1.381 10-23 J-K-1 h = 6.626 10-34 J-s e = 1.1602 10-19 C c = 2.998 108 m-s-1 me = 9.11 10-31 kg -12 C2 N-1 m-2
Ohio State - CHEM - 221
QUANTITATIVE CHEMICAL ANALYSIS (revised - March 28, 2008)Chemistry 221 MWF, 8:30 AM, 1015 McPherson Laboratory Spring Quarter 2008 Instructor: Philip Grandinetti Office: 0044B McPherson Lab Phone: (614) 292-6818 Contact: www.grandinetti.org/Contact
Ohio State - CHEM - 221
Problem Set Chemistry 2211. Define absorption, absorbance, percent transmission, and transmittance. 2. Define absorptivity and molar absorptivity. 3. Describe radiation sources and detectors for the ultraviolet, visible, and infrared regions of the
Ohio State - CHEM - 221
Ohio State - CHEM - 520
Equationsax2 + bx + c = 0 P V = nRT PT = P+iandx=-b b2 - 4ac 2aQuadratic Equation Ideal Gas Law Dalton's law of partial pressuresPi (V - nb) = nRTan2 V2van der Waals equationZ =1+B C D + 2 + 3 + Vm Vm VmVirial Equation Viri
Ohio State - CHEM - 520
Chemistry 520 Midterm Exam, Winter Quarter 7:30 - 9:00 PM, 7 February 2008NAME Useful Constants R = 8.314 J-K-1 -mol-1 N0 = 6.022 1023 mol-1 1 J = 1 kg m2 s-2 kB = 1.381 10-23 J-K-1 -12 C2 N-1 m-2 0 = 8.854 10 1 amu = 1.661 10-27 kg e = 1.1602
Ohio State - CHEM - 520
Ohio State - CHEM - 520
Chemistry 520 Winter Quarter 2008 Take Home Quiz 1 This quiz is due no later than 4 pm on Wednesday, Jan. 16, 2008 in room 0048 McPherson lab. Submit your answers on single sided paper. 1. A system receives 50 J of electrical energy and delivers 150
Ohio State - CHEM - 520
Chemistry 520 Winter Quarter 2008 Take Home Quiz 2 This quiz is due no later than 4 pm on Friday, Feb. 1, 2008 in room 0048 McPherson lab. Write down first the mathematical equation with symbols and leave the numerical substitution at the end. Submit
Ohio State - CHEM - 520
Chemistry 520 Winter Quarter 2008 Take Home Quiz 3 This quiz is due no later than 4 pm on Friday, Feb. 18, 2008 in room 0048 McPherson lab. Write down first the mathematical equation with symbols and leave the numerical substitution at the end. Submi
Ohio State - CHEM - 520
Chemistry 520 Winter Quarter 2008 Take Home Quiz 4 This quiz is due no later than 4 pm on Wednesday, Feb. 27, 2008 in room 0048 McPherson lab. Write down first the mathematical equation with symbols and leave the numerical substitution at the end. Su
Ohio State - CHEM - 520
Chemistry 520 Winter Quarter 2008 Take Home Quiz 5 This quiz is due no later than 4 pm on Wednesday, Mar. 5, 2008 in room 0048 McPherson lab. Write down first the mathematical equation with symbols and leave the numerical substitution at the end. Sub