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Course: CHEM 1420, Spring 2010School: North Texas
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North Texas - CHEM - 1420
Practice Exam 1 Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. B D D E C D C D E10. A 11. B [second sentence of Question should be: To determine the Activation Energy,] 12. A 13. B 14. A 15. B 16. Ask me to draw it in recitation 17. Rate = k[NO][H2]2 - k = 9.6x104 M
North Texas - CHEM - 1420
North Texas - CHEM - 1420
Practice Exam 2Answers1.E - Was on our first test2.B- Was on our first test3.A- Was on our first test4.E- Was on our first test5.B- Was on our first test6.C7.D- Was on our first test8.B9.D- Was on our first test10. Not on either test1
North Texas - CHEM - 1420
North Texas - CHEM - 1420
Practice Exam 3 Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. B D B A B A C B D10. A 11. C 12. E 13. A 14. A 15. B [should read: A Bronsted-Lowry acid is a substance that can donate a proton and . - i.e. not donate a proton donor] 16. 8.0x10-5 17. 3.37 3.4 18. 10.2
North Texas - CHEM - 1420
North Texas - CHEM - 1420
Practice Exam 4Answers1.D2.C3.D - However, you are not responsible for this question4.C5.E6.D7.D8.A9.B10. B - but you are not responsible for entropy of solution of ionic solids11. D12. B13. B - However, you are not responsible for
North Texas - CHEM - 1413
COVALENT BONDINGChapter 8 OutlineProbs:Sect.12, 17, 32, 33, 34, 36, 42, 44, 46, 51+ supplementary questions (attached)Title and CommentsRequired?1.Covalent BondingYES2.Single Covalent Bonds and Lewis StructuresYES3.Multiple Covalent BondsY
North Texas - CHEM - 1413
North Texas - CHEM - 1413
CHEM 1413.001 - Exam 1 Sept. 22, 2010 - Part 1Name_(51)MULTIPLE CHOICE (Circle the ONE correct answer)NOTE: All required Molar Masses and non-metric conversion factors aregiven at the end of the test.1. 25. g/cm3 = _ g/m3 ?(A) 2.5x10-11(C) 2.5x10-
North Texas - CHEM - 1413
North Texas - CHEM - 1413
CHEM 1413.001 - Exam 1 Sept. 21, 2011 - Part 1Constants and Conversion FactorsAvogadro's Number: NA = 6.02x1023 mol-1Conversion Factors:1 cm3 = 1mLMolar Masses: C3H8O3 - 92.PF5 - 126.P - 31.1 lb = 454 gF - 19.1 Qt. = 1.06 LCHEM 1413.001 - Exam
North Texas - CHEM - 1413
CHEM 1413.001 - Exam 2 October 13, 2010 - Part 1Name_(60)MULTIPLE CHOICE (Circle the ONE correct answer)NOTE: All required Molar Masses and non-metric conversion factors aregiven at the end of the test.1. The reaction, 2 Al(s) + 3 O2(g) Al2O3(s), is
North Texas - CHEM - 1413
CHEM 1413.001 - Exam 3 November 3, 2010 - Part 1Name_(27)MULTIPLE CHOICE (Circle the ONE correct answer)NOTE: All required Molar Masses and non-metric conversion factors aregiven at the end of the test.1. For a process in which the internal energy c
North Texas - CHEM - 1413
CHEM 1413.001 - Exam 3 November 2, 2011 - Part 1Molar Masses: C6H14 - 86.H2O - 18.C2H4 - 28.H2 - 2.Li - 6.9NH4Cl - 53.5CH3OH - 32.C2H6 - 30.CHEM 1413.001 - Exam 3 November 2, 2011 - Part 1Name_(27)MULTIPLE CHOICE (Circle the ONE correct answer
North Texas - CHEM - 1413
North Texas - CHEM - 1413
CHEM 1413.001 - Exam 4 November 30, 2011 - Part 1Avogadro's Number: NA = 6.02x1023 mol-1Conversion Factors:1 atm. = 760 torr ,Molar Masses: CO2 - 44.CH4 - 16.C2H6 - 30.SO2 - 64.1CHEM 1413.001 - Exam 4 November 30, 2011 - Part 1Name_(42)MULTIPLE
North Texas - CHEM - 1413
North Texas - CHEM - 1413
CHEM 1413.001 - Final Exam Dec. 13, 2010If you would like to have your grade posted on the course web site, using a fourdigit number, then please sign in the space below and put the number.If you do not wish to have your grade posted, you can still lea
North Texas - CHEM - 1413
CHEM 1413.001 - Final Exam Dec. 14, 2011If you would like to have your grade posted on the course web site, using a fourdigit number, then please sign in the space below and put the number.If you do not wish to have your grade posted, you can still lea
North Texas - CHEM - 1413
CHEM 1413Chapter 6Homework SolutionsTEXTBOOK HOMEWORK29. Answer: 5.00 105 JStrategy and Explanation: Given the mass and initial temperature of a liquid substance, theboiling point of the liquid, the final temperature of the gaseous substance, the sp
North Texas - CHEM - 1413
CHEM 1413Chapter 8Homework SolutionsTEXTBOOK HOMEWORK12. Answer: (a). .l.C.. .Cl. ..O..O.(b) H_H.O.(c)HHBHH+H(d)HPHH. .. Cl ... Cl. . P . .Cl .(e) . ... Cl. .Cl.. ...F . .F ..17. Answer: (a)..F. .CC.
North Texas - CHEM - 1413
CHEM 1413Chapter 10Homework SolutionsTEXTBOOK HOMEWORK17. Answer: CH2Cl2 < Kr < N2 < CH4Strategy and Explanation: Given the formulas of various atoms and molecules and theircommon temperature, put their gases in order of increasing average molecular
North Texas - CHEM - 1413
CHEM 1413Chapter 11Homework SolutionsTEXTBOOK HOMEWORK10. Answer: Reduce the pressureStrategy and Explanation: A liquid can be converted to a vapor without changing thetemperature by reducing the pressure above it. This can be accomplished by puttin
North Texas - CHEM - 1413
North Texas - CHEM - 1413
CHAPTER 7SUPPLEMENTARY HOMEWORK QUESTIONS1.The maximum wavelength of light which can break the CO bond is 112 nm. Calculate the CObond strength, in kJ/mol.2.A lamp rated at 40 W (1 W = 1 J/s) emits blue light at a wavelength of 470 nm. How manyphot
North Texas - CHEM - 1413
CHAPTER 8SUPPLEMENTARY HOMEWORK QUESTIONS1.Without using a table of electronegativies, predict which order is correct for increasing polarity(least polar first)?(A) Si-Cl (B) S-Cl(A) C<D<B<A(D) C<B<D<A2.(C) P-Cl(D) Si-Si(B) D<C<B<A(E) A<B<C<D
North Texas - CHEM - 1413
North Texas - CHEM - 1413
North Texas - CHEM - 1413
ENERGY AND CHEMICAL REACTIONSChapter 6 OutlineProbs:Sect.29, 36, 37, 41, 47, 50, 54, 58, 63, 70, 79+ Supplementary Questions (attached)Title and CommentsRequired?1.The Nature of EnergyYES2.Conservation of EnergyYESIn addition to discussing w
North Texas - CHEM - 1413
ELECTRON CONFIGURATIONS AND THE PERIODIC TABLEChapter 7 OutlineProbs:Sect.8, 23, 30, 38, 45, 47, 55, 57, 62, 65, 73, 88+ supplementary questions (attached)Title and CommentsRequired?1.Electromagnetic Radiation and MatterYES2.Plancks Quantum Th
North Texas - CHEM - 5200
Chapter 2The First Law1DefinitionsSYSTEM AND SURROUNDINGSThe system is the part of the universe we are interested in.The rest of the universe is the surroundings.System2DefinitionsTYPES OF SYSTEMSOpen:Exchange Matter and Energy withsurroundin
North Texas - CHEM - 5200
Chapter 3The Second Law1Spontaneous Processes2Spontaneous ProcessesHot BarCold BarRoom25 oCIce Cube0 oCMid-temperatureBarsRoom24.9 oCPuddle of water24.9 oC3Observations All four processes are spontaneous only in thedirection shown (le
North Texas - CHEM - 5200
Chapter 6Chemical Equilibrium1Spontaneous Chemical ReactionsThe Gibbs Energy MinimumConsider the simple equilibrium reaction: A BThe equilibrium concentrations (or pressures) will be at the extent ofreaction at which the Gibbs function of the syste
North Texas - CHEM - 5200
Chapter 21The Rates ofChemical Reactions1Spontaneous Reactions Dont Always OccurConsider:H2(g) + O2(g) H2O(l)At 298 K, Go = -237.1 kJ/molK = 4x1041Therefore, this reaction proceeds 100% to completion.But, how long does it take?Forever!Without
North Texas - CHEM - 5200
CHEM 5200 - Exam 1 - September 29, 2010Name_(21) MULTIPLE CHOICE [3 points per question] (Circle the ONE correct answer)1. A certain gas at 50 oC and 50 atm. has a Molar Volume of 0.65 L/mol. Under theseconditions, the Compression Factor (Z) is _, and
North Texas - CHEM - 5200
CHEM 5200 - Exam 1 - September 22, 2011INFORMATION PAGE (Use for reference and for scratch paper)Constants and Conversion Factors:R = 0.082 L-atm/mol-K = 8.31 J/mol-K = 8.31 kPa-L/mol-K1 L-atm = 101 J1 L-bar = 100 J1 kPa-L = 1 J1 bar = 100 kPa1 at
North Texas - CHEM - 5200
North Texas - CHEM - 5200
CHEM 5200 - Exam 2 - October 27, 2010Name_(33) MULTIPLE CHOICE [3 points per question] (Circle the ONE correct answer)1. The entropy change is +237 J/K for the reaction: 2 HgO(s) 2 Hg(l) + O2(g). Thestandard molar entropies of HgO(s) and O2(g) are 70
North Texas - CHEM - 5200
North Texas - CHEM - 5200
CHEM 5200 - Exam 2 - October 20, 2011INFORMATION PAGE (Use for reference and for scratch paper)Constants and Conversion Factors:R = 0.082 L-atm/mol-K = 8.31 J/mol-K = 8.31 kPa-L/mol-K1 L-atm = 101 J1 L-bar = 100 J1 kPa-L = 1 J1 bar = 100 kPa1 bar
North Texas - CHEM - 5200
CHEM 5200 - Final Exam Dec. 15, 2010If you would like to have your grade posted on the course web site, using a four digitnumber, then please sign in the space below and put the number.If you do not wish to have your grade posted, you can still learn y
North Texas - CHEM - 5200
North Texas - CHEM - 5200
CHEM 5200 - Final Exam - December 15, 2011INFORMATION PAGE (Use for reference and for scratch paper)Constants and Conversion Factors:F = 96,500 C/mol1 C-Volt = 1 JR = 8.31 J/mol-K = 8.31 C-V/mol-K = 8.31 kPa-L/mol-K1 L-atm = 101 J1 L-bar = 100 J1
North Texas - CHEM - 5200
Page 1CHEM 5200FALL 2011Lecture:Instructor:Office:Off. Hrs:Thursdays - 6:00 PM to 8:50 PM - Room 253Martin SchwartzRm 272M, W, F - 1:00 PM - 2:00 PM + ANYTIMEOffice Ph.:Cell/Home Ph.:565-3542382-1370E-mail:marty@unt.eduWeb Site:or:Chem
North Texas - CHEM - 5200
THE SECOND LAWChapter 3 OutlineHOMEWORKExercises (part a): 3, 4, 7, 8, 10, 11, 13, 16, 17, 21, 22Problems:7Text Examples:1, 2, 6Supplementary HW Questions BelowSect.Title and CommentsRequired?1.The Dispersal of EnergyYES2.EntropyWe wont b
North Texas - CHEM - 5200
SIMPLE MIXTURESChapter 5 OutlineHOMEWORKExercises (part a): 2, 5, 6, 8, 9Problems:NoneSupplementary Questions (see below)Sect.Title and CommentsRequired?1.Partial Molar QuantitiesYES2.The Thermodynamics of MixingYES3.The Chemical Potentia
North Texas - CHEM - 5200
CHEMICAL EQUILIBRIUMChapter 6 OutlineHOMEWORKExercises (part a): 3-5, 8, 9, 17, 23Problems:NoneText Examples:1, 2Supplementary Questions below.Sect.Title and CommentsRequired?1.The Gibbs Energy MinimumYES2.The Description of EquilibriumSk
North Texas - CHEM - 5200
THE RATES OF CHEMICAL REACTIONSChapter 21 OutlineHOMEWORKExercises (part a): 3, 5, 8, 16(1st. part only), 21, 22, 23Problems:NoneSupplementary Questions below.Sect.Title and CommentsRequired?1.Experimental TechniquesYES2.The Rates of Reactio
North Texas - CHEM - 5200
Chapter 2 Homework SolutionsTextbook HomeworkE2.3a V1 = 22.4 L , V2 = 44.8 L , T = 273 K (constant) , n = 1.00 mol(a) ReversibleU nCV ,m T 0H nC p ,m T 0w nRT ln(V2 / V1 ) 1 mol (8.31 J / mol K )(273 K ) ln(44.8 / 22.4) 1570 Jq = U - w = +1570 J(b
North Texas - CHEM - 5200
Chapter 4 Homework SolutionsTextbook HomeworkE4.8a103 LVm ( sol ) 161 mL / mol 0.161 L / mol1 mL103 LVm (liq ) 163.3 mL / mol 0.1633 L / mol1 mLVm Vm (liq ) Vm ( sol ) 0.1633 0.161 0.0023 L / molp1 = 1. atmp2 = 100. atmp p2 p1 100 1 99 atm
North Texas - CHEM - 5200
Chapter 5 Homework SolutionsTextbook HomeworkE5.2a Initial: Determination of nW (moles water) and nE (moles ethanol).Assume 1 L = 1000 cm3 of solution.0.914 gmtot 1000 mL 914 g1 mLmw = 0.50 x 914 = 457 gmE = 0.50 x 914 = 457 g1 molnw 457 g 2
North Texas - CHEM - 5200
Chapter 6 Homework SolutionsTextbook HomeworkE6.3a (a) Q = 0.01 (same equation used for (b)G Go RT ln Q 32.9 kJ / mol (8.31x103 kJ / mol K )(298 K ) ln(0.01) 44.3 kJ / mol(a)(b)(c)(d)(e)Q0.011.010.01x1051x106G-44.3 kJ/mol-32.9-27.2-4.4
North Texas - CHEM - 5200
North Texas - CHEM - 5200
North Texas - CHEM - 5200
North Texas - CHEM - 1423
North Texas - CHEM - 1423
CHEM 1423 - Exam 1 February 10, 2011Constants and Conversion FactorsR = 8.31 J/mol-KCHEM 1423 - Exam 1 February 10, 2011Name_Note: There are only 92 points on this test (because the amount of material was cut down dueto snow/ice days). Don't be conc