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Ex-3-1413-F11

Course: CHEM 1413, Fall 2011
School: North Texas
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1413.001 CHEM - Exam 3 November 2, 2011 - Part 1 Molar Masses: C6H14 - 86. H2O - 18. C2H4 - 28. H2 - 2. Li - 6.9 NH4Cl - 53.5 CH3OH - 32. C2H6 - 30. CHEM 1413.001 - Exam 3 November 2, 2011 - Part 1 Name______________________________ (27) MULTIPLE CHOICE (Circle the ONE correct answer) NOTE: Required Molar Masses and non-metric conversion factors are given on the cover sheet. 1. Which of the following...

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1413.001 CHEM - Exam 3 November 2, 2011 - Part 1 Molar Masses: C6H14 - 86. H2O - 18. C2H4 - 28. H2 - 2. Li - 6.9 NH4Cl - 53.5 CH3OH - 32. C2H6 - 30. CHEM 1413.001 - Exam 3 November 2, 2011 - Part 1 Name______________________________ (27) MULTIPLE CHOICE (Circle the ONE correct answer) NOTE: Required Molar Masses and non-metric conversion factors are given on the cover sheet. 1. Which of the following statements is/are true. (i) (ii) (iii) (A) Enthalpy is defined by: H = E + PV q = H for constant volume reactions Hesss law is valid becauses work (w) is a state function i only (B) ii only (C) i and iii (D) None 2. The Enthalpy of Fusion of hexane, C6H14, is 12.4 kJ/mol, and the constant pressure molar heat capacity is 190 J/mol-oC. What is the heat involved in cooling 129 grams of liquid hexane from 25 oC to its melting point, -95 oC, and freezing the sample? (A) -34.2 kJ (B) -38.6 kJ (C) -15.6 kJ (D) -52.8 kJ For #3 - #4: Consider the thermochemical equation for the decomposition of ethene: 2 C2H4(g) 4 C(graphite) + 4 H2(g) H = -104 kJ 3. The heat involved in the production of 42 grams of ethene from the elements carbon and hydrogen, is (A) +156 kJ (B) +78 kJ (C) -78 kJ (D) -208 kJ 4. If ethene is reacted, and 520 kJ of heat is released, how many grams of H2 have been formed? (A) 10 g (B) 1.25. g (C) 40. g (D) 20. g For #5 - #6: Consider the reaction of Lithium with water: 2 Li(s) + 2 H2O(l) 2 LiOH(aq) + H2(g) Ho = -160 kJ The Enthalpy of Fusion of H2O(s) is 6.0 kJ/mol and the specific heat capacity of H2O(l) is 4.18 J/g-oC. 5. When 10 grams of Li(s) is dropped in a container containing ice and liquid water at 0 oC, approximately how many grams of ice will melt? (A) 78.5 g (B) 350 g (C) 19.3 g (D) 700 g 6. When 10 grams of Li(s) is dropped into 900 grams of H2O(l), originally at 23 oC, what will be the approximate final temperature of the water? (A) 62 oC (B) 85 oC (C) 54 oC (D) 31 oC 7. H for the combustion (reaction with O2) of 2 (two) moles of propane, C3H8, to form CO2 and H2O is -4446. kJ. From this information and the enthalpies of formation of CO2 (-394 kJ/mol), and H2O (-286 kJ/mol), what is the Enthalpy of Formation of propane? (A) +103 kJ/mol (B) -206 kJ/mol (C) -103 kJ/mol (D) There is insufficient data given to perform this calculation 8. When 8. grams of NH4Cl(s) is dissolved in water, the water cools down and the NH4Cl absorbs 5.6 kJ of heat. What is H for dissolving 1 mol of NH4Cl in water? (A) +37.5 kJ (B) +44.8 kJ (C) -37.5 kJ (D) The specific heat capacity, Cs, and mass, m, of water is required. 9. The Fuel Value of methanol, CH3OH, is 23 kJ/g. Therefore, H for combusting 3 (three) moles of methanol in O2 is approximately: (A) -100 kJ (B) -2200 kJ (C) -740 kJ (D) +2200 kJ -------------------------------------------------------------------------------------------------------------------------Two problems on following pages. PROBLEMS (You MUST show your work to receive credit) (10) 1. A 120 g sample of ethane gas, C2H6(g), is initially at 250 oC in a 200. Liter container (the pressure in the container is 0.86 atm.). The sample is cooled at constant volume to 50 oC (and the pressure decreases to 0.53 atm). The constant volume molar heat capacity of ethane is 44.6 J/mol-oC. Calculate q, w and E for this process, all in kJ (kJoules). (10) 2. From the following of enthalpies reaction: 2 B2O3(s) 4 B(s) + 3 O2(g) H = +2510 kJ 4 H2(g) + 2 O2(g) 4 H2O(l) H = -1144 kJ B2O3(s) + 3 H2O(l) B2H6(g) + 3 O2(g) H = +2148 kJ Calculate the enthalpy change for the reaction, B2H6(g) 2 B(s) + 3 H2(g) CHEM 1413.001 - Exam 3 November 2, 2011 - Part 2 Constants: NA = 6.02x1023 mol-1 h = 6.63x10-34 J-s Conversion Factors: 1 eV = 1.60x10-19 J Molar Masses: NaCl - 58.5 c = 3.00x108 m/s 1 hr = 3600 s Ca3(PO4)2 - 310.3 CHEM 1413.001 - Exam 3 November 2, 2011 - Part 2 Name______________________________ (39) MULTIPLE CHOICE (Circle the ONE correct answer) NOTE: Required Molar Masses and non-metric conversion factors are given on the cover sheet. 10. A stellar object is emitting photons with a frequency of 2.5x1014 s-1. If a detector is capturing 5x108 photons/s, what is the total energy of the photons detected in one hour? (A) 3.0x10-7 J (B) 2.2x106 J (C) 140 J (D) 1.4x10-18 J 11. Which of the following sets of quantum numbers are allowed for an electron in a hydrogen atom? (i) n = 4, l = 3, ml = -2, ms = - (ii) n = 3, l = -1, ml = 0, ms = + (iii) n = 5, l = 1, ml = -1, ms = - (A) i & ii (B) ii & iii (C) i & iii (D) i, ii & iii 12. How many orbitals are contained in the n = 4 shell of an atom? (A) 9 (B) 16 (C) 18 (D) 32 13. What is the energy of a photon with a wavelength of 6500. nm? (A) 3.1x10-20 J (B) 3.1x10-29 J (C) 1.0x10-28 J (D) 4.6x1013 J 14. Of the four atoms, K, Ca, As and Se, ___ has the most unpaired electrons and ____ has the least unpaired electrons. (A) Se , K (B) As , Ca (C) Se , Ca (D) As , K 15. The complete electron configuration of the Co3+ ion is: (A) 1s22s22p63s23p64s23d7 (B) 1s22s22p63s23p64s23d4 (C) 1s22s22p63s23p63d6 (D) 1s22s22p63s23p64s23d10 16. Arrange the following atoms in order of increasing atomic radius: Pb , Br , Te , Sb (A) Te < Br < Sb < Pb (B) Br < Te < Pb < Sb (C) Pb < Sb < Te < Br (D) Br < Te < Sb < Pb 17. Of the four atoms, As, Cl, Br, Ga, ____ has the largest radius and ____ has the largest First Ionization Energy. (A) Cl , As (B) Be , Ga (C) Cl , Ga (D) Ga , Cl 18. The Electron Affinity of an atom is defined as the energy change for which one of the following processes? + - + (A) M M + e - (B) X X + e - - (C) X + e X- (D) X+ + e X 19. Which of the following atoms would have a fourth ionization energy much larger than its third ionization energy? (A) Mg (B) Al (C) Si (D) P 20. The total number of lone pairs of electrons in the PO-1 ion is/are: (A) 4 (B) 3 (C) 2 (D) 1 21. The number of lone pairs of electrons around the central atom in TeBr2 is: (A) 0 (B) 1 (C) 2 (D) 3 22. In the AsS-1 ion (As=Arsenic), the Arsenic-Sulfur bond is a ____ bond and there is/are _____ lone pairs of electrons around the Sulfur atom. (A) double, 1 (B) double , 2 (C) triple , 1 (D) triple , 2 _________________________________________________________________________ TWO PROBLEMS (You MUST show your work to receive credit) (05) 3. Write both the Complete and Condensed electron configuration of Astatine (At). (09) 4. When photons of light with a wavelength of 140 nm strike a platinum (Pt) electrode, the ejected electrons have a kinetic energy of 5.2x10-19 J. Calculate the Binding Energy of an electron in platinum, in eV (electron-volts).
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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 &lt; Kr &lt; N2 &lt; 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&lt;D&lt;B&lt;A(D) C&lt;B&lt;D&lt;A2.(C) P-Cl(D) Si-Si(B) D&lt;C&lt;B&lt;A(E) A&lt;B&lt;C&lt;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
k1 = 5x10+4
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
North Texas - CHEM - 1423
North Texas - CHEM - 1423
CHEM 1423 - Exam 1 February 9, 2012Constants and Conversion FactorsR = 8.31 J/mol-KCHEM 1423 - Exam 1 February 9, 2012Name_(60) PART I.MULTIPLE CHOICE (Circle the ONE correct answer)1. Consider the hypothetical reaction, 2A + B C + 3D. The rate of
North Texas - CHEM - 1423
CHEM 1423 - Exam 2 March 3, 2011Constants and Conversion FactorsR = 0.082 L-atm/mol-K1 atm. = 760 torrMolar MassesC6H12O6 - 180.C2H6O2 - 62.H2O - 18.HCl - 36.5NaOH - 40.HNO3 - 63.CHEM 1423 - Exam 2 March 3, 2011Name_(66) PART I.MULTIPLE CHOI
North Texas - CHEM - 1423
North Texas - CHEM - 1423
North Texas - CHEM - 1423
CHEM 1423 - Final Exam May 10, 2011Name_If you wish to have your final exam and course grade posted on the Web site, pleaseprovide me with a four (4) digit number which will be the ID number for your grade._Four (4) digit number for posting.PART I:
North Texas - CHEM - 1423
Chapter 12 Homework SolutionsTextbook HomeworkT20.2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g)Expt. #123[NO](mol/L)5.00 10315.0 10315.0 103Ro = k[NO]x[H2]y[H2](mol/L)2.50 1032.50 10310.0 103Initial rate(mol L1s1)3.0 1039.0 1033.6 102(a) Comp
North Texas - CHEM - 1423
Chapter 13 Homework SolutionsTextbook HomeworkH2O O2 Kc 2H2O2 2T11.(a)(b) Kc PCl5 PCl3 Cl2 (c) Kc = [CO]2(d) Kc H2SH2 T13.To obtain the second equation, one must multiply the first equation by 2 and reverse reactantsand products. Therefo
North Texas - CHEM - 1423
CHEM 1423Chapter 14Homework SolutionsTEXTBOOK HOMEWORK6. As shown in Figure 14.11, generally the solubility of ionic compounds in water increases as thetemperature increases. Increased temperature causes an increase in kinetic energy. The increasedm
North Texas - CHEM - 1423
CHEM 1420Chapter 17Homework SolutionsTEXTBOOK HOMEWORK16. Answer: (a) negative (b) positive (c) positiveStrategy and Explanation: Use the qualitative guidelines for entropy changes described in Section17.3.(a) H2O(g)H2O(s)The solid product has lo
North Texas - CHEM - 1423
CHEM 1423Chapter 18Homework SolutionsTEXTBOOK HOMEWORK6. Answer: see chart belowStrategy and Explanation: Follow the methods described in the answers to Questions 27-33 inChapter 5.Substance Substance Oxidizing ReducingReactantProductoxidizedre
North Texas - CHEM - 1423
CHEM 1423Chapter 19Homework SolutionsTEXTBOOK HOMEWORK12. Answer: (a) alpha emission (b) beta emission (c) electron capture or positron emission (d)beta emissionStrategy and Explanation: The mass numbers and atomic numbers must balance. Use that to
North Texas - CHEM - 1423
CHAPTER 12SUPPLEMENTARY HOMEWORK QUESTIONSS1. Consider the hypothetical reactionA + 3B 2C + D .The rate of reaction istimes [B]/t andtimes [C]/t.a. -2; -3b. -1/3; 1/2c. 1/2; - 1/3d. -1/2; -1/3e. -1; 2/3S2. In a reaction that is first-order wit
North Texas - CHEM - 1423
North Texas - CHEM - 1423
North Texas - CHEM - 1423
North Texas - CHEM - 1423
CHEMICAL KINETICS: RATES OF REACTIONSChapter 12 OutlineProbs:Sect.20, 25, 27, 29, 37, 39, 43, 47, 52*, 66, 67, 78 [*Also determine the rate equation]+ Supplementary Questions (attached)Title and CommentsRequired?1.Reaction RateYES2.Effect of C