#Chem 162-2008 homework 6th week
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#Chem 162-2008 homework 6th week

Course Number: CHEM 162 162, Spring 2009

College/University: Rutgers

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Chem 162-2008 Hill & Petrucci Homework Chapter 15 (Note that my algebra-solving calculator requires my setting up equations involving ",X" and lots of parentheses. Ignore these equations.) Brnsted-Lowry Acids and Bases *21. For each of the following pairs, write an equation in which the first species given acts as a BrnstedLowry acid. (a) HIO4, NH3 (b) H2O, NH2OH (c) H3BO3, NH2 (a) HIO4 + NH3 IO4- +...

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162-2008 Chem Hill & Petrucci Homework Chapter 15 (Note that my algebra-solving calculator requires my setting up equations involving ",X" and lots of parentheses. Ignore these equations.) Brnsted-Lowry Acids and Bases *21. For each of the following pairs, write an equation in which the first species given acts as a BrnstedLowry acid. (a) HIO4, NH3 (b) H2O, NH2OH (c) H3BO3, NH2 (a) HIO4 + NH3 IO4- + NH4+ (b) H2O + NH2OH OH- + NH3OH+ (c) H3BO3 + NH2- H2BO3- + NH3 *23. For each of the following, identify the conjugate acid-base pairs. (a) HOClO3 + H2O H3O+ + OClO2 (b) HSeO4- + NH3 NH4+ + SeO42 (c) HCO3- + OH- CO32- + H2O (d) C5H5NH+ + H2O C5H5N + H3O+ (a) HOClO3 and OClO2- is a conjugate acid-base pair. H2O and H3O+ is a conjugate acid-base pair. (b) HSeO4- and SeO42- is a conjugate acid-base pair. NH3 and NH4+ is a conjugate acid-base pair. (c) HCO3- and CO32- is a conjugate acid-base pair. OH- and H2O is a conjugate acid-base pair. (d) C5H5NH+ and C5H5N is a conjugate acid-base pair. H2O and H3O+ is a conjugate acid-base pair. *25. Identify the species that is amphiprotic, and write two equations for its reaction with H2O(l) that illustrate its amphiprotic character: HI, H2PO4-, NH4+, H2CO3, CO32-. HI is not amphoteric because it can lose a proton, but can't accept a proton (due to the I- being an extremely weak base). H2PO4- is amphoteric. H2PO4- + H2O HPO42- + H3O+ H2PO4- + H2O H3PO4 + OHNH4+ isn't amphoteric, because it can lose a proton to become NH3, but it has no electrons available to gain another proton becoming NH52+. H2CO3 is not amphoteric, because it can lose a proton, but it cannot gain a proton to become H3CO3+. CO32- is not amphoteric, because it can gain a proton, but it has no proton to lose. Molecular Structure and the Strengths of Acids and Bases *31. Identify the stronger acid in each pair, and explain your choice. (a) H2S or H2Se (b) HClO3 or HIO3 (c) H3AsO4 or H2PO4(d) H2Se or HBr (e) HN3 or HCN (f) HNO3 or HSO4(a) H2Se is stronger, because in vertical groups (binary groups) the lower acid has a weaker H-X bond, which makes it a stronger acid. (b) HClO3 is a stronger acid. The strength of the H-O bonds are approximately equal, but the greater electronegativity of the Cl vs the I makes HClO3 a stronger acid. (c) H3AsO4 is a stronger acid, because removing another hydrogen from H2PO4- results in the formation of a doubly negatively charged substance, which doesn't form easily due to it being high energy. (d) HBr is stronger, because due to the higher electronegativity of Br vs. Se, the HBr is more polarized, resulting in a greater interaction with H2O, which results in it being a stronger acid. (e) HN3 is stronger, because the hydrogen is directly connected to the nitrogen, which is strongly electronegative, whereas in HCN the hydrogen is directly connected to the carbon, which is weakly electronegative. (f) HNO3 is stronger, because for the HSO4- to lose a proton it would become a high energy HSO42- ion. *35. Phenol, C6H5OH, used as a disinfectant and in the manufacture of plastics, dyes, and indicators, ionizes as an acid. C6H5OH + H2O C6H5O- + H3O+ Ka = 1.0 x 10-10 Arrange the following substituted phenols in the order in which you would expect their Ka values to increase. Where would you expect phenol itself to fit into this ranking? (a) 3-chlorophenol (b) 2,4-dichlorophenol (c) 2,4,6-trichlorophenol (d) 4-chlorophenol C6H5OH < 4-chlorophenol < 3-chlorophenol < 2,4-dichlorophenol < 2,4,6-trichlorophenol (d) (a) (b) (c) Chlorines are electron withdrawing; hence, the more chlorines, and the closer they are to the acidic proton, the greater the acidity. The positions of "d" vs "a" are controversial, since resonance is also a factor, but we are not responsible for resonance in this course. The pH Scale *37. What is the [H3O+] in (a) lemonade, pH 2.91; (b) shampoo, pH 9.26; (c) tomato puree, pH 4.35; (d) tea, pH 3.94? pH = -log[H+] [H+] = 10-pH (a) [H+] = 10-pH [H+] = 10-2.91 = 1.23 x 10-3 (b) [H+] = 10-pH [H+] = 10-9.26 = 5.50 x 10-10 (c) [H+] = 10-pH [H+] = 10-4.35 = 4.47 x 10-5 (d) [H+] = 10-pH [H+] = 10-3.94 = 1.15 x 10-4 *39. What is the pH of each of the following aqueous solutions? (a) 0.039 M HCl (b) 0.070 M KOH (c) 0.65 M HBr (d) 2.5 x 10-4M Ca(OH)2 pH = -log[H+] (a) pH = -log[0.039] = 1.41 (b) pOH = -log[0.070] = 1.15; pH = 14 1.15 = 12.85 (c) pH = -log[0.65] = 0.19 (d) Ca(OH)2 Ca2+ + 2OH2.5 x 10-4MCa(OH)2 x ((2OH-)/(1Ca(OH)2) = 5.0 x 10-4M[OH-] pOH = -log[OH-] = -log[(5.0 x 10-4)] = 3.30; pH = 14 3.30 = 10.70 *41. What is the pOH of each of the following aqueous solutions? (a) 0.073 M LiOH (b) 1.75 M NaOH (c) 0.045 M Ba(OH)2 (d) 9.1 x 10-2 M HClO4 pOH = -log[OH-] (a) pOH = -log[0.073] = 1.14 (b) pOH = -log[1.75] = -0.24 (c) pOH = -log[2 x 0.045] = 1.05 (d) pH = -log[9.1 x 10-2] = 1.04; pOH = 14 1.04 = 12.96 Equilibria in Solutions of Weak Acids and Weak Bases *47. Calculate the pH of (a) a 0.22 M aqueous solution of phenylacetic acid, HC8H7O2 and (b) an aqueous solution containing 32.9 g formic acid, HCOOH, per liter (a) HC8H7O2 + H2O H3O+ + C8H7O2HC8H7O2 + Initial Change Equilibrium 0.22M H2O H3O+ + C8H7O2- HC8H7O2 + Initial Change Equilibrium 0.22M -X 0.22-X H2O H3O+ + 0 +X +X C8H7O20 +X +X ([H3O+][C8H7O2-])/[HC8H7O2] = (4.9 x 10-5) ([X][X])/[0.22-X] = (4.9 x 10-5) (((([X][X])/[0.22-X]) = (4.9 x 10-5)),X) X = 3.26 x 10-3 M pH = -log(3.26 x 10-3) = 2.49 (b) mol = 32.9g/46.03gmol-1 = 0.715M formic acid HCOOH + H2O H3O+ + HCO2HCOOH Initial Change Equilibrium 0.715M + H2O H3O+ + HCO2- HC8H7O2 + Initial Change Equilibrium 0.715M -X 0.715-X H2O H3O+ + 0 +X +X C8H7O20 +X +X ([H3O+][HCO2-])/[HCOOH] = (1.8 x 10-4) ([X][X])/[0.715-X] = (1.8 x 10-4) (((([X][X])/[0.715-X]) = (1.8 x 10-4)),X) X = 1.13 x 10-2 M pH = -log(1.13 x 10-2) = 1.95 *48. Calculate the pH of the following: (a) a 0.084 M aqueous solution of quinoline, C9H7N, and (b) an aqueous solution containing 5.65 g hydroxylamine, NH2OH, in 226 mL of solution. (a) C9H7N + H2O C9H7NH+ + OHC9H7N Initial Change Equilibrium 0.084 M + H2O C9H7NH+ + OH- C9H7N + Initial Change Equilibrium 0.084 M -X 0.084-X H2O C9H7NH+ + 0 +X +X OH0 +X +X ([C9H7NH+][OH-])/[C9H7N] = (6.3 x 10-10) ([X][X])/[0.084-X] = (6.3 x 10-10) (((([X][X])/[0.084-X]) = (6.3 x 10-10)),X) X = 7.27 x 10-6 M = [OH-] pOH = -log(7.27 x 10-6) = 5.14 pH = 14 5.14 = 8.86 (b) NH2OH + H2O NH3OH+ + OH(5.65g/33.04gmol-1)/0.226L = 0.757M NH2OH Initial Change Equilibrium 0.757 M + H2O NH3OH+ + OH- NH2OH Initial Change Equilibrium 0.757 M -X 0.084-X + H2O NH3OH+ 0 +X +X + OH0 +X +X ([NH3OH+][OH-])/[NH2OH] = (9.1 x 10-9) ([X][X])/[0.757-X] = (9.1 x 10-9) (((([X][X])/[0.757-X]) = (9.1 x 10-9)),X) X = 8.30 x 10-5 M = [OH-] pOH = -log(8.30 x 10-5) = 4.08 pH = 14 4.08 = 9.92 *51. A 1.00 g sample of aspirin (acetylsalicylic acid) is dissolved in 0.300 L of water at 25oC, and its pH is found to be 2.62. What is the Ka of aspirin? o-C6H4(OCOCH3)COOH + H2O H3O+ + o-C6H4(OCOCH3)COOKa = ? (1.00g/(180.17gmol-1))/0.300L = 0.0185 M pH = 2.62 10-2.62 = [H+] = 0.00240 HA + Initial Change Equilibrium 0.0185M 0.00240 HA + Initial Change Equilibrium X = 0.00240 HA + Initial Change Equilibrium 0.0185M -0.00240 0.0161 H2O H3O+ + 0 +0.00240 0.00240 A0 +0.00240 +0.00240 0.0185M -X 0.0185-X H2O H3O+ + 0 +X 0.00240 A0 +X +X H2O H3O+ + A- ([H3O+][A-]/[HA] = Ka ([0.00240][0.00240]/[0.0161] = Ka Ka = 3.58 x 10-4 *52. Codeine, C18H21NO3, a commonly prescribed painkiller, is a weak base. A saturated aqueous solution contains 1.00 g codeine in 120 mL of solution and has a pH = 9.8. What is the Kb of codeine? C18H21NO3 + H2O [C18H21NHO3]+ + OHKb = ? B + H2O HB+ + OH- (1g/(299.4gmol-1))/0.120L = 0.0278M pH = 9.8 pOH = 4.2 [OH-] = 10-4.2 = 6.31 x 10-5 B Initial Change Equilibrium + H2O HB+ + 0 OH0 6.31 x 10-5 B Initial Change Equilibrium X = 6.31 x 10-5 B Initial Change Equilibrium + H2O HB+ + 0 6.31 x 10-5 +6.31 x 10-5 OH0 +6.31 x 10-5 +6.31 x 10-5 0.0278M -6.31 x 10-5 0.0277 + HB+ + 0 +X +X OH0 +X 6.31 x 10-5 0.0278M H2O 0.0278M -X 0.0278 X ([HB+][OH-]/[B] - = Kb ([6.31 x 10-5][6.31 x 10-5])/[0.0277] = Ka Ka = 1.44 x 10-7 Polyprotic Acids *57. Without doing detailed calculations, determine which of the following polyprotic solutions will have the lower pH: 0.0045 M H2SO4 or 0.0045M H3PO4. 0.0045 M H2SO4 will have the lower pH. H2SO4 is a strong acid. The first step of the dissociation will be complete, providing 0.0045 M H3O+. The second step will be weak, increasing the H3O+ concentration slightly. H3PO4 is a weak acid. The first step will provide very low H3O+ concentration, well below 0.0045 M. The second step will barely add to that. *59. Oxalic acid, HOOCCOOH, is a weak dicarboxylic acid found as its potassium or calcium salt in the cell sap of many plants (for example, rhubarb leaves). The solubility of oxalic acid at 25oC is about 83 g/ L. What are the (a) pH and (b) [-OOCCOO-] in the saturated solution? HOOCCOOH + H2O H3O+ + HOOCCOO HOOCCOO- + H2O H3O+ + -OOCCOO83g/(90.04gmol-1)/L = 0.922 M HOOCCOOH HOOCCOOH Initial Change Equilibrium 0.922 M -X 0.922 - X + H2O H3O+ 0 +X +X + HOOCCOO0 +X +X Ka1 = 5.4 x 10-2 Ka2 = 5.3 x 10-5 ([H3O+][HOOCCOO-])/[HOOCCOOH] = ([X][X])/(0.922-X) = 5.4 x 10-2 (((([X][X])/(0.922-X)) = (5.4 x 10-2)),X) X = 0.1978M HOOCCOOH Initial Change Equilibrium 2nd ionization: HOOCCOO- + Initial Change Equilibrium 0.1978 M -X 0.1978 - X H2O H3O+ + - + H2O H3O+ 0 +X 0.1978M + HOOCCOO0 +X 0.1978M OOCCOO- 0.922 M -X 0.7242M 0.1978M +X 0.1978 + X 0 +X +X ([H3O+][-OOCCOO-])/[HOOCCOO-] = ([0.1978 + X][X])/(0.1978-X) = 5.4 x 10-2 (((([0.1978 + X][X])/(0.1978-X)) = (5.3 x 10-5)),X) X = 5.3 x 10-5M = [-OOCCOO-] pH = -log(0.1978 + (5.3 x 10-5)) = -log(0.1978) = 0.70 We could have made the calculations a lot simpler by dropping the X and +X in the calculations of the second ionization step since the X's were likely to be very small due to the very small second ionization Ka. Hydrolysis *63. Predict whether each of the following solutions is acidic, basic, or neutral. Write balanced equations for any hydrolysis reaction that occur. (a) RbClO4(aq) (b) CH3CH2NH3Br(aq) (c) HCOONH4(aq) (a) RbClO4 Rb+ + ClO4Rb+ + H2O NR ClO4- + H2O NR Therefore, neutral solution. (b) CH3CH2NH3Br CH3CH2NH3+ + BrCH3CH2NH3+ + H2O CH3CH2NH2 + H3O+ Br- + H2O NR Acidic + no reaction = acidic solution (c) HCOONH4 HCOO- + NH4+ HCOO- + H2O HCOOH + OHNH4+ + H2O NH3 + H3O+ Formation of hydroxide and acid is approximately a neutral solution. To be certain, calculations need to be done with the respective equilibrium constants. The Solutions Manual solves this problem quantitatively, doing a better job than I did. *67. For a solution that is 0.080M NaOCl, (a) write an equation for the hydrolysis reaction that occurs, and determine (b) the equilibrium constant for this hydrolysis and (c) the pH. (a) NaOCl Na+ + OClNa+ + H2O NR OCl- + H2O HOCl + OHInitial Change Equilibrium OCl0.080M -X 0.080 - X + H2O HOCl 0 +X +X + OH0 +X +X Ka for HOCl = 2.9 x 10-8 But this reaction is not an acid reacting with water. It is a conjugate base reacting with water. Kw = Ka x Kcb Kcb = Kw/Ka = (1 x 10-14)/(2.9 x 10-8) = 3.45 x 10-7 ([HOCl][OH-])/[OCl-] = 3.45 x 10-7 X2/(0.080 X) = 3.45 x 10-7 Drop the X because it is very small. X = 1.66 x 10-4 = [OH-] [H+] = (1 x 10-14)/(1.66 x 10-4) = 6.02 x 10-11 pH = -log(6.02 x 10-11) = 10.2 *69. What is the molarity of a sodium acetate solution that has a pH = 9.05? NaA Na+ + ANa+ + H2O NR A- + H2O HA + OHpH = 9.05 pOH = 14 9.05 = 4.95 [OH-] = 10-4.95 = 1.12 x 10-5 AY -X Y-X + H2O HA 0 +X +X + OH0 +X 1.12 x 10-5 OH0 +1.12 x 10-5 1.12 x 10-5 Initial Change Equlibrium Initial Change Equlibrium A+ Y -X Y-(1.12 x 10-5) H2O HA + 0 +1.12 x 10-5 +1.12 x 10-5 This is not a Ka calculation. It is a conjugate base calculation. Kcb = Kw/Ka = (1 x 10-14)/(1.8 x 10-5) = 5.56 x 10-10 ([HA][OH-])/[A-] = 5.56 x 10-10 ([1.12 x 10-5][1.12 x 10-5])/[(Y (1.12 x 10-5))] = 5.56 x 10-10 (((([1.12 x 10-5][1.12 x 10-5])/[(Y (1.12 x 10-5))] ) = (5.56 x 10-10)),X) Y = 0.226M ACommon Ion Effect *71. Which of the following solutions and substances will suppress the ionization of formic acid, HCOOH(aq); (a) NaCl, (b) KOH(aq), (c) HClO4, (d) (HCOO)2Ca, (e) Na2CO3? Explain HCOOH + H2O H3O+ + HCOO(a) NaCl would have no effect. It doesn't affect the equilibrium. (b) KOH would react preferentially with the H3O+, decreasing its concentration, consequently shifting the reaction to the right in order to maintain equilibrium. (c) HClO4 would increase the H3O+ concentration. Therefore, it would shift the equilibrium to the left. (d) (HCOO)2Ca provides additional HCOO- concentration. Therefore, it would shift the reaction to the left. (e) Na2CO3 would react preferentially with the H3O+, decreasing its concentration, consequently shifting the reaction to the right in order to maintain equilibrium. *75. Calculate the pH of a solution that is 0.350 M CH3CH2COOH and 0.0786 M in the salt CH3CH2COOK. (For CH3CH2COOH, Ka = 1.3 x 10-5.) CH3CH2COOH + H2O H3O+ + CH3CH2COOCH3CH2COOH + Initial Change Equilibrium 0.350M -X 0.350 - X H2O H3O+ 0 +X +X + CH3CH2COO0.0786M +X 0.0786 + X Ka = ([H3O+][CH3CH2COO-])/[CH3CH2COOH] = 1.3 x 10-5 ([X][0.0786 + X])/[0.350 - X] = 1.3 x 10-5 (((([X][0.0786 + X])/[0.350 - X]) = (1.3 x 10-5)),X) X = 5.78 x 10-5M pH = -log(5.78 x 10-5) = 4.24 More Acid-Base Equilibria *77. Without doing detailed calculations, determine which of the following solutions should have a pH higher than 7.00, and which lower than 7.00: (a) 0.0050 M C6H5COOH, (b) 1.0 x 10-5M NH3, (c) 0.022 M CH3NH3Cl, (d) 0.10M KH2PO4, (e) saturated Ba(OH)2(aq), (f) 0.25 M NaNO2. (a) C6H5COOH is benzoic acid, which is an acid. HB + H2O H3O+ + BBenzoic acid should provide an acidic solution, with a pH lower than 7.00. (b) NH3 + H2O NH4+ + OHBasic solution, with pH > 7.00. (c) CH3NH3Cl CH3NH3+ + ClCl- + H2O NR CH3NH3+ + H2O CH3NH2 + H3O+ Acidic solution, with pH < 7.00. (d) KH2PO4 K+ + H2PO4K+ + H2O NR H2PO4- is amphoteric. It can act as a conjugate base: H2PO4- + H2O H3PO4 + OHKcb = Kw/Ka1 = (1 x 10-14)/(7.1 x 10-3) = 1.41 x 10-12 or it can act as an acid: H2PO4- + H2O HPO42- + H3O+ Ka2 = 6.3 x 10-8 The equilibrium constant is much larger when it acts as an acid. Therefore it will provide an acidic solution, with a pH < 7.00. (e) Ba(OH)2 Ba2+ + 2OHBa2+ + H2O NR This leaves OH-, which is basic, with pH > 7.00 (f) NaNO2 Na+ + NO2Na+ + H2O NR NO2- + H2O HNO2 + OHBasic solution, with pH > 7.00. *79. Write two equations--one to show the ionization of HPO42- as an acid and one to show its ionization (hydrolysis) as a base. Would you expect an aqueous solution of Na2HPO4 to be acidic or basic? (Hint: What are the relevant Ka and Kb values?) HPO42- + H2O PO43- + H3O+ Ka3 = 4.3 x 10-13 HPO42- + H2O H2PO4- + OHKcb = Kw/Ka = (1 x 10-14)/(6.3 x 10-8) = 1.59 x 10-7 The Kcb value is so much larger than the Ka3 value that I would expect Na2HPO42- to act as a base. *110. Determine [H3O+] in 0.020 M H2SO4(aq), and compare your result with that obtained in Exercise 15.12B. H2SO4 + H2O H3O+ + HSO4K= The Ka1 of H2SO4 is so large that the reaction can be considered to go to completion. Initial Exchange Equilibrium H2SO4 0.020 M -0.020 0 + H2O H3O+ 0 +0.020M +0.020M + HSO40 +0.020M +0.020M 2nd ionization: HSO4- + H2O H3O+ + SO42HSO4Initial Exchange Equilibrium 0.020 M -X 0.020 - X + H2O Ka2 = 1.1 x 10-2 H3O+ 0.020M +X 0.020 + X + SO420 +X +X ([H3O+][SO42-])/[HSO4-] = 0.011 ([0.020 + X][X])/[0.020 - X] = 0.011 (((([0.020 + X] x [X])/[0.020 - X]) = (0.011)),X) X = 5.96 x 10-3M Therefore, the [H3O+] = 0.020 + 0.00596 = 0.02596 M (which compares very well with the estimated 0.025). Apply Your Knowledge *129. [Environmental] A batch of sewage sludge is dewatered to 28% solids; the remaining water has a pH of 6.0. The sludge is made quite alkaline to kill disease-causing microorganisms and make the sludge acceptable for spreading on farmland. How much quicklime (CaO) is needed to raise the pH of 1.0 ton of this sludge to 12.0? CaO(s) + H2O(l) Ca(OH)2(aq) I'm not bothering to do this problem.

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CHEMISTRY 162-2007 3rd WEEK RECITATION ANNOUNCEMENTS E-MAIL ATTENDANCE Sign in QUIZ Recitation quiz this week Wed Chapter 12.1 13.3 Thu Chapter 12.7 13.5 (Mon Chapter 12.8 13.7) Will run late due to quiz. Quiz will be at end of recitation. Pick up graded
Rutgers - CHEM 162 - 162
CHEMISTRY 162-2008 4th WEEK RECITATION ANNOUNCEMENTS E-MAIL ATTENDANCE Sign in EXAMS 1st hourly: Wed., Feb. 20th, 9:4011:00 PM. Review session: Thursday, Feb. 14th, 8:0010:00 PM, CDL 102 Chapter 12 Solutions Sunday, Feb. 17th, 8:3010:30 PM, Hck 138 Chapte
Rutgers - CHEM 162 - 162
Hill &amp; Petrucci Chapter 12 Homework Problems Chem 162-2008 Homework 1st &amp; 2nd Week 21. Describe how you would prepare 2.30 kg of an aqueous solution that is 4.85% NaNO3 by mass. 4.85 g NaNO3/100g solution gNaNO3/2300g solution Numerator conversion: Done.
Rutgers - CHEM 162 - 162
PRACTICE PROBLEMS CHEM 162-2007 EXAM I CHAPTER 13 - KINETICS RATES, RATE CONSTANTS, REACTION ORDERS, HALF-LIFES CONCEPTS RATES,RATE CONSTANTS, REACTION ORDERS, HALF-LIFES CALCULATIONS REACTION MECHANISMS, ACTIVATION ENERGY &amp; CATALYSTS CONCEPTS REACTION M
Rutgers - CHEM 162 - 162
PRACTICE PROBLEMS CHEM 162-2007 EXAM I CHAPTER 12 - PROPERTIES OF SOLUTIONS SOLUTION CONCENTRATIONS CONCEPTS (Molarity, mole fraction, etc.) SOLUTION CONCENTRATIONS CALCULATIONS (Molarity, mole fraction, etc.) ENERGETICS OF SOLUTIONS AND SOLUBILITY CONCE
San Jose State - CMPE - 202
Beauty and the BeastFebruary 2009 CmpE 202-05AbstractA prince had no love or affection, was cursed to become a beast by an Enchantress. Learning to love another and earn her love in return is the only way he could break the spell, but only before the l
San Jose State - CMPE - 202
CmpE 202 Software Systems EngineeringTeam Project #1The requirements analysis and design specifications part of the projects will be specified using the Unified Modeling Language (UML) on Rational Rose or any other suitable modeling tool, such as Visio
San Jose State - CMPE - 202
I qbalFebruary 2009 CmpE 202-05TABLE OF CONTENTS Iqbal 1. Abstract .3 2. Domain Description.4 3. Description of the program that is wanted .6 4. Block Diagram .7 5. Detailed Requirement.8-9 6. Use Cases &amp; User Context .9-12 6.1. Milk, the buffalo .9 6.2
San Jose State - CMPE - 202
CmpE 202 Software Systems EngineeringTeam Project #3Pattern Assignments Due Date: Check the due date schedule for pattern assignments due date. Major Tasks: 1. Use traditional or meta model to model the concepts definitions 2. Use software stability mod
San Jose State - CMPE - 202
1Chronicles of Narnia The Lion, the Witch and the WardrobeFebruary 2009 CmpE 202-05TABLE OF CONTENTS 1. ABSTRACT.3 2. DESCRIPTION OF DOMAIN.3 3. BLOCK DIAGRAM.6 4. DESCRIPTION OF THE PROGRAM THAT IS WANTED.7 5. DETAILED REQUIREMENTS.7 6. USE CASES AND
San Jose State - CMPE - 202
San Jose State - CMPE - 202
UNDERGROUND WATER DEPRECIATION AND POLLUTION Fall 2008, CmpE 202-05 ABSTRACT: Goals: To provide solution for effective and minimal use of Underground water resources To minimize water pollution by using various techniques and methods To discuss differen
San Jose State - CMPE - 202
EFFECTS OF HURRICANE Fall 2008, CmpE 202-051. ABSTRACT Goal: To provide the precautionary measures that could be taken to reduce the effect of disaster and to provide economical, physical, spiritual, and emotional help to the victims of this calamity. Mo
San Jose State - CMPE - 202
Design Session (01) PP1 [1] InvitationYou invited and hosted a group of people for dinner. Would you think of the invitation and dinner as a modeling problem and answer the following questions? (1) Document two of the Use Cases in Your Problem: (a) Ident
San Jose State - CMPE - 202
Team Projects (PP2)You are working on team projects for CmpE 202. Would you think of working on team projects as a modeling problem and answer the following questions? (1) Document three of the Use Cases in Your Problem: (a) Identify two of the use cases
San Jose State - CMPE - 202
Exercises (PP4)Prepare a class diagram for each of the above modeling problems: 1. Invitation (PP01) 2. Hunting Trip (PP01) 3. Camping Trip (PP01) 4. Team Project (PP02) 5. Seeking a Loan
San Jose State - CMPE - 202
CmpE 202-05 Software System Engineering Fall 2009Due Dates (Thursdays)Team Projects Your Contact Info Your Team Name &amp; Contact Info Team Projects' Reqs. Statements Starts at Extra Team Project Sign up Team Problem Statements Submission Project 1 Submiss
San Jose State - CMPE - 202
Exercises (PP5)Prepare traditional models for the following problems showing at least 10 relationships among the following object classes, including associations, aggregations, and generalizations. Show multiplicities in your diagrams. You don't need to
San Jose State - CMPE - 202
Exercises (PP6)Prepare traditional models for the following problems showing at least 10 relationships among the following object classes, including associations, aggregations, and generalizations. Show multiplicities in your diagrams. You don't need to
University of Texas - M - 57420
Version 050 Exam 2 Louiza Fouli (59590) This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. V1:1, V2:1, V3:3, V4:1, V5:1. 001 (part 1 of 1) 10 points If f is the fun
University of Texas - M - 57420
Version 050 Exam 2 Louiza Fouli (59590) This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. V1:1, V2:1, V3:3, V4:1, V5:1. 001 (part 1 of 1) 10 points If f is the fun
University of Texas - M - 57420
Martinez, Gorge Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Diane Radin This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1
University of Texas - M - 57420
Tamayo, Nubia Exam 1 Due: Feb 20 2007, 11:00 pm Inst: Gary Berg6 5 4 3 2 1 0 -1 -2 -3 -4 -51This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Ce
University of Texas - M - 57420
Martinez, Gorge Exam 2 Due: Oct 31 2007, 1:00 am Inst: Diane Radin This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1
University of Texas - M - 57420
Martinez, Gorge Exam 3 Due: Dec 4 2007, 11:00 pm Inst: Diane Radin This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1
University of Texas - M - 57420
Department of Mathematics M 408L Integral Calculus Test #1 Review TA: Chris Mirabito October 5, 2006Test #1 will be held on Tuesday, October 10, 2006 at 7:00pm in FAC 21 (this is the Peter T. Flawn Academic Center, located next to the Main Building). The
University of Texas - M - 57420
M 408L - Integral CalculusFinal Exam Review May 6, 2009 Toics: Applications: Motion, Area, Volume (of solids, of paperweights, as double integrals) 1. Jim decides to go BASE jumping off an Austin building. Unfortunately, he hits the ground with a speed o
University of Texas - M - 57420
Martinez, Gorge Final 1 Due: Dec 14 2007, 10:00 pm Inst: Diane Radin This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of
University of Texas - M - 57420
Garza, Ernesto Final 1 Due: May 11 2007, 11:00 pm Inst: JEGilbert This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1)
University of Texas - M - 57420
badruddin (ssb776) HW01 Radin (57410) This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. HW01 is a review of M408K material. It overlaps HW02 and HW03. 001 Determin
University of Texas - M - 57420
badruddin (ssb776) HW02 Radin (57410) This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points i.e., D = -3. Consequently, f (t) = t3 - 4t2 + 8t - 3 . 002
University of Texas - M - 57420
badruddin (ssb776) HW03 Radin (57410) This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 pointsn1But in this case xi = i . , f (x) = tan x, [a, b] = 0,
University of Texas - M - 57420
badruddin (ssb776) HW04 Radin (57410) This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 4) 10.0 points A Calculus student begins walking in a straig
University of Texas - M - 57420
badruddin (ssb776) HW05 Radin (57410) This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points -2 For which one of the following shaded regions is its are
University of Texas - M - 57420
moorea8 Review Exam01 Gilbert (58430) This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points 1. increase in Mira's weight from age 2 to 9 2. Mira's weig
University of Texas - M - 57420
Create assignment, 57611, Homework 14, Mar 23 at 9:27 pm This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. version 888 CalC8e58s 54:0
University of Texas - M - 57420
Create assignment, 57611, Homework 14, Mar 23 at 9:27 pm This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. version 888 CalC8e58s 54:0
University of Texas - M - 57420
Create assignment, 93205, Homework 3, Apr 26 at 12:49 am This print-out should have 40 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. version 728 CalC12c04a 55:
University of Texas - M - 57420
Create assignment, 93205, Homework 3, Apr 26 at 12:50 am This print-out should have 40 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. version 728 CalC12c04a 55:
University of Texas - M - 57420
Version 040 EXAM 2 Odell (58340) This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 Determine ifx1Consequently, the limit exists and lim (ln x)2 = 0 . 3x + 7
University of Texas - M - 57420
Version 040 Exam 3 Odell (58340) This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points and if it does, find its limit 1. sequence diverges 2. limit = e
University of Texas - M - 57420
Version 004 FINAL Odell (58340) This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 Find f (/2) when 1 2 f (t) = cos t + 4 sin t 3 3110.0 pointsR2 a R1 b can
University of Texas - M - 57420
Version 040 EXAM 1 Odell (58340) This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 0016 5 4 3 2 1 0 -1 -2 -3 -4 -5 5 4 3 2 1 0 -1 -2 -314 2 2 -2 -4-2 -1 0 1 2
University of Texas - M - 57420
M 408L - Integral CalculusWorksheet - Homework 3 February 2, 2009Group 1: Problem like #3. Consider the following Riemann sum: 2 n 4+ 2 n2+2 n4+4 n2+2 n4+6 n2+ +2 n4+2n n21. Rewrite the sum using sigma notation starting with i = 1. How
University of Texas - PHY - 59335
HW#1 Solution#1. Question E2 p.14 12 cupcakes for 900ml 12 cupcakes : 900ml = 8cupcakes : x 900ml 8 cupcakes x= = 600ml 12 cupcakes #2. Question E6 p.16 1cm = 10mm 1m = 100cm 1cm 220mm 10mm 1cm 220mm 10mm20cm 1m 100cm0.22m#3. Question E10, p15 1 mile
University of Texas - PHY - 59335
1. question 21 page 35 a. Velocity is vector quantity. If we set the forward movement to the positive vector, the car moves forward because velocity is everywhere positive on the graph. b. ans: A , The change of velocity is greatest, or slope is deepest.