CHEM 104 Module 1, 2, 3, 4, 5, 6 Exam (New, 2023-2024): General Chemistry:Portage Learning, Exams of Chemistry

Download CHEM 104 Module 1, 2, 3, 4, 5, 6 Exam (New, 2023-2024): General Chemistry:Portage Learning and more Exams Chemistry in PDF only on Docsity! Module 1: Question 1 In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below, the following data table is obtained: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) Data Table #2 Time (sec) [N2O5] [O2] 0 0.300 M 0 300 0.272 M 0.014 M 600 0.224 M 0.038 M 900 0.204 M 0.048 M 1200 0.186 M 0.057 M 1800 0.156 M 0.072 M 2400 0.134 M 0.083 M 3000 0.120 M 0.090 M 1. Using the [O2] data from the table, show the calculation of the instantaneous rate early in the reaction (0 secs to 300 sec). 2. Using the [O2] data from the table, show the calculation of the instantaneous rate late in the reaction (2400 secs to 3000 secs). 3. Explain the relative values of the early instantaneous rate and the late instantaneous rate. Your Answer: 1. rate = (0.014 - 0) / (300 - 0) = 4.67 x 10-5 mol/Ls 2. rate = (0.090 - 0.083) / (3000 - 2400) = 1.167 x 10-5 mol/Ls 3. The late instantaneous rate is smaller than the early instantaneous rate. Question 2 The following rate data was obtained for the hypothetical reaction: A + B → X + Y Experiment # [A] [B] rate 1 0.50 0.50 2.0 2 1.00 0.50 8.0 3 1.00 1.00 64.0 1. Determine the reaction order with respect to [A]. 2. Determine the reaction order with respect to [B]. 3. Write the rate law in the form rate = k [A]n [B]m (filling in the correct exponents). 4. Show the calculation of the rate constant, k. Your Answer: rate = k [A]x [B]y rate 1 / rate 2 = k [0.50]x [0.50]y / k [1.00]x [0.50]y 2.0 / 8.0 = [0.50]x / [1.00]x 0.25 = 0.5x x = 2 rate 2 / rate 3 = k [1.00]x [0.50]y / k [1.00]x [1.00]y 8.0 / 64.0 = [0.50]y / [1.00]y 0.125 = 0.5y y = 3 rate = k [A]2 [B]3 2.0 = k [0.50]2 [0.50]3 k = 64 Question 3 ln [A] - ln [A]0 = - k t 0.693 = k t1/2 An ancient sample of paper was found to contain 19.8 % 14C content as compared to a present-day sample. The t1/2 for 14C is 5720 yrs. Show the calculation of the decay constant (k) and the age of the paper. H2O = [0.0380], CH4 = [0.0380], CO = [0.0620] and H2 = [0.186] with Kc = 3.62. If the concentration of CO at equilibrium is increased to [0.200], how and for what reason will the equilibrium shift? Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. CO (g) + 3 H2 (g) CH4 (g) + H2O (g) Your Answer: Kc = 3.62 Qc = [0.200] [0.0380] / [0.0620] [0.186]3 Qc = 19.0 Qc is greater than Kc and thus the equilibrium will shift to the left towards the direction of the reactants. Kc = 3.62 when H2O = [0.0380], CH4 = [0.0380], CO = [0.0620] and H2 = [0.186 M] When CO = [0.200], Q = [0.0380] [0.0380] = 1.14 [0.200] [0.186]3 The reaction must shift briefly in the forward direction to decrease the [CO] to come back to equilibrium.This is in agreement with Qc < Kc which also predicts the reaction will proceed to the right. new CO concentration is 0.2 Question 9 The equilibrium reaction below has the Kc = 3.93. If the volume of the system at equilibrium is decreased from 6.00 liters to 2.00 liters, how and for what reason will the equilibrium shift? Be sure to calculate the value of the reaction quotient, Q, and use this to confirm your answer. CO (g) + 3 H2 (g) CH4 (g) + H2O (g) Your Answer: Initial volume = 6.00 L Final volume = 2.00 L Volume is reduced by 3. Pressure is inversely proportional to volume, thus pressure is tripled and concentration of gases is also tripled. Final concentration = Initial concentration x 3 Qc = [3CH4] [3H2O] / [3CO] [3H2]3 Qc = Kc / 9 Qc is lesser than Kc and thus the equilibrium will shift to the right towards the direction of the products. When volume decreases from 6.00 to 2.00, the pressure triples and the concentration of all gases (CO, H2, CH4, and H2O) triples so: (at equilibrium) Qc =Kc = [CH4] [H2O] = 3.93 [CO] [H2]3 (volume 1/3 = pressure tripled = conc tripled) Qc = [3 CH4] [3 H2O] = Kc [3 CO] [3 H2]3 9 The reaction must shift briefly in the direction that decreases the pressure by going toward the side with the lesser moles of gas (forward direction : 4 moles of gas yields 2 moles of gas) to come back to equilibrium. This is in agreement with Qc < Kc: the reaction will proceed to the right (in the direction of the products). - Qc should be used to confirm your answer - for what reason will it shift this way? Question 10 The equilibrium reaction below has the Kc = 3.93 at 25oC. If the temperature of the system at equilibrium is decreased to 10oC, how and for what reason will the equilibrium shift? Also show and explain how and why the Kc value will change. CO (g) + 3 H2 (g) CH4 (g) + H2O (g) ∆H0 = -206.2 kJ Your Answer: Temperature is decreased in the reaction at equilibrium that has a - ∆H0, thus reaction shifts in the forward direction. The reaction shift in the direction that produces some heatIn this reaction, the concentration of the product increases and the concentration of the reactants decreases and thus increases the value of Kc. Module 2: Question 1 Identify each of the compounds below as ACID, BASE or SALT on the basis of their formula and explain your answer. (1) Cr(OH)3 (2) HAsO4 (3) CoCO3 Your Answer: 1. base - contains Cr metal + OH polyatomic group 2. acid - contains H + polyatomic group (AsO4) 3. salt - contains Co metal + polyatomic group (CO3) Question 2 For the Brønsted-Lowry acid base reactions shown below, list the stronger acid, stronger base, weaker acid and weaker base in the answer blanks provided: NH4 + + H2PO4 - NH3 + H3PO4 Stronger acid: Stronger base: Weaker acid: Weaker base: Your Answer: Stronger acid: H3PO4 Stronger base: NH3 Weaker acid: NH4 + Weaker base: H2PO4 - Question 3 For the compounds shown below, choose the stronger acid, explain why it is the stronger acid and write the formula of its conjugate base in the answer blank: pH = - log [H+] = - log (3.41x10-3) pH = 2.47 % ionization = ([C2H3O2 -]/ [HC2H3O2]) x 100 % ionization = (3.41x10-3 / 0.645) x 100 % ionization = 0.53% Question 7 Predict and explain whether a solution of LiF is acidic, basic or neutral. Your Answer: Basic since Li hydrolyzes to form strong base [LiOH]. LiF: Basic since F- hydrolyzes to form a weak acid (HF) and OH- Question 8 Show calculation of the pH of a buffer prepared by mixing 0.100 M NH4Cl and 0.0750 M NH3. NH3 + H2O (liq) NH4 + + OH- Ka = 1.8 x 10-5 Your Answer: NH3 + H2O (liq) NH4 + + OH- 0.0750 0.100 0 -x +x +x 0.0750-x 0.100+x x Ka = 1.8 x 10-5 Ka = [NH4 +] [OH-] / [NH3] 1.8 x 10-5 = (0.100+x) (x) / (0.0750-x) 1.8 x 10-5 = (0.100) (x) / (0.0750) x = 1.35x10-5 = [OH-] pOH = - log [OH-] = - log (1.35x10-5) pOH = 4.87 pH = 14 - pOH = 14 - 4.87 pH = 9.13 Question 9 Show the calculation of the pH of a solution obtained by adding 0.0100 mole of OH- to a buffer of pH 4.74 which originally contains 0.100 M NaC2H3O2 and 0.100 M HC2H3O2. HC2H3O2 H+ + C2H3O2 - Ka = 1.8 x 10-5 Your Answer: HC2H3O2 H+ + C2H3O2 - 0.100 0 0.100 +0.0100 mole OH- 0.090 0 0.110 -x +x +x 0.090-x x 0.110+x Ka = 1.8 x 10-5 Ka = [H+] [C2H3O2 -] / [HC2H3O2] 1.8 x 10-5 = (x) (0.110+x) / (0.090-x ) 1.8 x 10-5 = (x) (0.110) / (0.090) x = 1.47x10-5 = [H+] pH = - log [H+] = - log (1.47x10-5) pH = 4.83 pH went up slightly fron 4.74 to 4.83 when OH- was added. Question 10 List and explain what type of acid and base are involved in the titration curve below and select an appropriate indicator to be used in this titration from the list below. pH Range of Indicator Color Change Methyl violet 0.0 - 1.6 Phenolphthalein 8.0 - 9.8 Alizarin yellow R 10.1 - 12.0 Your Answer: Titration of a strong acid with a strong base because the location of the equivalence point is at pH 7. The rapid rise occurs approximately between pH 3 and pH 11. Indicator that can be used is limited to phenolphthalein because its range of color change is at 8.0 - 9.8. Methyl violet and Alizarin yellow R cannot be used for this titration. Module 3: Question 1 Show the calculation of the molar solubility (mol/L) of BaF2, Ksp of BaF2 = 1.0 x 10-6. Your Answer: Ksp = [Ba+2] [F-]2 = 1.0x10-6 ΔGf 0 = ΔGf 0 products - ΔGf 0 reactants ΔGf 0 = [(4 ΔGf 0 CO2 ) + (6 ΔGf 0 H2O)] - [(2ΔGf 0 C2H6) + (7 ΔGf 0 O2)] ΔGf 0 = [(4x-394.4) + (6x-228.6)] - [(2x-32.9) + (7x0)] ΔGf 0 = -2883.4 kJ/mol Reaction is spontaneous since ΔG is a larger negative number than -20 kJ. Question 8 Using the values of ΔH0 and ΔS0 given, determine the value of ΔG0 for the reaction below at 25oC. ΔH0 = 311.4 kJ/mole and ΔS0 = 232.6 J/mole. Explain whether this reaction is spontaneous as written. C2H6 (g) → C2H2 (g) + 2 H2 (g) What would be the value of ΔG0 for the reaction at 900oC? Explain whether this reaction is spontaneous as written at 900oC. Your Answer: ΔG0 = ΔH0 - TΔS0 ΔG0 = 311.4 kJ/mol - (25+273K)(232.6/1000 kJ/mol) ΔG0 = 242.1 kJ/mol Reaction is nonspontaneous at 25oC since ΔG is a larger positive number than +20 kJ. ΔG0 = ΔH0 - TΔS0 ΔG0 = 311.4 kJ/mol - (900+273K)(232.6/1000 kJ/mol) ΔG0 = 38.56 kJ/mol Reaction is nonspontaneous at 900oC since ΔG is a larger positive number than +20 kJ. Question 9 The formation of Mn from MnS is nonspontaneous as written in reaction 1 below but can be accomplished by coupling it with the oxidation of S to SO2 (shown in reaction 2 below). Show the overall spontaneous reaction resulting from the coupling of these two reactions and calculate the ΔG0 of the overall reaction. (1) MnS (s) → Mn (s) + S (g) ΔG0 = + 214.2 kJ (2) S (s) + O2 (g) → SO2 (g) ΔG0 = - 300.1 kJ Your Answer: MnS (s) + O2 (g) → Mn (s) + SO2 (g) ΔG0 = -85.9 kJ Question 10 Using the table and images (A - F) below, predict, and explain by doing a Qc vs Ksp calculation, which tube shows the result when colorless 1 x 10-6 M AgNO3 solution was added to colorless 1 x 10-6 M NaBr solution. A B C D E F Substance Formula Color Ksp Copper (II) carbonate CuCO3 Blue 1.4 x 10-10 Lead (II) chromate PbCrO4 Yellow 2.8 x 10-13 Silver bromide AgBr White 5.0 x 10-13 Your Answer: Ksp = [Ag+] [Br-] = 5 .0 x 10-6 Qc = [Ag+]initial [Br-]initial = [1.0 x 10-6] [ 1.0 x 10-6] Qc = 1.0x10-12 Qc (1.0x10-12) < Ksp (5 .0 x 10-6) thus the reaction should go in the forward direction (solution) and AgBr will not precipitate. Image B shows the result of when colorless AgNO3 solution was added to colorless NaBr solution because it does not precipitate, its color is still colorless. Qc = (1 x 10-6) x (1.0 x 10-6) = 1 x 10-12 Ksp for AgBr = 5.0 x 10-13 Since Qc > Ksp a white precipitate will form = E Module 4: Question 1 For the cell described by the following cell diagram. Cr (s) | Cr+3 (aq, 1 M) || Al+3 (aq, 1 M) | Al (s) Al+3 + 3e- → Al E0 = - 1.66 v Cr+3 + 3e- → Cr E0 = - 0.73 v (1) The anode half reaction is (2) The cathode half reaction is (3) The overall cell reaction is (4) Show the calculation for the total cell potential (5) State and explain whether the cell is voltaic or electrolytic Your Answer: 1) Cr (s) → Cr+3 (aq) + 3e- 2) Al+3 (aq) + 3e- → Al (s) 3) Cr (s) + Al+3 (aq) → Cr+3 (aq) + Al (s) 4) E0 cell = E0 cathode - E 0 anode E0 cell = -1.66 - (-0.73) E0 cell = -0.93 v 5) Since the total cell potential is negative, cell will not occur spontaneously and would have to be electrolytic. Question 2 In the following reaction, identify the following and explain your answers Zn + 2 H+ + 2 Cl- → Zn+2 + 2 Cl- + H2 Question 6 Answer the following multiple choice questions: 1. The ground state of _________ has an electron configuration of 1s1. A. Hydrogen B. Nitrogen C. Carbon D. Sodium E. Iron 2. According to the course materials, which of the following planets has a significant amount of hydrogen in the atmosphere due to its large gravitational pull? A. Venus B. Mars C. Earth D. Jupiter E. Uranus 3. Which of the following is known as protium? A. 1P B. 1H C. 2H D. 3H E. 2P 4. Which of the following is a mixture of metals? A. Mineral B. Ore C. Alloy D. Metallurgy E. None of the above Question 7 Answer the following multiple choice questions: 1. Smelting is a type of _________ processing of metals. A. Electrometallurgy B. Hydrometallurgy C. Pyrometallurgy D. Thermometallurgy E. None of the above 2. Which of the following indicates that metals can be pounded into shape? A. Ductile B. Malleable C. Luster D. Electrical conductivity E. Thermal conductivity 3. Which of the following is not an alkali metal? A. Mg B. Na C. Li D. K E. All of the above are 4. What gas is evolved when alkali metals react with water? A. Br2 B. O2 C. Cl2 D. N2 E. None of the above Question 8 Answer the following multiple choice questions: 1. Which alkali compound is used in water softening? A. MgF2 B. BaCO3 C. Na2CO3 D. LiNO3 E. All of the above 2. Gunpowder contains which alkali compound? A. KNO3 B. KCl C. NaF D. NaNO3 E. LiNO3 3. Which compound is the crucial ingredient in milk of magnesia? A. CaCO3 B. MgCl2 C. Mg(OH)2 D. Ca(OH)2 E. CaO 4. When magnesium burns in air, what is one compound produced? A. MgH2 B. MgI2 C. Mg2N3 D. Mg2P3 E. None of the above Question 9 Answer the following multiple choice questions: 1. Tin and lead tend to form compounds with ____ and _____ charges. A. +1, +2 B. +2, +3 C. +2, +4 D. -2, -4 E. -1, -2 2. Which of the following is not a characteristic of nonmetals? A. Brittle B. Poor electrical conductors C. Covalent bonding D. Forms only – charges E. None of the above 3. Carbon exists as two allotropes, _______ and _______. A. graphite; ozone B. Coal; oil C. Carbon monoxide; carbon dioxide Question 6 The reaction shown below is a/an ____________________ type reaction. CH3CH2CH=CH2 + 6 O2 → 4 CO2 + 4 H2O Your Answer: oxidation Question 7 In what type of reaction is a hydrocarbon converted to CO2? Your Answer: oxidation Question 8 The following is a ________________ spectrum. Your Answer: mass Question 9 The compound most likely to have given the SEM/EDS shown below is __________. 1. NaNO3 2. NaHCO3 3. NSO2 4. NaHSO4 The compound most likely to have given the SEM/EDS shown below is NaHSO4 H is not detected by SEM/EDS Question 10 Which of the structural formulas below is likely for an organic compound, C4H8O, whose IR spectrum is shown below? Note and assign at least 3 absorptions as part of your answer. O O O ║ ║ ║ 1. CH3-C-O-CH2CH3 b. CH3CH2-C-CH3 c. CH3-CH2CH2-C-H The organic compound is ____________________ (letter and type) The 3 most important absorption peaks and the bond deformations they describe are: Approximate peak location Bond ______________________ __________________ ______________________ __________________ ______________________ __________________ Your Answer: b. ketone Peak location and bond: 1. 3000-2800 cm-1 peak for C-H bond 2. 1750-1650 cm-1 peak for C=O bond 3. no peak at 2900-2700 cm-1 so no O=C-H bond, not an aldehyde The organic compound is B (ketone) The 3 most important absorption peaks and the bond deformations they describe are: Approximate peak location Bond 3000-2800 cm-1 alkane C-H (stretch) 1720 cm-1 C=O (stretch) 1460 cm-1 C-H (bend) or 1375 cm-1 C-H (bend) Question 8 Label the following nuclear reaction as fusion or fission and explain your answer. 238U92 + 14N7 → 248Es99 + 4 1n0 Your Answer: fusion a larger nuclei is formed from two smaller nuclei Question 9 Balance the following nuclear reaction and calculate the energy yield from the reaction: 7Be4 + ? → 7Li3 Your Answer: 7Be4 + 0e-1 → 7Li3 7Be4 = 7.01693 0e-1 = 0.000549 7Li3 = 7.01600 m = (7.01693 + 0.000549) - 7.01600 = 1.479x10-3 g/mol = 1.479x10-6 kg/mol E = mc2 = 1.479x10-6 (3.00x108)2 = 1.331x1011 J/mol Question 10 1. What is a nucleon? 2. Radioactivity nuclear reactions generally have ______ reactant(s) and ______ product(s). Your Answer: 1. Nucleon is a particle found in the nucleus and can be a proton or a neutron. 2. one ; more than one