CHEM 1411 Zumdahl & Zumdahl PRACTICE EXAM II (Chapters 4, 5, 6) Multiple Choices: Select one best answer. 1. Which of the following is a weak electrolyte? (a) barium hydroxide solution (b) ammonia solution (c) argon gas (d) liquid aluminum chloride (e) water th Hint: For 9 ed.: Section 4.2. Strong electrolytes: (1) All ionic compound and (2) Molecular compounds which are strong acids: HNO3, H2SO4, HCl, HBr, HI, HClO3, HBrO3, HlO3, HClO4, HBrO4, HlO4. Note: Strong bases are strong electrolytes because they are IA and IIA metal oxides and metal hydroxides. Weak electrolytes: Molecular compounds that are weak acids and weak bases. Common weak acids: H2CO3, HNO2, H2S, CH3COOH, HF, H3PO4, etc. Common weak base: NH3 (some people write it as NH4OH. Nonelectrolytes: Molecular compounds exclude strong and weak electrolytes. Common nonelectrolytes: sugar ( -ose), acetone ( -one), alcohol ( -ol), ether ( -O- ) etc. 2. How many grams of KOH are present in 35.0 ml of a 5.50 M solution? (a) 5.5 (b) 10.8 (c) 15.7 (d) 17.8 (e) 21.3 th Hint: For 9 ed., see Section 4.3: p.p. 145 150. From M = mole solute/liter solution and 1 L = 1000 ml Then 5.50 = mole KOH/0.035 mole KOH = 5.50x0.035 = 0.1925 From mole = mass in gram/molar mass Mass in gram = mole x molar mass = 0.1925x(39.10+16+1) = 0.1925x56.1 = 10.799 g 3. Which of the following solutions DOES contain 1.0 M nitrate ion? (a) 1.0 M Co(NO3)2 (b) 0.5 M Al(NO3)3 (c) 0.25 M Mn(NO3)4 (d) 0.4 M Mg(NO3)2 (e) 0.5 M AgNO3 Hint: For 9 th ed. See Section 4.3: p. 147: Interactive Example 4.3. 4. What is the final concentration (in M) of a solution when water is added to 25.0 ml of a 0.866 M KNO3 solution until the volume of the solution is exactly 500.0 ml? (a) 0.0252 (b) 0.0368 (c) 0.0117 (d) 0.0534 (e) 0.0433 Hint: 9 th ed., see Section 4.3: p.p. 150 153: Interactive Example 4.7. Another type of dilution: see in CHEM 1412: acid-base titration. What is the concentration (in M) of the final solution when a 46.2 ml, 0.568 M Ca(NO3)2 solution is mixed with 80.5 ml of 1.396 M Ca(NO3)2 solution? (a) 0.568 (b) 1.09 (c) 0.953 (d) 1.545 (e) 1.874 Hint: Mixing solutions is diluting solution. 1
New concentration = total mole of solute/ total volume of solutions. Thus, M = (0.568x46.2+1.396 x 80.5) mmol/(46.2+80.5) ml = 1.09 M. 5. Which of the following pair when mixed will NOT produce a precipitate or solid? (a) MgSO 4 (aq) and Pb(NO3) 2 (aq) (b) NaCl(aq) and AgNO3(aq) (c) Ca(OH) 2 (aq) and K2CO3(aq) (d) Hg(NO 3 )2(aq) and HBr(aq) (e) NH4OH(aq), ammonia and CH3COOH(aq) vinegar Hint: For 9 th ed. See Section 4.5: p.p. 153-158, Interactive Example 4.8. Be sure to use Table 4.1: Simple Rules for the Solubility of Salts in Water to answer this question. 6. Which of the following is wrong concerning a net ionic equation? 2 2+ (a) S (aq) + Zn (aq) ZnS (s) (b) 2Na(s) + Mg 2+ (aq) Mg(s) + 2Na + (aq) (c) Pb(NO 3 )2(aq) + 2Na(s) 2NaNO 3 (aq) + Pb(s) 2 + (d) CO 3 (aq) + 2H (aq) H 2 O(l) + CO 2 (g) (e) Fe(s) + Ni 2+ (aq) Fe 2+ + Ni(s) Hint: For 9 th ed. See Section 4.6: p.p. 158 160. Check with the solubility table for the products, which must be solids (s), liquids (l) or gases (g). For (e): see Fig 4.16. 7. When aqueous solutions of sodium sulfate and lead (II) nitrate are mixed, lead (II) sulfate precipitates. How many grams of lead (II) sulfate formed when 1.25 L of 0.0500 M lead (II) nitrate and 2.00 L of 0.0250 M sodium sulfate are mixed? (a) 15.2 g (b) 19.0 g (c) 34.1 g (d) 34.2 g (e) 68.3 g Hint: For 9 th ed. See Section 4.7: p.p. 160-162: Interactive Example 4.11. This is a comprehensive question involving chapter 2 (naming and chemical formula) and balancing equation (chapter 3) and limiting reagent (chapter 3) and obtaining mole from molarity and volume (chapter 4). This is a solution-stoichiometry-with-limiting-reagent question (chapter 3): Na2SO 4 (aq) + Pb(NO 3 )2 (aq) PbSO 4 (s) + 2NaNO 3 (g) Assume lead (II) nitrate Pb(NO 3 )2 is the limiting reagent: (1.25 L Pb(NO 3 )2 x {0.0500 mole Pb(NO 3 )2 /1 L Pb(NO 3 )2 } x (1 mol PbSO 4/1 mol Pb(NO 3 )2 ) x (303.2 g PbSO 4/ 1 mol PbSO 4) = 18.95 g PbSO 4 Assume sodium sulfate Na2SO 4 is the limiting reagent: (2.00 L Na2SO 4 ) x {0.0250 mole Na2SO 4 /1 L Na2SO 4 } x (1 mol PbSO 4/1 mol Na2SO 4 ) x (303.2 g PbSO 4/ 1 mol Na2SO 4 ) = 15.16 g PbSO 4 So sodium sulfate Na2SO 4 is the limiting reagent and lead (II) nitrate Pb(NO3) 2 is the excess reagent and there are 15.16 g lead (II) sulfate PbSO 4 produced. 2
8. What volume (in ml) of a 0.500 M HCl solution is needed to neutralize 10.0 ml of a 0.2000 M Ba(OH)2 solution? (a) 8.00 (b) 4.00 (c) 2.00 (d) 1.00 (e) 0.50 Hint: For 9 th ed., see Section 4.8: p.p. 167 169; also see Interactive Example 4.14: Solution Stoichioimetry. Short cut without balancing equation: Formula for acid base neutralization: ia x Ma x Va = ib x Mb x Vb where a refers to acid and b refers to base. ia refers to number of H in the chemical formula of acid and ib refers to the number of OH in the chemical formula of base. 9. Which of the following underlined atoms contains the oxidation number as 1? (a) Cs2O (b) CaC2 (c) SO4 2 (d) PtCl4 2 (e) NaO2 Hint: For 9 th ed. See Section 4.9: p.p. 170 174: Interactive Example 4.16. Be sure to memorize Table 4.2 Rules for Assigning Oxidation States. 10. What volume (in L) does a sample of air occupy at 6.6 atm when 1.2 atm, 3.8 L of air is compressed? (a) 0.34 (b) 0.57 (c) 0.69 (d) 0.77 (e) 0.86 Hint: For 9 th ed., see Section 5.2: Boyle s law: Interactive Example 5.2. P1V1 = P2V2 1.2x3.8 = 6.6xV V = 0.69 L. 11.What is the final temperature (in K), under constant pressure condition, when a sample of hydrogen gas initially at 88 o C and 9.6 L is cooled until its finial volume is 3.4 L? (a) 31 (b) 68 (c) 94 (d) 128 (e) 261 Hint: For 9 th ed. See Section 5.2: Charles s Law: Interactive Example 5.4. T is in Kelvin. V1/T1 = V2/T2 9.6/(88+273.15) = 3.4/T2 9.6/361.15 = 3.4/T2 T2 = 3.4/0.02658 = 127.907 Kelvin 12. What is the volume (in L) of 88.4 g of carbon dioxide gas at STP? (a) 45.1 (b) 53.7 (c) 62.1 (d) 74.6 (e) 83.2 Hint: For 9 th ed., see Section 5.3: p.p. 198-203: Ideal gas law. From PV = nrt where P is pressure in atm, V is volume in liter, n is mole, R is ideal gas constant 0.082 atm.l/mol.k and T is temperature in Kelvin; so V = nrt/p = {(88.4/44)x0.082x(0+273.15)}/1 = 45.1 Liters. Application: A gas evolved during the fermentation of glucose (wine making) has a volume of 0.78 L at 20.1 o C and 1.00 atm. What was the volume (L) of this gas at the fermentation temperature of 36.5 o C and 2.00 atm pressure? (a) 0.41 (b) 0.82 (c) 1.43 (d) 2.67 (e) 3.54 Hint: For 9 th ed., see Section 5.3: Combined Gas Law: Interactive Example 5.9: p. 202. P1V1/T1 = P2V2/T2 1x0.78/(20.1+273.15) = 2xV/(36.5+273.15) 0.00266 = 0.00646xV V = 0.00266/0.00646 = 0.412 L 3
13. What is the molar mass (g/mol) of 7.10 grams of gas whose volume is 5.40 L at 741 torr and 40 o C? (a) 35.0 (b) 70.3 (c) 86.2 (d) 94.6 (e) 102.3 Hint: For 9 th ed., see Section 5.4: p.p. 207-208: Formula (5.1); Interactive Example 5.14. From PM = drt where P is in unit of atm and T is in unit of Kelvin Here, density is given indirectly as density = mass/volume M= drt/p = [(7.10/5.40)x0.082x(40+273.15)]/(741/760) = 35.0 g/mol Application: What is most likely the unknown gas according to Q 13? (a) C2H4 (b) HCl (c) C3H4 (d) CO2 Hint: From the molar mass comparison, the closest one will be most likely the unknown. Application: What is the density of HBr gas in grams per liter at 733 mmhg and 46 o C? (a) 0.54 (b) 1.36 (c) 2.24 (d) 2.97 (e) 3.57 Hint: 9 th ed., see Section 5.4: Formula (5.1) From PM= drt d = PM/RT = (733/760)x(1.008+79.90)/[0.082x(46+273.15)] = 2.97 g/l 14. The combustion process for methane, major component of natural gas, is CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) If 15.0 moles of methane are reacted, what is the volume of carbon dioxide (in L) produced at 23.0 o C and 0.985 atm? (a) 370 (b) 430 (c) 510 (d) 630 (e) 720 Hint: Gas stoichiometry: For 9 th ed., see Section 5.4:203-208; Interactive Examples 5.12 and 5.13. From the given information, 15.0 mole of methane, we can calculate the theoretical yield in mole of carbon dioxide. Then apply PV = nrt to calculate volume in liter for carbon dioxide. Here, mole of carbonxide = mole of methane = 15 moles and thus V = nrt/p = 15x0.082x(23+273.15)/0.985 = 369.81 L. 4
15. Application: Comprehensive Question: In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide: C6H12O6(s) C2H5OH(l)) + 2 CO2(g) If 5.97 g of glucose are reacted and 1.44 L of carbon dioxide gas are collected at 293 K and 0.984 atm, what is the percent yield of the reaction? (a) 88.9 % (b) 76.3% (c) 65.9% (d) 56.2% (e) 47.6% Hint: Hint: Gas stoichiometry: For 9 th ed., see Section 5.4:203-208; Interactive Examples 5.12 and 5.13. From mass of glucose, we can calculate the theoretical yield of carbon dioxide in mole; from 293 K and 0.984 atm, we can calculate the actual yield of carbon dioxide in mole. Thus, the theoretical yield of carbon dioxide = 2 x mole of glucose = 2 x (5.97/12x6+1x12+16x6) = 2 X (5.97/180) = 0.0663 mole. From PV = nrt n = PV/RT we can calculate the actual yield = 0.984x1.44/(0.082x293) = 0.05898 = 0.0590 mole. So the percent yield = (actual yield / theoretical yield) x 100% = (0.0590/0.0663)x100% = 88.95 %. 5
16. What is the total pressure (in atm) of the mixture when a 2.5 L flask at 15 o C contains a mixture if nitrogen, helium, and neon gases at partial pressure of 0.32 atm for nitrogen, 0.15 atm for helium, and 0.42 atm for neon? (a) 0.49 (b) 0.51 (c) 0.64 (d) 0.73 (e) 0.89 Hint: Dalton s Law of Partial Pressures: For 9 th ed., see Section 5.5. The total pressure is the sum of all partial pressures of component gases: Total pressure = 0.32+0.15+0.42= 0.89 atm. To use this formula, all the gases in the same container cannot react to each other (i.e. non reactive gases). Application: See Interactive Example 5.18: When collecting gas over the water, the gas must be insoluble or slightly soluble in water: for acidic gases, HCl(g), SO2(g), etc., and basic gas, NH3(s), they cannot be collected over the water because they dissolve in water easily. From Dalton s partial pressure, Ptotal = Pwater vapor + Pgas, then P gas = P total - P water vapor 17. Nickel forms a gaseous compound of the formula Ni(CO)x. What is the value of x given the fact that under the same conditions of temperature and pressure, methane (CH4) effuses 3.3 times faster than the compound? (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 Hint: Graham s law of effusion: For 9 th ed., see Section 5.7: p.p. 222-224: Application of Figure 5.24. See End-of-Chapter Exercises 111, 112, and 113. {Rate of effusion for CH4 / Rate of effusion for Ni(CO)x} = Square root of {molar mass of CH4 / molar mass of Ni(CO)x} = {molar mass of CH4 / molar mass of Ni(CO)x} 1/2 = 3.3/ 1 Take square at the both side of equation, then Rate of effusion for CH4 / Rate of effusion for Ni(CO)x = 10.89 Rate of effusion for Ni(CO)x = 16 x 10.89 = 174.24 = 58.69 + 28x x = {(174.24 58.69)/28} = 4.1267 requiring rounding to whole number thus x = 4 18. What is the change in energy (J) of the gas when a gas expends and does P V work on the surroundings equal to 325 J and at the same time absorbs 127 J of heat from the surroundings? (a) 198 (b) + 198 (c) + 157 (d) 157 (e) + 452 Hint: For 9 th ed., see Section 6.3: Example 6.1. 19. What is the heat absorbed (in kj) when 250 grams of water is heated from 22 o C to 98 o C? The specific heat of water is 4.18 J/g.K (or 4.18 J/g. o C) (a) 79 (b) 88 (c) 97 (d) 102 (e) 137 6
Hint: For 9 th ed., see Section 6.2: p. 255. 20. A sheet of gold weighing 10.0 g and at a temperature of 18.0 o C is placed flat on a sheet of iron weighing 20.0 g and at a temperature of 55.6 o C. What is the final temperature ( o C) of the combined metals? Assume that no heat is lost to the surroundings. The specific heat of iron and gold are 0.444 and 0.129 J/g. o C, respectively. (Hint: the heat gained by the gold must be equal to the heat lost by the iron.) (a) 50.7 (b) 63.1 (c) 72.4 (d) 47.2 (e) 36.8 Hint: Heat absorbed by the colder object gold + heat released by the hotter object iron = 0 or Hear absorbed by the colder object gold = heat released by the hotter object iron. For 9 th ed., see Section 6.2: End-of-Chapter Exercises 53 & 55. From the Law of Conservation of Energy, qhot + qcold = 0 10.0 x 0.129 x (T 18) + 20.0 x 0.444 x (T 55.6) = 0 T = 50.7 o C. Prior to introduce Hess s Law, learn the characteristics of Thermodynamics: 1. When reverse the reaction, the Hnew = - Hold 2. When the reaction coefficient is multiplied by a constant, the Hnew = constant x H old 3. When the third chemical equation is the sum of the other two equations, then H3 = H1 + H2 21. Consider the reaction below: 2CH3OH(l) + 3O2(g) 4H2O(l) + 2CO2(g) H = 1452.8 kj (g)? What is the value of H for the reaction of 8H2O(l) + 4CO2(g) 4 CH3OH (l) + 6O2 (a) +1458.2 (b) 1458.2 (c) 3.46 (d) 2905.6 (e) +2905.6 Hint: This question is a combination of Rules 1 and 2. Thus, H = 2x(+1452.8 kj) = +2905.6 kj 22. Calculate the standard enthalpy change (kj) for the reaction Given that 2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s) 2Al(s) + 3/2O2(g) Al2O3(s) 2Fe(s) + 3/2O2(g) Fe2O3(s) H = 1669.8 kj H = 822.2 kj (a) 637.1 (b) 847.6 (c) 984.6 (d) 1120.3 (e) +847.6 Hint: Hess s Law: the indirect method for calculating the Hrxn. For 9 th ed., see Section 6.3: 7
p.p. 260-264: Interactive Examples 6.2 & 6.8; End-of-Chapter Exercises 110, 111, and 112. Terms used and procedures: *Identity refers to either react or product; *Quantity refers to coefficient; *Given reactions: Two or more reactions with H given; *Target reaction/equation: The one that is asked; Must use it as reference. Any given reaction that is different from it must be reversed or multiplied by a number (whole or fraction); *Resultant reaction/equation: The one that was obtained by adding the given reactions that have been manipulated. Conclusion: If the resultant reaction is identical to the target reaction, you are solving it correctly. So add all the given reactions that have been manipulated together to get the H. 23. From the following data, C(graphite) + O2(g) CO2(g) H2(g) + 1/2O2(g) H2O(l) 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l) H = 393.5 kj H = 285.8 kj H = 3119.6 kj What is the enthalpy change (kj) for the reaction 2C(graphite) + 3H2(g) C2H6(g)? (a) 84.6 (b) 42.3 (c) +84.6 (d) +42.3 (e) 67.2 Hint: The Indirect Method (Hess Law). For 9 th ed., see Section 6.3. 24. Calculate the standard enthalpy change (kj) for the reaction Given that 2HCl(g) + F2(g) 2HF(l) + Cl2(g) 4HCl(g) + O2(g) 2H2O(l) + 2Cl2(g) 1/2H2(g) + 1/2F2(g) HF(l) H2(g) + ½O2(g) H2O(l) H = -202.4 kj H = -600.0 kj H = -285.8 kj (a) -637.1 (b) -847.6 (c) -984.6 (d) -1015.4 (e) +847.6 8
25. Calculate the heat of combustion (kj) for the following reaction: 2H2S(g) + 3O2(g) 2H2O(l) + 2SO2(g) The standard enthalpies of formation ( Hf) for H2S(g), H2O(l), SO2(g) and O2(g) are Hf 20.15, 285.8, 296.4 and 0.0, respectively. kj/mol (a) +1124 (b) 1124 (c) +562 (d) 562 (e) +281 Hint: The Direct Method: Using the standard entropy of formation ( Hf) to calculate the entropy of reaction ( Hrxn). For 9 th ed., see Section 6.4: 264-271: Use formula (6.1) in p. 267. H rxn = {2 H f, H2O(l) + 2 H f, SO2(g) } {2 H f, H2S(g) + 3 H f, O2(g) } = {2x(-285.8) +2x(-296.4)} {2x(-20.15) + 3x0} = 1124 kj 9
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