CHEM 1411 PRACTICE EXAM II (Chapters 4, 5, 6): 25 questions. Multiple Choices: Select one best answer. Q1‐9: Chapter 4; Q10‐18: Chapter 5; Q19‐25: Chap 6 Page numbers include both 9th and 10th editions of the textbooks. 1. Which of the following is a weak electrolyte? (a) barium hydroxide solution (b) argon gas (c) ammonia solution (d) water (e) liquid aluminum chloride Hint: 11th ed.: p.p. 119‐120; 10th ed. p. 122; 9th ed. p.p. 120‐121. Weak electrolytes include molecular compounds that are weak acids and weak bases. Strong electrolytes include ionic compounds and molecular compounds that are strong acids. You need to memorize all the strong acids as strong electrolytes. Nonelectrolytes include all molecular compounds except strong acids, weak acids and weak bases. 2. Which of the following is wrong concerning a net‐ionic equation? (a) S2‐(aq) + Zn2+(aq)  ZnS(s) (b) 2NaOH(aq) + MgSO4(aq) Mg(OH)2(s) + Na2SO4(aq) (d) CO32‐(aq) + 2H+(aq)H2O(l) + CO2(g) (c) SO42‐(aq) + Pb2+(aq) PbSO4(s) (e) Fe(s) +Ni2+(aq)  Fe2+ + Ni(s) Hint: 11th ed.: p.p. 123‐125; 10th ed. p.p. 126‐127. Example 4.2; 9th ed. p.p. 124‐126 & Examples 4.1 & 4.2. Check with the solubility table for the products, which must be solids (s), liquids (l) or gases (g). For (e): see Fig 4.16. 3. Which of the following is a Brǿnsted acid? (b) CH3COO‐ (c) SO42‐ (d) ClO4‐ (e) NO3‐ (a) HSO4‐ th
th
th
Hint: 11 ed.: p.p. 127‐128; 10 ed. p. 130; 9 ed. p. 128. Example 4.3. You need to memorize the definition. Definition: Brǿnsted acids must contain available (releasable) hydrogen ions while the Brǿnsted bases must be able to accept hydrogen ions. To obtain Bronsted acid, simply add H+; to obtain Bronsted base, simply subtract H+. 4. Which of the following underlined atoms contains the oxidation number as ‐1? (b) CaC2 (c) SO42‐ (d) PtCl42‐ (e) NaO2 (a) Cs2O th
Hint: 11 ed.: p.p. 135‐136 and see Examples 4.5 and 4.6; 10th ed. p.p. 136‐139; 9th ed. p.p. 134‐137. You need to Memorize the rules. See Example 4.5. 5. Which of the following redox reactions will occur according to activity series? (a) Cu(s) + 2HCl(aq)  CuCl2(aq) + H2(g) (b) 2Ag(s) + 2H2O(l)  AgOH(s) + H2(g) (c) Mg(s) + CuSO4(aq)  MgSO4(aq) + Cu(s) (d) Pt(s) + FeSO4(aq)  PtSO4(aq) + Fe(s) (e) 2Al(s) + 3Ca(NO3)2(aq)  2Al((NO3)3(aq) + 3Ca(s) Hint: 11th ed.: Activity table: 11th ed.: p. 141, fig. 4.16; 10th ed. Fig 4.16, p. 143; 9th ed. Fig. 4.16, p. 139. The one on the top of the Table is more reactive then the one at the bottom. So in order to have a reaction, the metal on top must be in the neutral atomic form and the one below must be in the compound or ionic form. Or you can simply memorize that the reaction direction always goes from top metal to the bottom metal. 6. 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 Hint: 11th ed.: p.p. 145‐147 and Example 4.7; 10th ed. p.p. 147‐148, Example 4.6; 9th ed. p.p. 142‐146. Example 4.6. 1
7. 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? (d) 0.0534 (e) 0.0433 (a) 0.0252 (b) 0.0368 (c) 0.0117 Hint: 11th ed.: p.p. 147‐148 and Example 4.9; 10th ed. p.p. 149‐151 Similar to Example 4.8; 9th ed. p.p. 146‐148. Similar to the reverse question of Example 4.8. 8. 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: 11th ed.: p.p. 147‐148 and p. 162: 4.77 and 4.78; 10th ed. p.p. 149‐151; 9th ed. p. 146‐148. p. 160: 4.74. Mixing solutions is diluting solution. New concentration = mole of solute/ total volume of solutions. Thus, M = (0.568x46.2+1.396 x 80.5)/(46.2+80.5) = 1.09. 9. 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: 11th ed.: p.p. 151‐154 and Example 4.12; 10th ed. p.p. 153‐156 Example 4.11 ; 9th ed. p.p. 150‐
153. Solution stoichioimetry. Example 4.11. Short‐cut formula of 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. 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: Boyle’s law: 11th ed.: p.p. 178‐181 and p. 216: 5.19; 10th ed. p.p. 179‐182 Equation (5.2); 9th ed. p.p. 175‐178. Equation (5.2). 11. What is the final temperature (in K), under constant‐pressure condition, when a sample of hydrogen gas initially at 88oC and 9.6 L is cooled until its finial volume is 3.4 L? (a) 31 (b) 68 (c) 94 (d) 130 (e) 260 Hint: Charles’s Law: 11th ed. 183 and p. 216: 5.24; 10th ed. p.p. 183‐184 Equation (5.4); 9th ed. p.p. 178‐180. Equation (5.4). 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: Ideal gas law. 11th ed.: p. 184 and the Practice Exercise of Example 5.3 (p. 186); Equation (5.8). 10th ed. p.p. 185‐187 Examples 5.3 & 5.4; 9th ed. p.p. 181‐182. Equation (5.8). Example 5.4. 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 20.1oC and 1.00 atm. What was the volume (L) of this gas at the fermentation temperature of 36.5oC and 2.00 atm pressure? (a) 0.41 (b) 0.82 (c) 1.43 (d) 2.67 (e) 3.54 Hint: Combined gas law. p. 184 equation (5.10). 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 oC? (a) 35.0 (b) 70.3 (c) 86.2 (d) 94.6 (e) 102.3 Hint: 11th ed.: p. 191, equation (5.12) and Example 5.9; 10th ed. p.p. 192‐194 Examples 5.9 & 5.10; 9th ed. p. 187. Equation (5.12). Example 5.9. From PM = dRT where P is in unit of atm and T is in unit of Kelvin  M = dRT/P = [(7.10/5.40)x0.082x(40+273.15)]/(741/760) = 35.0 g/mol 2
Application: What is the density of HBr gas in grams per liter at 733 mmHg and 46 oC? (a) 0.54 (b) 1.36 (c) 2.24 (d) 2.97 (e) 3.57 Hint: 11th ed.: p. 190 and Example 5.8; 10th ed. p.p. 190‐191 Example 5.8; 9th ed. p. 186. Equation (5.11). Example 5.8. From PM= dRT  d = PM/RT = (733/760)x(1.008+79.90)/[0.082x(46+273.15)] = 2.97 g/L 14. A compound has the empirical formula as SF4. At 20 oC, 0.100 gram of the gaseous compound occupies a volume of 22.1 mL and exerts a pressure of 1.02 atm. What is the molecular formula of the gas? (b) SF6 (c) S2F10 (d) S4F16 (e) S5F20 (a) SF4 Hint: 11th ed. p.p. 191‐193 Example 5.10; 10th ed. p.p. 193‐194 Example 5.10; 9th ed. p. 211 (5.50) or p. 187. Equation (5.12). Example 5.9. To find molecular formula, we need to find the molar mass from PM = dRT and convert 22.1 mL to 0.0221 L  M = dRT/P = (0.100/0.0221)x0.082x(20+273.15)/1.02 = 106.64 g/mol. Ratio = molar mass/empirical molar mass = 106.64/(32.07+19x4) = 0.99 round to whole number as 1. So the molecular formula contains one empirical formula unit (see Example 3.11 in Chapter 3) and thus the molecular formula is identical to the empirical formula as SF4. 15. 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 oC and 0.985 atm? (a) 370 (b) 430 (c) 510 (d) 630 (e) 720 Hint: Gas stoichiometry: 11th ed. p.p. 193‐195 Examples 5.11 and 5.12; 10th ed. p.p. 194‐196 Examples 5.12 & 5.13; 9th ed. p.p. 190‐192. Example 5.11. This question is taken from p. 211 (5.52). From 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. 16. In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide: C6H12O6(s)  2 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: Gas stoichiometry: 11th ed. p.p. 193‐195 Examples 5.11 and 5.12; 10th ed. p.p. 194‐196. This question is taken from p. 217 (5.54); 9th ed. p.p. 190‐192. This question is taken from p. 212 (5.54). 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 %. 17. What is the total pressure (in atm) of the mixture when a 2.5‐L flask at 15 oC 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: Total pressure is the sum of all partial pressures of component gases. To use this formula, all the gases in the same container can not react to each other (i.e. non‐reactive gases). 3
11th ed. p.p. 195‐198; 10th ed. p.p. 196‐199. This question is taken from p. 218 (5.64 (a)); 9th ed. p.p. 192‐195.This question is taken from p. 212 (5.64 (a)). Total pressure = 0.32+0.15+0.42= 0.89 atm. 18. 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: 11th ed. p.p. 204‐209 Example 5.17; 10th ed. p.p. 208‐209 Example 5.17. This question is taken from p. 219 (5.84); 9th ed. p.p. 203‐204. Example 5.17. This question is taken from p. 213 (5.84). {rate of CH4 / rate of Ni(CO)x} = Square root of {molar mass of CH4 / molar mass of Ni(CO)x} = 3.3/ 1 =  Take square at the both side  rate of CH4 / rate of Ni(CO)x = 10.89  rate of Ni(CO)x = 16 x 10.89 = 174.24 = 58.69 + 28x  x = {(174.24 – 58.69)/28} = 4.1267  Round to whole number = 4 19. 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: 11th ed. p. 239 Example 6.2 table 6.1; 10th ed. p. 238: Example 6.2; 9th ed. p. 232: Example 6.2. 20. Consider the reaction below: 2CH3OH(l) + 3O2(g)  4H2O(l) + 2CO2(g) ∆H = ‐1452.8 kJ What is the value of ∆H for the reaction of 8H2O(l) + 4CO2(g)  4CH3OH (l) + 6O2 (g)? (a) +1458.2 (b) ‐1458.2 (c) 3.46 (d) ‐2905.6 (e) +2905.6 Hint: 11th ed. Application of rules in p.p. 242‐244 or see p. 265 End‐of‐Chapter Questions & Probems # 6.24; 10th ed. p.p. 241‐243; 9th ed. p. 236: Guidelines or p. 249: Examle 6.9: concepts to use in multiple or reversing equation.. 21. What is the heat absorbed (in kJ) when 250 grams of water is heated from 22oC to 98oC? The specific heat of water is 4.18 J/g.K (or 4.18 J/g. oC) (a) 79 (b) 88 (c) 97 (d) 102 (e) 137 Hint: 11th ed. p.p.246‐247 Example 6.5; 10th ed. p.p. 245‐246 Example 6.5; 9th ed. p. 240: Example 6.5. 22. A sheet of gold weighing 10.0 g and at a temperature of 18.0oC is placed flat on a sheet of iron weighing 20.0 g and at a temperature of 55.6oC. What is the final temperature (oC) 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.oC, 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. 11th ed. p. 252 Example 6.7; 10th ed. p. 249: Example 6.7 ; 9th ed. p. 243: Example 6.7. p. 256: problem 6.36. So 10.0x0.129x(T‐18)+20.0x0.444x(T‐55.6) = 0  T = 50.7 oC. 23. Calculate the heat of combustion (kJ) for the following reaction: 2H2S(g) + 3O2(g)  2H2O(l) + 2SO2(g) The standard enthalpies of formation for H2S(g), H2O(l), SO2(g) and O2(g) are 4
‐20.15, ‐285.8, ‐296.4 and 0.0, respectively. (a) +1124 (b) ‐1124 (c) +562 (d) ‐562 (e) +281 Hint: The Direct Method: 11th ed. p. 256 and Example 6.10; 10th ed. 252‐254 Equation (6.18), Example 6.10; 9th ed. p. 246: Equation 6.17 or 6.18, p. 247: in‐text example, and p. 250: Example 6.10. 24. 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 Calculate 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). 11th ed. p.p. 256‐259 and Example 6.9; 10th ed. p.p. 255‐
258 Example 6.9; 9th ed. p.p. 248‐250: in‐text example and Example 6.9. Terms used and procedures: (a) Identity refers to either react or product; (b) Quantity refers to coefficient; (c) Given reactions: Two or more reactions with ∆H given; (d) 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); (e) Resultant reaction/equation: The one that was obtained by adding the given reactions that have been manipulated. (f) 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. 25. Calculate the standard enthalpy change (kJ) for the reaction 2Al(s) + Fe2O3(s)  2Fe(s) + Al2O3(s) Given that ∆H = ‐1669.8 kJ 2Al(s) + 3/2O2(g)  Al2O3(s) ∆H = ‐822.2 kJ 2Fe(s) + 3/2O2(g)  Fe2O3(s) (a) ‐637.1 (b) ‐847.6 (c) ‐984.6 (d) ‐1120.3 (e) +847.6 Hint: Indirect Method (Hess’ Law). 11th ed. p.p. 256‐259 and Example 6.9;10th ed. p.p. 255‐258 Example 6.9; 9th ed. p.p. 248‐250: in‐text example and Example 6.9. 5