Chem 111 Key Review Final

Review - Chem 111 1) Which of the following compounds will produce an acidic solution when dissolved in water? a) NO2 (nonmetal oxide) b) NaClO4 c) K2SO3 d) N2O e) NaCN 2) Which of the following compounds will produce a basic solution when dissolved in water? a) K2O (metal oxide) b) HNO3 c) NH4Cl d) HBr e) KBr 3) For the equilibrium given below, list the two pairs of conjugate acid/base: HSO3- + HF <--> H2SO3 + F- Answer: a) F-/HF b) HSO3-/H2SO3 4) Consider a saturated solution of CaF2 (s). Which of the following may take place upon the addition of Ca(NO3)2? Circle all correct answers.   CaF2 (s) <--> Ca2+ (aq) + 2F- (aq) a) More CaF2 (s) dissolves. b) More CaF2 will precipitate. c) The concentration of the fluoride ions will decrease. d) The concentration of Ca2+ ions will increase. e) The concentration of NO3- will have no effect on the solubility of CaF2 (s). f)   None of (a) to (e) will happen. All of the above will happen. 5) The equilibrium concentration of HSO3- is much higher than the equilibrium conc. of SO32- in the reaction: HSO3- + H2O   <--> SO32- + H3O+ List two conjugate acid/base pairs and label each species as stronger or weaker acid or base. a) HSO3-/SO32- (weaker acid/stronger base) b) H3O+/H2O (stronger acid/weaker base) 6) A) Write the equilibrium equation for the solubility of CaCO3 (s). B) Circle all correct answers: The molar solubility of CaCO3 (s) in a saturated solution can be decreased by: a) Adding Na2CO3 b) Adding a strong acid c) Adding CaCl2 d) Adding more CaCO3 (s) 7) Fill in the table given below: | Unit cell | Simple cube | Body-centered cube | Face-centered cube | Hexagonal unit cell | |---|---|---|---|---| | Number of particles inside the unit cell   | 1 | 2 | 4 | 4 | | The coordination number | 6 | 8   | 12 | 12 | | Relative packing efficiency | lowest | high | highest | highest | | Relative Density | lowest | high | highest | highest | **8. a) What are the structural components that exist in a compound for hydrogen bonding to take place?** Ans: H-F, H-O, or H-N **b) The intermolecular forces that exist between nonpolar molecules are   called London forces.** **c) The intermolecular forces that exist between polar molecules are called dipole-dipole forces.** **9. a) MgCl2** **Ionic bond** **b) SO2 (bent geometry)** **Dipole-dipole forces** **c) Copper** **Metallic bond** **d) NH3 (pyramidal)** **Hydrogen bonding** **e) CO2 (linear)** **London forces** **f) KB** **Ionic bond** **10. a) What is the mass of one mole of cobalt atoms in grams?** Ans: 58.93 g **b) What is the mass of one cobalt atom in amu?** Ans: 58.93 amu **c) What is the mass of one cobalt atom in grams? Show the set-up:** 58.93 g Co * (1 mol Co / 6.022 x 10^23 atoms Co) = 9.79 x 10^-23 g/atom **11. a) Define ‘solution’.** A homogeneous mixture. **b) Is air a compound, an element, or a solution?** Solution **c) If you combine sand and water, are you preparing a new element, a new compound, or a solution? Explain your answer.**   Heterogeneous mixture. The mixture can be separated into components by physical means. **12. Write the chemical formulas of the following compounds:** a) sodium nitride: Na3N b) cobaltous phosphide: Co3P2 c) nickel (II) bisulfide: Ni(HS)2 d) Antimony (III) bisulfite: Sb(HSO3)3 e) lead (II) thiocyanate: Pb(SCN)2 f) Aluminum thiosulfate: Al2(S2O3)3 **13. How many moles of C6H6O contain 7.03 x 10^24 carbon atoms?** Setup: 7.03 x 10^24 atoms C * (1 mol C6H6O / 6 mol C) = 1.17 x 10^24 mol C6H6O **14. a) Explain how particles of a hydrophobic sol remain dispersed without precipitating.** **b) Heating may cause a hydrophobic sol to coagulate. Why?** **15. List three methods for coagulating a hydrophobic colloid.** a) b) c)   **54. What would be the ionization constant, Ka, of a weak monoprotic acid, HX, if it is 5.0% ionized in a 0.18 M solution?** Setup: HX + H2O ⇌ H3O+ + X- Ka = [H3O+][X-] / [HX] [H3O+] = [X-] = 5% of 0.18 M = 0.009 M [HX] = 0.18 - 0.009 = 0.171 M Ka = (0.009)² / 0.171 = 4.7 x 10^-4 **55. A compound contains 1.55 g Phosphorus and 1.20 g oxygen. Calculate the simplest formula of the compound. (molar mass: P = 31.00, O = 16.00)** Setup: P: 1.55 g / 31.00 g/mol = 0.050 mol O: 1.20 g / 16.00 g/mol = 0.075 mol Simplest formula: P2O3 **56. Calculate the molarity of a solution made by diluting 8.00 mL of 15.00 M H3PO4 to a 0.500 L solution.** Setup: M1V1 = M2V2 15.00 mol/L * 0.008 L = M2 * 0.500 L M2 = 0.240 mol/L **57. What would be the H+ concentration of a solution resulting from mixing 35.0 mL of 0.20 M HCl and 35.0 mL of 0.15 M NaOH? HCl + NaOH → NaCl + H2O** Setup: moles HCl = 0.20 mol/L * 0.035 L = 7.0 x 10^-3 mol moles NaOH = 0.15 mol/L * 0.035 L = 5.3 x 10^-3 mol moles HCl left over = 7.0 x 10^-3 - 5.3 x 10^-3 = 1.7 x 10^-3 mol [H+] = (1.7 x 10^-3 mol) / 0.070 L = 0.024 M **58. What is the mass of CO2 (g) collected in a 580 mL flask at 50°C and 1.50 atm? (R = 0.0821 L.atm/mol.K)** Setup: PV = nRT n = PV/RT = (580 mL * 1.50 atm) / (0.0821 L.atm/mol.K * 323 K) = 33.8 mol mass = moles * molar mass = 33.8 mol * 44.0 g/mol = 1484 g **59. What is the density of NH3 (g) at 100°C and 1.35 atm? (R = 0.0821 L.atm/mol.K)** Setup: D = P * molar mass / RT = (1.35 atm * 17.0 g/mol) / (0.0821 L.atm/mol.K * 373 K) = 0.749 g/L **54. What would be the ionization constant, Ka, of a weak monoprotic acid, HX, if it is 5.0% ionized in a 0.18 M solution?** Setup: HX + H2O ⇌ H3O+ + X- Ka = [H3O+][X-] / [HX] [H3O+] = [X-] = 5% of 0.18 M = 0.009 M [HX] = 0.18 - 0.009 = 0.171 M Ka = (0.009)² / 0.171 = 4.7 x 10^-4 **55. A compound contains 1.55 g Phosphorus and 1.20 g oxygen. Calculate the simplest formula of the compound. (molar mass: P = 31.00, O = 16.00)** Setup: P: 1.55 g / 31.00 g/mol = 0.050 mol O: 1.20 g / 16.00 g/mol = 0.075 mol Simplest formula: P2O3 **56. Calculate the molarity of a solution made by diluting 8.00 mL of 15.00 M H3PO4 to a 0.500 L solution.** Setup: M1V1 = M2V2 15.00 mol/L * 0.008 L = M2 * 0.500 L M2 = 0.240 mol/L **57. What would be the H+ concentration of a solution resulting from mixing 35.0 mL of 0.20 M HCl and 35.0 mL of 0.15 M NaOH? HCl + NaOH → NaCl + H2O** Setup: moles HCl = 0.20 mol/L * 0.035 L = 7.0 x 10^-3 mol moles NaOH = 0.15 mol/L * 0.035 L = 5.3 x 10^-3 mol moles HCl left over = 7.0 x 10^-3 - 5.3 x 10^-3 = 1.7 x 10^-3 mol [H+] = (1.7 x 10^-3 mol) / 0.070 L = 0.024 M **58. What is the mass of CO2 (g) collected in a 580 mL flask at 50°C and 1.50 atm? (R = 0.0821 L.atm/mol.K)** Setup: PV = nRT n = PV/RT = (580 mL * 1.50 atm) / (0.0821 L.atm/mol.K * 323 K) = 33.8 mol mass = moles * molar mass = 33.8 mol * 44.0 g/mol = 1484 g **59. What is the density of NH3 (g) at 100°C and 1.35 atm? (R = 0.0821 L.atm/mol.K)** Setup: D = P * molar mass / RT = (1.35 atm * 17.0 g/mol) / (0.0821 L.atm/mol.K * 373 K) = 0.749 g/L