Name Date Period key Change Font to white CHEMISTRY FINAL REVIEW CHAPTER 10: THE MOLE - Atomic Mass and Formula Mass - What is a Mole? - Avogadro s Number - Molar Mass - Mole Conversions - Mass to Moles - Particles to Moles - Volume to Moles - Empirical and Molecular Formulas - % Composition - Determining Empirical Formula - Determining Molecular Formula CHAPTER 11: THE MATHEMATICS OF CHEMICAL EQUATIONS (STOICHIOMETRY) - Interpreting Balanced Chemical Equations - Moles to Moles - Verifying the Law of Conservation of Matter - Solving Problems - Mass to? - Volume to? - Particles to? - Limiting Reactant Problems - Identifying the Limiting Reactant - Percent Yield CHAPTER 13: GASES - Kinetic Molecular Theory - The Nature of Gases - Properties of Gases (have mass, fill containers etc ) - Measuring Gases (P,V,n & T) - Atmospheric Pressure and the Barometer - Enclosed Gases (Manometers) - STP - The Gas Laws - Boyle s Law (P vs V) - Charles s Law (T vs V) - Avogadro s Law (n vs V) - Dalton s Law of Partial Pressures - The Ideal Gas Law - Relating the Ideal Gas Law to the Kinetic Molecular Theory - Gas Effusion CHAPTER 14: LIQUIDS AND SOLIDS (KNOW HOW THE INTERMOLECULAR FORCES PLAY A ROLE IN ALL THESE TOPICS) - Condensed States of Matter - Kinetic Molecular Theory Applied to Liquids and Solids - Intermolecular Forces (3 Types) - Intramolecular Forces (3 Types) - Properties of Liquids - Viscosity - Surface Tension - Unusual Properties of water Last Updated: 6/11/12 Page 1/9
- Nature of Solids - Changes of State - Energy and Change of State - Vaporization and Condensation - Freezing and Melting - Sublimation and Deposition - Heating Curves - Phase Diagrams CHAPTER 22: KINETICS - Reaction Rates - Reaction Mechanisms - Rate Laws - Collision Theory - Energy in Chemical Reactions (exo vs endo, Pot. E. graphs) - Factors Affecting Reaction Rates - Naturre of the Reactants - Temperature - Concentration - Surface Area - Catalysts CHAPTER 16: EQUILIBRIUM - Reversible Reactions - Chemical Equilibrium - The Equilibrium Constant (K eq ) - Homogeneous and Heterogeneous Equilibria - The Reaction Quotient (Q) - Le Chatelier s Principle - Changes in Concentration - Changes in Pressure - Effect of Changing Temperature - The Haber Process CHAPTER 18: ACIDS, BASES AND SALTS - Defining Acids and Bases - Properties of Acids and Bases - The Arrhenius Definition - The Bronsted-Lowry Definition - The Hydronium Ion - Conjugate Acid-Base Pairs - Determining the Strengths of Acids and Bases - Strong and Weak Acids - Strong and Weak Bases - Strength of Conjugate Acid-Base Pairs - The Acid Dissociation Constant - The Base Dissociation Constant - Calculating Dissociation Constants (Bikini Problems) - Acid-Base Properties of Salts - Naming and Identifying Acids and Bases CHAPTER 19: THE SELF IONIZATION OF WATER AND PH - The Self Ionization of Water - The ph Scale Last Updated: 6/11/12 Page 2/9
CHEMISTRY FINAL REVIEW PROBLEMS s 3.5 x 10 23 0.78 1.8 x 10-5 13.83 98.78 0.72 50.97 18,550 0.30 0.97 3.73 5.3 x 10 22 78.22 5.25 10-2 0.902 310.17 39.82 0.96 1.08 852 6.7 1. Calculate the molar mass of calcium phosphate Ca 3 (PO 4 ) 2 (3 40.078) + (2 30.974) + (8 15.999) 310.17 g/ mol 2. Find the mass of 0.89 mol of calcium chloride. 0.89 mol CaCl 2 110.984 g CaCl 2 = 98.78g CaCl 2 1 mol CaCl 2 3. Determine the number of atoms that are in 0.58 mol of selenium. 0.58 mol Se 6.02 10 23 atoms = 3.5 10 23 atoms 1 mol Se 4. A container with a volume of 893 L contains how many mole of air at STP? P = 1 atm V = 893 L N =? T = 0 C = 273.15 K PV = nrt ; n = PV = (1 atm)(893 L) = 39.82 mol Air RT (0.0821)(273.15 K) 5. How many formula units are in 3.5 g of sodium hydroxide? 98.78 g CaCl 2 3.5 10 23 atoms 39.82 mol Air 3.5 g NaOH 1 mol NaOH 6.02 10 23 formula units = 5.3 10 22 f.u. of NaOH 40 g NaOH 1 mol NaOH 5.3 10 22 f.u.naoh 6. What is the molarity of the solution produced when 145 g of sodium chloride is dissolved in enough water to make 2.75 L of solution? 145 g NaCl 1 mol NaCl = 0.902 mol NaCl = 0.902 M NaCl 2.75 Liter 58.443 g NaCl Liter 0.902 M NaCl 7. Lead will react with hydrochloric acid to produce lead(ii) chloride and hydrogen gas. How many moles of hydrochloric acid are needed to completely react with 0.36 mol of lead Mole - Mole Pb + 2HCl PbCl 2 + H 2 0.36 mol? mol 0.36 mol Pb 2 mol HCl = 0.72 mol HCl 1 mol Pb 0.72 mol HCl 8. Determine the mass of lithium hydroxide produced when 0.38 g of lithium nitride reacts with water to make ammonia (NH 3 ) and lithium hydroxide. Mass Mass Li 3 N + 3H 2 O NH 3 + 3 LiOH 0.38 g? g 0.38 g Li 3 N 1 mol Li 3 N 3 mol LiOH 23.9479 g LiOH = 0.78 g LiOH 34.83 g Li 3 N 1 mol Li 3 N 1 mol LiOH 0.78 g LiOH 9. Find the mass of glucose (C 6 H 12 O 6 ) required to produce 1.82 g of carbon dioxide from the following unbalanced reaction: C 6 H 12 O 6 (s) C 2 H 6 O (l) + CO 2 (g) Mass - Volume? g 1.82 g 1.82 g CO 2 1 mol CO 2 1 mol C 6 H 12 O 6 180.15 g C 6 H 12 O 6 = 7.32 g C 6 H 12 O 6 44 g CO 2 2 mol CO 2 1 mol C 6 H 12 O 6 C 6 H 12 O 6 (s) 2 C 2 H 6 O (l) + 2 CO 2 (g) 3.73 g C 6 H 12 O 6 Last Updated: 6/11/12 Page 3/9
10. How many moles of carbon dioxide gas are produced when 2.8 g of oxygen gas is used to burn propane (C 3 H 8 )? Volume - Volume C 3 H 8 + 5 O 2 3 CO 2 + 4 H 2 O 2.8 L? L 2.8 g O 2 1 mol O 2 3 mol CO 2 = 5.25 10-2 32.0 g O 2 5 mol O 2 5.25 10-2 mol 11. Determine the percent yield for the reaction between 15.87 g of ammonia (NH 3 ) and excess oxygen to produce 21.87 g of NO gas and an unknown amount of water. NH3 is the limiting reactant because O2 is in excess 21.87 g actually produced 4 NH 3 + 5 O 2 4 NO + 6 H 2 O 15.87 g excess? g 15.87 g NH 3 1 mol NH 3 4 mol NO 30.01 g NO = 27.96 g NO theoretically produced 17.03 g NH 3 4 mol NH 3 1 mol NO 21.87 g = 0.782 = 78.22 % Yield 78.22 % 27.96 g 12. If 14.1 g of chromium is heated with 9.3 g of chlorine gas, what mass of CrCl 3 will be produced? Limiting reactant is C2 2 Cr + 3 Cl 2 2 CrCl 3 14.1 g 9.3 g? g 14.1 g Cr 1 mol Cr 2 mol CrCl 3 158.355 g CrCl 3 = 42.94 g CrCl 3 51.996 g Cr 2 mol Cr 1 mol CrCl 3 9.3 g Cl 2 1 mol Cl 2 2 mol CrCl 3 158.355 g CrCl 3 = 13.83 g CrCl 3 theoretically produced 70.906 g Cl 2 3 mol Cl 2 1 mol CrCl 3 13. The air pressure for a certain tire is 109 kpa. What is this pressure in atmospheres? 109 kpa 1 atm = 1.08 atm 101.325 kpa Unit conversion 13.83 g CrCl 3 : 1.08 atm 14. What volume would be occupied by 100 g of oxygen gas at a pressure of 1.50 atm and a temperature of 25 C? P = 1.50 atm V = X n = 100 g O 2 1 mol O 2 = 3.125 molo 2 32 g O 2 T = 25 C + 273 C = 298 K V = nrt = (3.125 mol)(0.0821 L atm)(298 K) = P (mol K)(1.50 atm) : 50.97 L Last Updated: 6/11/12 Page 4/9
15. A gas occupies a volume of 458 ml at a pressure of 1.01 kpa and a temperature of 295 K. When the pressure is changed, the volume becomes 477 ml. If there has been no change in temperature, what is the new pressure? V 1 = 458 ml V 2 = 477 ml P 1 = 1.01 kpa P 2 = X P 2 V 2 = n 2 RT 2 P 1 V 1 = n 1 RT 1 P 1 V 1 = P 2 V 2 P 2 = P 1 V 1 V 2 = (1.01 kpa)(458 ml) = (477 ml) If given in atm, the : answer is 0.0096 atm 0.97 kpa 16. What is the pressure of a mixture of helium, nitrogen and oxygen if their partial pressures are 600 mm Hg, 150 mm Hg, 102 mm Hg? P Tot = X P He = 600 mm Hg P N2 = 150 mm Hg P Hg = 102 mm Hg P Total = P He + P N2 + P Hg = 600 mm Hg + 150 mm Hg + 102 mm Hg = 852 mm Hg : 17. If 25 g of solid ice at -20 C is heated to gaseous water a 125 C how, much energy is required? Specific heat of water = 1.00 cal/ g ºC Heat of fusion ( H fus ) = 80 cal/ g Specific heat of ice = 0.50 cal/ g ºC Specific heat of steam = 0.48 cal/g ºC This problem is a combination of all the problems below: from 20 C to 0 C, you need 10 cal of energy 25 = 250 to melt the ice you need(no temp change)..80 cal of energy 25 = 2000 for 0 C to 100 C you need..100 cal of enery 25 = 2500 to vaporize the water you need 540 cal of energy 25 = 13500 from 100 C to 125 C you need. 12 cal of energy 25 = 300 Heat of vaporization ( H vap ) = 540 cal/ g 852 mm Hg or 742 25 742 is how much energy it takes for 1g, then multiply by 25 º : 18,550 Cal Trial # Initial [A] Initial [B] Initial [C] Rate M/s 1 0.1 0.2 0.1 4 2 0.2 0.2 0.1 8 3 0.2 0.1 0.1 8 4 0.1 0.2 0.2 16 18. Use the table above to answer the following: a. What is the order with respect to: A: 1 st order b. What is the overall order? Comparing Trials #1 & #2 B: 0 order or Independent Comparing Trials #2 & #3 C: 2 nd order Comparing Trials #1 & #4 Adding up all the orders, we get: 1 + 0 + 2 = 3 rd order c. What is the value of k? Randomly choosing trial #1 to determine k, we get: : 3 rd order 4 = k(0.1)(0.1) 2 = 4000 4000 d. If [A] = 0.3, [B] = 0.4, and [C] = 0.5. What is the rate? rate = 4000(0.3)(0.5) 2 = 300 e. Write the overall rate expression: 300 rate = k[a][c] 2 Last Updated: 6/11/12 Page 5/9
19. Write out the equilibrium expression for the following reaction. Gaseous hydrogen reacts with gaseous oxygen to form water vapor. Keq = [H 2 H 2 (g) + O 2 (g) 2 H 2 O (g) 2 O] 2. [H 2 ] 2 [O 2 ] 20. At 740 C, K eq = 0.0060 for solid calcium carbonate decomposing into solid calcium oxide and carbon dioxide gas. Find the reaction quotient (Q) and predict how the reaction will proceed if [CO 2 ] = 0.0004 M. CaCO 3 (s) CaO (s) + CO 2 (g) K eq = [CO 2 ] = Q = 0.0004, so the reaction will shift RIGHT Q < K eq, too many reactants, not enough products so the reaction will shift right, forward, toward the products 21. For the following reaction 2 HI (g) H 2 (g) + I 2 (g) + heat In which direction will the equilibrium position be shifted for each of the following changes? a. H 2 (g) is added. The reaction would shift LEFT b. I 2 (g) is removed. The reaction would shift RIGHT c. HI (g) is removed. The reaction would shift LEFT d. Some Ar (g) is added. No effect e. The volume of the container is doubled. This would decrease the pressure, for this reaction, that would have no effect. f. The temperature is decreased. Since the reaction is exothermic, decreasing the temp causes the reaction to shift RIGHT 22. Identify the roles of the following (acid, base, conjugate acid, conjugate base) HOCl + C 6 H 5 NH 2 OCl - + C 6 H 5 NH 3 + Acid Base C. Base C. Acid 23. Acetic acid is a weak monoprotic acid. It is the active ingredient in vinegar. If the initial concentration of acetic acid is 0.200 M and the equilibrium concentration of H 3 O + is 1.9 x 10-3 M, calculate K a for acetic acid. [H 3 O + ] = 0.0019 M HA + H 2 O H 3 O + + A - [A - ] = 0.0019 M [HA] = 0.200 0.0019 = 0.1981 M K a = [H 3 O + ][A - ] = (0.0019)(0.0019) = 1.8 x 10-5 [HA] (0.1981) 1.8 10-5 24. Analysis of a sample of maple syrup reveals that the concentration of OH - ions is 5.0 x 10-8 M. What is the ph of this syrup? Is it acidic, basic or neutral? K w = [H 3 O + ][OH - [H ] 3 O + ] = K w. [OH - ] = (1.0 x 10-14 ) = 2.0 x 10-7 (5.0 x 10-8 ) - log (2.0 x 10-7 ) = 6.70 ph = -log[h 3 O + Acidic or ] [ph] 6.70 Basic? Acidic 25. A volume of 90-mL of 0.2 M HBr neutralizes a 60-mL sample of NaOH solution. What is the concentration of the NaOH solution? 0.018 mol NaOH = 0.30 M NaOH 0.090 L HBr 0.2 mol HBr 1 mol NaOH = 0.018 mol NaOH 0.060 L 1 L HBr 1 mol HBr 0.30 M NaOH 1. What are some things that intermolecular forces are responsible for? Assuming the forces get stronger how do they affect these things? Increase Surfaces tension, Increase viscosity, Increase boiling point, Increase melting point, Lower the evaporation Last Updated: 6/11/12 Page 6/9
rate 2. Describe the behavior of the molecules in a liquid. Explain this behavior in terms of intermolecular forces. In a liquid, the molecules can move relatively freely, the intermolecular forces keep them close, but not locked in place. As the intermolecular forces get stronger the molecules are less free to move until they become locked in place, this is a solid. 3. What is the most likely type of IMF s present in a substance if the substance has a high vapor pressure? dispersion forces 4. As you go down the nitrogen group of the Periodic Table (N, P, As, Sb), the vapor pressures go down and the boiling points increase. Account for this based on IMF s. As go down, MM goes up, so dispersion forces go up, so VP down and BP up. 5. Why does water have an unusually high boiling point? Water forms lots of hydrogen bonds with itself. 6. What is the force between SO 2 molecules? SO2 is a polar molecule, therefore it has dispersion forces (everything does) and more importantly it has dipoledipole forces. 7. What is the force that accounts for HF being a liquid, while H 2 and F 2 are gases? H2 and F2 are both completely non-polar, so they only have dispersion forces, which are the weakest. HF is polar and is an F stuck to H therefore, it forms hydrogen bonds which are the strongest intermolecular force, this causes it to be a liquid at room temperature. 8. How do intermolecular forces affect the viscosity of a liquid? The stronger the intermolecular forces in a liquid the more viscous the liquid is. 9. Which is likely to be more viscous, H3C-OH or HO-CH2-OH? WHY? (Note these are structural formulas which partially show the Lewis structures of the molecules) HO-CH2-OH b/c it has 2 sites for H-bonding, vs only 1 for H3C-OH. 10. Which of the following will a bug be more likely to be able to stand on HCl (l) or HF (l)? WHY? HF b/c it H-bonds and therefore has stronger IMF s than HCl which only has dipole-dipole IMF s 11. For the following compounds, which would you pick to have the highest boiling point? The lowest? What intermolecular forces are present in each? H2O CO2 SO2 Since water is bent and it has hydrogen bonding it will have the highest boiling point. Since CO2 is linear it is limited to having only dispersion forces, which are the weakest, therefore, it has the lowest boiling point. SO2 is bent; therefore it has dipole-dipole, which puts it roughly in the middle. 12. When you taste something, it has been attracted to & hooked to your tongue. Is this permanent or temporary? Temporary 13. Based on your answer to the question above, is the taste of food due to an intramolecular (i.e. covalent) or intermolecular (IMF) force? IMF s 14. Fill in the following table for the six changes of state States Involved Name of the Change Exothermic or Endothermic a. Solid Liquid Melting Endothermic b. Liquid Solid Freezing Exothermic c. Liquid Gas Vaporization (boiling, Evaporating) Endothermic d. Gas Liquid Condensation Exothermic e. Solid Gas Sublimation Endothermic f. Gas Solid Deposition Exothermic Last Updated: 6/11/12 Page 7/9
A Melting Pressure Solid Triple Point C Liquid B Gas Critical Point Pressure Deposition Freezing Vaporization Condensation Sublimation Temperature Temperature 15. In either of the phase diagrams what is happening if the temperature and the pressure fall exactly on one of the lines? If the temperature and pressure are at points that meet directly on a line you have dynamic equilibrium between the two corresponding states. 16. What is happening at the triple point? If the temperature and pressure are at the correct combination, then achieve conditions that support dynamic equilibrium between all three states of matter. The substance is undergoing sublimation, deposition etc 17. What happens if a substance follows the path shown by Arrow A? Liquid Solid, freezing 18. What happens if a substance follows the path shown by Arrow B? Gas Liquid, condensation 19. What happens if a substance follows the path shown by Arrow C? Solid Gas, sublimation Last Updated: 6/11/12 Page 8/9
Name Date Period 1. 2. 3. Molecule Lewis Structure Shape Polar? Carbon tetrahydride Boron trichloride selenium dioxide H H C H H Cl B Cl Cl Exception, but not taught.. O Se =O Tetrahedral Trigonal Planar Bent (Trigonal Planar) No No Yes Corrected Lewis Structure with partial charges if polar H C / \ H H H Cl Cl \ / B Cl..(δ + ) Se = O (δ - ) / O (δ - ) Approximate Bond Angles Resonace Structures Possible? # Unshared pairs on Central Atom? 109.5º No 0 120º Yes 0 120º Yes 1 4. Sulfur difluoride.. F S F.. Bent (Tetrahedral) Yes (δ + ) S F (δ - ) / F (δ - ) 105º No 2 5. Nitrate ion O N O O Trigonal Planar No O \ / O N = O 120º Yes 0 Last Updated: 6/11/12 Page 9/9