1. Page 12

2.  2K + Cl 2  2KCl Type: Synthesis  2AlBr 3 + 3 Na 2 (CO 3 )  Al 2 (CO 3 ) 3 +6 NaBr Type: Double replacement  2C 2 H 6 + 7O 2  4 CO 2 +6 H 2 O Type:Combustion  3Cu 2 (CrO 4 ) + 2 Fe  Fe 2 (CrO 4 ) 3 + 6Cu Type: Single replacement  Mg(CO 3 )  MgO + CO 2 Type: decomposition  3H 2 + Fe 2 S 3  3 H 2 S + 2 Fe Type: Single replacement

3.  2AlBr 3 (aq)+ 3 Na 2 (CO 3 ) (aq)  Al 2 (CO 3 ) 3 (s) +6 NaBr(aq) Precipitate! (The insoluble product)

4.  3H 2 + Fe 2 S 3  3 H 2 S + 2 Fe will not occur. Iron is more reactive than hydrogen. (Refer to the activity series on the back of your periodic table). The more reactive element will be in the compound.

5.  Copper(II) chloride reacts with iron to produce iron(III) chloride and copper metal. Skeleton: CuCl 2 + Fe  FeCl 3 +Cu Balanced: 3CuCl 2 + 2Fe  2FeCl 3 +3Cu Type of reaction : Single replacement

6.  Hydrogen gas and bromine liquid react to yield hydrogen bromide. Skeleton: H 2 + Br 2  HBr Balanced: H 2 + Br 2  2HBr Type of reaction : Synthesis

7.  Carbon tetrahydride reacts with oxygen to produce carbon dioxide and water vapor. Skeleton: CH 4 + O 2  CO 2 + H 2 O Balanced: CH 4 + 2O 2  CO 2 + 2H 2 O Type of reaction : Combustion

8. Pages 13 and 14

9. ___HCl + ___Ca(OH) 2  ___CaCl 2 + ___H 2 O Balance: 2HCl + 1Ca(OH) 2  1CaCl 2 +2H 2 O Use the mole ratio from the balanced equation to convert from moles of HCl to moles of CaCl 2 : 2HCl + 1Ca(OH) 2  1CaCl 2 +2H 2 O 1.53 mol HCl1 mol CaCl 2 =.765 mol CaCl 2 2 mol HCl

10. ___HBr + ___Fe  ___FeBr 3 + ___H 2 Balance: 6HBr + 2Fe  2FeBr 3 + 3H 2  Use the mole ratio from the balanced equation to convert from moles of Fe to moles of FeBr 3 :  Use molar mass to convert between moles & grams. 6HBr + 2Fe  2FeBr 3 + 3H 2 2.05 mol Fe 2 mole FeBr 3 295.557g = 606g FeBr 3 2 mole Fe 1 mole FeBr 3 Molar mass of FeBr 3

11. ___Na + ___Cl 2  ___NaCl Balance: 2Na + 1Cl 2  2NaCl  Use the mole ratio from the balanced equation to convert from moles of Na to moles of Cl 2 :  Use molar mass to convert between moles & grams. 2Na + 1Cl 2  2NaCl 2.3g Na 1 mol Na 2 mol NaCl 58.44g NaCl = 5.8g NaCl 22.990g Na 2 mol Na 1 mol NaCl Molar mass of NaClMolar mass of Na

12. A. If a sample containing 18.1 grams of NH 3 reacted with 90.4 grams of copper(II) oxide, which is the limiting reactant? 2NH 3 +3CuO  1N 2 + 3Cu + 2H 2 O 18.1g NH 3 1 mol NH 3 1 mol N 2 28.014 g N 2 = 14.9 g N 2 17.031g NH 3 2 mol NH 3 1 mol N 2 90.4g CuO 1 mol CuO 1 mol N 2 28.014 g N 2 = 10.6 g N 2 79.545g CuO 3 mol CuO 1 mol N 2 A. CuO is the limiting reactant. B. How many grams of N2 will theoretically be formed? 10.6g N 2

13. 1C 7 H 6 O 3 + 1 C 4 H 6 O 3  1 C 9 H 8 O 4 + 1 C 2 H 4 O 2 200.0g C 7 H 6 O 3 1 mol C 7 H 6 O 3 1 mol C 9 H 8 O 4 180.069g C 9 H 8 O 4 = 138.052g C 7 H 6 O 3 1 mol C 7 H 6 O 3 1 mol C 9 H 8 O 4 Answer = 260.9g theoretical Actual x 100 = % yield Theoretical 231x100 = 88.5% 260.9

14. 1C 6 H 6 + 1Br 2  1C 6 H 5 Br + 1HBr 30.0g C 6 H 6 1 mol C 6 H 6 1 molC 6 H 5 Br 156.95g C 6 H 5 Br = 60.32gC 6 H 5 Br 78.054g C 6 H 6 1 mol C 6 H 6 1 mol C 6 H 5 Br 65.0g Br 2 1 mol Br 2 1 mol C 6 H 5 Br 156.95g C 6 H 5 Br = 63.8 C 6 H 5 Br 159.808g Br 2 1 mol Br 2 1 mol C 6 H 5 Br C 6 H 6 was the limiting reactant, so only 60.32 g of C 6 H 5 Br would be theoretically produced.

15. Actual x 100 = % yield Theoretical 42.3x100 = 65.82% 60.32

16. Pages 15,16,17, 18

17. Matter Pure SubstancesMixtures ElementsCompounds Homogeneous Mixtures Heterogeneous Mixtures

18. a. matter – anything that has mass or takes up space b. physical property – property of matter that can be observed or measured without changing the substance  Examples: color, density, mass, volume

19. c. extensive physical property – depends on amount of substance present  Examples: mass, volume, length d. intensive physical property – does not depends on the amount of a substance  Examples: density, color, melting point, boiling point

20. e. Chemical property- used to describe ability Examples: reactivity, flammability, separating mixtures f. Physical change- alters the appearance but does not change the composition of the substance. Examples: phase change, separating mixtures, dissolving, evaporating

21.  a. Peas and carrots- chromatography to separate by color.  b. charcoal powder and iron powder- chromatography to separate by color.  c. salt water- salt dissolves in water so you would use crystallization  d. pigments in green food coloring- if the pigments are dissolved use crystallization, if not dissolved use filtration  e. rubbing alcohol and water- distillation to separate by boiling points

22.  57. Define chemical change – change occurs when one or more substances undergoes a chemical reaction to form a new substance Examples: cooking, combustion, oxidation, fizzes  58. List AND EXPLAIN the four indicators of a chemical change. a.Color changec. gas evolution b. formation of a precipitated. odor

23.  a. A student pours hydrochloric acid into a test tube containing a white, crystalline powder. The mixture begins to bubble and the test tube begins to feel cold. ENDOTHERMIC  b. A student pours HCl at 25.2 C into a test tube containing a small metal strip. The mixture begins to fizz and the temperature of the mixture rises to 38.6 C. EXOTHERMIC

24. SolidsLiquidsGases CompressibilityIncompressible Compressible StructureTightly packedLoosely packed particles No attraction between particle MotionParticles vibrate Ability to flowMove freely ability to flow ShapeFixed shapeTakes shape of container Spread out in container VolumeFixed shapeFixed volumeDepends on size of container

25. A. melting- solid to liquid B. Freezing- liquid to solid C. Vaporization- liquid to gas D. condensation- gas to liquid E. Sublimation- solid to gas F. deposition- gas to solid

26. Solid Liquid Gas Triple point- point where all 3 phases coexist critical point- anything above this point will be a gas

27.  a. Pure substance – uniform unchanging composition, ex: elements and compounds  b. Element- single type of atom ex: gold (Au), Hydrogen (H)  c. Compound- more than one element combined ex: NaCl

28.  d. Mixture- combination of two or more pure substances ex: salt water  e. Homogeneous mixture- constant composition throughout and are always in one phase  f. Solution- homogeneous mixture

29.  g. Heterogeneous mixture- mixtures do not blend together smoothly and the individual substances remain distinct ex: colloid, suspension  h. Colloid- one substance is suspended evenly throughout another substance ex: milk, fog, jello  i. Suspension-large substance particles are suspended in another substance. Ex: muddy water, paint

31. 64. 85 grams 65. 130 grams 66. 50 o C

32. Pages 19,20

34.  67. unsaturated solution  68. 30g will dissolve, 20 grams will remain at the bottom of the beaker  69. 25 more grams will dissolve

35.  Molarity = moles of solute Liters of solution Molarity units = M (concentration) You may have to covert mL to L Given mL1 L 1000 mL You may have to covert grams to moles in order to solve for Molarity Given grams1 mole molar mass

36.  M 1 V 1 = M 2 V 2  M= Molarity  V= Volume

37.  Molarity = moles of solute Liters of solution  Convert Grams  moles 7.20g1 mole=.087 moles 83 g  Convert mL  L 500mL 1L =.500L 1000 mL  Plug into equation: Molarity =.087=.18M.500

38.  Molarity = moles of solute Liters of solution Plug into equation: 2.5 = X= 3.375 mol 1.35

39.  Formula: M 1 V 1 =M 2 V 2  M = Molarity, V = volume  Solving for V 1 (5.0M)(V 1 )= (.25M)(100mL) (5.0M)(V 1 )= 25 (V 1 )= 5mL Multiply.25 x 100 Divide both sides by 5.0 to get (V 1 ) by itself

40.  Formula: M 1 V 1 =M 2 V 2  M = Molarity, V = volume  Solving for M 2 (3.5M)(20mL)= (M 2 )(100mL) 70 = (M 2 )(100mL).7M= (M 2 ) Multiply 3.5 x 20 Divide both sides by 100 to get (M 2 ) by itself