Page 12

 2K + Cl 2  2KCl Type: Synthesis  2AlBr 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

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

 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.

 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

 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

 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

Pages 13 and 14

___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

___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 mol Fe 2 mole FeBr g = 606g FeBr 3 2 mole Fe 1 mole FeBr 3 Molar mass of FeBr 3

___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 g Na 2 mol Na 1 mol NaCl Molar mass of NaClMolar mass of Na

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 g N 2 = 14.9 g N g NH 3 2 mol NH 3 1 mol N g CuO 1 mol CuO 1 mol N g N 2 = 10.6 g N g 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

1C 7 H 6 O C 4 H 6 O 3  1 C 9 H 8 O C 2 H 4 O g C 7 H 6 O 3 1 mol C 7 H 6 O 3 1 mol C 9 H 8 O g C 9 H 8 O 4 = g 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

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 g C 6 H 5 Br = 60.32gC 6 H 5 Br g 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 g C 6 H 5 Br = 63.8 C 6 H 5 Br g Br 2 1 mol Br 2 1 mol C 6 H 5 Br C 6 H 6 was the limiting reactant, so only g of C 6 H 5 Br would be theoretically produced.

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

Pages 15,16,17, 18

Matter Pure SubstancesMixtures ElementsCompounds Homogeneous Mixtures Heterogeneous Mixtures

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

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

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

 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

 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

 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

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

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

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

 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

 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

 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

grams grams o C

Pages 19,20

 67. unsaturated solution  g will dissolve, 20 grams will remain at the bottom of the beaker  more grams will dissolve

 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

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

 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

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

 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

 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