44. BALANCE THESE EQUATIONS AND CLASSIFY AS BEING A SYNTHESIS, DECOMPOSITION, SINGLE OR DOUBLE REPLACEMENT, OR COMBUSTION REACTION:
2K + Cl2 2KCl Type: Synthesis
2AlBr3 + 3 Na2(CO3) Al2(CO3)3 +6 NaBr Type: Double replacement
2C2H6 + 7O2 4 CO2 +6 H2O Type:Combustion
3Cu2(CrO4) + 2 Fe Fe2(CrO4)3 + 6Cu
Type: Single replacement
Mg(CO3) MgO + CO2 Type: decomposition
3H2 + Fe2S3 3 H2S + 2 Fe Type: Single replacement
45. IDENTIFY THE SOLUBILITY OF EACH COMPOUND IN THE DOUBLE REPLACEMENT REACTION ABOVE
2AlBr3 (aq)+ 3 Na2(CO3) (aq) Al2(CO3)3 (s) +6 NaBr(aq)
Precipitate!(The insoluble product)
46. WHICH SINGLE REPLACEMENT REACTION ABOVE WILL NOT OCCUR? 3H2 + Fe2S3 3 H2S + 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.
47A. WRITE THE SKELETON EQUATION THEN BALANCE Copper(II) chloride reacts with iron to
produce iron(III) chloride and copper metal.
Skeleton: CuCl2 + Fe FeCl3 +Cu
Balanced: 3CuCl2 + 2Fe 2FeCl3 +3Cu
Type of reaction : Single replacement
47B. WRITE THE SKELETON EQUATION THEN BALANCE Hydrogen gas and bromine liquid react
to yield hydrogen bromide.
Skeleton: H2 + Br2 HBr
Balanced: H2 + Br2 2HBr
Type of reaction : Synthesis
47C. WRITE THE SKELETON EQUATION THEN BALANCE Carbon tetrahydride reacts with oxygen
to produce carbon dioxide and water vapor.
Skeleton: CH4 + O2 CO2+ H2O
Balanced: CH4 + 2O2 CO2 + 2H2O
Type of reaction : Combustion
48. HOW MANY MOLES OF CALCIUM CHLORIDE CAN BE THEORETICALLY YIELDED FROM 1.53 MOLES OF HYDROGEN CHLORIDE?
___HCl + ___Ca(OH)2 ___CaCl2 + ___H2O
Balance: 2HCl + 1Ca(OH) 2 1CaCl2 +2H2O
Use the mole ratio from the balanced equation to convert from moles of HCl to moles of CaCl2 :
2HCl + 1Ca(OH) 2 1CaCl2 +2H2O
1.53 mol HCl 1 mol CaCl2 = .765 mol CaCl2 2 mol HCl
49. HOW MANY GRAMS OF IRON(III) BROMIDE WILL BE THEORETICALLY YIELDED FROM THE REACTION OF 2.05 MOLES OF IRON?
___HBr + ___Fe ___FeBr3 + ___H2
Balance: 6HBr + 2Fe 2FeBr3 + 3H2 Use the mole ratio from the balanced equation
to convert from moles of Fe to moles of FeBr3: Use molar mass to convert between moles &
grams. 6HBr + 2Fe 2FeBr3 + 3H2
2.05 mol Fe 2 mole FeBr3 295.557g = 606g FeBr3
2 mole Fe 1 mole FeBr3 Molar mass of FeBr3
50. IN A REACTION OF SODIUM WITH CHLORINE, 2.30 GRAMS OF CHLORINE ARE CONSUMED. HOW MANY GRAMS OF PRODUCT CAN BE THEORETICALLY YIELDED?
___Na + ___Cl2 ___NaCl
Balance: 2Na + 1Cl2 2NaCl
Use the mole ratio from the balanced equation to convert from moles of Na to moles of Cl2:
Use molar mass to convert between moles & grams.
2Na + 1Cl2 2NaCl
2.3g Na 1 mol Na 2 mol NaCl 58.44g NaCl = 5.8g NaCl
22.990g Na 2 mol Na 1 mol NaClMolar mass of NaCl
Molar mass of Na
51. NITROGEN GAS CAN BE PREPARED BY PASSING GASEOUS AMMONIA (NH3) OVER SOLID COPPER(II) OXIDE AT HIGHTEMPERATURES. THE OTHER PRODUCTS OF THE REACTION ARE SOLID COPPER AND WATER VAPOR.
A. If a sample containing 18.1 grams of NH3 reacted with 90.4 grams of copper(II) oxide, which is the limiting reactant?
2NH3 +3CuO 1N2 + 3Cu + 2H2O
18.1g NH3 1 mol NH3 1 mol N2 28.014 g N2 = 14.9 g N2
17.031g NH3 2 mol NH3 1 mol N2
90.4g CuO 1 mol CuO 1 mol N2 28.014 g N2 = 10.6 g N2
79.545g CuO 3 mol CuO 1 mol N2
A. CuO is the limiting reactant.
B. How many grams of N2 will theoretically be formed? 10.6g N2
52. A STUDENT ADDS 200.0G OF C7H6O3 TO AN EXCESS OF C4H6O3, THIS PRODUCES C2H4O2 AND C2H4O2. CALCULATETHE PERCENT YIELD IF 231 G OF ASPIRIN (C9H8O4) IS PRODUCED.
1C7H6O3 + 1C4H6O3 1C9H8O4 + 1C2H4O2
200.0g C7H6O3 1 mol C7H6O3 1 mol C9H8O4 180.069g C9H8O4 =
138.052g C7H6O3 1 mol C7H6O3 1 mol C9H8O4
Answer = 260.9g theoretical
Actual x 100 = % yield
Theoretical
231 x100 = 88.5%
260.9
53. BENZENE (C6H6) REACTS WITH BROMINE TO FORM BROMOBENZENE (C6H5BR ) IN THE REACTION BELOW. A. WHAT IS THE THEORETICAL YIELD OF BROMOBENZENE IN THIS REACTION WHEN 30.0 GRAMS OF BENZENE REACTS WITH 65.0 GRAMS OF BROMINE?
1C6H6 + 1Br2 1C6H5Br + 1HBr
30.0g C6H6 1 mol C6H6 1 molC6H5Br 156.95g C6H5Br = 60.32gC6H5Br
78.054g C6H6 1 mol C6H6 1 mol C6H5Br
65.0g Br2 1 mol Br2 1 mol C6H5Br 156.95g C6H5Br = 63.8 C6H5Br
159.808g Br2 1 mol Br2 1 mol C6H5Br
C6H6 was the limiting reactant, so only 60.32 g of C6H5Br would be theoretically produced.
53.B IF THE ACTUAL YIELD OF BROMOBENZENE WAS 42.3 GRAMS, WHAT WAS THE PERCENT YIELD?C6H6 was the limiting reactant, so only 60.32 g of
C6H5Br would be theoretically produced.
Actual x 100 = % yield
Theoretical
42.3 x100 = 65.82%
60.32
54. CLASSIFICATION OF MATTER
Matter
Pure Substances Mixtures
Elements CompoundsHomogeneous
MixturesHeterogeneous
Mixtures
55. DEFINE:
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
55. DEFINE: 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
55. DEFINEe. Chemical property- used to describe
abilityExamples: 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
56. WHICH PHYSICAL SEAPARATION TECHNIQUE? 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 AND 58 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 change c. gas evolution
b. formation of a precipitate d. odor
59. INDICATE ENDOTHERMIC OR EXOTHERMIC 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
60. STATES OF MATTERSolids Liquids Gases
Compressibility
Incompressible
Incompressible
Compressible
Structure Tightly packed
Loosely packed particles
No attraction between particle
Motion Particles vibrate
Ability to flow Move freely ability to flow
Shape Fixed shape Takes shape of container
Spread out in container
Volume Fixed shape Fixed volume Depends on size of container
61.LIST THE SIX PHASE CHANGES, DEFINE THEM, AND GIVE AT LEAST ONE EXAMPLE OF EACH:
A. melting- solid to liquidB. Freezing- liquid to solid C. Vaporization- liquid to gasD. condensation- gas to liquidE. Sublimation- solid to gas F. deposition- gas to solid
62.
Solid
Liquid
Gas
Triple point- point where all 3 phases coexist
critical point- anything above this point will be a gas
63. DEFINE 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
63. DEFINE 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
63. DEFINE 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
67. unsaturated solution
68. 30g will dissolve, 20 grams will remain at the bottom of the beaker
69. 25 more grams will dissolve
70. MOLARITY FORMULA
Molarity = moles of soluteLiters of solution
Molarity units = M (concentration)You may have to covert mL to LGiven mL 1 L
1000 mLYou may have to covert grams to
moles in order to solve for MolarityGiven grams 1 mole
molar mass
71. WHAT FORMULA IS USED FOR DILUTING A CONCENTRATED SOLUTION TO A GIVEN MOLARITY?
M1V1= M2V2
M= Molarity
V= Volume
72. DETERMINE THE MOLARITY OF 500 ML OF A SOLUTION CONTAINING 7.20 G OF SODIUM ACETATE Molarity = moles of solute
Liters of solution Convert Grams moles
7.20g 1 mole = .087 moles
83 g Convert mL L
500mL 1L = .500L
1000 mL Plug into equation:
Molarity = .087 = .18M
.500
73. HOW MANY MOLES OF CALCIUM CHLORIDE ARE IN 1.35 L OF A 2.5 M SOLUTION OF CALCIUM CHLORIDE?
Molarity = moles of solute Liters of solution
Plug into equation:
2.5 = X = 3.375 mol
1.35
74. HOW MANY MILLILITERS OF A 5.0 M H2SO4 STOCK SOLUTION WOULD YOU NEED TO PREPARE 100.0 ML OF 0.25 MH2SO4??
Formula: M1V1=M2V2 M = Molarity, V = volume Solving for V1
(5.0M)(V1)= (.25M)(100mL)
(5.0M)(V1)= 25
(V1)= 5mL
Multiply .25 x 100
Divide both sides by 5.0 to get (V1) by itself