QUESTION

1
QUESTION

• A forensic chemist in a crime lab weighs a shard
  of colored thread removed from a suspicious
  vehicle and records the weight as 0.002500 g.
• This number contains how many significant
  figures?
• 2
• 4
• 6
• 7

2
ANSWER

• Choice 2 is the correct answer. The first three
  zeroes are insignificant because they are
  leading zeroes, which can be thought of as
  place holders. The last two zeroes would not be
  recorded unless they were significant. Remember
  that you should always assume that you are
  looking at a measured number in which the last
  digit recorded is estimated and is therefore
  significant.
• Section 2.5 Significant Figures

3
QUESTION

• While driving in London you encounter a speed
  limit sign that reads 80 km/hr. How fast can you
  legally travel in miles/hr?
• 80
• 70
• 60
• 50

4
ANSWER

• Choice 4
• If you use the conversions given in the chapter,
  your setup might look like this
• 80 km/hr x 1000m/km x 1.094 yd/m x 1 mile/1760 yd
  50 mph
• If you had available the conversion 0.62 miles
  1 km, you could have done the problem in one
  step
• 80 km/hr x 0.62 miles/km 50 miles/hr
• Section 2.6 Problem Solving and Dimensional
  Analysis

5
QUESTION

• If ethanol has a density of 0.785 g/mL, calculate
  the volume of 82.5g ethanol
• 105 L
• 105 mL
• .009 mL
• 202 mL

6
ANSWER

• Choice 2 is correct.
• Densityg/mL 0.78582.5/mL
• 0.785mL 82.5
• 82.5/.785105 mL

7
QUESTION

• When you dissolve sugar in a hot cup of coffee in
  the morning,
• this is an example of a
• chemical change
• separation
• chemical reaction
• physical change

8
ANSWER

• Choice 4 is the correct answer because the sugar
  still retains its chemical identity. It just
  changes from a pure solid form into a dissolved
  form.
• Section 3.2 Physical and Chemical Properties and
  Changes

9
QUESTION

• Pure substances consist of either
• elements or compounds
• elements or mixtures
• elements or solutions
• elements or energy

10
ANSWER

• Choice 1 is correct, as shown in the figure
  here
• Section 3.3 Elements and Compounds
• Section 3.5 Separation of Mixtures

11
QUESTION

• The normal body temperature for a dog is
  approximately 102oF.
• What is this equivalent to on the Kelvin
  temperature scale?
• 373 K
• 312 K
• 289 K
• 202 K

12
ANSWER

• Choice 2 is correct.
• (102 32)oF x 5oC/9oF 273 312 K
• Section 2.7 Temperature Conversions

13
QUESTION

• The symbol for the element Z15 is
• P
• Sn
• K
• Au

14
ANSWER

• Choice 1, P is for phosphorus, while Sn and Au
  are the symbols for tin and gold respectively.
• Section 4.2 Symbols for the Elements

15
QUESTION

• The chemical properties of a particular element
  are largely determined by the number of
  __________ an atom of that element possesses.
• protons
• neutrons
• electrons

16
ANSWER

• Choice 3 is correct. The protons and neutrons
  are buried within the atom in the nucleus. The
  electrons occupy most of the volume of the atom
  and affect the ways in which atoms interact, or
  react, with each other.
• Section 4.6 Introduction to the Modern Concept
  of Atomic Structure

17
QUESTION

• Lead-210 is used in radioactive dating of
  sediment cores. The number of protons, electrons,
  and neutrons in an atom of lead-210 respectively
  is
• variable
• 82, 82, 128
• 210, 82, 128
• 128, 128, 82

18
ANSWER

• Choice 2 is correct. Lead has an atomic number
  of 82 hence every lead atom has 82 protons and
  electrons. To find the number of neutrons, we
  subtract the number of protons from the mass
  number (210), which yields 128 neutrons. Lead-210
  has a characteristic common to radioactive
  isotopes it is relatively rich in neutrons.
• Section 4.7 Isotopes

19
QUESTION

• When aluminum foil is dropped into liquid bromine
  the ionic compound aluminum bromide forms. This
  compounds formula is
• AlBr
• AlBr2
• AlBr3
• Al3Br

20
ANSWER

• Choice 3 provides the correct formula. Note that
  on the periodic table Al is in Group 3, and hence
  forms a 3 ion. Bromine, a halogen, forms an
  anion with a charge of -1. We need three Br-1 to
  balance the Al3 and yield an overall charge of
  zero for the compound.
• Section 4.11 Compounds That Contain Ions

21
QUESTION

• Calomel is the common name for Hg2Cl2, which was
  used in the 1800s as a treatment for dysentery
  and is now used in making reference electrodes.
  It is a Type ___ compound and is named
• ______________.
• II, mercury(I) chloride
• II, mercury(II) chloride
• I, mercury chloride
• I, mercury(I) chloride

22
ANSWER

• Choice 1 provides the two correct responses.
  Mercury exists as a 1 or 2 cation, so it is a
  Type II binary compound when combined with a
  chloride ion. In this compound we have the Hg(I)
  ion that exists as Hg22, which is two Hg(I) ions
  chemically bound together.
• Section 5.2 Naming Compounds That Contain a
  Metal and a Nonmetal (Types I II)

23
QUESTION

• One of the primary functions of a catalytic
  converter is to convert CO (named _________), a
  major component of smog, into CO2 (named
  _________), which is unfortunately a greenhouse
  gas.
• monocarbon monoxide, monocarbon dioxide
• carbon monoxide, carbon dioxide
• carbon oxide, carbon dioxide
• monocarbon oxide, monocarbon dioxide

24
ANSWER

• Choice 2 correctly names these two Type III
  compounds, both of which consist of two
  nonmetals. Notice that although these compounds
  use prefixes, the prefix mono- is not used when
  the first element of the compound has only one
  atom present.
• Section 5.3 Naming Binary Compounds That Contain
  Only Nonmetals (Type III)

25
QUESTION

• The chlorine that is used to disinfect water
  actually consists largely of hypochlorite
  compounds like sodium hypochlorite, which has the
  formula
• NaCl
• NaClO4
• NaClO2
• NaClO

26
ANSWER

• Choice 4. Hypochlorite is the oxyanion ClO-,
  which combines with Na to form NaClO.
• Section 5.5 Naming Compounds That Contain
  Polyatomic Ions

27
QUESTION

• Write a balanced equation for the reaction of
  sulfuric acid with the base potassium hydroxide
  to form potassium sulfate.
• H2S(aq) NaOH(aq) NaS(aq)
  H2O(l)
• H2SO4(aq) KOH(aq) KSO4(aq)
  H2O(l)
• H2SO4(aq) KOH(aq) K2SO4(aq)
  H2O(l)
• H2SO4(aq) 2KOH(aq) K2SO4(aq)
  2H2O(l)

28
ANSWER

• Choice 4 represents the balanced equation.
  Notice first that the formulas of each component
  must be correct, and that once that is
  accomplished we then use coefficients to achieve
  a mass balance.
• Section 6.3 Balancing Chemical Equations

29
QUESTION

• The fuel in small portable lighters is butane
  (C4H10). The coefficient in front of oxygen when
  the reaction below is balanced is
• C4H10 O2 CO2
  H2O
• 3
• 6
• 13
• 15

30
ANSWER

• Choice 3 2C4H10 13O2
  8CO2 10H2O
• One would probably begin balancing the equation
  by placing a 4 in front of the carbon dioxide and
  a 5 in front of the water. The carbons and
  hydrogens are now balanced. The right side of
  the equation now shows 13 oxygens, 8 in the
  carbon dioxide and 5 in the water. The oxygen on
  the left side is diatomic and any whole number
  coefficient placed in front of oxygen would
  generate an even number of oxygens. This problem
  can be solved by doubling the coefficients in
  front of the the butane, carbon dioxide, and
  water.
• Section 6.3 Balancing Chemical Equations

31
QUESTION

• When we study reactions in aqueous solutions, we
  will learn in detail about reactions between
  soluble ionic compounds to form an insoluble
  product (precipitate). Write the balanced
  chemical equation for the reaction between
  soluble calcium nitrate and sodium fluoride to
  form the insoluble precipitate calcium fluoride
  and soluble sodium nitrate.

32
ANSWER

• Ca(NO3)2(aq) 2NaF(aq) CaF2(s)
  2NaNO3(aq)
• It is crucial in this problem to get the formulas
  correct by balancing ionic charges. The 2 is then
  placed in front of NaNO3 in order to balance the
  nitrate ions, which prompts us to place a 2 in
  front of NaF to balance the Na atoms as well as
  the fluorides. Note how it is easier to balance
  the nitrate as one unit since it stays intact
  during the reaction.
• Section 6.2 Chemical Equations
• Section 6.3 Balancing Chemical Equations

33
QUESTION

• Based on solubility rules, which of the compounds
  below is insoluble?
• Ag2CO3
• Na2CO3
• Ag2SO4
• Ba(NO3)2

34
ANSWER

• Choice 1 correctly selects Ag2CO3 as the
  insoluble compound according to Rule 6 in Table
  7.1. Note that carbonates are considered
  slightly soluble in Rule 6, but that the
  footnote also tells us that this means
  insoluble in practical terms. Why? Because if
  we mix Ag and CO32- ions together, a very little
  bit will remain dissolved, but a precipitate will
  form.
• Section 7.2 Reactions in Which a Solid Forms

35
QUESTION

• Mixing which of the pairs of solutions listed
  below will result in the formation of a
  precipitate?
• NaCl K2CO3
• NaOH Mg(NO3)2
• KCl Ba(NO3)2
• Na2S KNO3

36
ANSWER

• Choice 2 correctly predicts the formation of
  insoluble Mg(OH)2 according to 5 of the
  solubility rules.
• Section 7.2 Reactions in Which a Solid Forms

37
QUESTION

• Write the net ionic equations for the reaction
  that occurs when mixing aqueous solutions of lead
  (II) nitrate and ammonium sulfate.

38
ANSWER

• First we determine the ions we have from correct
  formulas of the compounds we are mixing. Upon
  dissolving Pb(NO3)2 (NH4)2SO4 we get the ions
  and switch the anions to yield the total
  ionic equation Pb2 2NO3- 2NH4 SO42-
• 2NH4 2NO3- PbSO4(s).
• Note that the formation of the insoluble product,
  lead(II) sulfate, is the driving force in this
  reaction (solubility rule 4).
• Since nitrate and ammonium ions are the spectator
  ions, we exclude them to yield the
• net ionic equation Pb2 SO42-
  PbSO4(s)
• Section 7.3 Describing Reactions in Aqueous
  Solutions

39
QUESTION

• Diamond is an allotrope of carbon. If 1.0 carat
  has a mass of 0.20 grams, how many carbon atoms
  are in a 3.0 carat diamond?
• 6.0 x 1015
• 7.2 x 1010
• 3.0 x 1026
• 3.0 x 1022

40
ANSWER

• Choice 4 correctly gives the atomic make up of
  the 3.0 carat diamond
• Section 8.2 Atomic Masses Counting Atoms by
  Weighing

41
QUESTION

• The metal gallium (Ga) has most of the properties
  you would expect in a metal, but it is so soft it
  will melt in your hand. If you had 20.0 grams of
  solid gallium in your palm and watched it turn
  into liquid, how many moles of liquid gallium
  would you soon have?
• 20.0 moles
• 0.287 moles
• 4.32 moles
• 3.6 ? 1023 moles

42
ANSWER

• Choice 2 represents the mole equivalent for 20.0
  grams of gallium.
• Section 8.3 The Mole

43
QUESTION

• The compound Cr2O3 (chromium (III) oxide) is one
  of the key components responsible for the red
  color of rubies. Calculate the molar mass of
  chromium(III) oxide.

44
ANSWER

• (52.0 x 2) (16.0 x 3) 152.0 g/mol
• Section 8.4 Molar Mass

45
QUESTION

• Propane, C3H6, is a common fuel used in heating
  homes in rural areas. What is the mole ratio of
  reactants?
• C3H6 6O2 3CO2
  6H2O
• 16
• 13
• 63
• 66

46
ANSWER

• Choice 1 provides the correct response
• It takes 1 mole of propane to react with 6 moles
  of oxygen
• Section 9.2 Mole-Mole Relationships

47
QUESTION

• How many grams of oxygen are needed to react with
  42.8 g of chromium in the production of Cr2O3, a
  common paint pigment?
• 4Cr 3O2 2Cr2O3
• 42.8 g
• 33.4 g
• 19.8 g
• 10.5 g

48
ANSWER

• Choice 3 can be easily arrived at if we first
  write the balanced equation and then use
  dimensional analysis.
• 4Cr 3O2
  2Cr2O3
• Section 9.3 Mass Calculations

49
QUESTION

• Ammonium nitrate decomposes when heated to form
  laughing gas, N2O, and water. If this reaction
  forms 36.0 g H2O, how many grams of N2O are also
  formed?
• NH4NO3 ? N2O 2H2O
• 36.0 g
• 40.0 g
• 44.0 g
• 48.0 g

50
ANSWER

• Choice 3. First we write a balanced equation and
  then we use dimensional analysis
• NH4NO3 N2O
  2H2O
• Section 9.3 Mass Calculations

51
QUESTION

• One calorie of energy is equivalent to
• 1.0 kilojoule
• 2.35 millijoules
• 4.184 joules
• a temperature change of 1 K

52
ANSWER

• Choice 3 shows the proper relationship between
  calories and joules.
• Section 10.5 Measuring Energy Changes

53
QUESTION

• The heat specific capacity of lead is 0.13 J/g
  ºC. How many joules of heat would be required to
  raise the temperature of 150.0 g of Pb from 25ºC
  to 100ºC?
• 130 J
• 1.0 x 103 J
• 1.5 x 103 J
• 7.5 x 102 J

54
ANSWER

• Choice 3 can be correctly found by using
• Q s x m x ?T
• Q 0.13 J/g ºC x 150.0 g x
  75ºC 1.5 x 103 J
• Section 10.5 Measuring Energy Changes

55
QUESTION

• Calculate ?H for the reaction SO2 ½O2
  SO3 ?H ?
• Given (1) S O2 SO2
  ?H -297 kJ (2) 2S
  3O2 2SO3 ?H -792
  kJ
• -693 kJ
• 101 kJ
• 693 kJ
• -99 kJ

56
ANSWER

• Choice 4 is correct. We have to rearrange the
  equations so that they add up to the desired
  equation. First, reverse equation (1) to give
  SO2 S O2 ?H 297
  kJ
• Then add ½ equation (2) S 3/2O2 SO3
  ?H -396 kJ
• The two new eqns add up SO2 ½O2 SO3
  ?H -99 kJ
• Section 10.7 Hesss Law

57
QUESTION

• Write out the electron configuration for an atom
  of phosphorus and indicate the number of unpaired
  electrons
• Ne3s23p3 3 unpaired electrons
• Ne2s22p63s23p3 3 unpaired electrons
• Ne3s23p3 5 unpaired electrons
• Ne3p5 5 unpaired electrons

58
ANSWER

• Choice 1 is correct. Each of the three 3p
  orbitals has three unpaired electrons in it. The
  electrons in the 3s orbital are paired.
• Section 11.9 Electron Arrangements in the First
  Eighteen Atoms on the Periodic Table

59
QUESTION

• The number of valence electrons in an atom of Cl
  is
• 7
• 5
• 3
• 1

60
ANSWER

• Choice 1 is correct. Cl has 2 valence electrons
  in the 3s orbital and 5 in the 3p subset.
• Section 11.9 Electron Arrangements in the First
  Eighteen Elements on the Periodic Table

61
QUESTION

• Which element would be likely to have the lowest
  ionization energy?
• Na
• Al
• Cl
• Cs

62
ANSWER

• Choice 4 should be selected. Ionization energy
  increases from left to right, but decreases as
  from top to bottom.
• Section 11.11 Atomic Properties and the Periodic
  Table

63
QUESTION

• Most chemical bonds consist of electrostatic
  attractive forces and are called ____ bonds, or
  of shared electrons and are called _____bonds.
• electric shared
• ionic covalent
• ionic molecular
• electronic coordinate

64
ANSWER

• Choice 2 properly identifies the bonds as ionic
  and covalent.
• Section 12.1 Types of Chemical Bonds

65
QUESTION

• When considering a bond between two atoms, the
  greater the difference in ____________, the more
  __________is the bond.
• polarity divided
• atomic weight nonpolar
• electronegativity polar
• electronegativity nonpolar

66
ANSWER

• Choice 3 correctly expresses the idea that bonds
  become more polar as the difference in the
  electronegativity of the atoms increases.
• Section 12.2 Electronegativity

67
QUESTION

• Lewis structures show the arrangement of
  ________electrons in an atom or ion.
• all
• core
• valence
• missing

68
ANSWER

• Choice 3 correctly indicates that only valence
  electrons are included in Lewis structures.
• Section 12.6 Lewis Structures

69
QUESTION

• In the Lewis structure for H2S there are a total
  of ______ electrons and ____ pair(s) of
  nonbonding electrons.
• 9 2
• 9 1
• 8 2
• 8 1

70
ANSWER

• Choice 3 correctly describes the Lewis structure
  for hydrosulfuric acid (6 electrons from S) (2
  x 1 2 electrons from H) 8 total electrons.
  The Lewis structure is the same as that for H2O,
  with S in place of O.
• Section 12.6 Lewis Structures

71
QUESTION

• The vapor pressure over a beaker of hot water is
  measured as 656 torr. What is this pressure in
  atmospheres?
• 1.16 atm
• 0.863 atm
• 0.756 atm
• 0.500 atm

72
ANSWER

• Choice 2 provides the correct value, which is
  easily obtained by dimensional analysis
• Section 13.1 Pressure

73
QUESTION

• A gas sample with a volume of 800.0 mL and a
  pressure of 1.00 atm is reduced in volume to
  250.0 mL at constant temperature. The new
  pressure of the gas is
• 0.100 atm
• 0.313 atm
• 1.50 atm
• 3.20 atm

74
ANSWER

• Choice 4 correctly gives the response arrived at
  using Boyles Law
• P1V1 P2V2
• (1.00 atm) (800.0 ml) (P2) (250.0 ml)
• P2 3.20 atm
• Section 13.2 Pressure and Volume Boyles Law

75
QUESTION

• If a 1.00 L sample of CO2 at 27oC is cooled to
  15oC at constant atmospheric pressure of 1.00
  atm, what is the new volume of the gas?
• 1.50 L
• 1.33 L
• 1.04 L
• 0.960 L

76
ANSWER

• Choice 4 is arrived at by using Charless Law
• V2 0.960 L
• YOU MUST USE KELVIN !
• Section 13.3 Volume and Temperature Charless
  Law

77
QUESTION

• Carbon monoxide is a toxic gas that competes very
  successfully for active sites on hemoglobin, and
  can thus cause suffocation. If 22.0 g of pure CO
  is in a 20.0 L container at a temperature of
  25oC, what pressure (atm) does the gas exert?
• 0.961 atm
• 0.881 atm
• 1.04 atm
• 1.68 atm

78
ANSWER

• Choice 1. Use the Ideal Gas Law, PV nRT, and
  rearrange
• Note that due to units for R, we must use Kelvin!
• Section 13.5 The Ideal Gas Law

79
QUESTION

• It is increasingly recognized that the key to
  smog reduction is the control of NO emissions
  from vehicles. If 50.0 ml of the NO emitted from
  an exhaust pipe at 200oC exerts a pressure of
  1.80 atm, what pressure will it exert when it is
  transferred into a 100.0 mL container and cooled
  to 25oC?
• 2.14 atm
• 0.567 atm
• 1.95 atm
• 0.877 atm

80
ANSWER

• Choice 2 correctly predicts the pressure. First
  of all, we expect the pressure to decrease
  because the temperature is decreasing and the
  volume is being expanded. To solve numerically we
  use the combined form of the Ideal Gas Law for
  changing conditions
• P2 0.567 atm
• Section 13.5 The Ideal Gas Law

81
QUESTION

• If 1.00 mole of NaCl is added to 1.00 liter of
  H2O and dissolved, what is the mass composition
  of the solution?
• 5.85 NaCl
• 5.53 NaCl
• 4.29 NaCl
• 1.00 NaCl

82
ANSWER

• Choice 2 reflects the correct by mass of NaCl
  for the solution.
• (58.5g NaCl)/(58.5g NaCl 1000.0g H2O) x 100
  5.53 NaCl
• Section 15.3 Solution Composition Mass Percent

83
QUESTION

• Calculate the molarity of a solution prepared by
  dissolving 10.5g of baking soda (NaHCO3) in water
  and diluting to a total volume of 500.0 mL.
• 21.0 M
• 0.021 M
• 0.250 M
• 0.125 M

84
ANSWER

• Choice 3 is arrived at by dimensional analysis
• Section 15.4 Solution Composition Molarity

85
QUESTION

• Concentrated nitric acid is sold commercially as
  15.9 M HNO3. How many mL of the concentrated
  reagent are needed to prepare 200.0 mL of 0.050 M
  HNO3?
• 318 mL
• 32 mL
• 3.2 mL
• 0.63 mL

86
ANSWER

• Choice 4 correctly predicts the volume of
  concentrated HNO3
• M1 x V1 M2 x
  V2
• (0.050) x (200.0) (15.9) x V2 V2
  0.63 mL
• Section 15.5 Dilution

87
QUESTION

• If 25.5 mL of 0.100 M NaOH is diluted to 500.0
  mL, what is the molarity of the diluted solution?
• 1.27 M
• 0.050 M
• 0.00510 M
• 0.00260 M

88
ANSWER

• Choice 3 is correct. Use M1 x V1 M2 x V2
• (0.100) x (25.5)
  M2 x (500.0)
• 
  M2 0.00510 M
• Section 15.5 Dilution