Name: ______________________
Class: _________________
Date: _________
2010 Honors Chemistry Semester 1 Practice Test
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. The
a.
b.
c.
d.
two most important properties of all matter are
the ability to carry an electric current well and to hold electric charge.
taking up space and having mass.
being brittle and hard.
being malleable and ductile.
____
2. A compound is
a. a pure substance that cannot be broken down into simpler, stable substances.
b. a substance, made of two or more atoms that are chemically bonded, that can be
broken down into simpler, stable substances.
c. the smallest unit of matter that maintains its chemical identity.
d. any substance, whether it is chemically bonded or not.
____
3. Matter includes all of the following except
a. air.
b. light.
c.
d.
smoke.
water vapor.
____
4. A student recorded the following while completing an experiment.
Color of substance: yellow, shiny powder
Effect of magnet: yellow, shiny powder was attracted
The student should classify the substance as a(n)
a. element.
c. mixture.
b. compound.
d. plasma.
____
5. Which of the following is not a physical change?
a. grinding
c. boiling
b. cutting
d. burning
____
6. A physical change occurs when a
a. peach spoils.
b. silver bowl tarnishes.
c. bracelet turns your wrist green.
d. glue gun melts a glue stick.
1
ID: A
Name: ______________________
____
ID: A
7. Which part of the illustration below shows the particles in a heterogeneous mixture?
a.
b.
a
b
c.
d.
c
d
____
8. If a mixture is uniform in composition, it is said to be
a. homogeneous.
c. heterogeneous.
b. chemically bonded.
d. a compound.
____
9. All
a.
b.
c.
d.
of the following are steps in the scientific method except
observing and recording data.
forming a hypothesis.
discarding data inconsistent with the hypothesis.
developing a model based on experimental results.
____ 10. Which of the following observations is qualitative?
a. A chemical reaction was complete in 2.3 seconds.
b. The solid had a mass of 23.4 grams.
c. The pH of a liquid was 5.
d. Salt crystals formed as the liquid evaporated.
____ 11. A theory is accepted as the explanation of an observed phenomenon until
a. one study contradicts the theory.
b. repeated observations conflict with the theory.
c. a new method is discovered.
d. a leading scientist declares that it is invalid.
____ 12. All of the following are SI units for density except
a. kg/m3 .
c. g/cm3 .
b. kg/L.
d. g/m2 .
____ 13. The density of aluminum is 2.70 g/cm3 . What is the mass of a solid piece of aluminum with a volume of
1.50 cm3 ?
a. 0.556 g
c. 4.05 g
b. 1.80 g
d. 4.20 g
2
Name: ______________________
ID: A
____ 14. What is the density of 37.72 g of material whose volume is 6.80 cm3 ?
a. 0.180 g/cm3
c. 30.9 g/cm3
3
b. 5.55 g/cm
d. 256. g/cm3
____ 15. How many minutes are in 1 week?
a. 168 min
b. 1440 min
c.
d.
10 080 min
100 800 min
____ 16. If repeated measurements agree closely but differ widely from the accepted value, these measurements are
a. neither precise nor accurate.
b. accurate, but not precise.
c. both precise and accurate.
d. precise, but not accurate.
____ 17. The measurement 0.035550 g rounded off to two significant figures would be
a. 0.03 g.
c. 0.036 g.
b. 0.35 g.
d. 3.5 × 102 g.
____ 18. The number of significant figures in the measurement 0.000 305 kg is
a. 2.
c. 6.
b. 3.
d. 7.
____ 19. The dimensions of a rectangular solid are measured to be 1.27 cm, 1.3 cm, and 2.5 cm. The volume
should be recorded as
a. 4.128 cm3 .
c. 4.13 cm3 .
3
b. 4.12 cm .
d. 4.1 cm3 .
____ 20. How many significant figures would the answer to the following calculation have?
3.475 × 1.97 + 2.4712 is
a.
b.
2
3
c.
d.
4
5
____ 21. Dalton's atomic theory agrees with modern atomic theory except for the statement that
a. all matter is made up of small particles.
b. atoms are not divided in chemical reactions.
c. atoms of the same element are chemically alike.
d. all atoms of the same element have the same mass.
____ 22. The deflection of cathode rays in Thomson's experiments was evidence of the ____ nature of electrons.
a. wave
c. particle
b. charged
d. spinning
____ 23. Because most particles fired at metal foil passed straight through, Rutherford concluded that
a. atoms were mostly empty space.
c. electrons formed the nucleus.
b. atoms contained no charged particles.
d. atoms were indivisible.
3
Name: ______________________
ID: A
____ 24. Because a few alpha particles bounced back from the foil, Rutherford concluded that they were
a. striking electrons.
b. indivisible.
c. repelled by densely packed regions of positive charge.
d. magnetic.
____ 25. The
a.
b.
c.
d.
____ 26. An
a.
b.
c.
d.
nucleus of an atom has all of the following characteristics except that it
is positively charged.
is very dense.
contains nearly all of the atom's mass.
contains nearly all of the atom's volume.
atom is electrically neutral because
neutrons balance the protons and electrons.
nuclear forces stabilize the charges.
the numbers of protons and electrons are equal.
the numbers of protons and neutrons are equal.
____ 27. Isotopes are atoms of the same element that have different
a. principal chemical properties.
c. numbers of protons.
b. masses.
d. numbers of electrons.
____ 28. All atoms of the same element have the same
a. atomic mass.
c.
b. number of neutrons.
d.
____ 29. The
a.
b.
c.
d.
mass number.
atomic number.
average atomic mass of an element
is the mass of the most abundant isotope.
may not equal the mass of any of its isotopes.
cannot be calculated.
always adds up to 100.
____ 30. Chlorine has atomic number 17 and mass number 35. It has
a. 17 protons, 17 electrons, and 18 neutrons.
b. 35 protons, 35 electrons, and 17 neutrons.
c. 17 protons, 17 electrons, and 52 neutrons.
d. 18 protons, 18 electrons, and 17 neutrons.
____ 31. The mass of a sample containing 3.5 mol of silicon atoms (atomic mass 28.0855 amu) is approximately
a. 28 g.
c. 72 g.
b. 35 g.
d. 98 g.
____ 32. A prospector finds 39.39 g of pure gold (atomic mass 196.9665 amu). She has
a. 1.204 × 102 3 atoms of Au.
c. 4.306 × 102 3 atoms of Au.
b. 2.308 × 102 3 atoms of Au.
d. 6.022 × 102 3 atoms of Au.
____ 33. Visible light, X rays, infrared radiation, and radio waves all have the same
a. energy.
c. speed.
b. wavelength.
d. frequency.
4
Name: ______________________
____ 34. The energy of a photon is related to its
a. mass.
b. speed.
ID: A
c.
d.
frequency.
size.
____ 35. A line spectrum is produced when an electron moves from one energy level
a. to a higher energy level.
b. to a lower energy level.
c. into the nucleus.
d. to another position in the same sublevel.
____ 36. The Bohr model of the atom was an attempt to explain hydrogen's
a. density.
c. mass.
b. flammability.
d. line-emission spectrum.
____ 37. For
a.
b.
c.
d.
an electron in an atom to change from the ground state to an excited state,
energy must be released.
energy must be absorbed.
radiation must be emitted.
the electron must make a transition from a higher to a lower energy level.
____ 38. Bohr's theory helped explain why
a. electrons have negative charge.
b. most of the mass of the atom is in the nucleus.
c. excited hydrogen gas gives off certain colors of light.
d. atoms combine to form molecules.
____ 39. Louis de Broglie's research suggested that
a. frequencies of electron waves do not correspond to specific energies.
b. electrons usually behave like particles and rarely like waves.
c. electrons should be considered as waves confined to the space around an atomic
nucleus.
d. electron waves exist at random frequencies.
____ 40. Which model of the atom explains the orbitals of electrons as waves?
a. the Bohr model
c. Rutherford's model
b. the quantum model
d. Planck's theory
____ 41. The region outside the nucleus where an electron can most probably be found is the
a. electron configuration.
c. s sublevel.
b. quantum.
d. electron cloud.
____ 42. The element with electron configuration 1s2 2s2 2p 6 3s2 3p 2 is
a. Mg (Z = 12).
c. S (Z = 16).
b. C (Z = 6).
d. Si (Z = 14).
____ 43. What is the electron configuration for nitrogen, atomic number 7?
a. 1s2 2s2 2p 3
b. 1s2 2s3 2p 2
c. 1s2 2s3 2p 1
d. 1s2 2s2 2p 2 3s1
5
Name: ______________________
____ 44. The
a.
b.
c.
d.
ID: A
periodic table
permits the properties of an element to be predicted before the element is discovered.
will be completed with element 118.
has been of little use to chemists since the early 1900s.
was completed with the discovery of the noble gases.
____ 45. Elements in a group or column in the periodic table can be expected to have similar
a. atomic masses.
c. numbers of neutrons.
b. atomic numbers.
d. properties.
____ 46. The group of soft, silvery, reactive metals, all of which have one electron in an s orbital, is known as the
a. alkaline-earth metals.
c. alkali metals.
b. transition metals.
d. metalloids.
____ 47. The
a.
b.
c.
d.
most characteristic property of the noble gases is that they
have low boiling points.
are radioactive.
are gases at ordinary temperatures.
are largely unreactive.
____ 48. In the alkaline-earth group, atoms with the smallest radii
a. are the most reactive.
b. have the largest volume.
c. are all gases.
d. have the highest ionization energies.
____ 49. A mutual electrical attraction between the nuclei and valence electrons of different atoms that binds the
atoms together is called a(n)
a. dipole.
c. chemical bond.
b. Lewis structure.
d. London force.
____ 50. As independent particles, most atoms are
a. at relatively high potential energy.
b. at relatively low potential energy.
c.
d.
very stable.
part of a chemical bond.
____ 51. If two covalently bonded atoms are identical, the bond is
a. nonpolar covalent.
c. dipole covalent.
b. polar covalent.
d. coordinate covalent.
____ 52. The B—F bond in BF3 (electronegativity for B is 2.0; electronegativity for F is 4.0) is
a. polar covalent.
c. nonpolar covalent.
b. ionic.
d. metallic.
____ 53. When a stable covalent bond forms, the potential energy of the atoms
a. increases.
c. remains constant.
b. decreases.
d. becomes zero.
____ 54. The substance whose Lewis structure shows three covalent bonds is
a. H2 O.
c. NH3 .
b. CH2 Cl2 .
d. CCl4 .
6
Name: ______________________
ID: A
____ 55. Compared with ionic compounds, molecular compounds
a. have higher boiling points.
c. have lower melting points.
b. are brittle.
d. are harder.
____ 56. Compared with nonmetals, the number of valence electrons in metals is generally
a. smaller.
c. about the same.
b. greater.
d. almost triple.
____ 57. Use VSEPR theory to predict the shape of the carbon tetraiodide molecule, CI4 .
a. tetrahedral
c. bent
b. linear
d. trigonal-planar
____ 58. A chemical formula includes the symbols of the elements in the compound and subscripts that indicate
a. atomic mass of each element.
b. number of atoms or ions of each element that are combined in the compound.
c. formula mass.
d. charges on the elements or ions.
____ 59. What is the formula for zinc fluoride?
a. ZnF
b. ZnF2
c.
d.
Zn2 F
Zn2 F3
____ 60. What is the formula for the compound formed by lead(II) ions and chromate ions?
a. PbCrO4
c. Pb2 (CrO4 ) 3
b. Pb2 CrO4
d. Pb(CrO4 ) 2
____ 61. What is the formula for barium hydroxide?
a. BaOH
b. BaOH2
c.
d.
Ba(OH)2
Ba(OH)
____ 62. Name the compound Zn3 (PO4 ) 2 .
a. zinc potassium oxide
b. trizinc polyoxide
c.
d.
zinc phosphate
zinc phosphite
____ 63. Name the compound KClO3 .
a. potassium chloride
b. potassium trioxychlorite
c.
d.
potassium chlorate
hypochlorite
____ 64. Name the compound SiO2 .
a. silver oxide
b. silicon oxide
c.
d.
silicon dioxide
monosilver dioxide
____ 65. What is the formula for dinitrogen trioxide?
a. Ni2 O3
c.
b. NO3
d.
N2 O6
N2 O3
____ 66. The molar mass of NO2 is 46.01 g/mol. How many moles of NO2 are present in 114.95 g?
a. 0.4003 mol
c. 2.498 mol
b. 1.000 mol
d. 114.95 mol
7
Name: ______________________
ID: A
____ 67. What is the mass of 0.240 mol glucose, C6 H1 2O6 ?
a. 24.0 g
c. 180.16 g
b. 43.2 g
d. 750. g
____ 68. How many molecules are there in 5.0 g of methyl alcohol, CH3 OH?
a. 9.4 × 102 2
c. 3.6 × 102 4
b. 3.0 × 102 4
d. 3.8 × 102 4
____ 69. What is the percentage composition of CO?
a. 50% C, 50% O
c.
b. 12% C, 88% O
d.
____ 70. The
a.
b.
c.
d.
25% C, 75% O
43% C, 57% O
empirical formula for a compound shows the symbols of the elements with subscripts indicating the
actual numbers of atoms in a molecule.
number of moles of the compound in 100 g.
smallest whole-number ratio of the atoms.
atomic masses of each element.
____ 71. A molecular compound has the empirical formula XY3 . Which of the following is a possible molecular
formula?
a. X2 Y3
c. X2 Y5
b. XY4
d. X2 Y6
____ 72. The molecular formula for vitamin C is C6 H8 O6 . What is the empirical formula?
a. CHO
c. C3 H4 O3
b. CH2 O
d. C2 H4 O2
____ 73. A compound's empirical formula is C2 H5 . If the formula mass is 58 amu, what is the molecular formula?
a. C3 H6
c. C5 H8
b. C4 H1 0
d. C5 H1 5
____ 74. In a chemical reaction
a. the mass of the reactants equals the mass of the products.
b. the mass of the products is greater than the mass of reactants.
c. the number of atoms in the reactants and products must change.
d. energy as heat must be added to the reactants.
____ 75. Which observation does not indicate that a chemical reaction has occurred?
a. formation of a precipitate
c. evolution of heat and light
b. production of a gas
d. change in total mass of substances
____ 76. The word equation solid carbon + oxygen gas → carbon dioxide gas + energy, represents a chemical
reaction because
a. the reaction releases energy.
b. CO2 has chemical properties that differ from those of C and O.
c. the reaction absorbs energy.
d. CO2 is a gas and carbon is a crystal.
8
Name: ______________________
ID: A
____ 77. Which of the following is a formula equation for the formation of carbon dioxide from carbon and
oxygen?
a. Carbon plus oxygen yields carbon
c. CO2 → C + O2
dioxide.
b. C + O2 → CO2
d. 2C + O → CO2
____ 78. Which coefficients correctly balance the formula equation
NH4 NO2 (s)→ N2 (g) + H2 O(l)?
a. 1, 2, 2
c. 2, 1, 1
b. 1, 1, 2
d. 2, 2, 2
____ 79. Which equation is not balanced?
a. 2H2 + O2 → 2H2 O
b. 4H2 + 2O2 → 4H2 O
c. H2 + H2 + O2 → H2 O + H2 O
d. 2H2 + O2 → H2 O
____ 80. In what kind of reaction do two or more substances combine to form a new compound?
a. decomposition reaction
c. double-displacement reaction
b. ionic reaction
d. synthesis reaction
____ 81. The equation AX + BY → AY + BX is the general equation for a
a. synthesis reaction.
c. single-displacement reaction.
b. decomposition reaction.
d. double-displacement reaction.
____ 82. The reaction represented by the equation Mg(s) + 2HCl(aq) → H2 (g) + MgCl2 (aq) is a
a. composition reaction.
c. single-displacement reaction.
b. decomposition reaction.
d. double-displacement reaction.
____ 83. In a double-displacement reaction, hydrogen chloride and sodium hydroxide react to produce sodium
chloride. Another product is
a. sodium hydride.
c. water.
b. potassium chloride.
d. hydrogen gas.
____ 84. An element in the activity series can replace any element
a. in the periodic table.
c. above it on the list.
b. below it on the list.
d. in its group.
____ 85. The Haber process for producing ammonia commercially is represented by the equation N2 (g) + 3H2 (g)
→ 2NH3 (g). To completely convert 9.0 mol hydrogen gas to ammonia gas, how many moles of nitrogen
gas are required?
a. 1.0 mol
c. 3.0 mol
b. 2.0 mol
d. 6.0 mol
9
Name: ______________________
ID: A
Use the table below to answer the following questions.
Element
Bromine
Calcium
Carbon
Chlorine
Cobalt
Copper
Fluorine
Hydrogen
Iodine
Iron
Lead
Magnesium
Mercury
Nitrogen
Oxygen
Potassium
Sodium
Sulfur
Symbol
Br
Ca
C
Cl
Co
Cu
F
H
I
Fe
Pb
Mg
Hg
N
O
K
Na
S
Atomic Mass
79.90
40.08
12.01
35.45
58.93
63.55
19.00
1.01
126.90
55.85
207.2
24.30
200.59
14.01
15.00
39.10
22.99
32.01
____ 86. For the reaction represented by the equation
required to produce 4.00 mol of sulfuric acid
a. 80.0 g
b. 160. g
SO3 + H2 O → H2 SO4 , how many grams of sulfur trioxide are
in an excess of water?
c. 240. g
d. 320. g
____ 87. For the reaction represented by the equation Cl2 + 2KBr → 2KCl + Br2 , how many grams of potassium
chloride can be produced from 300. g each of chlorine and potassium bromide?
a. 98.7 g
c. 188 g
b. 111 g
d. 451 g
____ 88. What is the ratio of the actual yield to the theoretical yield, multiplied by 100%?
a. mole ratio
c. Avogadro yield
b. percentage yield
d. excess yield
____ 89. What is the maximum possible amount of product obtained in a chemical reaction?
a. theoretical yield
c. mole ratio
b. percentage yield
d. actual yield
____ 90. For the reaction represented by the equation SO3 + H2 O → H2 SO4 , calculate the percentage yield if 500.
g of sulfur trioxide react with excess water to produce 575 g of sulfuric acid.
a. 82.7%
c. 91.2%
b. 88.3%
d. 93.9%
____ 91. According to the kinetic-molecular theory, particles of matter
a. are in constant motion.
c. have different colors.
b. have different shapes.
d. are always fluid.
10
Name: ______________________
ID: A
____ 92. According to the kinetic-molecular theory, which substances are made of particles?
a. gases only
c. all matter
b. liquids only
d. all matter except solids
____ 93. According to the kinetic-molecular theory, particles of an ideal gas
a. attract each other but do not collide.
b. repel each other and collide.
c. neither attract nor repel each other but collide.
d. neither attract nor repel each other and do not collide.
____ 94. By which process do gases take the shape of their container?
a. evaporation
c. adhesion
b. expansion
d. diffusion
____ 95. The volume of a gas is 400.0 mL when the pressure is 1.00 atm. At the same temperature, what is the
pressure at which the volume of the gas is 2.0 L?
a. 0.5 atm
c. 0.20 atm
b. 5.0 atm
d. 800 atm
____ 96. The volume of a gas is 5.0 L when the temperature is 5.0°C. If the temperature is increased to 10.0°C
without changing the pressure, what is the new volume?
a. 2.5 L
c. 5.1 L
b. 4.8 L
d. 10.0 L
____ 97. Why does the air pressure inside the tires of a car increase when the car is driven?
a. Some of the air has leaked out.
b. The air particles collide with the tire after the car is in motion.
c. The air particles inside the tire increase their speed because their temperature rises.
d. The atmosphere compresses the tire.
____ 98. On a cold winter morning when the temperature is –13°C, the air pressure in an automobile tire is 1.5
atm. If the volume does not change, what is the pressure after the tire has warmed to 15°C?
a. –1.5 atm
c. 3.0 atm
b. 1.7 atm
d. 19.5 atm
____ 99. The principle that under similar pressures and temperatures, equal volumes of gases contain the same
number of molecules is attributed to
a. Boyle.
c. Avogadro.
b. Graham.
d. Dalton.
____100. If the temperature of a container of gas remains constant, how could the pressure of the gas increase?
a. The mass of the gas molecules increase.
b. The diffusion of the gas molecules increases.
c. The size of the container increases.
d. The number of gas molecules in the container increases.
11
ID: A
2010 Honors Chemistry Semester 1 Practice Test
Answer Section
MULTIPLE CHOICE
1. ANS:
OBJ:
2. ANS:
OBJ:
3. ANS:
OBJ:
4. ANS:
OBJ:
5. ANS:
OBJ:
6. ANS:
OBJ:
7. ANS:
OBJ:
8. ANS:
OBJ:
9. ANS:
OBJ:
10. ANS:
OBJ:
11. ANS:
OBJ:
12. ANS:
OBJ:
13. ANS:
B
1
B
1
B
1
C
1
D
2
D
2
A
4
A
5
C
1
D
2
B
3
D
2
C
Solution:
D=
PTS: 1
DIF: I
REF: 2
PTS: 1
DIF: II
REF: 2
PTS: 1
DIF: I
REF: 2
PTS: 1
DIF: III
REF: 2
PTS: 1
DIF: II
REF: 2
PTS: 1
DIF: II
REF: 2
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
DIF: II
REF: 2
DIF: II
REF: 2
DIF: I
REF: 1
DIF: II
REF: 1
DIF: I
REF: 1
DIF: I
REF: 2
1
PS.9.15
1
PS.9.9
1
SI.12.1
1
SI.12.4
1
SWK.12.3
1
SI.12.5
m
V
3
3
m = DV = 2.70g/cm × 1.50cm = 4.05g
PTS: 1
STA: SI.12.2
14. ANS: B
Solution:
DIF: III
D=
D=
PTS: 1
STA: SI.12.2
m
V
37.72 g
6.80 cm
3
=5.55 g/cm
DIF: III
REF: 2
OBJ: 4
REF: 2
OBJ: 4
3
1
ID: A
15. ANS: C
Solution:
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
PTS:
STA:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
1
SI.12.2
D
1
C
2
B
2
D
3
B
3
B
2
B
1
A
2
C
2
D
3
C
3
B
1
D
2
B
2
A
3
D
Solution:
PTS: 1
32. ANS: A
1 week ×
7 days
24 h
60 min
×
×
=10 080min
1 week 1 day
1h
DIF: II
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
3.5 mol Si +
REF: 2
1
DIF:
SI.12.5
1
DIF:
SI.12.5
1
DIF:
SI.12.5
1
DIF:
SI.12.5
1
DIF:
SI.12.5
1
DIF:
PS.9.26
1
DIF:
PS.12.15
1
DIF:
PS.12.15
1
DIF:
PS.12.15
1
DIF:
PS.9.1
1
DIF:
PS.9.1
1
DIF:
PS.11.1
1
DIF:
PS.9.2 | PS.11.1
1
DIF:
PS.9.2 | PS.11.1
1
DIF:
PS.9.2 | PS.11.1
OBJ: 5
II
REF: 3
II
REF: 3
II
REF: 3
III
REF: 3
III
REF: 3
I
REF: 1
I
REF: 2
II
REF: 2
I
REF: 2
II
REF: 2
I
REF: 2
I
REF: 3
I
REF: 3
I
REF: 3
III
REF: 3
28.0855 g Si
= 98.3 or 98 g Si
1 mol Si
DIF: III
REF: 3
OBJ: 5
23
Solution:
PTS: 1
39.39 g Au ×
1 mol Au
6.022 × 10 atoms of Au
23
×
= 1.204 + 10 atoms of Au
196.9665 g Au
1 mol Au
DIF: III
REF: 3
2
OBJ: 5
ID: A
33. ANS:
OBJ:
34. ANS:
OBJ:
35. ANS:
OBJ:
36. ANS:
OBJ:
37. ANS:
OBJ:
38. ANS:
OBJ:
39. ANS:
OBJ:
40. ANS:
OBJ:
41. ANS:
OBJ:
42. ANS:
OBJ:
43. ANS:
OBJ:
44. ANS:
OBJ:
45. ANS:
OBJ:
46. ANS:
OBJ:
47. ANS:
OBJ:
48. ANS:
OBJ:
49. ANS:
OBJ:
50. ANS:
OBJ:
51. ANS:
OBJ:
52. ANS:
OBJ:
53. ANS:
OBJ:
54. ANS:
OBJ:
55. ANS:
OBJ:
C
1
C
3
B
3
D
4
B
4
C
4
C
1
B
2
D
2
D
3
A
3
A
2
D
4
C
4
D
4
D
2
C
1
A
2
A
3
B
5
B
2
C
5
C
4
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
1
DIF:
PS.9.18
1
DIF:
PS.12.13
1
DIF:
PS.12.13
1
DIF:
PS.12.14
1
DIF:
PS.12.14
1
DIF:
PS.12.14
1
DIF:
PS.12.14
1
DIF:
PS.12.14
1
DIF:
PS.12.14
1
DIF:
PS.12.12
1
DIF:
PS.12.12
1
DIF:
PS.9.4
1
DIF:
PS.9.4
1
DIF:
PS.9.4
1
DIF:
PS.9.4
1
DIF:
PS.9.4
1
DIF:
PS.9.7 | PS.12.1
1
DIF:
PS.9.7 | PS.12.1
1
DIF:
PS.9.7 | PS.12.1
1
DIF:
PS.9.7 | PS.12.1
1
DIF:
PS.9.7 | PS.12.1
1
DIF:
PS.9.7 | PS.12.1
1
DIF:
PS.9.7 | PS.12.1
3
I
REF: 1
I
REF: 1
II
REF: 1
I
REF: 1
II
REF: 1
II
REF: 1
I
REF: 2
I
REF: 2
I
REF: 2
II
REF: 3
II
REF: 3
I
REF: 1
I
REF: 1
I
REF: 2
I
REF: 2
II
REF: 3
I
REF: 1
I
REF: 1
I
REF: 1
III
REF: 1
I
REF: 2
III
REF: 2
I
REF: 3
ID: A
56. ANS: A
OBJ: 1
57. ANS: A
OBJ: 2
58. ANS: B
OBJ: 1
59. ANS: B
OBJ: 2
60. ANS: A
OBJ: 2
61. ANS: C
OBJ: 2
62. ANS: C
OBJ: 3
63. ANS: C
OBJ: 3
64. ANS: C
OBJ: 4
65. ANS: D
OBJ: 5
66. ANS: C
Solution:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
114.95 g NO 2 +
PTS: 1
STA: PS.12.1
67. ANS: B
Solution:
1 mol NO 2
46.01 g NO 2
5.0 g CH 3 OH +
= 9.4 + 10
22
I
REF: 4
III
REF: 5
I
REF: 1
III
REF: 1
III
REF: 1
III
REF: 1
III
REF: 1
III
REF: 1
II
REF: 1
II
REF: 1
REF: 3
OBJ: 2
= 2.498 mol NO 2
DIF: III
0.240 mol C 6 H 12 O 6 +
PTS: 1
STA: PS.12.1
68. ANS: A
Solution:
1
DIF:
PS.9.7 | PS.12.1
1
DIF:
PS.12.1
1
DIF:
PS.12.1
1
DIF:
PS.12.1
1
DIF:
PS.12.1
1
DIF:
PS.12.1
1
DIF:
PS.12.1
1
DIF:
PS.12.1
1
DIF:
PS.12.1
1
DIF:
PS.12.1
180.18 g C 6 H 12 O 6
1 mol C 6 H 12 O 6
DIF: III
1 mol CH 3 OH
32.05 g CH 3 OH
= 43.2 C 6 H 12 O 6
REF: 3
+
6.022 + 10
23
OBJ: 2
molecules of CH 3 OH
1 mol CH 3 OH
molecules of CH 3 OH
PTS: 1
STA: PS.12.1
DIF: III
REF: 3
4
OBJ: 3
ID: A
69. ANS: D
Solution:
percentage composition =
percentage carbon =
70.
71.
72.
73.
mass of element
× 100
total mass
12.01 g
× 100 = 42.88% C, 100% − 42.88% = 57.12% O
28.01 g
PTS: 1
DIF: III
STA: PS.12.1
ANS: C
PTS: 1
OBJ: 1
STA: PS.12.1
ANS: D
PTS: 1
OBJ: 3
STA: PS.12.1
ANS: C
PTS: 1
OBJ: 3
STA: PS.12.1
ANS: B
Solution:
molecular mass
58 amu
=
= 2.0
empirical mass
29.07 amu
REF: 3
OBJ: 4
DIF: I
REF: 4
DIF: II
REF: 4
DIF: III
REF: 4
DIF: III
REF: 4
OBJ: 4
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
DIF: I
REF: 1
DIF: I
REF: 1
DIF: I
REF: 1
DIF: II
REF: 1
DIF: III
REF: 1
DIF: III
REF: 1
DIF: I
REF: 2
DIF: II
REF: 2
DIF: II
REF: 2
DIF: II
REF: 2
DIF: I
REF: 3
Molecular formula is C 4 H 10 .
74.
75.
76.
77.
78.
79.
80.
81.
82.
83.
84.
PTS:
STA:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
1
PS.12.1
A
1
D
1
B
2
B
3
B
4
D
4
D
1
D
1
C
2
C
4
B
1
1
PS.9.7
1
PS.9.7
1
PS.9.7
1
PS.9.7
1
PS.9.7
1
PS.9.7
1
PS.9.7
1
PS.9.7
1
PS.9.7
1
PS.9.7
1
PS.9.4
5
ID: A
85. ANS: C
OBJ: 3
86. ANS: D
Solution:
PTS: 1
STA: PS.9.7
4.00 mol H 2 SO 4 ×
1 mol SO 3
1 mol H 2 SO 4
DIF: III
×
80.07 g SO 3
1 mol SO 3
REF: 1
= 320.28 or 320. g SO 3
PTS: 1
DIF: III
REF: 2
OBJ: 2
STA: PS.9.7
87. ANS: C
Solution:
1 mol Cl 2
2 mol KCl 74.55 g KCl
300. g Cl 2 ×
×
×
= 630.89 g KCl
70.90 g Cl 2
1 mol Cl 2
1 mol KCl
1 mol KBr
2 mol KCl 74.55 g KCl
×
×
= 187.9 g KCl
119.00 g KBr 2 mol KBr
1 mol KCl
Since Cl2 would produce the most KCl, KBr is the limiting reactant, thus 188 G KCl is produced.
300. g KBr ×
PTS: 1
STA: PS.9.7
88. ANS: B
OBJ: 3
89. ANS: A
OBJ: 3
90. ANS: D
Solution:
500. g SO 3 ×
91.
92.
93.
94.
DIF: III
REF: 2
OBJ: 4
PTS:
STA:
PTS:
STA:
DIF: I
REF: 3
DIF: I
REF: 3
1 mol SO 3
80.07 g SO 3
×
1
PS.9.7
1
PS.9.7
1 mol H 2 SO 4
1 mol SO 3
×
98.09 g H 2 SO 4
1 mol H 2 SO 4
percentage yield =
ex p erimental yield
× 100
actual yield
percentage yield =
575 g
× 100 = 93.88 or 93.9%
612.5 g
PTS:
STA:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
ANS:
OBJ:
1
PS.9.7
A
1
C
1
C
2
B
3
= 612.5 g H 2 SO 4
DIF: III
REF: 3
OBJ: 4
PTS:
STA:
PTS:
STA:
PTS:
STA:
PTS:
STA:
DIF: I
REF: 1
DIF: I
REF: 1
DIF: I
REF: 1
DIF: II
REF: 1
1
PS.9.11
1
PS.9.11
1
PS.9.11
1
PS.9.11
6
ID: A
95. ANS: C
Solution:
P 1 V1
1.00 atm × 400.0 mL
P2 =
=
= 0.20 atm
V2
2000 mL
PTS: 1
DIF: III
REF: 2
96. ANS: C
Solution:
V1 T 2
5.0 L × 283 K
V2 =
=
= 5.09 or 5.1L
T2
278 K
OBJ: 2
PTS: 1
DIF: III
REF: 2
97. ANS: C
PTS: 1
DIF: II
OBJ: 4
98. ANS: B
Solution:
P 1 T2
1.5 atm × 288 K
P2 =
=
= 1.66 or 1.7 atm
T1
260 K
OBJ: 3
REF: 2
PTS:
99. ANS:
OBJ:
100. ANS:
OBJ:
1
C
2
D
2
DIF: III
PTS: 1
REF: 2
DIF: I
OBJ: 4
REF: 3
PTS: 1
DIF: I
REF: 3
7