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AP Chemistry
Worksheet Fall Semester Review
Name ________________________
Period _____
Write the number next to the word that fits the paragraph.
Measurement in Chemistry
Science knowledge is advanced by observing patterns, (1), and constructing explanations, (2); which
are supported by repeatable (3) evidence.
Measurements are made using the metric system, where the standard units are called (4) units, which
are based on the meter, kilogram, and second as the basic units of length, mass, and time, respectively.
The SI temperature scale is the (5) scale, although the (6) scale is frequently used in chemistry. The metric
system employs a set of prefixes to indicate decimal fractions or multiples of the base units; k (10 -3), c (10-2),
m (10-3), m (10-4) and n (10-9).
All measured quantities are inexact to some extent. The (7) of a measurement indicates how closely
different measurements of a quantity agree with one another. The (8) of a measurement indicates how well
a measurement agrees with the accepted value. Significant figures indicate the level of (9) in a
measurement. Significant figures in a measured quantity include one (10) digit, the last digit of the
measurement. Calculations involving measured quantities are reported with the appropriate number of
significant figures. In multiplication and division, the (11) of significant figures is used. In addition and
subtraction, the (12) of the least accurate significant figure is used. Relative difference between a measured
value and an accepted (true) value is called (13). Relative spread of multiple measurements is called (14).
Mass and volume measure (15) of matter. Density relates mass to volume in the equation, (16).
Chemical processes involve interaction of (17), which are measured in (18). The number of particles in a
mole is called Avogadro's number, (19). This number is based on using periodic table masses to be equal
to mass of formula unit labeled in (20). (21) is the sum of atomic masses in the chemical formula. The
mass of one molecule of H2O, for example, is 18.0 u, so the molar mass of H2O is 18.0 g.
In the dimensional analysis approach to problem solving, we keep track of units as we carry
measurements through calculations. The given units are multiplied by a series of conversion factors, which
are (22) of equivalent quantities. After canceling out units algebraically, what remain are the target units.
Atomic Nature of Matter
Atoms are the basic building blocks of matter; they are the smallest units of an element that can
combine with other elements. Atoms are composed of even smaller subatomic particles. Experiments led to
the discovery and characterization of subatomic particles. (1) experimented with cathode rays in magnetic
and electric fields, which led to the discovery of the electron and its (2) ratio. (3), working with oil-drops in a
vacuum to determined the (4) of the electron. (5) observed the scattering of  particles by gold metal foil and
concluded that atoms have a dense, (6) nucleus.
The atom's nucleus contains (7) and (8), whereas (9) move in the space around the nucleus. The
charges of subatomic particles in terms of the charge of a proton, +1, are: electron (10) and neutron (11).
Masses in terms of the mass of a proton, 1, are: neutron (12), and electron  0.
Elements are classified by atomic number or Z value equals the number of (13). The mass number or A
value is the sum of protons and (14). Atoms of the same element that differ in mass number are called (15).
In a neutral atom, the number of protons equals the number of electrons. An (16) is formed when the
number of electrons and protons are not equal.
The unified atomic mass unit, u, is 1/12 the mass of a C-12 atom. The average atomic mass of an
element is calculated using the formula: 100m av = (17).
The two kinds of pure substances are (18), which are identified by a chemical symbol and (19), which
are composed of two or more elements joined chemically and identified by a chemical formula, which shows
the composition. Molecular compounds have a defined size, whereas crystalline compounds are
unbounded, where their formula shows the (20) of atoms in the compound.
(21) are composed of multiple pure substances in an object or container and have variable composition.
They can be homogeneous or heterogeneous. Homogeneous mixtures are also called (22) and are uniform
throughout.
Radioactivity
There are four kinds of radioactive decay: emission of (1) particles ( or 42He), (2) particle ( or 0-1e), (3)
particle (, 01e), and (4) radiation (00).
In nuclear equations, reactant and product nuclei are represented by AZX , called a (5) symbol. In a
balanced equation the sum of reactant A values equals the sum of product A values and the sum of reactant
Z values equals the sum of product Z values.
Modes of decay can be predicted by comparing the number of neutrons with the average. In general,
neutron-rich nuclei tend to emit (6) particles; neutron-poor nuclei tend to emit (7) particles and nuclei above
Z = 83 tend to emit (8) particles.
Nuclear (9), induced conversions of one nucleus into another, can be brought about by bombarding
nuclei with either charged particles or neutrons.
The rate of decay (radioactivity) is rate = (10). The time for half of the radioactive atoms to decay is
constant, t½ = (11). The time it takes radioactive atoms to decay is determined by using the formula kt = (12).
Electron Structure—Bohr Model
The electronic structure of an atom describes the energies and arrangement of electrons around the
atom. Much of what is known about the electronic structure of atoms was obtained by observing atomic (1),
which is the radiant energy emitted or absorbed by matter. Radiant energy equals: Ephoton = (2).
14
13
19
8
15
6
9
16
10
3
20
5
1
21
18
11
17
12
7
22
4
2
% deviation
% difference
6.02 x 1023
accuracy
amount
Celsius
certainty
d = m/V
estimated
experimental
grams
Kelvin
laws
Molar mass
moles
number
particles
position
precision
ratios
SI
theories
17
12
11
10
4
2
19
9
18
16
15
3
21
8
14
6
7
13
20
5
22
1
1
8
2
6
4
12
5
3
7
10
11
9
%1m1 + %2m2
1
0
-1
charge
charge/mass
compounds
electrons
elements
ion
isotopes
Millikan
Mixtures
neutrons
neutrons
positive
protons
protons
ratio
Rutherford
solutions
Thomson
alpha
alpha
beta
beta
gamma
ln(No/Nt)
nuclear
positron
positron
kNt
(ln2)/k
transmutations
10
3
9
11
absorbed
Bohr
emitted
En-final – En-initial
(3) analyzed the wavelengths of light emitted by hydrogen atoms and proposed a model that explains
its atomic spectrum. In this model the energy of the hydrogen atom depends on its quantum number, (4).
The value of n must be a positive integer (1, 2, 3, . . .) and each value of n corresponds to a different energy;
En = (5). As n increases, the energy of the electron (6) until it reaches a value of 0 J, where n equals infinity
and the electron leaves the atom or ionizes. The lowest energy is n = 1; this is called the (7) state. Other
values of n correspond to (8) states. Light is (9) when the electron drops from a higher energy state to a
lower energy state and light is (10) when excited from a lower energy state to a higher one. The energy of
light emitted or absorbed equals the difference in energy between the two transition states, Ephoton = (11).
Quantum Mechanical Model
In the quantum mechanical model each electron has a precisely known energy, but according to the (1)
Uncertainty Principle, the (2) of the electron cannot be determined exactly; rather, the 90 % probability of it
being at a particular point in space is given by its (3). An orbital is described by a combination of four
quantum numbers. The principal quantum number, n, is indicated by the integers 1, 2, 3. . . 7. This
quantum number relates to the (4) and energy of the orbital. The sublevel quantum number, l, is indicated
by the letters (5), corresponding to the values of (6). The l quantum number defines the shape of the orbital.
For a given value of n, l can have integer values from (7). The orbital quantum number, ml, relates to the
orientation of the orbital in space. For a given value of l, ml can have integral values ranging from (8). The
spin quantum number, ms, defines the orientation of the electron's (9) field and has two possible values,
(10). The (11) Exclusion Principle states that no two electrons in an atom can have the same spin in the
same orbital. This principle limits the number of electrons that can occupy any one atomic orbital to (12),
which differ in their value of ms.
Electron Arrangements in Atoms and Ions
Energy (1) as n increases (1 < 2 < 3, etc) and within the same value of n, energy increases as the
sublevel progresses from letters (2). Orbitals within the same sublevel are (3), meaning they have the same
energy.
The energies of the (4) sublevels are less than the energy of the next higher s sublevel, whereas the
energies of the (5) sublevels are greater than the next higher s sublevel. This restricts the outermost
occupied sublevels for any atom to s and p. Electrons that occupy the outermost sublevels are involved in
chemical bonding and are called (6) electrons. Non-valence electrons are called (7) electrons.
The periodic table is partitioned into different types of elements, based on their electron arrangement.
Elements with the same valence energy level form a row or (8). Elements with the same number of valence
electrons form a column or (9). The elements in which an s or p sublevel is being filled are called the (10)
elements, which include group 1—(11) metals, group 2—(12) metals, group 17—(13) and group 18—(14).
Transition metals are where the (15) sublevel is filling. The (16) sublevel filling regions are called lanthanide
and actinide.
Electron (17) show how electrons are distributed among the atom's sublevels. In (18) state
configurations electrons occupy the lowest sublevel available until its capacity is reached. Additional
electrons fill the next lowest sublevel until its filled, etc. (19) state configurations have gaps.
(20) diagrams show how the electrons fill the specific orbitals, where arrows are used to represent
electrons; () for ms = +½ and () for ms = –½. When electrons occupy a sublevel with more than one
degenerate orbital, (21) rule applies. The rule states that the lowest energy is attained by maximizing the
number of electrons with the same (22).
Transition metals in columns (23) have a half-filled s sublevel in order to have a Half-filled or fully
occupied d sublevel, which is more stable than other arrangements.
The electron arrangement for monatomic ions is the same as the element with the same number of
electrons. Elements within (24) squares on the periodic table of a noble gas form ions with the same
electron arrangement as the noble gas and are (25). Transition metals form ions by losing all s level
electrons first.
Elements with (26) electrons have reinforcing magnetic fields, which makes the atom (27). If all of the
electrons are paired, then the atom is (28).
8
7
6
4
1
5
excited
ground
increases
n
spectra
-2.18 x 1018 J/n2
2.00
x 10-25 J•m/
2
10
7
6
8
1
2
9
3
11
4
5
12
23
11
12
17
7
15
5
3
28
19
16
18
9
13
21
1
25
10
14
20
27
8
2
4
22
24
26
6
Periodic Properties—Main Groups
12
Many properties of atoms are due to the distance of the outer electrons from the nucleus and to the (1) 19
nuclear charge experienced by these electrons. The (2) electrons are very effective in screening the outer
13
electrons from the full charge of the nucleus, whereas electrons in the valence shell do not screen each
20
other very effectively at all. As a result, the effective nuclear charge experienced by valence electrons
18
increases as we move (3) across a period, except for (4) metals where the increase in Zeff is minimum since 14
the added electrons enter the core and cancel the added protons. As a result, atomic radii (5) as we go
2
down a group and (6) as we proceed left to right across a period of main group elements.
6
Cations are (7) than their parent atoms; anions are (8) than their parent atoms, but group and period
9
trends are the same as atomic radius. For an isoelectronic series the radius (9) with increasing nuclear
1
charge since the electrons are attracted more strongly to the nucleus.
5
The ionization energy is the energy needed to (10) an electron from a gaseous atom; forming a cation.
11
Successive ionization energies show a sharp (11) after all the valence electrons have been removed,
because of the much higher effective nuclear charge experienced by the core electrons. For the main group 8
3
elements, the first ionization energy trend is generally opposite the atomic radii trend, with smaller atoms
having higher first ionization energies, except for columns (12) (removing p orbital electron) and column (13) 15
17
(removing an electron from a full orbital).
The electron affinity measures the energy change when (14) an electron to a gaseous atom; forming an 16
10
+½ and -½
0 to (n – 1)
0, 1, 2, and 3
–l to +l
Heisenberg
location
magnetic
orbital
Pauli
radius
s, p, d, and f
Two
6 and 11
alkali
alkaline earth
configurations
core
d
d and f
degenerate
diamagnetic
Excited
f
ground
group
halogens
Hund's
increases
isoelectronic
main-group
noble gases
Orbital
paramagnetic
period
spdf
s and p
spin
three
unpaired
Valence
13
15
16
18
2
adding
core
decrease
decreases
effective
increase
increase
larger
left to right
negative
negative
positive
remove
anion. A (15) electron affinity means that the anion is stable; a (16) electron affinity means that the anion is
not stable relative to the separate atom and electron. In general, electron affinities become more (17) as we
proceed from left to right across the main groups except for column (18) (adding an electron to a p orbital),
column (19) (adding an electron to a half-filled orbital) and column (20) (adding an electron to the next
higher energy level).
Bonding
Bonds are classified into three broad groups: (1) bonds, which are the electrostatic forces that exist
between ions of opposite charge; (2) bonds, which result from the sharing of electrons by two atoms; and (3)
bonds, which bind together the atoms in metals.
The formation of bonds involves interactions of the (4) electrons of atoms. The (5) electrons of an atom
can be represented by electron-dot symbols, called (6) symbols. The tendencies of atoms to gain, lose, or
share their valence electrons often follow the (7) rule, which can be viewed as an attempt by atoms to
achieve a (8) electron configuration.
Ionic bonding results from the complete (9) of electrons from one atom to another, with formation of a
three-dimensional lattice of charged particles. The strength of the electrostatic attractions between ions is
measured by the (10) energy, which (11) with ionic charge and (12) with distance between ions.
A covalent bond results from the sharing of electrons. The sharing of one pair of electrons produces a
(13) bond; whereas the sharing of two or three pairs of electrons produces double or triple bonds,
respectively. The bond length (14) as the number of bonds between the atoms increases whereas the bond
strength (15).
(16) measures the ability of an atom to attract bonding electrons. Electronegativity generally (17) from
left to right in the periodic table, and (18) going down a column. The difference in electronegativity of bonded
atoms is used to determine the (19) of a bond; the (20) the difference, the more polar the bond. A polar
molecule has a positive side (+) and a negative side (–). This separation of charge produces a dipole, the
magnitude of which is given by the dipole moment.
7
4
smaller
transition
2
covalent
12
decreases
14
decreases
18
decreases
16 Electronegativity
20
greater
11
increases
15
increases
17
increases
1
ionic
10
lattice
6
Lewis
3
metallic
8
noble gas
7
octet
4
outermost
19
polarity
13
single
9
transfer
5
valence
Lewis Structures
3
eight
Electron distribution in molecules is shown with (1) structures, which indicate how many valence
7
formal
electrons are involved in forming bonds and how many remain as (2) electron pairs. If we know which atoms 8
half
are connected to one another, we can draw Lewis structures for molecules and ions by a simple procedure,
1
Lewis
where (3) electrons are placed around each atom. When there are too few valence electrons, then it will be
9
low
necessary to add bonds. When there are too many valence electrons (and the central atom has at least 10
negative
(5) energy level electrons), then it will be necessary to place additional electrons (up to 10 or 12) around the 6
odd
central atom. When the total number of valence electrons is an (6) number, then it will be necessary to place 5
third
seven electrons around the atom with the odd number of valence electrons.
2
unshared
When there are multiple valid Lewis structures for a molecule or ion, we can determine which is most
4

likely by assigning a (7) charge to each atom, which is the sum of (8) the bonding electrons and all the
unshared electrons. Most acceptable Lewis structures will have (9) formal charges with any (10) formal
charge residing on the more electronegative atom.
VSEPR Model
5
domain
The valence-shell electron-pair repulsion (VSEPR) model rationalizes molecular geometries based on
1
domains
the repulsions between electron (1), which are regions about a central atom where electrons are likely to be
2
far apart
found. Pairs of electrons, bonding and non-bonding, create domains around an atom, which are as (2) as
3
greater
possible. Electron domains from non-bonding pairs exert slightly (3) repulsions, which leads to (4) bond
6
molecular
angles than idealized values. The arrangement of electron domains around a central atom is called the
7
nonpolar
electron (5) geometry; the arrangement of atoms is called the (6) geometry.
8
polar
Certain molecular shapes, such as linear and trigonal planar, have cancelling bond dipoles, producing a 4
smaller
(7) molecule, which is one whose dipole moment is zero. In other shapes, such as bent and trigonal
pyramidal, the bond dipoles do not cancel and the molecule is (8) (a nonzero dipole moment).
Valence-Bond Theory
15
blend
Valence-bond theory is an extension of Lewis's notion of electron-pair bonds. In valence-bond theory,
4
bonding
covalent bonds are formed when atomic (1) on neighboring atoms overlap. The bonding electrons occupy
16
delocalized
the overlap region and are attracted to both (2) simultaneously, which bonds the atoms together.
3
hybrid
To extend valence-bond theory to polyatomic molecules, s, p, and sometimes d orbitals are blended to
2
nuclei
form (3) orbitals, which overlap with orbitals on another atom to make a bond. Hybrid orbitals also hold non- 12
one
(4) pairs of electrons. A particular mode of hybridization can be associated with each of the five common
1
orbitals
electron-domain geometries (linear = (5); trigonal planar = (6); tetrahedral = (7); trigonal bipyramidal = (8);
17
order
and octahedral = (9)).
11
pi
Covalent bond in which the electron density lies along the line connecting the atoms is called (10), ,
14
resonance
bond. Bonds can also be formed from the sideways overlap of p orbitals. Such a bond is called a (11) ,
18
share
bond. A double bond, such as that in C2H4, consists of one  bond and (12)  bond; a triple bond, such as
10
sigma
that in C2H2, consists of one  and (13)  bonds. The formation of a  bond between carbons requires that
5
sp
molecules adopt a specific orientation; the two CH2 groups in C2H4, for example, must lie in the same plane. 6
sp2
As a result, the presence of  bonds introduces rigidity into molecules.
7
sp3
Sometimes a  bond can be placed in more than one location. In such situations, we describe the
8
sp3d
molecule by using two or more (14) structures. The molecule is envisioned as a (15) of these multiple
9
sp3d2
resonance structures and the  bonds are (16); that is, spread among several atoms. The bond (17) value
13
two
represents the actual bond strength and is sum of the  bond plus a (18) of the  bond(s).
Naming Binary Molecules
The procedures used for naming two-element, (1), molecular compounds follow the rules below.
1. The (2) electronegative element is written (3) in the formula and named as an element.
2. The name of the second element is given an (4) ending.
3. Greek prefixes are used to indicate the (5) of atoms of each element; (6) is not used with the first.
Molecular compounds that contain hydrogen and one other element are an important exception. The
formula for pure hydrogen sulfide is (7), but when dissolved in water H2S(aq) is (8) acid.
1
binary
3
first
7
H2S
8
hydrosulfuric
4
ide
2
lower
6
mono
5
number
Hydrocarbons
5
carbon
The simplest types of (1) compounds are hydrocarbons. There are four major kinds of hydrocarbons:
10
cis-trans
alkanes, alkenes, alkynes, and aromatic hydrocarbons. (2) are substances that possess the same molecular 7
cyclo
formula, but differ in the arrangements of atoms. In (3) isomers the bonding arrangements of the atoms
12 functional group
differ. Different isomers are given different (4). The naming of hydrocarbons is based on the longest
2
Isomers
continuous chain of (5) atoms in the structure. The locations of alkyl groups, which branch off the chain, are
9
location
specified by (6) along the carbon chain. Ring structures have the prefix (7). The names of alkenes and
8
multiple
alkynes are based on the longest continuous chain of carbon atoms that contains the (8) bond, and the (9)
4
names
of the multiple bond is specified by a numerical prefix. Alkenes exhibit not only structural isomerism but
6
numbering
geometric (10) isomerism as well. In geometric isomers the bonds are the same, but the molecules have
1
organic
different geometries. Geometric isomerism is possible in alkenes because rotation about the C=C double
11
restricted
bond is (11).
structural
3
The chemistry of an organic compound is dominated by the presence of the (12).
Gas State
18
1
Substances that are gases at room temperatures tend to be (1) with (2) molar mass. Air, a mixture
9
101
composed mainly of (3), is the most common gas we encounter. Some liquids and solids can also exist in
17
273
the gaseous state, where they are known as vapor. Gases' volume can change because they are (4) and
10
760
they mix in all proportions because their component molecules are far apart.
11
760
To describe the state, or condition, of a gas, we must specify four variables: pressure (P), volume (V),
15
atm
temperature (T), and quantity (n). Volume is measured in (5), temperature in (6), and quantity of gas in (7).
39
attractive
(8) is the force per unit area. In chemistry, pressure is measured in atmospheres (atm), torr (named after
12
barometer
Torricelli), millimeter of mercury (mm Hg) and kilopascals (kPa). One atmosphere of pressure equals (9)
25
chaotic
kPa = (10) torr = (11) mm Hg. A (12) is used to measure atmospheric pressure and a (13) is used to
4
compressible
measure the pressure of enclosed gases.
37
decreases
The ideal-gas law equation is (14). The term R is the gas constant, which is 0.0821 when P is in (15) or
35
diffusion
8.31 when P is in (16). Most gases at pressures of about 1 atm and temperatures near 273 K and above
33
effusion
obey the ideal-gas law reasonably well. The conditions of (17) K and (18) atm are known as the standard
38
finite
temperature and pressure and abbreviated as (19).
40
ideal-gas
Using the ideal-gas law equation, we can relate the density of a gas to its molar mass: MM = (20). In all
36
increases
applications of the ideal-gas law we must remember to convert temperatures to the (21) scale; the only
21
Kelvin
scale where 0 equals no warmth. In gas mixtures the total pressure is the sum of the (22) pressures that
kelvins
each gas would exert if it were present alone under the same conditions (Dalton's law). The partial pressure 6
16
kPa
of a component of a mixture is equal to its mole fraction times the total pressure, PA = (23). The mole
5
liters
fraction is the ratio of the moles of one component of a mixture to the (24) moles of all components. In
2
low
calculating the quantity of a gas collected over water, correction must be made for the partial pressure of
water vapor in the gas mixture.
13
manometer
The kinetic-molecular theory accounts for the properties of an ideal gas in terms of a set of statements
20
dRT/P
about the nature of gases. Briefly, these statements are as follows: Molecules are in continuous, (25)
31
molar mass
motion; the volume of gas molecules is (26) compared to the volume of their container; the gas molecules
34
molar mass
have (27) attraction for one another; their collisions are elastic; and the average kinetic energy of the gas
1
molecular
molecules is proportional to the absolute (28).
7
moles
Molecules of a gas do not all have the same kinetic energy at a given instant. Their (29) are distributed
3
N2 and O2
over a wide range; the distribution varies with the molar mass of the gas and with temperature. The root26
negligible
mean-square speed, u, varies in proportion to the square root of the absolute (30) and in inverse proportion 27
no
with the square root of the (31): u = (32).
23
XAPtot
It follows from kinetic-molecular theory that the rate at which a gas undergoes (33) (escapes through a
22
partial
tiny hole into a vacuum) is inversely proportional to the square root of its (34) (Graham's law). The (35) of
8
Pressure
one gas through the space occupied by a second gas is another phenomenon related to the speeds at
14
PV = nRT
which molecules move. Collisions between molecules limit the rate at which a gas molecule can diffuse.
29
speeds
Departures from ideal behavior increase in magnitude as pressure (36) and as temperature (37). Real
19
STP
gases depart from ideal behavior because the molecules possess (38) volume and because the molecules
28
temperature
experience (39) forces for one another. The van der Waals equation is an equation that modifies the (40)
30
temperature
law equation to account for molecular volume and intermolecular forces.
24
total
32
(3RT/MM)½
Phase Change
23
1
Substances that are gases or liquids at room temperature are usually composed of (1). In gases the
22
atmospheric
intermolecular attractive forces are (2) compared to the kinetic energies of the molecules; thus, the
19 condensation
molecules are widely separated and undergo constant, chaotic motion. In liquids the intermolecular forces
16
critical
are strong enough to keep the molecules in close proximity; nevertheless, the molecules are free to move
4
dipole-dipole
with respect to one another. In solids the inter-particle attractive forces are strong enough to restrain
molecular (3) and to force the particles to occupy specific locations in a three-dimensional arrangement.
Three types of intermolecular forces exist between neutral molecules: (4) forces, London (5) forces, and
(6) bonding. London dispersion forces operate between all molecules. The relative strengths of the dipoledipole and dispersion forces depend on the polarity, polarizability, size, and shape of the molecule. Dipoledipole forces increase in strength with (7) polarity. Dispersion forces increase in strength with increasing
molecular (8), although molecular shape is also an important factor. Hydrogen bonding occurs in
compounds containing (9) bonded to H. Hydrogen bonds are (10) than dipole-dipole or dispersion forces.
The stronger the intermolecular force, the greater is the viscosity, or resistance to flow, of a liquid. The
surface tension of a liquid also increases as intermolecular forces increase in strength. Surface tension is a
measure of the tendency of a liquid to maintain a minimum surface area. The adhesion of a liquid to the
walls of a narrow tube and the cohesion of the liquid account for capillary action and the formation of a
meniscus at the surface of a liquid.
A substance may exist in more than one state of matter, or (11). Phase changes are transformations
from one state to another. Changes of a solid to liquid, (12), solid to gas, (13), and liquid to gas, (14), absorb
energy. The reverse processes (15) energy. A gas cannot be liquefied by application of pressure if the
temperature is above its (16) temperature. The pressure required to liquefy a gas at its critical temperature
is called the critical pressure.
The vapor pressure of a liquid indicates the tendency of the liquid to (17). The vapor pressure is the
partial pressure of the vapor when it is in dynamic equilibrium with the liquid. At equilibrium the rate of (18),
transfer of molecules from the liquid to the vapor, equals the rate of (19), transfer from the vapor to the
liquid. The higher the vapor pressure of a liquid, the more readily it evaporates and the more (20) it is. Vapor
pressure increases nonlinearly with temperature. Boiling occurs when the (21) pressure equals the (22)
pressure. The normal boiling point is the temperature at which the vapor pressure equals (23) atm.
The equilibria between the solid, liquid, and gas phases of a substance as a function of temperature
and pressure are displayed on a phase diagram. Equilibria between any two phases are indicated by a (24).
The line through the melting point usually slopes slightly to the right as pressure increases, because the
solid is usually (25) dense than the liquid. The melting point at 1 atm is the (26) melting point. The point on
the diagram at which all three phases coexist in equilibrium is called the (27) point.
Crystalline Solids
In a crystalline solid, particles are arranged in a regularly repeating pattern. An amorphous solid or
glass is one whose particles show no such order.
The properties of solids depend both on the type of particles and on the attractive forces between them.
(5) solids, which consist of atoms or molecules held together by intermolecular forces, are soft and (6)
melting. Covalent (7) solids, which consist of atoms held together by covalent bonds that extend throughout
the solid, are hard and (8) melting. (9) solids are hard and brittle and have (10) melting points. (11) solids,
which consist of metal (12) held together by a sea of (13), exhibit a wide range of properties.
Solubility
(1) form when one substance disperses uniformly throughout another. The dissolving medium of the
solution (usually in the greater amount) is called the (2). The substance dissolved in a solvent (usually the
smaller amount) is called the (3). The attractive interaction of solvent molecules with solute is called (4).
When the solvent is water, the interaction is called (5). The dissolution of ionic substances in water is
promoted by hydration of the separated ions by the polar water molecules. The overall change in energy
upon solution formation may be either positive (6) or negative (7), depending on the relative value of (8)
energy (positive) and (9) energy (negative).
The equilibrium between a saturated solution and undissolved solute is dynamic; the process of solution
and the reverse process, (10), occur simultaneously. In a solution in equilibrium with undissolved solute, the
two processes occur at equal rates, giving a (11) solution. The amount of solute needed to form a saturated
solution at any particular temperature is the (12) of that solute at that temperature.
The solubility of one substance in another depends on the tendency of systems to become more
random, by becoming more dispersed in space, and on the relative intermolecular solute-solute and solventsolvent energies compared with solute-solvent interactions. Polar and (13) solutes tend to dissolve in polar
solvents such as water and alcohol, and nonpolar solutes tend to dissolve in nonpolar solvents ("like
dissolves like"). Liquids that mix in all proportions are (14); those that do not dissolve significantly in one
another are immiscible. Hydrogen-bonding interactions between solute and solvent often play an important
role in determining solubility; for example, ethanol and water, whose molecules form hydrogen bonds with
each other, are miscible. The solubilities of gases in a liquid are generally proportional to the (15) of the gas
over the solution, as expressed by Henry's law: Sg = (16). The solubilities of most solid solutes in water (17)
as the temperature of the solution increases. In contrast, the solubilities of gases in water generally (18) with
increasing temperature.
Concentrations of solutions can be expressed quantitatively by several different measures, including
mass (19) [(mass solute/mass solution) x 102], parts per (20) (ppm) [(mass solute/mass solution) x 106],
parts per (21) (ppb) [(mass solute/mass solution) x 109], and mole (22) [mol solute/(mol solute + mol
solvent)]. (23), M, is defined as moles of solute per liter of solution; (24), m, is defined as moles of solute per
kg of solvent. Conversions between concentration units is possible if (25) mass of solute and solvent are
known and/or the (26) of the solution is known.
5
17
18
6
7
24
8
12
1
25
3
9
2
26
11
15
10
13
27
21
14
20
dispersion
evaporate
evaporation
hydrogen
increasing
line
mass
melting
molecules
more
motion
N, O or F
negligible
normal
phase
release
stronger
sublimation
triple
vapor
vaporization
volatile
12
13
8
10
9
6
11
5
7
21
10
18
26
6
7
22
5
9
17
13
8
20
14
24
25
23
19
15
11
16
12
3
1
4
2
cations
electrons
high
high
Ionic
low
Metallic
Molecular
network
billion
crystallization
decrease
density
endothermic
exothermic
fraction
hydration
hydration
increase
ionic
lattice
million
miscible
molality
molar
Molarity
percent
pressure
saturated
kPg
solubility
solute
Solutions
solvation
solvent
Colligative Properties
A physical property of a solution that depends on the concentration of solute particles present,
regardless of the nature of the solute, is a (1) property. Colligative properties include vapor-pressure
lowering, freezing-point lowering, boiling-point elevation, and osmotic pressure. The (2) of vapor pressure is
expressed by Raoult's law, Pvap = (3). A solution containing a nonvolatile solute possesses a (4) boiling
point than the pure solvent. The molal boiling-point constant, (5), represents the increase in boiling point for
a 1 m solution of solute particles as compared with the pure solvent. Similarly, the molal freezing-point
constant, (6), measures the (7) of the freezing point of a solution for a 1 m solution of solute particles. The
temperature changes are given by the equations Tb = (8) and Tf = (9). When NaCI dissolves in water,
(10) moles of solute particles are formed for each mole of dissolved salt. The boiling point or freezing point
is thus elevated or depressed, respectively, approximately (11) as much as that of a nonelectrolyte solution
of the same concentration. The multiplier is called the van't Hoff factor, (12). Similar considerations apply to
other strong electrolytes. (13) is the movement of solvent molecules through a semipermeable membrane
from a (14) concentrated to a (15) concentrated solution. This net movement of solvent generates an
osmotic pressure, (16), which can be measured in units of gas pressure, such as atm. The osmotic pressure
of a solution as compared with pure solvent is proportional to the solution molarity:  = (17). Osmosis is a
very important process in living systems, in which cell membranes act as semipermeable walls, permitting
the passage of (18), but restricting the passage of ionic and macromolecular components.
1
4
12
5
6
14
2
7
15
13
3
11
10
18
8
9
16
17
Chemical Reactions
24
The study of the quantitative relationships between chemical formulas and chemical equations is known 13
as (1). One of the important concepts of stoichiometry is the law of conservation of (2), which states that the 4
total mass of the products of a chemical reaction is the same as the total mass of the reactants. Likewise,
3
the same numbers of atoms of each type are present before and after a (3) reaction. A (4) chemical
9
equation shows equal numbers of atoms of each element on each side of the equation. Equations are
5
balanced by placing (5) in front of the chemical formulas for the reactants and products of a reaction, not by 16
changing the (6) in chemical formulas.
10
Among the reaction types described in this unit are (a) combination reactions, in which two reactants
19
combine to form (7) product; (b) decomposition reactions, in which a (8) reactant forms two or more
12
products; (c) combustion reactions in oxygen, in which a hydrocarbon or related compound reacts with O2 to
11
form (9); (d) electron exchange reactions in which an element exchanges (10) with an ion in a compound;
22
(e) ion exchange reactions, in which (11) exchange "partners"; and (f) proton exchange reactions, in which
20
an acid gives its (12) to a hydroxide ion, OH-, to form water. The last three reactions will be discussed in
2
more detail later in the year.
17
(13) reactions (electron exchange, ion exchange and proton exchange) usually involve ions, where only
one of the two ions in a ionic compound or acid react, the other is non-reacting and is called a (14) ion. In a 15
7
net ionic equation, those ions that go through the reaction unchanged are (15).
25
The (16) in a balanced equation give the relative numbers of moles of the reactants and products. To
18
calculate the number of grams of a product from the number of grams of a reactant, therefore, first convert
8
grams of reactant to (17) of reactant. We then use the coefficients in the balanced equation to convert the
14
number of moles of reactant to moles of (18). Finally, we convert moles of product to (19) of product. Many
reactions occur in water solutions, aqueous. One of the common ways to express the concentration of a
1
solute in a solution is in terms of molarity. The molarity of a solution is the number of moles of solute per
6
(20) of solution. Molarity makes it possible to interconvert solution (21) and number of moles of solute.
23
A (22) reactant is completely consumed in a reaction. When it is used up, the reaction stops, thus
21
limiting the quantities of products formed. The (23) yield is the quantity of product(s) calculated to form when
all of the limiting reagent is consumed. The (24) yield of a reaction is always less than the theoretical yield.
The (25) yield compares the actual and theoretical yields.
Gravimetric Analysis
4
The (1) formula of any substance can be determined from its (2) composition by calculating the relative
1
number of moles of each atom in 100 g of the substance. Similarly, the empirical formula can be determined 3
from the (3) of each element in the compound, or if it is a (4) reaction, from the mass of CO 2 and H2O
6
produced. If the substance is molecular in nature, its (5) formula can be determined from the empirical
5
formula if the molecular (6) is also known.
2
Volumetric Analysis
4
Solutions of known molarity can be formed either by adding a measured (1) of solute and diluting it to a
2
known volume or by the dilution of a more (2) solution of known concentration (a stock solution). Adding
10
solvent to the solution (the process of dilution) decreases the concentration of the solute without changing
8
the number of (3) of solute in the solution (Mstandard)(Vstandard) = (4).
7
In the process called (5), we add a solution of known molarity, (6), to a buret and an unknown solution
1
to a flask. The titrant is slowly added to the flask until the (7) changes color; which is called the (8) point,
9
where stoichiometrically equivalent quantities of reactants are brought together (also called the end point).
3
The moles of unknown are calculated from the volume and molarity of titrant (moles = (Mstandard)(Vstandard). By 6
knowing moles of unknown, the (9) of unknown or (10) of unknown solution can be determined.
5
21
Change in Enthalpy (H)
Chemical reactions typically involve (1) some bonds between reactant atoms and forming new bonds.
11
Breaking bonds (2) energy, therefore the chemical system gains bond energy and the surroundings lose
19
energy, typically in the form of heat. In contrast, forming bonds (3) energy; resulting in lose of energy by the
2
colligative
higher
i
Kb
Kf
less
lowering
lowering
more
Osmosis
XPo
twice
two
water
Kbm
Kfm

MRT
actual
Aqueous
balanced
chemical
CO2 and H2O
coefficients
coefficients
electrons
grams
H+
ions
limiting
liter
mass
moles
omitted
one
percent
product
single
spectator
stoichiometry
subscripts
theoretical
volume
combustion
empirical
mass
mass
molecular
percent
(Mstock)(Vstock)
concentrated
concentration
equivalence
indicator
mass
molar mass
moles
titrant
titration
(C + mc)T
+H
4.18 J/g•K
absorbs
chemical system and a gain in energy by the surroundings (also in the form of heat).
When energy required to break bonds is greater than the energy released to form new bonds (+H),
then products are at a (4) energy state than reactants (making the product bonds (5) than the reactant
bonds) and energy of the system (6), which is described as (7) because the surroundings typically (8) heat
energy and (9). Alternatively, when energy required to break bonds is (10) the energy released to form new
bonds (11), then (12) are at a lower energy state than reactants (making the product bonds stronger than
the reactant bonds) and energy of the system decreases, which is described as exothermic because the
(13) typically gain heat energy and warm up. The change in (14), H, is listed to the right of a balanced
chemical equation. H can be treated in the same way as a (15) when using dimensional analysis.
The amount of heat transferred between the system and the surroundings is measured experimentally
by (16). A calorimeter measures the (17) change accompanying a process. The temperature change of a
calorimeter depends on its heat capacity, the amount of heat required to raise its temperature by 1 K. The
heat capacity for one mole of a pure substance is called its molar heat capacity; for one gram of the
substance, we use the term (18). Water has a very high specific heat, c = (19). The exchange of heat, Q,
with the surroundings is the product of the surrounding medium's specific heat (c), mass (m), and change in
temperature (T), such that Q = (20). If a Bomb calorimeter is used, then the bomb constant, C. is
incorporated in the equation: Q = (21).
(22), B.E., measures the energy needed to break a covalent bond in a diatomic, gaseous molecule.
The bond energy is approximately the same for any gaseous molecule. Change in enthalpy is estimated by
adding the bond energies of all bonds that are broken (B.E.R)and subtracting the bond energies of all bonds
formed (B.E.P): H = (23).
Change in Entropy (S)
All chemical systems have an inherent amount of (1), S, because of the (2) of the atoms in the
molecules, the (3) of molecules with respect to each other; and the overall (4).
An increase in S for (5) changes can be predicted based on whether the molecules spread out.
Evaporation, diffusion and effusion have (6) values. Dissolving is more complicated although most
dissolving is +S. Chemical reactions that result in more moles of (7) products compared to reactants have
+S.
Thermodynamic Data
The standard enthalpy of formation, (1), of a substance is the enthalpy change for the reaction in which
one mole of substance is formed from its constituent elements under standard conditions of (2) atm and (3)
K. For any element in its most stable state under standard conditions, Hfo = (4). Most compounds have (5)
values of Hfo. Large negative Hfo indicate a (6) bond and stable compound. The standard entropy, (7), is
based on H+ having So = (8) (although the AP exam often lists the values in (9)). The thermodynamic data
chart lists the Hfo and So for common substances.
H depends only on the initial and final states of the system. Thus, the enthalpy change of a process is
the same whether the process is carried out in one step or in a (10) of steps. (11) law states that if a reaction
is carried out in a series of steps, H for the reaction will be equal to the (12) of the enthalpy changes for the
steps. We can therefore calculate H for any process, as long as we can write the process as a series of
steps for which H is known. In addition, Hfo applies to situations involving more than one mole, where Hfo
is (13) by the number of moles, and involving decomposition, where H = (14). An important use of Hfo and
So is calculate H and S for a wide variety of reactions under laboratory conditions, where H  Ho = (15)
and S  So = (16).
Change in Free Energy (G)
The Gibbs free energy (or just free energy), G, combines enthalpy and entropy. For processes that
occur at constant temperature, G = (1). The sign of G relates to the spontaneity of the process. When G
is negative, the process is (2). When G is positive, the process is (3); the reverse process is spontaneous.
When G = 0, then the reaction is at a threshold between spontaneous and nonspontaneous.
The values of H and S generally do not vary much with (4). As a consequence, the dependence of
G with temperature is governed mainly by the value of T in the expression G = H –TS. The threshold
temperature, T = (5), is when a reaction goes from spontaneous to nonspontaneous. This only occurs when
H and S are both positive or both negative. When they are both positive, the reaction is spontaneous at
all temperatures (6) the threshold. When they are both negative, the reaction is spontaneous at all
temperatures (7) the threshold. When H and S have opposite signs, then the reaction is spontaneous at
all temperatures (8) or spontaneous at no temperature (9).
Reaction Rate
Chemical kinetics is the area of chemistry that studies the rates of chemical reactions and the factors
that affect them, namely, concentration, temperature, and catalysts.
Reaction (1) are usually expressed as changes in concentration per unit time: Typically, for reactions in
solution, rates are given in units of molarity per second, (2). For most reactions, a plot of molarity versus
time shows that the rate (3) as the reaction proceeds. The instantaneous rate is the slope of a line drawn
tangent to the concentration-versus-time curve at a specific time. Rates can be written in terms of products,
which are (4) rates, or in terms of reactants, which are (5) rates. The (6) in the balanced equation are
proportional to the various rates for the same reaction.
The quantitative relationship between rate and concentration is expressed by a (7), which has the form:
23
22
1
16
15
9
7
14
4
6
10
8
20
12
3
18
13
17
5
B.E.r – B.E.p
Bond energy
breaking
calorimetry
coefficient
cool down
endothermic
enthalpy
higher
increases
less than
lose
mcT
products
releases
specific heat
surroundings
temperature
weaker
6
+ S
1
disorder
7
gas
2
number
5
physical
3
spacing
4
speed
6
strong
7
So
9
J/mol•K
10
series
11
Hess's
12
sum
13
multiplied
1
Hfo
14
-Hfo
15
Hfop – Hfor
16
Sop – Sor
5
negative
4
0 kJ/mol
8
0 kJ/mol•K
2
1
3
298
6
above
7
below
3 nonspontaneous
2
spontaneous
4
temperature
1
H – T S
5
H/S
9
+H and –S
8
–H and +S
16
[A]o
17
[A]t
13 1 – overall order
20
1/[A]t
6
coefficients
21
half-life
15
In([A]o/[A]t)
18
In[A]
19
k[A]2
8
k[A]m[B]n
rate = (8), where A and B are (9), k is called the (10), and the exponents m and n are called reaction (11).
The sum of the reaction orders gives the (12) reaction order. Reaction orders must be determined
experimentally. The unit of the rate constant depends on the overall reaction order. The unit for k is Mxt-1,
where x = (13).
Rate laws can be used to determine the concentrations of reactants or products at any time during a
reaction. In a first-order reaction, rate = (14) and kt = (15), where (16) is the initial concentration of A, (17) is
the concentration of A at time t, and k is the rate constant. Thus, for a first-order reaction, a graph of (18)
versus time yields a straight line of slope -k. In a second-order reaction, rate = (19), and kt = 1/[A]t – 1/[A]o.
In this case a graph of (20) versus time yields a straight line. The (21) of a reaction, t½, is the time required
for the concentration of a reactant to drop to one-half of its original value. For a first-order reaction, t½ = (22).
14
k[A]t
22
ln2/k
2
M/s
5
negative
11
orders
12
overall
4
positive
10
rate constant
7
rate law
1
rates
9
reactants
3
slows down
7
8.31 J/mol•K
4 activated complex
3
activation
6
-Ea/R
2
kinetic
5
orientation
1
temperature
Collision Model
The collision model, which assumes that reactions occur as a result of collisions between molecules,
helps explain why the rate constant increases with increasing (1). At a higher temperature, reactant
molecules have more (2) energy and their collisions are more energetic. The minimum energy required for a
reaction to occur is called the (3) energy, Ea. A collision with energy Ea or greater can cause the atoms of
the colliding molecules to reach the (4), which is the highest energy arrangement in the pathway from
reactants to products. Even if a collision is energetic enough, it may not lead to reaction; the reactants must
also have correct (5) for a collision to be effective. Because the kinetic energy depends on temperature, the
rate constant is dependent on temperature. The relationship between k and temperature is given by the
Arrhenius equation lnk = -Ea/R(1/T) + InA, which is an equation of a straight line (y = mx + b) so that the
slope of lnk versus 1/T equals (6), where R = (7). The two point form is ln(k1/k2) = (Ea/R)(1/T2 – 1/T1).
Reaction Mechanism
1
1 or 2
Many reactions occur by a multistep mechanism, involving two or more elementary reactions, or steps.
7 activation energy
A reaction mechanism details the individual steps that occur in the course of a reaction. Each of these steps 6
catalyst
has (1) reactants and (2) activation energy. The rate law for each step corresponds exactly to the number of 3
exponents
reactant molecules, so that reactant coefficients become (3) in the rate law. An (4) is produced in one
12 heterogeneous
elementary step and is consumed in a later elementary step and therefore does not appear in the overall
8
homogeneous
equation for the reaction. When a mechanism has several elementary steps, the overall rate is limited by the 4
intermediate
slowest elementary step, called the (5) step.
2
low
A (6) is a substance that increases the rate of a reaction without undergoing a net chemical change
10
overall
itself. It does so by providing a different mechanism for the reaction, one that has lower (7). A (8) catalyst is 11
rate law
one that is in the same phase as the reactants. It is consumed in the (9) step and reappears in a later step.
5 rate-determining
As a result, it is not included in the (10) reaction, but is included in the (11). A (12) catalyst has a different
9
slow
phase from the reactants and is written above the reaction arrow.
1. A student measures the mass of an object to be 75.011 g.
mav = (206)(0.255) + (207)(0.221) + (208)(0.524) = 207
The true mass is 77.500 g. What is the percent error?
6. Place the forms of hydrogen (H, H+, H-, H2, NaH) in the table.
100(77.500 g – 75.011 g)/77.500 = 3.212 %
ionic
Atom
Cation
Anion
Molecule
2. An empty graduated cylinder (25.044 g) is filled with 50.0 mL
Compound
of liquid. The total mass of cylinder and liquid is 69.886 g.
H
H+
HH2
NaH
a, What is the density of the liquid in g/mL?
7. Predict the product(s) for the following nuclear processes.
218 Po  4  + 214 Pb
D = m/V = (69.886 g – 25.044 g)/50.0 mL = 0.897 g/mL
alpha emission of Po-218
84
2
82
3.
b, What is the density of the liquid in kg/m3?
0.897 g x 1 kg x 1 mL x 1 cm3 = 897 kg/m3
1 mL 103 g 1 cm3 (10-2)3 m3
Consider the gas, UF6, which has a density of 15.7 g/L.
a. What is the molar mass of UF6?
beta emission of Pb-211
8.
What is the volume of 25.0 g of UF6?
5.
Write the nuclear symbol that has 26 p and 30 n.
56 Fe
26
Calculate the average atomic mass of Pb, which consists
of four stable isotopes whose atomic masses and
abundances are as follows:
206Pb
207Pb
208Pb
Isotope
Atomic Mass
206 u
207 u
208 u
Abundance
25.5%
22.1%
52.4%
Calculate the rate constant.
k = ln2/t½ = ln2/35.5 days = 0.0195 d-1
25.0 g x 1 L/15.7 g = 1.59 L
4.
 0-1 + 21183Bi
neutron bombardment of S-32 3216S + 10n  11H + 3215P
Consider a radioactive sample (t½ = 35.5 days).
a. How long will it take for 50 % of the sample to decay?
b.
How many moles of UF6 have a volume of 1.00 L?
1.00 L x 15.7 g/1 L x 1 mol/353 g = 0.0446 mol
c.
82Pb
1 half-life 35.5 days
238 g + 6(19 g) = 352 g
b.
211
9.
c. What percent of the sample will remain after 100. days?
ln(No/Nt) = kt
ln(100/Nt) = (0.0195)(100)  Nt = 14.2 %
d. How long will it take for 17 % of the sample to decay?
ln(No/Nt) = kt
ln(100/17) = (0.0195)(t)  t = 91 days
A hydrogen electron transitions from n = 4 to n = 3.
a. What is the transition energy?
E = E2 – E6
E = -2.18 x 10-18 J/32 – -2.18 x 10-18 J/42 = -1.06 x 10-19 J
b. What is the wavelength of emitted light?
Ephoton = hc/
-6 m
1.06 x 10-19 J = (2.00 x 10-25 J•m)/
10. Complete the following chart for magnesium.
symbol
period
group atomic number metal/nonmetal
Mg
3
2
12
metal
11. For each element:
a. Write the electron configuration for energy levels 3 & 4.
b. Circle the valence electrons on the configuration.
c. Write the abbreviated electron configuration.
d. Write the abbreviated electron configuration for the ion.
e. Write the orbital diagram for energy levels 3 & 4.
f. Circle the electron that has the quantum numbers.
g. Label the element paramagnetic or diamagnetic
Ca
Al
S
2
6
2
2
1
2
3s 3p 4s
3s 3p
3s 3p4
a/b
[Ar]4s2
c
d
e/f
g
12. a.
Ca2+
[Ar]
[Ne]3s23p1
Al3+
[Ar]3d10
[Ne]3s23p4
S2-
[Ne]3s23p4



3s 3p 
3p
4s
3s
3p
(4, 0, 0, -½)
(3, 1, -1, +½)
(3, 1, 1, +½)
diamagnetic
paramagnetic
paramagnetic
Write all the ions that are "isoelectronic" (contain the
same electron structure) as Neon.
3s
N3-, O2-, F-, Na+, Mg2+, Al3+
b.
List the ions from part (a) from smallest radius to
largest.
Al3+, Mg2+, Na+, F-, O2-, N3c. Explain your reasoning for your answer to part (b).
All ions have the same # of e-, but the # of p+
decreases from Sc to P. The smallest ion has the
most p+ (greatest net charge (# p+ – # core e-)
attracting the valence electrons and the largest has
the least p+.
13. Which of the elements, N, O or F, has the
N
largest atomic radius
largest ionic radius
N
largest ionization energy
F
most negative electron affinity value
F
most negative ionic charge
N
F
greatest electronegativity
14. Complete the following
Atomic radius increases down a group. Explanation:
The distance from the nucleus increase as the number
of electron energy levels increases, therefore the
attraction for valence electrons decreases down a
group and the atoms' radii increase.
Ionization energy increases across a period except for
columns 13 and 16. Explanation:
Generally, the reduced distance between nucleus and
valence electrons strengthens the electron attraction
and increases the ionization energy except for 13,
where the electron comes from a higher energy
sublevel and 16, where the electron comes from a
filled orbital.
The numerical value of electron affinity decreases across
a period except for columns 2, 15 and 18. Explanation:
The amount of energy that an electron release as it
enters an orbital increases for smaller atoms except
for columns 2, where the electron enters a higher
sublevel, or 15, where electron enters a half-filled
orbital rather than an empty orbital, or 18, where the
electron enters a higher energy level.
15. Rank the ionic bonds from greatest lattice energy to least.
Na-F
Al-N
K-Cl
Mg-O
Ca-S
4
1
5
2
3
16. Consider N2O, NO, NO2-, NO3-.
a. Draw the Lewis structures where N is the center atom.
N2O
NO2
NO2••
• ••
•• •• ••
:N=O:
:O=N–O:
:NN–O:
••
••
b. Which molecule "violates" the octet rule?
NO2
c. Draw resonance structures of N2O.
••
••
••
••
: N N – O :  : N = N = O :  : N – N O :
••
••
d. Determine the mostly likely Lewis structure N2O.
N N – O  N = N = O  N – N O
5 5 6
5 5 6
5 5 6
5 4 7
6 4 6
7 4 5
0 1 -1 -1 1 0 -2 1 1
e. Which side of the N–O bond is more negative?
The O side is more negative.
f. Rank the molecules/ions from high to low.
N2O
NO2
NO23
1
2
N–O Bond strength
1
3
2
N–O Bond length
17. Complete the chart for the molecules: SF2, SF4, and SF6.
SF2
SF4
SF6
F
F F
••/
\/
••
F–S–F
F–S–F
Lewis Structure
F–S–F
|
/\
••
F
F F
Trigonal
Electron
Tetrahedral
Octahedral
bipyramidal
Domain
Molecular
Bent
See saw
Octahedral
Geometry
Atomic radius decreases across a period. Explanation:
The number of core electrons is constant, but the
number of protons increases from left to right,
increasing the # of unshielded protons, which
increases the attraction for valence electrons and
decreases atomic radius
Ionization energy decreases down a group. Explanation:
The increased distance between nucleus and valence
electrons weakens the attraction; therefore it takes
less energy to remove a valence electron as you go
down a group.
Hybridization
sp3
Bond Angle
109.5o
sp3d
90o,
120o
sp3d2
90o
polar
polar
nonpolar
Polarity
18. Rank bond angles from largest to smallest (NH3, NH2-, NH4+).
NH4+ > NH3 > NH219. Draw semi-condensed Lewis structures and name the
following organic molecules.
Molecule
Lewis Structure
Name
C2H4
CH2=CH2
ethene
C3H7OH
CH3–CH2–CH2–OH
propanol
••
CH3–N–CH3
|
H
CH3-C-OH
||
O
(CH3)2NH
CH3COOH
dimethylamine
ethanoic acid
B
C
A
Heat Added (J)
Which phase takes more energy to
solid
liquid
raise the temperature 1oC?
Which process takes more energy
melting
boiling
per gram substance?
Which interval is there a mixture of
B-C
D-E
solid and liquid?
Which temperature is vaporization?
40oC
120oC
25. Determine the following using the phase diagram below.
solid
Densest phase
Phase change from
20. Consider the condensed structure CH2CHCOOH
a. Draw the full Lewis structure.
H H
| |
H–C=C–C–O–H
||
O
b. Determine the number of sigma and pi bonds.
8 sigma and 2 pi
21. Consider the alkyl halide, 1,2-dichloroethene.
a. Draw the structural formulas of the cis and trans
configurations of this molecule (label cis and trans).
Cl Cl
Cl H
\
/
\ /
C=C
C=C
/ \
/
\
H H
H
Cl
cis 1,2-dichloroethene trans 1,2-dichloroethene
b. Draw the semi-condensed structural formula and
name the structural isomer of this molecule.
Cl
|
C=CH2
1,1-dichloroethene
|
Cl
22. Rank butane, ethane, methane and propane from lowest
boiling point to highest.
methane < ethane < propane < butane
23. In addition to dispersion, what type of forces would you
expect in the following molecules?
C6H14
C6H12OH
NH3
CH3F
none
H-bond
H-bond
dipole
24. Consider the following heating profile for substance X.
D
E
120oC
40oC
68oC
Normal boiling point
0 oC
to
40oC
at 0.5 atm
sublimation
boiling
Phase change from 1.0 to 0.05 atm at 60oC
26. Explain why a helium balloon expands when the
temperature is increased.
Hotter atoms move faster, which strike the walls more
often and more forcefully; increasing the internal
pressure, which causes the balloon to expand to
equalize pressure.
27. 1.00 g of dry ice (solid CO2) is placed in a flexible balloon
and allowed to warm to room temperature (22oC). What is
the volume of the balloon if the room pressure is 1.04 atm?
1.00 g CO2 x 1 mol/44.0 g CO2 = 0.0227 mol CO2
PV = nRT
(1.04 atm)V = (0.0227 mol)(0.0821)(22.0 + 273) 
V = 0.529 L
28. a. 0.416 g of an unknown gas occupies 10.0-L at 25oC
and 103 KPa. What is the molar mass of the gas?
MM = mRT/PV
MM = (8.30 g)(8.31)(298 K)/(103 kPa)(10.0 L) = 20.0 g
b. What is the average (root mean square) speed?
vrms = (3RT/MM)½ = [3)(8.31)(298)/0.0200]½ = 609 m/s
c.
What is the density of this gas at STP (standard
temperature—0oC, and pressure—1 atm)?
d = MM•P/RT
d = (20.0 g/mol)(1 atm)/(0.0821)(273 K) = 0.892 g/L
d. Which noble gas (He, Ne, Ar, or Kr) would have an
effusion rate that is ½ that of this gas?
rateA/rateB = (MMB/MMA)½
2/1 = (MMA/20)½ MMA = 80 g/mol  Kr
29. A mixture containing 0.250 mol N2(g), 0.150 mol CH4(g),
and 0.100 mol O2(g) is confined in a 1.00-L vessel at 35oC.
a. Calculate the total pressure (in kPa) of the mixture.
PV = nRT
P(1.00 L) = (.250+.150+.100)(8.31)(308 K)  P = 1280 kPa
b. Calculate the partial pressure of O2(g) in the mixture.
PO2 = XO2Ptotal
PO2 = (0.100/0.500)(1280 kPa) = 256 kPa
30. a. Why do gases under high pressure deviate from ideal?
At high pressure, molecules fit in smaller volume
leaving less open space  the volume of container
(Vobserved) > volume of empty space (Videal).
b. Why do gases near their boiling point deviate from ideal?
Close to the boiling point, gas molecules begin
clumping because of molecular bonding  fewer
collision = less pressure (Pobserved < Pideal).
31. 5.00 g of water is added to a 10.0-L container filled with dry
air at 25oC (PH2O = 23.8 torr). The container is sealed.
a. How many grams of the water will evaporate?
PV = nRT (20 x 0.133)(10.0) = n(8.31)(20 + 273)
n = 0.0109 moles x 18.0 g/1 mole H2O = 0.197 g
b. How would the following change the amount of water
that evaporates? (increase, decrease, no change)
decrease
Use a 5.0 L container
Fill with humid air
decrease
Raise the temperature to 25oC
increase
Add 10 g of water
no change
32. Complete the chart for each type of solid.
Covalent
Metallic
Molecular
Network
ion
atom
molecule
Structural Unit
Bond name
metallic
Bond strength variable
Ionic
ion
covalent molecular
ionic
strong
weak
strong
Melting point
variable
high
low
high
Solubility
low
low
variable
high
Conductivity
high
low
low
low
Malleability
high
low
variable
low
copper diamond
water
salt
Example
33. Rank the following compounds in order of increasing
melting point. (C, KCl, CO2, CaS, NH3, Cu)
CO2 < NH3 < Cu < KCl < CaS < C
34. Which solutes are soluble in water?
NaCl
NO2
CH3OH
SiO2
35. When NaOH is added to water, the temperature of the
solution increases. Check the correct box
Hydration
Lattice
Bond that breaks
Bond that forms
Stronger bond
You would expect NaOH to be (more/less) soluble in warm
water compared to cold water.
36. Gases are more soluble at (high/low) temperature and
(high/low) pressure.
37. What is the solubility of CO2 in a closed soft drink at 25oC
where the partial pressure of CO2 is 4.00 atm?
(kCO2 = 3.1 x 10-2 mol/L•atm).
Mg = kPg = (3.1 x 10-2 mol/L•atm)(4.00 atm)
Mg = 0.124 mol/L
38. An aqueous solution is made by adding 100. g of glucose
(C6H12O6) to 985 g of water to make a final volume of 1.00 L.
a. What is the mass percent of glucose in the solution?
(100. g/985 g + 100. g)100 = 9.22 %
b. What is the mole fraction of glucose in the solution?
100. g x 1 mol/180 g = 0.556 mol C6H12O6
985 g x 1 mol/18.0 g = 54.7 mol H2O
(0.556 mol)/(0.556 mol + 54.7 mol) = 0.0101
c. What is the molality of glucose in the solution?
0.556 mol/0.985 kg = 0.564 mol/kg
d.
What is the molarity of glucose in the solution?
0.556 mol/1.00 L = 0.556 mol/L
e.
What is the vapor pressure (VPH2Oo = 20.0 torr)
VP = XH2OPH2O = (1 – 0.0101)(20.0 torr) = 19.8 torr
f.
What is the freezing point (Kf-H2O = 1.86oC/m)?
Tf = Kfmi = (1.86 oC/m)(0.564 m)(1) = 1.05oC
Tf = 0oC – 1.05oC = -1.05oC
c. What is the osmotic pressure?
 = MRTi = (0.556)(0.0821)(293)(1) = 13.4 atm
39. The freezing point decreases by 3.8oC when 0.500 g of an
unknown is dissolved in 5.00 g of BHT (Kf = 9.3oC/m).
What is the molar mass of the unknown?
Tf = Kfmi  3.8oC = (9.3 oC/m)m(1)  m = 0.41 mol/kg
m = molsolute/msolvent(kg)
0.41 mol/kg = molsolute/0.00500 kg  molsolute = 0.0021 mol
MM = msolute/molsolute = 0.500 g/0.0021 mol = 240 g/mol
40. Balance the following chemical equations.
a. _1_C3H8(g) + _5_O2(g)  _3_CO2(g) + _4_H2O(g)
b.
45. Consider aspirin, C9H8O4.
a. What is the mass percent of carbon?
_1_C3H9OH(l) + _5_O2(g)  _3_CO2(g) + _5_H2O(g)
_4_ Al(s) + _3_ O2(g)  _2_ Al2O3(s)
Zn(s) + 2 HCl(aq)  ZnCl2(aq) + H2(g)
What volume of 3.00 M HCl is needed to react 25.0 g Zn?
25.0 g Zn x 1 mol Zn x 2 mol HCl x
1L
= 0.255 L
65.4 g Zn 1 mol Zn 3.00 mol HCl
42.
6 Li(s) + N2(g)  2 Li3N(s)
A mixture of 1.00 g of Li and N2 react.
a. What is the limiting reactant?
1.00 g Li x 1 mol Li x 2 mol Li3N = 0.0480 mol Li3N
6.94 g Li 6 mol Li
1.00 g N2 x 1 mol N2 x 2 mol Li3N = 0.0714 mol Li3N
28.0 g N2
1 mol N2
b. How much excess reactant is there?
0.0480 mol Li3N x 1 mol N2 x 28.0 g N2 = 0.672 g N2
2 mol Li3N 1 mol N2
 1.00 g N2 – 0.672 g N2 = 0.328 g N2 excess
c. How many grams of Li3N are formed?
0.0480 mol Li3N x 34.8 g Li3N = 1.67 g Li3N
1 mol Li3N
d. If the percent yield is 88.5%, how many grams of Li 3N
are produced?
100(9)(12.0)/[9(12.0) + 8(1.01) + 4(16.0)] = 60.0 %
b.
c.
41.
1.67 g Li3N x 0.885 = 1.48 g Li3N produced
43.
2 KOH(aq) + NiSO4(aq)  Ni(OH)2(s) + K2SO4(aq)
a. 100. mL of 0.200 M KOH and 200. mL of 0.150 M
NiSO4 are mixed. How many moles of each
compound are initially present?
0.100 L x 0.200 mol/1 L = 0.0200 mol KOH
0.200 L x 0.150 mol/1 L = 0.0300 mol NiSO4
b. What is the limiting reactant?
.0200 mol KOH x 1 mol Ni(OH)2/1 mol KOH = .0100 mol
.0300 mol NiSO4 x 1 mol Ni(OH)2/1 mol KOH = .0300 mol
c. How many grams of Ni(OH)2(s) are produced?
0.0100 mol Ni(OH)2 x 92.7 g Ni(OH)2 = 0.927 g Ni(OH)2
1 mol Ni(OH)2
d. How many moles of each ion are initially in solution?
0.0200 mol KOH: 0.0200 mol K+, 0.0200 mol OH0.0300 mol NiSO4 0.0300 mol Ni2+, 0.0300 mol SO42e. How many moles of Ni2+ and OH- are taken out of
solution to make Ni(OH)2?
0.0100 mol Ni(OH)2  0.0100 mol Ni2+, 0.0200 mol OH-
f. What are the molarities of each ion in solution?
[K+]: 0.0200 mol/0.300 L = .067 M,
[OH-]: 0.0200 mol OH- – 0.0200 mol OH- = 0 mol  0 M
[Ni2+] = (0.0300 mol – 0.0100 mol)/0.300 L = .067 M
[SO42-]: 0.0300 mol/0.300 L = .100 M
44. A 1.000-g sample of a carbon-hydrogen-oxygen compound
in oxygen yields 1.95 g CO2 and 1.00 g H2O. A separate
experiment shows that the molecular mass is 90 g/mol.
Determine the molecular formula.
1.95 g CO2 x 12.0 g C/44.0 g CO2 = 0.532 g C
1.00 g H2O x 2.02 g H/18.0 g = 0.112 g H
1.00 g X – 0.532 g C – 0.112 g H = 0.356 g O
.532 g C x 1 mol/12.0 = .0443 mol C/.0223 = 2 mol C
.112 g H x 1 mol/1.01 g H = .111 mol H/.0223 = 5 mol H
.356 g O x 1 mol/16.00 g O = .0223 mol O/.0223 = 1 mol O
 empirical formula is C2H5O
C2H5O = 2(12) + 5(1) + 1(16) = 45 g/mol
90/45 = 2  C4H10O2
How many grams of aspirin contain 25.0 g of carbon?
25.0 g C x 100 g Asp/60 g C = 41.7 g Asp
46.
Pb2+(aq) + SO42-(aq)  PbSO4(s)
0.100 g of powder that contains CuSO4•5 H2O mixed with
an impurity is dissolved in water. All the sulfate from the
copper compound reacts with 3.25 mL of 0.100 M
Pb(NO3)2 according to the equation above. What is the
mass percent of CuSO4•5 H2O in the powder?
.00325 L x .100 mol Pb... x 1 mol Pb2+ = 3.25 x 10-4 mol Pb2+
1L
1 mol Pb...
3.25 x 10-4 mol Pb2+ x 1 mol Cu... x 250 g Cu... = 0.0813 g
1 mol Pb2+ 1 mol Cu...
(0.0813 g Cu.../0.100 g powder) x 100 = 81.3%
47.
HX(aq) + NaOH-(aq)  H2O(l) + NaX(aq)
The molecular mass HX is determined by titration with NaOH.
a. What is the concentration of the NaOH solution if it takes
12.8 mL to neutralize 1.00 g of KPH (MM = 204 g/mol)?
1.00 g KHP x 1 mol KHP x 1 mol NaOH = 4.90E-3 mol = .383 M
L NaOH 204 g KHP 1 mol KHP
0.0128 L
b. What is the molar mass of the monoprotic acid if it
takes 23.7 mL of the standardized NaOH to neutralize
1.000 g of the unknown acid?
.0237 L NaOH x .383 mol NaOH x 1 mol HX = .00908 mol HX
1L
1 mol NaOH
1.000 g HX/0.00908 mol HX = 110 g/mol
48. Estimate H per mole of C2H2 for the reaction:
C2H2(g) + 2 H2(g)  C2H6(g).
a. Using the calorimetry data: 10.0 g of C2H2 reacts with
excess H2 in a calorimeter that contains 875 g H2O.
The temperature rises from 25.0oC to 57.6oC.
H = -Q = -mcT
H = -(875 g)(4.19 J/goC)(57.6oC – 25oC) = -120,00 J
120. kJ/10.0 g x 26 g/1 mol = 312 kJ/mol C2H2
b. Using the bond energy values.
C–H
C–C
H–H
CC
414 kJ/mol
347 kJ/mol
820 kJ/mol
436 kJ/mol
H = BE broken - BE formed
H = BECC + 2 BEC–H + 2 BEH–H – (BEC–C + 6 BEC–H)
H = 820 + 2(414) + 2(436) – (347 + 6(414)) kJ = -311 kJ
Using the Hfo values.
C2H2 (g)
H2 (g)
C2H6 (g)
226.7 kJ/mol
0 kJ/mol
-84.7 kJ/mol

Ho = Hof products -Hof reactants
H = -84.7 kJ – (226.7 kJ + 2(0 kJ)) = -311 kJ
49. Consider the reaction:
C2H5OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(l)
H2O(l)
C2H5OH
O2
CO2
-277.0
0
-393.5
-285.8
Hfo (kJ/mol)
So (kJ/mol•K) 0.1607
0.2050
0.2136
0.0699
a. Calculate H.
 Ho = Hof products – Hof reactants
H = 2(-393.5) + 3(-285.8) – (-277.6 + 3(0)) kJ = -1366.8 kJ
b. Calculate S.
 So = Soproducts – Soreactants
S = 2(0.2136 ) + 3(0.0699) – (0.1607 + 3(.2050)) kJ
S = -0.1388 kJ/K
c. Calculate G at 25oC.
GHo – TSo
G  -1366.8 kJ – (298 K)(-0.1388 kJ/K)  -1325.4 kJ
c.
d.
For what temperature range is the reaction
spontaneous?
Tthreshold = H/S = -1366.8 kJ/-0.1388 kJ/K = 9847 K
 spontaneous at all T below 9847 K
50. Consider the reaction: 2 HI(g)  H2(g) + I2(g), where the
rate of the reaction in terms of [HI] = -0.500 M•min-1.
a. What is the rate of the reaction in terms of H2(g)?
rate = 0.250 M•min-1
b. What is the rate of the reaction in terms of I2(g)?
rate = 0.250 M•min-1
51. Consider the reaction: A + B2  P. The following
experimental data at 22oC were obtained:
A (M)
B2 (M)
Rate (M•s-1)
0.100
0.100
0.080
0.500
0.100
0.41
0.100
0.500
0.079
a. What is the order of the reaction with respect to each
reactant?
rate = k[A]m[B2]n
0.41 = k[0.500]m[0.100]n
0.080 k[0.100]m[0.100]n
5 = 5m  m = 1
n = 0 because the rate didn't change when [B2] changed
b.
What is the rate constant for the reaction, including
units?
rate = k[A]
k = rate/[A] = 0.080 M•s-1/(0.100 M) = 0.80 M1•s-1•M-1
k = 0.80 s-1
c. Calculate the rate of the reaction when the initial
concentration of A is 0.400 M.
rate = k[A]
rate = (0.80 s-1)(0.400 M) = 0.32 M•s-1
d. What percent of A will be used up in 1 second?
ln[A]o/[A]t = kt
ln(100/[A]t = (0.80 s-1)(1 s) = 0.80
100/[A]t = 2.2  [A]t = 45 (100 – 45 = 55 % is used up)
e. What is the half-life of the reaction?
t½ = ln2/k = ln2/(0.80 s-1) = 0.87 s
f. What would cause an increase in the rate constant?
An increase in temperature
g. The activation energy for the reaction is 115 kJ/mol.
What is the rate constant of the reaction at 27oC?
ln(k27/k22) = (Ea/R)(1/T22 – 1/T27)
ln(k27/0.80) = (115,000/8.31)(1/295 – 1/300)
k27/0.80 = 2.19  k27 = 1.75 s-1
52. The gaseous reaction between H2(g) and N2O2(g) occurs
by the following mechanism.
N2O2 + H2  N2O + H2O
N2O + H2  N2 + H2O
a. What is the balanced equation for the overall reaction?
2 H2(g) + N2O2(g)  N2(g) + 2 H2O(g)
b. What molecule (if any) is a catalyst?
There is no catalyst for this reaction.
c. What molecule is an intermediate?
N2O
d.
The rate law for this reaction is rate = k[H2][N2O2].
Which step is the slow step in the mechanism?
N2O2 + H2  N2O + H2O