AP Chemistry Name _________________________Fall Semester Review Worksheet Period _____

Unit 1-7 SummaryMeasurement in Chemistry

Science knowledge is advanced by observing patterns (laws) and constructing explanations (theories), which are supported by repeatable experimental evidence.

Measurements are made using the metric system, where the standard units are called SI 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 Kelvin scale, although the Celsius 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), (10-6) and n (10-9).

All measured quantities are inexact to some extent. The precision of a measurement indicates how closely different measurements of a quantity agree with one another. The accuracy of a measurement indicates how well a measurement agrees with the accepted value. Significant figures indicate the level of certainty in a measurement. Significant figures in a measured quantity include one estimated 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 number of significant figures is used. In addition and subtraction, the position of the least accurate significant figure is used. Relative difference between an experiment value (E) and a true value (T) is % difference: % = 100|E – T|/T.

Mass and volume measure amount of matter. Density relates mass to volume, d = m/V.

Chemical processes involve interaction of particles, which are measured in moles. The number of particles in a mole is 6.02 x 1023

(Avogadro's number), which is the number of atoms in a sample of the element that has a mass equal to its atomic mass measured in grams.

Molar mass (MM) is the sum of atomic masses in the chemical formula. For example, the mass of one H2O molecule is 18.0 u, so the molar mass of H2O is 18.0 g.

Dimensional analysis is a calculation strategy, where units are conserved. The given units are multiplied by a series of conversion factors, which are ratios of equivalent quantities. After canceling out units algebraically, what remain are the target units.

Atomic Nature of MatterAtoms 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.

Scientist DiscoveryDalton Atomic Theory (atoms are conserved in reactions)Thomson Electron charge/mass ratio (cathode ray bending)Millikan Electron mass (charged oil drops in electric field)Rutherford Planetary atomic model (alpha rays and gold foil)

The atom's nucleus contains protons and neutrons, whereas electrons move in the space around the nucleus.

Particle Charge Mass SymbolProton +1 1 1

1p or 11H

Neutron 0 1 10n

Electron –1 0 0-1e

Elements are classified by their atomic number or Z value, which equals the number of protons. The mass number or A value is the sum of protons and neutrons. Atoms of the same element that differ in mass number are called isotopes.

In a neutral atom, the number of protons equals the number of electrons. Anions are formed when electrons are added to neutral atoms. Cations are formed when electrons are removed from neutral atoms.

The unified atomic mass scale (u) is 1/12 the mass of a C-12 atom. The average atomic mass of an element is calculated using the formula: 100mav = %1m1 + %2m2 ...

The two kinds of pure substances are elements and compounds. Elements are identified by a chemical symbol. Compounds 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 ratio of atoms in the compound.

Mixtures are composed of multiple pure substances in an object or container and have variable compositions. They can be homogeneous or heterogeneous. Homogeneous mixtures are also called solutions and are uniform throughout.

RadioactivityThere are four kinds of radioactive decay: emission of alpha

particles ( or 42He), beta particle ( or 0

-1e), positron particle (, 0

1e), and gamma radiation (00).

In nuclear equations, reactant and product nuclei are represented by A

ZX, which is its nuclear symbol. In a balanced equation the sum of reactant A and Z values equal the sum of product A and Z values.

Nuclear transmutations, induced conversions of one nucleus into another, can be brought about by bombarding nuclei with either charged particles or neutrons.

The decay rate (radioactivity) is proportional to the number of radioactive atoms, rate = kNt. The time for half of the radioactive atoms to decay is constant, t½ = (ln2)/k. The time interval t for No number of radioactive atoms to reduce to Nt is determined by the formula, kt = ln(No/Nt).

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 spectra, which is the radiant energy emitted or absorbed by matter.

Equations for radiant energy, Ephoton = hf and speed of light, c = f are combined in Ephoton = hc/ = 2 x 10-25 J•m/.

Bohr 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 the value its quantum number n, where En = -2.18 x 1018 J/n2. The value of n is a positive integer (1, 2, 3 . . .). As n increases, the energy of the electron increases until it reaches a value of 0 J, where n equals infinity and the electron is ionized (leaves the atom). The lowest energy state where n = 1 is called the ground state. Other values of n correspond to excited states. Light is emitted when the electron drops to a lower energy state and light is absorbed when the electron is excited to a higher one. The energy of light emitted or absorbed equals the difference in energy between the two states, Ephoton = En-final – En-initial = 2.00 x 10-25 J•m/.

Quantum Mechanical ModelIn the quantum mechanical model each electron has a known

energy, but according to the Heisenberg Uncertainty Principle, the location of the electron cannot be determined exactly; rather, the 90 % probability of it being at a particular point in space is given by its orbital. An orbital is described by a combination of four quantum numbers. The principal quantum number n is indicated by the integers 1, 2 . . . This quantum number relates to the radius and energy of the orbital. The sublevel quantum number l is indicated by the letters s, p, d, and f, which correspond to l = 0, 1, 2, and 3 respectively. The l quantum number defines the shape of the orbital. For a given value of n, l can have integer values from 0 to (n – 1). 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 –l to +l. The spin quantum number ms defines the orientation of the electron's magnetic field and has two possible values +½ and –½. The Pauli 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 occupy any one atomic orbital to two.

Electron Arrangements in Atoms and IonsEnergy increases as n increases (1 < 2 < 3, etc) and within

the same value of n, energy increases as the sublevel progresses from letters s p d f. Orbitals within the same sublevel are degenerate, meaning they have the same energy.

The energies of s and p sublevels are less than the energy of the next higher s sublevel, whereas the energies of d and f 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 valence electrons. Non-valence electrons are called core 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 period. Elements with the same number of valence electrons form a column or group. The elements in which an s or p sublevel is being filled are called the main-group elements, which include group 1—alkali metals, group 2—alkaline earth metals, group 17—halogens and group 18—noble gases. Transition metals are where the d-sublevel is filling.

Electron configurations show how electrons are distributed among the atom's sublevels. In ground state configurations electrons occupy the lowest sublevel available until its capacity is reached. Additional electrons fill the next lowest sublevel until its filled, etc. Excited state configurations have gaps.

Orbital 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, Hund's Rule applies. The rule states that the lowest energy is attained by maximizing the number of electrons with the same electron spin.

Transition metals in columns 6 and 11 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 three squares on the periodic table of a noble gas form ions with the same electron arrangement as the noble gas and are isoelectronic. Transition metals form ions by losing all s level electrons first.

Elements with unpaired electrons have reinforcing magnetic fields, which makes the atom paramagnetic. If all of the electrons are paired, then the atom is diamagnetic.

Periodic Properties—Main GroupsCore electrons are very effective at screening the outer

electrons from the full charge of the nucleus, whereas electrons in the valence shell do not screen each other very effectively. As a result, the effective nuclear charge (Zeff) experienced by valence electrons increases as we move left to right across the main-group elements because the number of protons in the nucleus increases, without a corresponding increase in core electrons. The increase in Zeff is less pronounced with transition metals because the added electrons enter the core and cancel the added protons.

As a result of the Zeff atomic radius decreases as we go from left to right across a period of main group elements. Atomic radii increase as we go down a group because electrons fill the next higher energy level.

Cations are smaller than their parent atoms; anions are larger than their parent atoms, but group and period trends are the same as atomic radius. For an isoelectronic series the radius decreases with increasing nuclear charge.

Ionization energy is the energy needed to remove an electron from a gaseous atom; forming a cation. Successive ionization energies show a sharp increase 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 elements, the first ionization energy trend is generally opposite the atomic radii trend, with smaller atoms having higher first ionization energies, except for columns 13 (removing p orbital electron) and column 16 (removing an electron from a full orbital).

Electron affinity measures the energy change when adding an electron to a gaseous atom; forming an anion. A negative electron affinity means that the anion is stable; a positive electron affinity means that the anion is not stable relative to the separate atom and electron. In general, electron affinities become more negative from left to right across the main groups except for column 2 (adding an electron to a p orbital), column 15 (adding an electron to a half-filled orbital) and column 18 (adding an electron to the next higher energy level).

BondingBonds are classified into three broad groups: ionic bonds are

the result of electrostatic forces between cations and anions; covalent bonds form when electrons are shared between non-metal atoms; and metallic bonds, which bind metal cations with mutually shared valence electrons.

Bonds involve the interaction of valence electrons, which are represented by electron-dot symbols, called Lewis symbols. The tendencies of atoms to gain, lose, or share their valence electrons often follow the octet rule, which can be viewed as an attempt by atoms to achieve a noble gas electron configuration, which is a lower energy state.

The strength of the electrostatic attractions between ions is measured by the lattice energy, which increases with ionic charge and decreases with distance between ions.

Electronegativity measures the ability of an atom to attract electrons in a covalent bond. Electronegativity generally increases from left to right in the periodic table and decreases down a column. The difference in atoms' electronegativities is used to determine the polarity of a covalent bond; the greater the difference, the more polar the bond. A polar molecule has a positive side (+) and a negative side (–). The separation of charge produces a dipole, the magnitude of which is given by the dipole moment. Polar bonds are stronger and shorter than non-polar bonds.

Bond strength and length is also affected by the number of shared electrons. Sharing of one pair of electrons produces a single bond; whereas the sharing of two or three pairs of electrons produces double or triple bonds, respectively. Multiple bonds are stronger and shorter than single bonds.

The procedures used for naming two-element, binary, molecular compounds follow the rules below.1. The lower electronegative element is written first in the formula

and named as an element.2. The name of the second element is given an –ide ending.3. Prefixes are used to indicate the number of atoms of each

element; mono is not used with the first element.Lewis Structures

Electron distribution in molecules is shown with Lewis structures, which indicate how many valence electrons are involved in forming bonds and how many remain as unshared electron pairs. If we know which atoms are connected to one another, we can draw Lewis structures for molecules and ions by a simple procedure, where eight electrons are placed around each atom. When there are too few valence electrons, then it will be necessary to add double or triple bonds. When there are too many valence electrons (and the central atom has at least third energy level electrons), then it will be necessary to place additional electrons (up to 10 or 12) around the central atom; forming an expanded octet. When the total number of valence electrons is an odd number, then

it will be necessary to place seven electrons around the atom with the odd number of valence electrons.

When there are multiple valid Lewis structures for a molecule or ion, we can determine which is most likely by assigning a formal charge to each atom, which is the sum of half the bonding electrons and all the unshared electrons. Most acceptable Lewis structures will have low formal charges with any negative formal charge on the more electronegative atom.

VSEPR Model The valence-shell electron-pair repulsion (VSEPR) model

rationalizes molecular geometries based on the repulsions between electron domains, which are regions about a central atom where electrons are likely to be found. Pairs of electrons, bonding and non-bonding, create domains around an atom, which are as far apart as possible. Electron domains from non-bonding pairs exert slightly greater repulsions, which leads to smaller bond angles than idealized values. The arrangement of electron domains around a central atom is called the electron domain geometry; the arrangement of atoms is called the molecular geometry.

Certain molecular shapes have cancelling bond dipoles, producing a nonpolar molecule, which is one whose dipole moment is zero. In other shapes the bond dipoles do not cancel and the molecule is polar (a nonzero dipole moment). In general non-bonding pairs of electrons around the central atom produce polar molecules.

The table below summarizes the Domain Geometry, Ideal bond angle, Molecular Geometry and Molecular polarity based on the number of bonded atoms (–) and non-bonding pairs of electrons (• •) surrounding the central atom. (–) (• •) Domain

GeometryBondAngle

Molecular Geometry Polar?

2 0 Linear 180o Linear no3 0

Trigonal planar 120o planar no

2 1 Bent yes4 0

Tetrahedral 109o

Tetrahedral no3 1 pyramidal yes2 2 Bent yes5 0

Trigonal bipyramidal

90o, 120o

bipyramidal no4 1 Seesaw yes3 2 T-shaped yes2 3 Linear no6 0

Octahedral 90o

Octahedral no5 1 pyramidal yes4 2 planar no

Valence-Bond TheoryValence-bond theory is an extension of Lewis's notion of

electron-pair bonds. In valence-bond theory, covalent bonds are formed when atomic orbitals on neighboring atoms overlap. The bonding electrons occupy the overlap region and are attracted to both nuclei simultaneously, which bonds the atoms together.

To extend valence-bond theory to polyatomic molecules, s, p, and sometimes d orbitals are blended to form hybrid orbitals, which overlap with orbitals on another atom to make a bond. Hybrid orbitals also hold non-bonding pairs of electrons. A particular mode of hybridization can be associated with each of the five common electron-domain geometries (linear = sp; trigonal planar = sp2; tetrahedral = sp3; trigonal bipyramidal = sp3d; and octahedral = sp3d2).

Covalent bonds formed between hybridized electrons are called sigma () bonds, where the electron density lies along the line connecting the atoms. Bonds that form between non-hybridized p orbitals are called pi () bonds. A double bond consists of one bond and one bond and a triple bond consists of one and two bonds.

Sometimes a bond can be placed in more than one location. In such situations, we describe the molecule by using two or more resonance structures. The molecule is envisioned as a blend of these multiple resonance structures and the bonds are delocalized; that is, spread among several atoms. The bond order value represents the actual bond strength and is sum of the bond plus a share of the bond(s).

HydrocarbonsCarbon molecules (except CO, CO2) are called organic.

Hydrocarbons are organic molecules that contain mostly carbon and hydrogen. The four groups of hydrocarbons are alkanes, alkenes, alkynes, and aromatic. The naming of hydrocarbons is based on the longest continuous chain of carbon atoms in the structure.

Prefix based on number of carbons in longest chain1 meth 6 hex2 eth 7 hept3 prop 8 oct4 but 9 non5 pent 10 dec

Hydrocarbon branches are named according to the number of carbons in the branch with a "yl" ending and located by the number of the main chain carbon to which the branch is attached.

Ring structures have the prefix cyclo. The names of alkenes (double bond) and alkynes (triple bond)

are based on the longest continuous chain of carbon atoms that contain the multiple bond; with the location of the multiple bond specified by a numerical prefix.

The chemistry of an organic compound is dominated by the presence of the functional group.

Alcohol Acid AmineR–C–R

| O-H

R–C–O-H || O

••

R–N–R|R

Water Soluble Acid (releases H+) Base (Absorbs H+)Isomers are substances that possess the same molecular

formula, but differ in the arrangements of atoms. In structural isomers the bonding arrangements differ. Different isomers are given different names. Alkenes exhibit not only structural isomerism but geometric isomerism (cis-trans) as well. In geometric isomers the bonds are the same, but the molecules have different geometries. Geometric isomerism is possible in alkenes because rotation about the C=C bond is restricted.

Gas StateGases at room temperatures tend to be molecular with low

molar mass. Air, a mixture composed mainly of N2 and O2. Some liquids and solids can also exist in the gaseous state, where they are known as vapor. Gases' volume can change because they are compressible and they mix in all proportions because their component molecules are far apart.

The gas state is characterized by four variables: pressure (P), volume (V), temperature (T), and quantity (n). Volume is measured in liters, temperature in kelvins, and quantity of gas in moles. Pressure is the force per unit area. In chemistry, pressure is measured in atmospheres (atm), torr (named after Torricelli), millimeter of mercury (mm Hg) and kilopascals (kPa).

1 atm = 101 kPa = 760 torr = 760 mm Hg.A barometer is used to measure atmospheric pressure and a

manometer is used to measure the pressure of enclosed gases.The ideal-gas law equation is PV = nRT, where V is in L, n is

in moles and T is in K. The term R is the gas constant, which is 0.0821 when P is in atm or 8.31 when P is in kPa. The conditions of 273 K and 1 atm are known as the standard temperature and pressure and abbreviated as STP, where the molar volume of all gases is 22.4 L/mol. Additional equations using molar mass (MM) are MM = mRT/PV and MM = dRT/P.

In gas mixtures the total pressure (Ptot) is the sum of the partial pressures (PA) that gas A would exert if it were present alone

under the same conditions: Ptot = PA + PB ... and the pressure of gas A is proportional to its mole fraction (XA): PA = XAPtot. In calculating the quantity of a gas collected over water, correction must be made for the partial pressure of water vapor.

The kinetic-molecular theory accounts for the properties of an ideal gas in terms of a set of statements about the nature of gases: Molecules are in continuous, chaotic motion; the volume of gas molecules is negligible compared to the volume of their container; the gas molecules have no attraction for one another; their collisions are elastic; and the molecules' kinetic energy is proportional to the absolute temperature: K = 3/2RT.

Molecules of a gas do not all have the same kinetic energy at a given instant. Their speeds are distributed over a wide range; the distribution varies with the molar mass of the gas and with temperature. The root-mean-square speed, u = (3RT/MM)½.

Effusion (rate of escape through a tiny hole into a vacuum) and diffusion (rate spreading of one gas through another) are related to molar mass by Graham's law: rA/rB = (MMB/MMA)½.

Departures from ideal behavior increase as pressure increases (real gases molecules occupy a significant fraction of container volume resulting in an observed volume that is greater than predicted) and as temperature decreases (real gases attract each other resulting in an observed pressure that is less than predicted).

Phase ChangeSubstances that are gases or liquids at room temperature are

usually composed of molecules. In gases the intermolecular attractive forces are negligible compared to the kinetic energies of the molecules; thus, the molecules are widely separated and undergo constant, chaotic motion. In liquids the intermolecular forces are strong enough to keep the molecules in close proximity; nevertheless, the molecules are free to move with respect to one another. In solids the inter-particle attractive forces are strong enough to restrain molecular motion and to force the particles to occupy specific locations in a three-dimensional arrangement.

Three types of intermolecular forces exist between neutral molecules: dipole-dipole forces, London dispersion forces, and hydrogen bonding. London dispersion forces operate between all molecules and are the result of temporary polarization of the molecule. Dispersion forces increase in strength with increasing molecular mass, although molecular shape is also an important factor. Dipole-dipole forces exist between polar molecules, where the positive pole of one molecule is attracted to the negative pole of its neighbor. Hydrogen bonding occurs in compounds containing N, O or F bonded to H. Hydrogen bonds are stronger than dipole-dipole or dispersion forces.

Phase changes are transformations from one state to another. Changes of solid to liquid (melting), solid to gas (sublimation) and liquid to gas (vaporization or boiling) absorb energy. The reverse processes (freezing, deposition, condensation, respectively) release energy. A gas cannot be liquefied by application of pressure if the temperature is above its critical temperature.

The vapor pressure is the partial pressure of the vapor when it is in equilibrium with the liquid. At equilibrium the rate of evaporation, transfer of molecules from the liquid to the vapor, equals the rate of condensation, transfer from the vapor to the liquid. The higher the vapor pressure of a liquid, the more readily it evaporates and the more volatile it is. Vapor pressure increases nonlinearly with temperature. Boiling occurs when the vapor pressure equals the atmospheric pressure. The normal boiling point occurs at 1 atm pressure.

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 line. The line through the melting point usually slopes slightly to the right as pressure increases, because the solid is usually more dense than the liquid. The melting point at 1 atm is the normal melting point. The point on the diagram at which all three phases coexist in equilibrium is called the triple point.

Crystalline SolidsIn 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. Molecular solids, which consist of atoms or molecules held together by intermolecular forces, are soft and low melting. Covalent network solids, which consist of atoms held together by covalent bonds that extend throughout the solid, are hard and high melting. Ionic solids are hard and brittle and have high melting points. Metallic solids, which consist of metal cations held together by a sea of electrons, exhibit a wide range of melting points.

SolubilitySolutions form when one substance disperses uniformly

throughout another. The dissolving medium of the solution (usually in the greater amount) is called the solvent. The substance dissolved in a solvent (usually the smaller amount) is called the solute. The attractive interaction of solvent molecules with solute is called solvation. When the solvent is water, the interaction is called hydration. 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 (endothermic) or negative (exothermic), depending on the relative value of lattice energy (positive) and hydration energy (negative).

The equilibrium between a saturated solution and undissolved solute is dynamic; the process of solution and the reverse process, crystallization, occur simultaneously. For a solution in equilibrium with undissolved solute, the two processes occur at equal rates, producing a saturated solution. The amount of solute needed to form a saturated solution at any particular temperature is the solubility 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 solvent-solvent energies compared with solute-solvent interactions. Polar and ionic 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 miscible; 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 pressure of the gas over the solution, as expressed by Henry's law: Sg = kPg. The solubilities of most solid solutes in water increase as the temperature increases. In contrast, the solubilities of gases in water generally decrease with increasing temperature.

Concentrations of solutions can be expressed quantitatively by several different measures, includingmass percent: % = 102(msolute/mtotal) or 102(msolvent/mtotal)mole fraction: Xsolute = molsolute/moltotal or Xsolvent = molsolvent/moltotal

molarity: M = molsolute/Vsolution(L)

molality: m = molsolute/msolvent(kg)

Conversions between concentration units is possible if molar mass of solute and solvent are known and/or the density of the solution is known.

Colligative PropertiesA physical property of a solution that depends on the

concentration of solute particles present, regardless of the nature of the solute, is a colligative property. Colligative properties include vapor-pressure lowering, freezing-point lowering, boiling-point elevation, and osmotic pressure. The lowering of vapor pressure follows the equation, Pvap = XsolventPo

solvent. A solution containing a nonvolatile solute possesses a higher boiling point than the pure solvent. The molal boiling-point constant, Kb, 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, Kf, measures the lowering of the freezing point of a solution for a 1 m solution of solute particles. The temperature changes are given by the equations Tb = Kbm and Tf = Kfm. When NaCI dissolves in water, two 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 twice as much as that of a nonelectrolyte solution of the same concentration. The multiplier is called the van't Hoff factor i. Similar considerations apply to other strong electrolytes. Osmosis is the movement of solvent molecules through a semipermeable membrane from a less concentrated to a more concentrated solution. This net movement of solvent generates an osmotic pressure 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: = MRT.

Chemical ReactionsOne of the important concepts of stoichiometry is the law of

conservation of mass, which states that the total mass of the products of a chemical reaction is the same as the total mass of the reactants. Likewise, the same numbers of atoms of each type are present before and after a chemical reaction. A balanced chemical equation shows equal numbers of atoms of each element on each side of the equation. Equations are balanced by placing coefficients in front of the chemical formulas for the reactants and products of a reaction, not by changing the subscripts in chemical formulas.

Among the reaction types described in this unit are (1) combination reactions, in which two reactants combine to form one product; (2) decomposition reactions, in which a single reactant forms two or more products; (3) combustion reactions in oxygen, in which a hydrocarbon or related compound reacts with O2 to form CO2 and H2O; (4) electron exchange reactions in which reactants exchange electrons; (5) ion exchange reactions, in which ions exchange "partners"; and (6) proton exchange reactions, in which atoms exchange H+. The last three reactions will be discussed in more detail later in the year.

Aqueous reactions (electron exchange, ion exchange and proton exchange) usually involve ions, where only one of the two ions in an ionic compound or acid react, the other is non-reacting and is called a spectator ion. In a net ionic equation, those ions that go through the reaction unchanged are omitted.

The coefficients in a balanced equation give the relative numbers of moles of reactants and products. To calculate the grams of a product from the grams of a reactant, first convert grams of reactant to moles of reactant, then use the coefficients to convert the number of moles of reactant to moles of product, and finally convert moles of product to grams of product. Many reactions occur in water (aqueous). Molarity makes it possible to convert solution volume to number of moles of solute.

A limiting reactant is completely consumed in a reaction. When it is used up, the reaction stops, thus limiting the quantities of products formed. The theoretical yield is the quantity of product calculated to form when all of the limiting reagent reacts. The actual yield is always less than the theoretical yield. The percent yield compares the actual and theoretical yields.

Gravimetric AnalysisThe empirical formula can be determined from its percent

composition by calculating the relative number of moles of each atom in 100 g of the substance. Similarly, the empirical formula can be determined from the mass of each element in the compound, or if it is a combustion reaction, from the mass of CO2 and H2O produced. If the substance is molecular in nature, its molecular formula can be determined from the empirical formula if the molecular mass is also known.

Volumetric Analysis

Solutions of known molarity can be formed either by adding a measured mass of solute and diluting it to a known volume or by the dilution of a more concentrated solution of known concentration (a stock solution). Adding solvent to the solution (the process of dilution) decreases the concentration of the solute without changing the number of moles of solute in the solution (Mstock)(Vstock) = (Mstandard)(Vstandard).

In titration a solution of known concentration (a standard solution) is combined with a solution of unknown concentration in order to determine the moles of unknown. By knowing moles of unknown, the molar mass of unknown or concentration of unknown solution can be determined. The point in the titration at which stoichiometrically equivalent quantities of reactants are brought together is called the equivalence point. An indicator can be used to show the end point of the titration, which coincides closely with the equivalence point.

Change in Enthalpy (H)Chemical reactions typically involve breaking some bonds

between reactant atoms and forming new bonds. Breaking bonds absorbs energy, therefore the chemical system gains bond energy and the surroundings lose energy, typically in the form of heat. In contrast, forming bonds releases energy; resulting in lose of energy by the 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, then products are at a higher energy state than reactants (making the product bonds weaker than the reactant bonds) and energy of the system increases (+H), which is described as endothermic because the surroundings typically lose heat energy and cool down. Alternatively, when energy required to break bonds is less than the energy released to form new bonds, then products are at a lower energy state than reactants (making the product bonds stronger than the reactant bonds) and energy of the system decreases (–H), which is described as exothermic because the surroundings typically gain heat energy and warm up. The change in enthalpy, H, is listed to the right of a balanced chemical equation. H can be treated in the same way as a coefficient when using dimensional analysis.

The amount of heat transferred between the system and the surroundings is measured experimentally by calorimetry. A calorimeter measures the temperature 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 specific heat. Water has a very high specific heat, c = 4.18 J/g•K. 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 = mcT. If a Bomb calorimeter is used, then the bomb constant (C) is in the equation: Q = (C + mc)T.

Bond energy, 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 and subtracting the bond energies of all bonds formed: H = B.E.reactants – B.E.products.

Change in Entropy (S)All chemical systems have an inherent amount of disorder

because of the complexity of the atomic arrangement within molecules, the spacing of molecules with respect to each other; and the overall motion of the system. Increases in complexity, spacing and overall motion result in increased disorder as measured by change in entropy, S. A positive S for physical changes can be predicted based on whether the molecules spread out. Evaporation, diffusion and effusion have +S values. Dissolving is more complicated because spreading out solute and solvent increases disorder, but formation of hydration bonds

between solute and solvent decreases disorder, therefore it is impossible to predict the sign for S (although most dissolving is +S. All chemical reactions that result in more moles of gas products compared to reactants have a +S.

Thermodynamic DataThe standard enthalpy of formation, Hf

o, 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 1 atm pressure and 25oC (298 K). For any element in its most stable state under standard conditions, Hf

o = 0 kJ/mol. Most compounds have negative values of Hf

o. Large negative Hfo

indicate a strong bond and stable compound. The standard entropy So is based on H+ having So = 0 kJ/mol•K (although the AP exam often lists the values in J/mol•K). The thermodynamic data chart lists the Hf

o and So for common substances. Hf

o applies to situations involving more than one mole, where Hf

o is multiplied by the number of moles, and involving decomposition, where H = -Hf

o. An important use of Hfo and So

is calculate H and S for a wide variety of reactions under laboratory conditions, where H Ho = Hf

oproducts – Hf

oreactants and

S So = Soproducts – So

reactants. 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 series of steps. Hess's law states that if a reaction is carried out in a series of steps, H for the reaction will be equal to the sum 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.

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 = H – TS. The sign of G relates to the spontaneity of the process. When G is negative, the process is spontaneous. When G is positive, the process is nonspontaneous; the reverse process is spontaneous. At equilibrium the process is reversible and G = 0 kJ.

The values of H and S generally do not vary much with temperature. As a consequence, the dependence of G with temperature is governed mainly by the value of T in the expression G = H –TS. The threshold temperature, Tthreshold = H/S, is when a reaction goes from spontaneous nonspontaneous. This only occurs when H and S are both positive or both negative. When H and S are both positive, the reaction is spontaneous at all temperatures above the threshold. When they are both negative, the reaction is spontaneous at all temperatures below the threshold. When H and S have opposite signs, then the reaction is spontaneous at all temperatures (-H and +S) or spontaneous at no temperature (+H and –S).

Reaction RateReaction rates are expressed as changes in molarity per unit

time, M/s. For most reactions, a plot of molarity versus time shows that the rate slows down 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 have positive rates, or in terms of reactants, which have negative rates. The reaction rates for the each reactant and product are proportional to the coefficients in the balanced equation.

The quantitative relationship between rate and concentration is expressed by a rate law, which has the form: rate = k[A]m[B]n, where A and B are reactants k is called the rate constant, and the exponents m and n are called reaction orders. The sum of the reaction orders gives the overall 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 = 1 – overall order.

The overall order for the reaction is used to determine which equation is used to determine the reaction rate, time needed to change from [A]o to [A]t, the half-life (t½) and the linear plot. Order 0 1 2rate = k k[A] k[A]2

kt = [A]o – [A]t ln([A]o/[A]t) 1/[A]t – 1/[A]o

t½ = [A]o/2k ln2/k 1/k[A]o

linear Plot [A] vs. t ln[A] vs. t 1/[A] vs. t(Note: radioactive decay is a first order process)

Collision ModelThe collision model, which assumes that reactions occur as a

result of collisions between molecules, helps explain why the rate constant increases with increasing temperature. At higher temperature, reactant molecules have more kinetic energy and their collisions are more energetic. The minimum energy required for a reaction to occur is called the activation energy Ea. A collision with energy Ea or greater can cause the atoms of the colliding molecules to reach the activated complex, 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 orientation for a collision to be effective. Because the kinetic energy depends on temperature, the rate constant is dependent on temperature. The two point equation is ln(k1/k2) = (Ea/R)(1/T2 – 1/T1), where R = 8.31 J/mol•K. The slope of lnk versus 1/T equals -Ea/R.

Reaction MechanismMany reactions occur by a multistep mechanism, involving two

or more elementary reactions, or steps. A reaction mechanism details the individual steps that occur in the course of a reaction. Each of these steps has 1 or 2 reactants and low activation energy. The rate law for each step corresponds exactly to the number of reactant molecules, so that reactant coefficients become exponents in the rate law. An intermediate is produced in one elementary step and is consumed in a later elementary step and therefore does not appear in the overall equation for the reaction. When a mechanism has several elementary steps, the overall rate is limited by the slowest elementary step, called the rate-determining step.

A catalyst is a substance that increases the rate of a reaction without undergoing a net chemical change itself. It does so by providing a different mechanism for the reaction, one that has lower activation energy. A homogeneous catalyst is one that is in the same phase as the reactants. It is consumed in the slow step and reappears in a later step. As a result, it is not included in the overall reaction, but is included in the rate law. A heterogeneous catalyst has a different phase from the reactants and is written above the reaction arrow.

Unit 1 Practice Multiple Choice1. Based on the data, the density of the solid in g/mL is

Mass of metal 19.611 gVolume of water 12.4 mLVolume of water + metal 14.9 mL

(A) 7.8444 (B) 7.844 (C) 7.84 (D) 7.8 2. Which scientist is correctly matched with the discovery?

(A) Millikan discovered the electron charge-to-mass ratio.(B) Thomson discovered the charge of an electron.(C) Bohr discovered the four quantum numbers.(D) Rutherford discovered the nucleus.

3. Which represents a pair of isotopes? (A) 14

6C and 147N (B) 18

9F and 3517Cl

(C) 5626Fe2+ and 56

26Fe3+ (D) 3517Cl and 36

17Cl4. Copper has two isotopes, 63Cu and 65Cu. What is the

abundance of 63Cu if the average atomic mass is 63.5?(A) 90% (B) 75% (C) 50% (D) 20%

5. Which of the following is correct about beta particles? I. mass number of 4 and a charge of +2

II. more penetrating than alpha particles III. electron

(A) I only (B) III only (C) I and II (D) II and III6. For the types of radiation given, which is the correct order of

increasing ability to penetrate a piece of lead?(A) < < (B) < < (C) < < (D) < <

7. 24996Cm is radioactive and decays by the loss of one beta

particle. The other product is (A) 245

94Pu (B) 24997Bk (C) 248

96Cm (D) 25096Cm

8. 25198Cf 2 1

0n + 13154Xe + . . .

What is the missing product in the nuclear reaction?(A) 118

42Mo (B) 11844Ru (C) 120

42Mo (D) 12044Ru

9. The radioactive decay of C-14 to N-14 occurs by (A) beta particle emission (B)alpha particle emission(C) positron emission (D) electron capture

10. What is the resulting nucleus after 21484Po emits 2 and 2 ?

(A) 20683Bi (B) 210

83Bi (C) 20682Pb (D) 208

82Pb11. 235

92U + 10n 141

55Cs + 3 10n + X

Neutron bombardment of uranium can induce the reaction represented above. Nuclide X is which of the following? (A) 92

35Br (B) 9435Br (C) 91

37Rb (D) 9237Rb

12. If 87.5 % of a sample of pure Pb-210 decays in 36 days, what is the half-life of Pb-210? (A) 6 days (B) 8 days (C) 12 days (D) 14 days

13. The half-life of isotope Y is 12 minutes. What mass of Y was originally present if 1 g is left after 60 minutes?(A) 8 g (B) 16 g (C) 24 g (D) 32 g

Unit 2 Practice Multiple ChoiceBriefly explain why the answer is correct in the space provided.Questions 1-3

(A) Heisenberg uncertainty principle (B) Pauli exclusion principle (C) Hund's rule(D) Shielding effect

1. Can be used to predict that a gaseous carbon atom in its ground state is paramagnetic.

2. States that an orbital can hold no more than two electrons.3. Predicts that it is impossible to determine simultaneously the

exact position and the exact velocity of an electron.4. Which set of quantum numbers describes the highest energy

valence electron in a ground-state gallium atom (Z = 31)?(A) 4,0,0,½ (B) 4,0,1,½ (C) 4,1,1,½ (D) 4,1,2,½

5. Which has the outer electronic configuration, s2p3? (A) Si (B) Cl (C) Se (D) As

Questions 6-9 refer to atoms with the atomic orbitals shown. (A) 1s 2s (B) [He] 2s 2p (C) [He] 2s 2p (D) [Ar] 4s3d

6. Represents an atom that is chemically unreactive. 7. Represents an atom in an excited state. 8. Represents an atom that has four valence electrons.9. Represents an atom of a transition metal. Questions 10-12

(A) 1s2 2s22p5 3s23p5

(B) 1s2 2s22p6 3s23p6

(C) 1s2 2s22p62d10 3s23p6

(D) 1s2 2s22p6 3s23p63d3 4s2

10. An impossible electronic configuration 11. The ground-state configuration for a halogen anion12. The ground-state configuration for an alkaline earth cation13. Which represents the ground state for the Mn3+ ion?

(A) 1s2 2s22p6 3s23p63d4

(B) 1s2 2s22p6 3s23p63d5 4s2

(C) 1s2 2s22p6 3s23p63d2 4s2

(D) 1s2 2s22p6 3s23p63d8 4s2

14. In which group are the three species isoelectronic? (A) S2-, K+, Ca2+ (B) Sc, Ti, V2+

(C) O2-, S2-, CI- (D) Mg2+, Ca2+, Sr2+ 15. The ionization energies for element X are listed in

the table below. On the basis of the data, element X is most likely

Ionization Energies for element X (kJ mol-1)First Second Third Fourth Fifth580 1,815 2,740 11,600 14,800

(A) Na (B) Mg (C) Al (D) Si16. In the periodic table, as the atomic number increases from 11

to 17, what happens to the atomic radius? (A) It remains constant. (B) It increases only. (C) It decreases only. (D) It increases, then decreases.

17. Which elements have most nearly the same atomic radius?(A) Be, B, C, N (B) Ne, Ar, Kr, Xe(C) Mg, Ca, Sr, Ba (D) Cr, Mn, Fe, Co

Questions 18-19 Use the following options.(A) O (B) Rb (C) N (D) Mg

18. What element has the most negative electron affinity?19. Which of the elements above has the smallest ionic radius for

its most commonly found ion?20. Ca, V, Co, Zn, As

Which gaseous atoms of the above are paramagnetic?(A) Ca and As only (B) Zn and As only(C) Ca, V, and Co only (D) V, Co, and As only

21. Which property generally decreases across the periodic table from sodium to chlorine? (A) 1st ionization energy (B) Atomic mass (C) Ionic radius (D) Atomic radius

Questions 22-23 Consider a ground state atom of each element. (A) S (B) Ca (C) Ga (D) Sb

22. The atom that contains exactly two unpaired electrons23. The atom that contains only one electron in the highest

occupied energy sublevel

Unit 3 Practice Multiple Choice1. Types of hybridization exhibited by the C atoms in propene,

CH3CHCH2 include which of the following? I. sp II. sp2 III. sp3

(A) I only (B) II only (C) III only (D) II and III2. Which molecule contains 1 sigma () and 2 pi () bonds?

(A) H2 (B) F2 (C) N2 (D) O2 3. Which molecule has the shortest bond length?

(A) N2 (B) O2 (C) Cl2 (D) Br24. Which molecule has only one unshared pair of valence

electrons?(A) Cl2 (B) NH3 (C) H2O2 (D) N2

5. The electron pairs in a molecule where the central atom exhibits sp3d2 hybrid orbitals are directed toward the corners of (A) a tetrahedron (B) a square pyramid(C) a trigonal bipyramid (D) an octahedron

6. The SbCl5 molecule has trigonal bipyramid structure. Therefore, the hybridization of Sb orbitals should be(A) sp2 (B) sp3 (C) dsp2 (D) dsp3

7. For which molecule are resonance structures necessary to describe the bonding satisfactorily?(A) H2S (B) SO2 (C) CO2 (D) OF2

8. Which molecules have planar configurations? I. BCl3 II. CHCl3 III. NCl3

(A) I only (B) II only (C) III only (D) II and III9. CCl4, CO2, PCl3, PCl5, SF6

Which does NOT describe any of the molecules above?(A) Linear (B) Octahedral (C) Square planar (D) Tetrahedral

10. The geometry of the SO3 molecule is best described as (A) trigonal planar (B) trigonal pyramidal(C) square pyramidal (D) bent

11. Pi () bonding occurs in each of the following EXCEPT(A) CO2 (B) C2H4 (C) CN- (D) CH4

12. According to the VSEPR model, the progressive decrease in the bond angles in the series of molecules CH4, NH3, and H2O is best accounted for by the (A) increasing strength of the bonds (B) decreasing size of the central atom (C) increasing electronegativity of the central atom (D) increasing number of unshared pairs of electrons

Questions 13-15 refer to the following molecules. (A) PH3 (B) H2O (C) CH4 (D) C2H4

13. The molecule with only one double bond 14. The molecule with the largest dipole moment15. The molecule that has trigonal pyramidal geometry16. Which molecule has a zero dipole moment?

(A) HCN (B) NH3 (C) SO2 (D) PF517. Which molecule has the largest dipole moment?

(A) CO (B) CO2 (C) O2 (D) HF18. Which molecule has a dipole moment of zero?

(A) C6H6 (benzene) (B) NO(C) SO2 (D) NH3

19. Which pair of atoms should form the most polar bond?(A) F and B (B) C and O(C) F and O (D) N and F

20. Which pair of ions should have the highest lattice energy?(A) Na+ and Br- (B) Li+ and F-

(C) Cs+ and F- (D) Li+ and O2-

21. Which compound has the greatest lattice energy?(A) BaO (B) MgO (C) CaS (D) MgS

22. Which molecule has the weakest bond?(A) CO (B) O2 (C) Cl2 (D) N2

23. How are the bonding pairs arranged in the best Lewis structure for ozone, O3?(A) O–O–O (B) O=O–O (C) OO–O (D) O=O=O

24. Which species has the shortest bond length?(A) CN- (B) O2 (C) SO2 (D) SO3

25. Which species has a valid non-octet Lewis structure?(A) GeCl4 (B) SiF4 (C) NH4

+ (D) SeCl426. The Lewis structure for SeS2 with zero formal charge has

(A) 2 bonding pairs and 7 nonbonding pairs of electrons.(B) 2 bonding pairs and 6 nonbonding pairs of electrons.(C) 3 bonding pairs and 6 nonbonding pairs of electrons.(D) 4 bonding pairs and 5 nonbonding pairs of electrons.

27. Which molecular shape cannot exhibit geometric isomerism?(A) tetrahedron (B)square planar(C) trigonal bipyramid (D)octahedron

28. Which molecule is NOT polar?(A) H2O (B) CO2 (C) NO2 (D) SO2

29. Which species has sp2 hybridization for the central atom?(A) C2H2 (B) SO3

2- (C) O3 (D) BrI3

30. In which species is the F-X-F bond angle the smallest? (A) NF3 (B) BF3 (C) CF4 (D) BrF3

31. For ClF3, the electron domain geometry of Cl and the molecular geometry are, respectively, (A) trigonal planar and trigonal planar. (B) trigonal planar and trigonal bipyramidal.(C) trigonal bipyramidal and trigonal planar.(D) trigonal bipyramidal and T -shaped.

32. The size of the H-N-H bond angles of the following species increases in which order? (A) NH3 < NH4

+ < NH2- (B) NH3 < NH2

- < NH4+

(C) NH2- < NH3 < NH4

+ (D) NH2- < NH4

+ < NH3

33. What is the molecular geometry and polarity of BF3?(A) trigonal pyramidal and polar(B) trigonal pyramidal and nonpolar(C) trigonal planar and polar(D) trigonal planar and nonpolar

34. Which set does not contain a linear species?(A) CO2, SO2, NO2 (B) H2O, HCN, BeI2 (C) OCN-, C2H2, OF2 (D) H2S, CIO2

-, NH2-

35. The hybrid orbitals of nitrogen in N2O4 are(A) sp (B) sp2 (C) sp3 (D) sp3d

36. How many sigma and how many pi bonds are inCH2=CH–CH2–C–CH3? ||

O (A) 5 sigma and 2 pi. (B) 8 sigma and 4 pi. (C) 11 sigma and 2 pi. (D) 13 sigma and 2 pi.

37. What is the best estimate of the H-O-H bond angle in H3O+?(A) 109.5o (B) 107o (C) 90o (D) 120o

38. Which of the following pairs of compounds are isomers? (A) CH3–CH2–CH2–CH3 and CH3–CH(CH3)–CH3 (B) CH3–CH(CH3)–CH3 and CH3–CH3–C=CH2

(C) CH3–O–CH3 and CH3–CO–CH3 (D) CH3–OH and CH3–CH2–OH

39. CH3–CH2–CH2BrWhich structural formulas represents an isomer of the compound that has the above structural formula? (A) CH2Br–CH2–CH2 (B) CH3–CHBr–CH3

(C) CH2Br–CH2–CH2Br (D) CH3–CH2–CH2–CH2Br

Unit 4 Practice Multiple ChoiceQuestions 1-2 The molecules have the normal boiling points.

Molecule HF HCl HBr HIBoiling Point, oC +19 -85 -67 -35

1. The relatively high boiling point of HF can be correctly explained by which of the following?(A) HF gas is more ideal.(B) HF molecules have a smaller dipole moment.(C) HF is much less soluble in water.(D) HF molecules tend to form hydrogen bonds.

2. The increasing boiling points for HCl, HBr and HI can be best explained because of the increase in(A) dispersion force (B) dipole moment(C) valence electrons (D) hydrogen bonding

3. A sample of an ideal gas is cooled from 50oC to 25oC in a sealed container of constant volume. Which of the following values for the gas will decrease?

I. The average kinetic energy of the molecules II. The average distance between the molecules III. The average speed of the molecules

(A) I only (B) II only (C) III only (D) I and IIIQuestions 4-7 refer to the phase diagram of a pure substance.

4. Which phase is most dense?(A) solid (B) liquid(C) gas (D) can't determine

5. Which occurs when the temperature increases from 0°C to 40°C at a constant pressure of 0.5 atm? (A) Sublimation (B) Condensation(C) Freezing (D) Fusion

6. Which occurs when the pressure increases from 0.5 to 1.5 atm at a constant temperature of 60°C? (A) Sublimation (B) Condensation(C) Freezing (D) Fusion

7. The normal boiling point of the substance is closest to(A) 20oC (B) 40oC (C) 70oC (D) 100oC

Questions 8-9The graph shows the temperature of a pure solid substance as it is heated at a constant rate to a gas.

8. Pure liquid exists at time(A) t1 (B) t2 (C) t3 (D) t4

9. Which of the following best describes what happens to the substance between t4 and t5? (A) The molecules are leaving the liquid phase. (B) The solid and liquid phases coexist in equilibrium. (C) The vapor pressure of the substance is decreasing. (D) The average intermolecular distance is decreasing.

10. Which actions would be likely to change the boiling point of a sample of a pure liquid in an open container? (A) Placing it in a smaller container (B) Increasing moles of the liquid in the container (C) Moving the container to a higher altitude(D) Increase the setting on the hot plate

11. Gas in a closed rigid container is heated until its absolute temperature is doubled, which is also doubled?(A) The density of the gas(B) The pressure of the gas (C) The average speed of the gas molecules (D) The number of molecules per liter

12. A 2.00-L of gas at 27oC is heated until its volume is 5.00 L. If the pressure is constant, the final temperature is(A) 68oC (B) 120oC (C) 477oC (D) 677oC

13. Under the same conditions, which of the following gases effuse at approximately half the rate of NH3? (A) O2 (B) He (C) CO2 (D) Cl2

14. What is the partial pressure (in atm) of N2 in a gaseous mixture, which contains 7.0 moles N2, 2.5 moles O2, and 0.50 mole He at a total pressure of 0.90 atm. (A) 0.13 (B) 0.27 (C) 0.63 (D) 0.90

Questions 15-16 refer to the following gases at 0°C and 1 atm. (A) Ne (B) Xe (C) O2 (D) CO

15. Has an average atomic or molecular speed closest to that of N2 molecules at 0°C and 1 atm

16. Has the greatest density 17. A 2-L container will hold about 4 g of which of the following

gases at 0oC and 1 atm? (A) SO2 (B) N2 (C) CO2 (D) C4H8

18. Which is the same for the structural isomers C2H5OH and CH3OCH3? (Assume ideal behavior.)(A) Gaseous densities at STP(B) Vapor pressures at the same temperature(C) Boiling points(D) Melting points

19. As the temperature is raised from 20oC to 40oC, the average kinetic energy of Ne atoms changes by a factor of(A) ½ (B) (313/293)½

(C) 313/293 (D) 220. The system shown above is at equilibrium at 28°C.

At this temperature, the vapor pressure of wateris 28 mm Hg. The partial pressure (in mm Hg) of O2(g) is (A) 28(B) 56 (C) 133(D) 161

21. What is the mole fraction of benzene in a mixture of toluene (pressure = 22 torr) and benzene (pressure = 75 torr)? (A) 0.23 (B) 0.29 (C) 0.50 (D) 0.77

22. In which of the processes are covalent bonds broken? (A) I2(s) I2(g) (B) CO2(s) CO2(g)(C) NaCl(s) NaCl(l) (D) C(diamond) C(g)

23. Of the following compounds, which is the most ionic? (A) SiCl4 (B) BrCl (C) PCl3 (D) CaCl2

24. Which of the following oxides is a gas at 25°C and 1 atm? (A) Rb2O (B) N2O (C) Na2O2 (D) SiO2

25. Which of the following has the highest melting point? (A) S8 (B) I2 (C) SiO2 (D) SO2

26. Under which conditions is O2(g) the most soluble in H2O? (A) 5.0 atm, 80oC (B) 5.0 atm, 20oC (C) 1.0 atm, 80oC (D) 1.0 atm, 20oC

27. Which is lower for a solution of a volatile solute compared to the pure solvent?(A) normal boiling point (B) vapor pressure(C) normal freezing point (D) osmotic pressure

Questions 28-30 Refer to 0.20 M solutions of the following salts.(A) NaBr (B) KI (C) MgCl2 (D) C6H12O6

28. Has the lowest freezing point29. Has the lowest conductivity 30. Has the lowest boiling point31. The mole fraction of ethanol in a 6 molal aqueous solution is

(A) 0.006 (B) 0.1 (C) 0.08 (D) 0.232. What additional information is needed to determine the

molality of a 1.0-M glucose (C6H12O6) solution?(A) Volume (B) Temperature (C) Solubility of glucose (D) Density of the solution

33. The mole fraction of toluene (MM = 90) in a benzene (MM= 80) solution is 0.2. What is the molality of the solution?(A) 0.2 (B) 0.5 (C) 2 (D) 3

Unit 5 Practice Multiple Choice1. _ Fe2O3 + _ CO _ Fe + _ CO2

When the equation is balanced and reduced to lowest terms, the coefficient for CO2 is (A) 1 (B) 2 (C) 3 (D) 4

2. 1 CH3CH2COOH + _ O2 _ CO2 + _ H2OHow many moles of O2 are required to oxidize 1 mole of CH3CH2COOH according to the reaction above?(A) 2 (B) 5/2 (C) 3 (D) 7/2

3. C3H8 burns in excess oxygen gas. What is the coefficient for O2 when the equation is balanced with lowest terms?(A) 4 (B) 5 (C) 7 (D) 10

4. CaCO3 + 2 HCl CaCl2 + CO2 + H2OWhat is the mass percent of CaCO3 in a 1.25-g rock that generate 0.44 g of CO2 when reacted with HCl? (A) 35 % (B) 44 % (C) 67 % (D) 80 %

5. 8.0 mol of F2 and 1.7 mol of Xe are mixed. When all of the Xe reacted, 4.6 mol of F2 remain. What is the formula? (A) XeF (B) XeF3 (C) XeF4 (D) XeF6

6. What mass of Ca(NO3)2 contains 24 g of oxygen atoms?(A) 164 g (B) 96 g (C) 62 g (D) 41 g

7. Compounds contain 38 g, 57 g, 76 g, and 114 g of element Q per mole compound. A possible atomic mass of Q is(A) 13 (B) 19 (C) 28 (D) 38

8. What is the percent nitrogen by mass in N2O3?(A) 18 % (B) 22 % (C) 36 % (D) 45 %

9. Which formula is 54 % water by mass?(A) CaCO3 • 10 H2O (B) CaCO3 • 6 H2O(C) CaCO3 • 2 H2O (D) CaCO3 • H2O

10. Which formula forms 88 g of carbon dioxide and 27 g of water when burned in excess oxygen?(A) CH4 (B) C2H2 (C) C4H3 (D) C4H6

11. How many moles of H2O are produced when 0.56 g of C2H4 (MM = 28 g) is burned in excess oxygen?(A) 0.04 (B) 0.06 (C) 0.08 (D) 0.4

12. BrO3- + 5 Br- + 6 H+ 3 Br2 + 3 H2O

How many moles of Br2 can be produced when 25 mL of 0.20 M BrO3

- is mixed with 30 mL of 0.45 M Br-?(A) 0.0050 (B) 0.0081 (C) 0.014 (D) 0.015

13. 3 Ag + 4 HNO3 3 AgNO3 + NO + 2 H2OIf 0.10 mole of silver is added to 10 mL of 6.0 M nitric acid, the number of moles of NO gas that can be formed is (A) 0.015 (B) 0.020 (C) 0.033 (D) 0.045

14. What is the simplest formula of a compound that contains 1.10 mol of K, 0.55 mol of Te, and 1.65 mol of O? (A) KTeO (B) KTe2O (C) K2TeO3 (D) K2TeO6

15. In which compound is the mass ratio of chromium to oxygen closest to 1.6 to 1.0? (A) CrO3 (B) CrO2 (C) CrO (D) Cr2O

16. In which is the mass percent of magnesium closest to 60 %. (A) MgO (B) MgS (C) MgF2 (D) Mg3N2

17. A student obtained a percent water in a hydrate that was too small. Which is the most likely explanation for this? (A) Hydrate spattered out of the crucible during heating(B) The anhydrous absorbed moisture after heating. (C) The amount of hydrate sample used was too small. (D) The amount of hydrate sample used was too large.

18. 2 N2H4 + N2O4 3 N2 + 4 H2OWhat mass of water can be produced when 8.0 g of N2H4 (MM = 32 g) and 9.2 g of N2O4 (MM = 92 g) react? (A) 9.0 g (B) 18 g (C) 36 g (D) 7.2 g

19. A student wishes to prepare 2.00 L of 0.100 M KIO3 (MM = 214 g). The proper procedure is to weigh out (A) 42.8 g of KIO3 and add 2.00 kg of H2O(B) 42.8 g of KIO3 and add H2O to a final volume of 2.00 L(C) 21.4 g of KIO3 and add H2O to a final volume of 2.00 L(D) 42.8 g of KIO3 and add 2.00 L of H2O

20. The volume of distilled water that is added to 10 mL of 6.0 M HCI in order to prepare a 0.50 M HCI solution is(A) 50 mL (B) 60 mL (C) 100 mL (D) 110 mL

21. What volume of 12 M HCl is diluted to obtain 1.0 L of 3.0-M?(A) 4.0 mL (B) 40 mL (C) 250 mL (D) 1,000 mL

22. 400 mL of distilled water is added to 200 mL of 0.6 M MgCI2, what is the resulting concentration of Mg2+? (A) 0.2 M (B) 0.3 M (C) 0.4 M (D) 0.6 M

23. When 70. mL of 3.0 M Na2CO3 is added to 30. mL of 1.0 M NaHCO3 the resulting concentration of Na+ is(A) 2.0 M (B) 2.4 M (C) 4.0 M d. 4.5 M

24. The mass of H2SO4 (MM = 98 g) in 50 mL of 6.0-M solution(A) 3.10 g (B) 29.4 g (C) 300. g (D) 12.0 g

25. What mass of CuSO4• 5 H2O (MM = 250 g) is required to prepare 250 mL of a 0.10 M solution? (A) 4.0 g (B) 6.3 g (C) 34 g (D) 85 g

26. 2 KOH + SO2 K2SO3 + H2OWhat mass of SO2 reacts with 1.0 L of 0.25-M KOH?(A) 4.0 g (B) 8.0 g (C) 16 g (D) 20. g

27. How many moles Ba(NO3)2 should be added to 300. mL of 0.20-M Fe(NO3)3 to increase the [NO3

-] to 1.0 M?(A) 0.060 (B) 0.12 (C) 0.24 (D) 0.30

28. What is the concentration of HC2H3O2 if it takes 32 mL of 0.50-M NaOH solution to neutralize 20. mL of the acid? (A) 1.6 M (B) 0.80 M (C) 0.64 M (D) 0.60 M

29. 2 HCl + Ba(OH)2 BaCl2 + 2 H2OWhat volume of 1.5-M HCI neutralizes 25 mL of 1.2-M Ba(OH)2? (A) 20. mL (B) 30. mL (C) 40. mL (D) 60. mL

30. What is the concentration of OH- in a mixture that contains 40. mL of 0.25 M KOH and 60. mL of 0.15 M Ba(OH)2? (A) 0.10 M (B) 0.19 M (C) 0.28 M (D) 0.40 M

31. What mass of Au is produced when 0.0500 mol of Au2S3 is reduced completely with excess H2? (A) 9.85 g (B) 19.7 g (C) 24.5 g (D) 39.4 g

Unit 6 Practice Multiple Choice

1. I2(g) + 3 Cl2(g) 2 ICl3(g) According to the data in the table below, what is the value of Ho , in kJ, for the reaction represented above?

Bond I–I CI–CI I–ClBond Energy (kJ/mole) 150 240 210

(A) - 870 (B) - 390 (C) +180 (D) + 4502. C2H4(g) + 3 O2(g) 2 CO2(g) + 2 H2O(g)

For the reaction, H is -1,300 kJ. What is the value of H, in kJ, if the combustion produced liquid water rather than water vapor? (H for H2O(l) H2O(g) is 45 kJ/mol) (A) -1,300 (B) -1,210 (C) -1,345 (D) -1,390

3. CH4 (g) + 2 O2(g) CO2(g) + 2 H2O(l) Ho = -900 kJWhat is the standard heat of formation of CH4, in kJ/mol?(Hf

oH2O = -300 kJ/mol, Hf

oCO2 = -400 kJ/mol)

(A) -200 (B) -100 (C) 100 (D) 200 4. H2(g) + ½ O2(g) H2O(l) Ho = x

2 Na(s) + ½ O2(g) Na2O(s) Ho = y Na(s) + ½ O2(g) + ½ H2(g) NaOH(s) Ho = z

What is H for the reaction: Na2O(s) + H2O(l) 2 NaOH(s)? (A) x + y + z (B) x + y – z(C) x + y - 2z (D) 2z - x - y

5. Which is true when ice melts at its normal melting point?(A) H < 0, S > 0, G = 0 (B) H < 0, S < 0, G > 0(C) H > 0, S < 0, G < 0 (D) H > 0, S > 0, G = 0

6. Which of the following reactions has the largest positive value of S per mole of Cl2?

(A) H2(g) + Cl2(g) 2 HCl(g)(B) Cl2(g) + O2(g) Cl2O(g)(C) Mg(s) + Cl2(g) MgCl2(s)(D) 2 NH4Cl(s) 4 H2(g) + Cl2(g)

7. Ice is added to hot water in an insulated container, which is then sealed. What has happened to the total energy and the total entropy when the system reaches equilibrium?(A) Energy and entropy remain constant (B) Energy remains constant, entropy decreases(C) Energy remains constant, entropy increases(D) Energy decreases, entropy increases

8. N2(g) + 3 H2(g) 2 NH3(g) The above reaction is thermodynamically spontaneous at 298 K, but becomes nonspontaneous at higher temperatures. Which of the following is true at 298 K? (A) G, H, and S are all positive.(B) G, H, and S are all negative.(C) G and H are negative, but S is positive.(D) G and S are negative, but H is positive.

9. 3 C2H2(g) C6H6(g)What is the standard enthalpy change, Ho, for the reaction represented above? (Hf

oC2H2 is 230 kJ•mol-1; Hf

oC6H6 is 80 kJ•mol-1)

(A) -610 kJ (B) 150 kJ (C) -770 kJ (D) 610 kJ10. When solutions of NH4SCN and Ba(OH)2 are mixed in a

closed container, the temperature drops and a gas is produced. Which of the following indicates the correct signs for G, H, and S for the process? (A) –G –H –S (B) –G +H –S(C) –G +H +S (D) +G –H +S

11. X(s) X(l)Which of the following is true for any substance undergoing the process represented above at its normal melting point? (A) S < 0 (B) H = 0(C) H = TG (D) H = TS

12. For a reaction, Ho = -150 kg/mol and So = -50 J/mol•K. Which statement is true about this reaction?(A) It is spontaneous at high temperature only.(B) It is spontaneous at low temperature only.(C) It is spontaneous at all temperatures.(D) It is non-spontaneous at all temperatures.

Unit 7 Practice Multiple Choice

Questions 1-2 refer to the following types of energy. (A) Activation energy (B) Free energy (C) Ionization energy (D) Lattice energy

1. The energy in a chemical or physical change that is available to do useful work

2. The energy required to form the transition state in a chemical reaction

3. For the reaction whose rate law is, Rate = k[X], a plot of which of the following vs. time is a straight line?(A) [X] (B) 1/[X] (C) ln[X] (D) [X]2

4. The initial-rate data in the table were obtained for the reaction: 2 NO(g) + O2(g) 2 NO2(g).

Exp [NO]o

(mol•L-1)[O2]o

(mol•L-1)Initial Rate of Formation

of NO2 (mol•L-1•s-1)1 0.10 0.10 2.52 0.20 0.10 5.03 0.20 0.40 80.

What is the experimental rate law? (A) Rate = k[NO][O2] (B) Rate = k[NO][O2]2 (C) Rate = k[NO]2[O2] (D) Rate = k[NO]2[O2]2

Questions 5-6 The oxidation of iodide ions by arsenic acid in acidic aqueous solution occurs according to the reaction.

H3AsO4 + 3 I- + 2 H3O+ H3AsO3 + I3- + H2O

The rate law is: Rate = k[H3AsO4][I-][H3O+]5. What is the order of the reaction with respect to I-?

(A) 0 (B) 1 (C) 2 (D) 36. According to the rate law for the reaction, an increase in the

concentration of H3O+ has what effect on this reaction?(A) The rate of reaction increases.(B) The rate of reaction decreases.(C) The value of the rate constant increases.(D) The value of the rate constant decreases.

Questions 7-8 The concentrations of X measured over a period of time for the reaction X + Y Z is graphed below.

7. What is the half-life for this reaction?(A) 2 minutes (B) 7 minutes(C) 10 minutes (D) half-life is not constant

8. What is the order of reaction with respect to X?(A) 0 (B) 1 (C) 2 (D) can't be determined

9. How long does it take in minutes for the partial pressure of the reactant in a first order reaction to decrease from 1.0 atm to 0.125 atm, where the half-life is 19 minutes?(A) 38 (B) 57 (C) 76 (D) 152

10. Which is associated with relatively slow reaction rates? (A) The presence of a catalyst(B) High temperature (C) High concentration of reactants (D) Strong bonds in reactant molecules

11. Which of the following best describes the role of a match in setting a fire?

(A) The match decreases Ea for the slow step. (B) The match increases the concentration of the reactant. (C) The match supplies Ea for the combustion reaction. (D) The match provides a more favorable activated complex

for the combustion reaction. 12. 2 A(g) + B(g) 2 C(g)

Which best explains the observation that there is no change in reaction rate when the concentration of B(g) is doubled?(A) The order of the reaction with respect to B is 1.(B) The overall order of the reaction is 0.(C) B is not involved in the rate-determined step.(D) substance B is a catalyst.

13. 2 NO + O2 2 NO2

The mechanism for the overall reaction is the following. (1) NO + NO N2O2 slow (2) N2O2 + O2 2 NO2 fast

Which rate law is consistent with this mechanism? (A) Rate = k[NO]2 (B) Rate = k[NO][O2]-1

(C) Rate = k[NO]2[O2]-1 (D) Rate = k[NO]2[O2]14. A reaction was observed for 20 days and the percentage of

the reactant remaining after each day was recorded below. time 0 1 2 3 4 5 6 7 10 20% 100 79 63 50 41 31 25 20 10 1Which best describes the order and half-life of the reaction? (A) First order with a 3 day half-life(B) First order with a 10 day half-life(C) Second order with a 3 day half-life(D) Second order with a 6 day half-life

Questions 15-17 refer to the proposed steps for the catalyzed reaction between Ce4+ and Tl+.

Step 1: Ce4+ + Mn2+ Ce3+ + Mn3+ Step 2: Ce4+ + Mn3+ Ce3+ + Mn4+ Step 3: Mn4+ + TI+ Tl3+ + Mn2+

15. The products of the overall catalyzed reaction are (A) Ce4+ and TI+ (B) Ce3+ and Tl3+

(C) Ce3+ and Mn3+ (D) Ce3+ and Mn4+ 16. The catalyst for the reaction is

(A) Ce4+ (B) Mn2+ (C) Ce3+ (D) Mn4+

17. Intermediates in the reaction are(A) Ce4+ and Ce3+ (B) Ce3+ and Mn3+

(C) Tl+ and Tl3+ (D) Mn3+ and Mn4+

18. Rate = k[M][N]2

If the concentrations of M and N are doubled, the reaction rate will increase by a factor of(A) 2 (B) 4 (C) 6 (D) 8

Questions 19-20 The energy diagram for the reaction X + Y Z is shown below.

19. Which of the following is true for this reaction?(A) The reaction is exothermic where Ea > |H|(B) The reaction is endothermic where Ea > |H|.(C) The reaction is exothermic where Ea < |H|.(D) The reaction is endothermic where Ea < |H|.

20. The addition of a catalyst to this reaction would cause a change in which of the indicated energy differences? (A) I only (B) II only (C) Ill only (D) I and II only

Unit 1 Practice Free Response15. Hydrogen atoms in the ground state are ionized by UV light.

a. Calculate the energy needed to ionize an electron from n = 1? (En = -2.18 x 10-18/n2 J).

b. Calculate the wavelength of UV light. (Ephoton = (2.00 x 10-25 J•m)/)

16. Answer the following questions regarding light and its interactions with molecules, atoms, and ions. a. The longest wavelength of light with enough energy to

break the CI-CI bond in CI2(g) is 495 nm. (1) Calculate the frequency in s-1 of the light.

(2) Calculate the energy in J of a photon of the light.

(3) Calculate the energy in kJ•mol-1 of the CI-CI bond.

b. A certain line in the spectrum of atomic hydrogen is associated with the electronic transition in the H atom from the sixth energy level (n = 6) to the second energy level (n = 2). (1) Indicate whether the H atom emits energy

or whether it absorbs energy during the transition. Justify your answer.

(2) Calculate the wavelength in nm of the radiation associated with the spectral line.

Unit 2 Practice Free Response6. The table shows the first three ionization energies in kJ/mol

for third period elements, which are numbered randomly. Use the information to answer the questions.Elemen

t First Second Third

1 1,251 2,300 3,8202 496 4,562 6,9103 738 1,451 7,7334 578 1817 2745

a. Which element is most metallic in character? Explain.

b. Identify element 3. Explain.

c. Complete the following.Electron configuration of element 3

Common ion charge of element 2

Chemical symbol of element 2

Element with the smallest atomic radius

Chemical symbol for element 4

7. Use the principles of atomic structure to explain each. a. The atomic radius of Li is larger than that of Be.

b. The electron affinity for K is less than Ca.

c. The first ionization energy of Se is less than As.

8. Consider the element strontium (Sr). Justify each answer.a. What is the outer electron configuration of Sr?

b. How does the atomic radius of Sr compare to Rb?

c. How does the atomic radius of Sr compare to Ca?

d. Compare Sr to Ca and Rb in first ionization energy.

e. How does the Sr2+ ion compare in size to the Sr atom?

f. How does the Sr2+ ion compare in size to the Br- ion?

g. As successive electrons are removed from the Sr atom, where does the largest jump in ionization energy occur?

h. Is strontium diamagnetic or paramagnetic?

Unit 3 Practice Free Response2. There are several oxides of nitrogen; among the more

common are N2O, NO2 and NO3-.

a. Draw the Lewis structures of these molecules.

N2O NO2 NO3-

b. Which of these molecules "violate" the octet rule? Explain.

c. Draw resonance structures of N2O.

d. For each resonance structure from question 2c, calculate the formal charge and evaluate which structure is most likely. Explain.

e. Which side of the N–O bond is +? Explain.

f. Rank the strength of the N–O bond in order of strongest (1) to weakest (3). Explain your answer.

N2O NO2 NO3-

4. Consider the ion SF3+.

a. Draw a Lewis structure.

b. Identify the type of hybridization exhibited by sulfur.

c. Identify the electron-domain and molecular geometries.Electron-domain geometry Molecular geometry

d. Predict whether the F-S-F bond angle is equal to, greater than or less than 109.5°. Explain

5. Consider the ion SF5-

a. Draw a Lewis structure.

b. Identify the type of hybridization exhibited by sulfur.

c. Identify the electron-domain and molecular geometries.Electron domain geometry Molecular geometry

6. Two Lewis structures can be drawn for the OPF3 molecule.Structure 1

..: O :

.. | ..: F – P – F :

.. | ..: F :

..

Structure 2

: O :.. || ..

: F – P – F :.. | ..

: F :..

Which Lewis structures best represents a molecule of OPF3? Justify your answer in terms of formal charge.

8. Compounds of Xe and F form molecules where the hybridization of Xe is sp3d and sp3d2. Write the formula and draw the Lewis structure for the two molecules.

sp3d sp3d2

Unit 4 Practice Free Response1. a. 3.327 g of an unknown gas occupies 1.00-L at 25oC and

103 kPa. What is the molar mass of the gas?

b. What is the density of this gas at STP (standard temperature—0oC, and pressure—1 atm)?

c. Which noble gas would have twice the effusion rate?

2. N2 with V = 200. mL, P = 99.7 kPa, and T = 27.0oC is mixed with O2 and transferred to a 750.-mL container at 27.0oC. The total pressure of the mixture is 90.4 kPa, at 27.0oC.a. Calculate the moles of N2.

b. Calculate the total moles of gas.

c. Calculate the partial pressure of each gas.

3. Explain why methane gas does not behave as an ideal gas at low temperatures and high pressures.

4. 10 g of water is added to a 10.0-L container filled with dry air at 20oC (PH2O = 20 torr). The container is sealed.a. How many grams of the water will evaporate?

b. Would the amount of water that evaporates increase (), remain the same (=) or decrease () for the following changes?

= Use a 5.0 L containerUse humid airRaise the temperature to 25oC

Add 20.0 g of water5. Consider the following solids.

a. Rank the solids from highest melting point (1) to lowest.CH4 H2O MgO Na NaCl SiO2

b. Justify your relative ranking of CH4 and H2O.

c. Justify your relative ranking of MgO and NaCl.

6. Explain the following observations.a. NH3 boils at 240 K, whereas NF3 boils at 144 K.

b. At 25°C and 1 atm, F2 is a gas, whereas I2 is a solid.

7. Hydrogen gas is produced when aluminum foil is added to a solution of hydrochloric acid.a. The hydrogen is collected over water at 25oC and a total

pressure of 756 torr. What is the mole fraction of H2(g) in the wet gas? (PH2O) at 25oC = 23.8 torr.

b. If 255 mL of wet gas is collected, what is the yield of hydrogen in grams?

c. What is the density of the wet gas?

8. When CaCl2 is added to water, the temperature of the solution decreases. a. Justify which bond is stronger; hydration bonds between

ions and water or ionic bonds between ions?

b. Justify why you expect CaCl2 to be more or less soluble in warm water compared to cold water?

11. Explain the following observations. Your responses must include specific information about all substances. a. When table salt (NaCI) and sugar (C12H22O11) are

dissolved in water, it is observed that (1) both solutions have higher boiling points

than pure water

(2) the boiling point of 0.10 M NaCI(aq) is higher than that of 0.10 M C12H22O11(aq).

b. Ammonia, NH3, is very soluble in water, whereas phosphine, PH3, is only moderately soluble in water.

12. Consider camphor, C10H16O, a substance obtained from the Formosa camphor tree. It has considerable use in the polymer and drug industry. A solution of camphor is prepared by mixing 30.0 g of camphor with 1.25 L of ethanol, C2H5OH (d = 0.789 g/mL). Assume no change in volume when the solution is prepared.a. What is the mass percent of camphor in the solution?

b. What is the molarity of the solution?

c. What is the molality of the solution?

d. The vapor pressure of pure ethanol at 25oC is 59.0 mm Hg. What is the vapor pressure of ethanol in the solution at this temperature?

e. What is the osmotic pressure of the solution at 25oC?

f. What is the boiling point of the solution? The normal boiling point of ethanol is 78.26oC (Kb = 1.22oC/m).

g. The molar mass of cortisone acetate is determined by dissolving 2.50 g in 50.0 g camphor (Kf = 40.0oC/m). The freezing point of the mixture is 173.44oC; that of pure camphor is 178.40oC. What is the molar mass of cortisone acetate?

Unit 5 Practice Free Response6. Answer the following questions about acetylsalicylic acid.

a. What is mass percent of acetylsalicylic acid in a 2.00-g tablet that contains 0.325 g acetylsalicylic acid?

b. Acetylsalicylic acid contains H, C and O. Combustion of 3.00 g yields 1.20 g H2O and 3.72 L of CO2 at 50oC and 1.07 atm, calculate the mass of each element.

c. Determine the empirical formula of acetylsalicylic acid.

d. 1.625 g of pure acetylsalicylic acid reacts with NaOH. The reaction requires 88.43 mL of 0.102 M NaOH.(1) Calculate the molar mass of the acid. (it takes one

mole of acid to react with each mole of NaOH)

(2) What is the molecular formula of the acid?

(3) Suppose the NaOH buret was rinsed with distilled water resulting in the first few drops of NaOH to be more dilute, how would this affect the calculated value for the equivalent mass of the acid?

(4) Suppose some of the solid acid was left in the weighing cup, how would this affect the calculated value for the equivalent mass of the acid?

7. Bismuth (Bi) reacts with fluorine to form BiF3.a. Calculate the mass percent of Bi and F in the compound.

b. Calculate the mass of fluorine required to form 16.5 g of compound.

c. Write a balanced equation for the reaction.

d. How many moles of F2 are required to react with 0.240 mol Bi?

e. How many grams of F2 are required to react with 1.60 g Bi?

f. If 5.00 g of Bi react with 2.00 g F2, what is the limiting reactant?

g. What is the theoretical yield of BiF3 when 5.00 g Bi and 2.00 g F2 react?

h. When BiF3 reacts with water, one of the products is a compound containing 85.65 % Bi, 6.56 % O, and 7.79 % F. What is the simplest formula of this compound?

i. Write a balanced equation for the reaction between BiF3 and H2O.

Unit 6 Practice Free Response1. Consider the combustion of butanoic acid at 25oC:

HC4H7CO2(l) + 5 O2(g) 4 CO2(g) + 4 H2O(l) Ho= -2,183.5 kJSubstance Hf

o (kJ/mol) So (kJ/mol•K)CO2(g) -393.5 0.2136H2O(l) -285.8 0.0699O2(g) 0.0 0.2050

C3H7COOH(l) ? 0.2263a. Calculate Hf

o, for butanoic acid.

b. Calculate So for the combustion reaction at 25oC.

c. Calculate Go for the combustion reaction at 25oC.

d. What is the spontaneous temperature range?

2. Consider the synthesis reaction: N2(g) + 3 F2(g) 2 NF3(g)(Ho

298 = -264 kJ mol-1, So298 = -278 J K-1 mol-1)

a. Calculate Go298 for the reaction.

b. For what temperature range is the reaction spontaneous?

c. Calculate the heat released when 0.256 mol of NF3(g) is formed from N2(g) and F2(g) at 1.00 atm and 298 K.

d. Calculate the F–F bond energy using the information above and the bond energies (NN = 946 kJ/mol, N–F = 272 kJ/mol).

3. The combustion of carbon monoxide is represented by the equation: CO(g) + ½O2(g) CO2(g) a. Determine Ho for the reaction above using the values.

C(s) + ½ O2(g) CO(g) Ho298 = -110.5 kJ•mol-1

C(s) + O2(g) CO2(g) Ho298 = -393.5 kJ•mol-1

b. Determine So for the reaction above using the tableSubstance CO(g) CO2(g) O2(g)So (J/mol•K) 197.7 213.7 205.1

c. Determine Go for the above reaction at 298 K.

4. The dissolving of AgNO3(s) in water is represented by the equation: AgNO3(s) Ag+(aq) + NO3

-(aq)a. Is G positive, negative, or zero? Justify your

answer.

b. The solution cools when AgNO3(s) is dissolved. Is H positive, negative or zero? Justify your answer.

c. Is S positive, negative, or zero? Justify your answer.

5. Consider the thermochemical equation:C2H6(g) + 7/2 O2(g) 2 CO2(g) + 3 H2O(l) H = -1559.7 kJa. Calculate H for the thermochemical equations:2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(l)2 CO2(g) + 3 H2O(l) C2H6(g) + 7/2 O2(g) b. The heat of vaporization of H2O(l) is +44.0 kJ/mol.

Calculate H for the equation:C2H6(g) + 7/2 O2(g) 2 CO2(g) + 3 H2O(g)

c. The heat of formation of CO2(g) and H2O(l) are -393.5 kJ/mol and -285.8 kJ/mol. Calculate Hf

o of C2H6(g).

d. How much heat is evolved when 1.00 g of C2H6(g) is burned to give CO2(g) + H2O(l) in an open container?

e. What is the bomb constant C if the change in temperature is 13.13oC when 1.00 g of C2H6 reacts in the bomb calorimeter that contains 250 g H2O?

Unit 7 Practice Free Response1. Consider the reaction: 2 HI(g) H2(g) + I2(g), where the rate

of the reaction in terms of [HI] = -0.100 M•min-1.a. What is the rate of the reaction in terms of H2(g)?

b. What is the rate of the reaction in terms of I2(g)?

2. Consider the reaction: A + B2 Products. The following experimental data at 22oC were obtained:

A (M) B2 (M) Rate (M•s-1)0.100 0.100 0.0800.500 0.100 0.400.100 0.500 2.0

a. What is the order of the reaction with respect to each reactant?

b. What is the rate constant for the reaction, including units?

c. What would cause an increase in the rate constant?

d. The activation energy for the reaction is 115 kJ/mol. What is the rate constant of the reaction at 27oC?

3. Consider the first-order decomposition of A. The rate constant is 1.7 x 10-2 min-1.a. Calculate the rate of the reaction when the initial

concentration of A is 0.400 M.

b. What percent of A will be used up in two hours?

c. What is the half-life of the reaction?

4. The redox reaction between Tl+ by Ce4+ occurs by the following mechanism.

Ce4+(aq) + Mn2+(aq) Ce3+(aq) + Mn3+(aq)Ce4+(aq) + Mn3+(aq) Ce3+(aq) + Mn4+(aq)

Mn4+(aq) + Tl+(aq) Mn2+(aq) + Tl3+(aq)a. What is the balanced equation for the overall

reaction?

b. What molecule acts as a catalyst for this reaction?

c. What molecules is an intermediate for this reaction?

d. The rate law for this reaction is rate = k[Ce4+][Mn2+]. Which step is the slow step in the mechanism?

5. I-(aq) + CIO-(aq) IO-(aq) + Cl-(aq) Three initial-rate experiments were conducted; the results are shown in the following table.Experiment [I-]

(M)[ClO-] (M)

[IO-]/t (M•s-1)

1 0.017 0.015 0.1562 0.052 0.015 0.4763 0.016 0.061 0.596

a. Determine the order of the reaction with respect to each reactant listed below. Show your work. (1) I-

(2) CIO-

b. For the reaction, (1) write the rate law for the reaction.

(2) calculate rate constant, k, and specify units.

c. The rate constant for this reaction is determined at various temperatures. The data is graphed in order to determine the activation energy, Ea. (1) What variables are graphed?

(2) Explain how to calculate the activation energy from this graph.