AP Chemistry Name ________________________Worksheet Fall Semester Review 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.

14 % deviation13 % difference19 6.02 x 1023

8 accuracy15 amount6 Celsius9 certainty

16 d = m/V10 estimated3 experimental

20 grams5 Kelvin1 laws

21 Molar mass18 moles11 number17 particles12 position7 precision

22 ratios4 SI2 theories

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. (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: 100mav = (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.

17 %1m1 + %2m2

12 111 010 -14 charge2 charge/mass

19 compounds9 electrons

18 elements16 ion15 isotopes3 Millikan

21 Mixtures8 neutrons

14 neutrons6 positive7 protons

13 protons20 ratio5 Rutherford

22 solutions1 Thomson

RadioactivityThere are four kinds of radioactive decay: emission of (1) particles ( or 4

2He), (2) particle ( or 0-1e), (3)

particle (, 01e), and (4) radiation (0

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

ZX , 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).

1 alpha8 alpha2 beta6 beta4 gamma

12 ln(No/Nt)5 nuclear3 positron7 positron

10 kNt

11 (ln2)/k9 transmutations

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).

10 absorbed3 Bohr9 emitted

11 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).

8 excited7 ground6 increases4 n 1 spectra5 -2.18 x 1018 J/n2

22.00 x 10-25 J•m/

Quantum Mechanical ModelIn 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.

10 +½ and -½7 0 to (n – 1)6 0, 1, 2, and 38 –l to +l1 Heisenberg2 location9 magnetic3 orbital

11 Pauli4 radius5 s, p, d, and f

12 Two

Electron Arrangements in Atoms and IonsEnergy (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).

23 6 and 1111 alkali12 alkaline earth17 configurations7 core

15 d5 d and f3 degenerate

28 diamagnetic19 Excited16 f18 ground9 group

13 halogens21 Hund's1 increases

25 isoelectronic10 main-group14 noble gases20 Orbital27 paramagnetic8 period2 s p d f4 s and p

22 spin24 three26 unpaired6 Valence

Periodic Properties—Main GroupsMany properties of atoms are due to the distance of the outer electrons from the nucleus and to the (1)

nuclear charge experienced by these electrons. The (2) electrons are very effective in screening the outer electrons from the full charge of the nucleus, whereas electrons in the valence shell do not screen each other very effectively at all. As a result, the effective nuclear charge experienced by valence electrons increases as we move (3) across a period, except for (4) metals where the increase in Zeff is minimum since the added electrons enter the core and cancel the added protons. As a result, atomic radii (5) as we go down a group and (6) as we proceed left to right across a period of main group elements.

Cations are (7) than their parent atoms; anions are (8) than their parent atoms, but group and period trends are the same as atomic radius. For an isoelectronic series the radius (9) with increasing nuclear charge since the electrons are attracted more strongly to the nucleus.

The ionization energy is the energy needed to (10) an electron from a gaseous atom; forming a cation. 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 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) (removing an electron from a full orbital).

12 1319 1513 1620 1818 214 adding2 core6 decrease9 decreases1 effective5 increase

11 increase8 larger3 left to right

15 negative17 negative16 positive

The electron affinity measures the energy change when (14) an electron to a gaseous atom; forming an 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).

10 remove7 smaller4 transition

BondingBonds 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.

2 covalent12 decreases14 decreases18 decreases16 Electronegativity20 greater11 increases15 increases17 increases1 ionic

10 lattice6 Lewis3 metallic8 noble gas7 octet4 outermost

19 polarity13 single9 transfer5 valence

Lewis StructuresElectron distribution in molecules is shown with (1) structures, which indicate how many valence

electrons are involved in forming bonds and how many remain as (2) 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 (3) electrons are placed around each atom. When there are too few valence electrons, then it will be necessary to add bonds. When there are too many valence electrons (and the central atom has at least (5) energy level electrons), then it will be necessary to place additional electrons (up to 10 or 12) around the central atom. When the total number of valence electrons is an (6) 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 (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.

3 eight7 formal8 half1 Lewis9 low

10 negative6 odd5 third2 unshared4

VSEPR Model The valence-shell electron-pair repulsion (VSEPR) model rationalizes molecular geometries based on

the repulsions between electron (1), 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 (2) as possible. Electron domains from non-bonding pairs exert slightly (3) repulsions, which leads to (4) bond angles than idealized values. The arrangement of electron domains around a central atom is called the electron (5) geometry; the arrangement of atoms is called the (6) geometry.

Certain molecular shapes, such as linear and trigonal planar, have cancelling bond dipoles, producing a (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).

5 domain1 domains2 far apart3 greater6 molecular7 nonpolar8 polar4 smaller

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 (1) on neighboring atoms overlap. The bonding electrons occupy the overlap region and are attracted to both (2) simultaneously, which bonds the atoms together.

To extend valence-bond theory to polyatomic molecules, s, p, and sometimes d orbitals are blended to form (3) orbitals, which overlap with orbitals on another atom to make a bond. Hybrid orbitals also hold non- (4) pairs of electrons. A particular mode of hybridization can be associated with each of the five common electron-domain geometries (linear = (5); trigonal planar = (6); tetrahedral = (7); trigonal bipyramidal = (8); and octahedral = (9)).

Covalent bond in which the electron density lies along the line connecting the atoms is called (10), , bond. Bonds can also be formed from the sideways overlap of p orbitals. Such a bond is called a (11) , bond. A double bond, such as that in C2H4, consists of one bond and (12) bond; a triple bond, such as that in C2H2, consists of one and (13) bonds. The formation of a bond between carbons requires that molecules adopt a specific orientation; the two CH2 groups in C2H4, for example, must lie in the same plane. As a result, the presence of bonds introduces rigidity into molecules.

Sometimes a bond can be placed in more than one location. In such situations, we describe the molecule by using two or more (14) structures. The molecule is envisioned as a (15) of these multiple resonance structures and the bonds are (16); that is, spread among several atoms. The bond (17) value

15 blend4 bonding

16 delocalized3 hybrid2 nuclei

12 one1 orbitals

17 order11 pi14 resonance18 share10 sigma5 sp6 sp2

7 sp3

8 sp3d9 sp3d2

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 binary3 first7 H2S8 hydrosulfuric4 ide2 lower6 mono5 number

HydrocarbonsThe simplest types of (1) compounds are hydrocarbons. There are four major kinds of hydrocarbons:

alkanes, alkenes, alkynes, and aromatic hydrocarbons. (2) are substances that possess the same molecular formula, but differ in the arrangements of atoms. In (3) isomers the bonding arrangements of the atoms differ. Different isomers are given different (4). The naming of hydrocarbons is based on the longest continuous chain of (5) atoms in the structure. The locations of alkyl groups, which branch off the chain, are specified by (6) along the carbon chain. Ring structures have the prefix (7). The names of alkenes and alkynes are based on the longest continuous chain of carbon atoms that contains the (8) bond, and the (9) of the multiple bond is specified by a numerical prefix. Alkenes exhibit not only structural isomerism but geometric (10) isomerism 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 double bond is (11).

The chemistry of an organic compound is dominated by the presence of the (12).

5 carbon10 cis-trans7 cyclo

12 functional group2 Isomers9 location8 multiple4 names6 numbering1 organic

11 restricted3 structural

Gas StateSubstances that are gases at room temperatures tend to be (1) with (2) molar mass. Air, a mixture

composed mainly of (3), is the most common gas we encounter. Some liquids and solids can also exist in the gaseous state, where they are known as vapor. Gases' volume can change because they are (4) and they mix in all proportions because their component molecules are far apart.

To describe the state, or condition, of a gas, we must specify four variables: pressure (P), volume (V), temperature (T), and quantity (n). Volume is measured in (5), temperature in (6), and quantity of gas in (7). (8) 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). One atmosphere of pressure equals (9) kPa = (10) torr = (11) mm Hg. A (12) is used to measure atmospheric pressure and a (13) is used to measure the pressure of enclosed gases.

The ideal-gas law equation is (14). The term R is the gas constant, which is 0.0821 when P is in (15) or 8.31 when P is in (16). Most gases at pressures of about 1 atm and temperatures near 273 K and above obey the ideal-gas law reasonably well. The conditions of (17) K and (18) atm are known as the standard temperature and pressure and abbreviated as (19).

Using the ideal-gas law equation, we can relate the density of a gas to its molar mass: MM = (20). In all applications of the ideal-gas law we must remember to convert temperatures to the (21) scale; the only scale where 0 equals no warmth. In gas mixtures the total pressure is the sum of the (22) pressures that each gas would exert if it were present alone under the same conditions (Dalton's law). The partial pressure of a component of a mixture is equal to its mole fraction times the total pressure, PA = (23). The mole fraction is the ratio of the moles of one component of a mixture to the (24) moles of all components. In calculating the quantity of a gas collected over water, correction must be made for the partial pressure of water vapor in the gas mixture.

The kinetic-molecular theory accounts for the properties of an ideal gas in terms of a set of statements about the nature of gases. Briefly, these statements are as follows: Molecules are in continuous, (25) motion; the volume of gas molecules is (26) compared to the volume of their container; the gas molecules have (27) attraction for one another; their collisions are elastic; and the average kinetic energy of the gas molecules is proportional to the absolute (28).

Molecules of a gas do not all have the same kinetic energy at a given instant. Their (29) 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, varies in proportion to the square root of the absolute (30) and in inverse proportion with the square root of the (31): u = (32).

It follows from kinetic-molecular theory that the rate at which a gas undergoes (33) (escapes through a tiny hole into a vacuum) is inversely proportional to the square root of its (34) (Graham's law). The (35) of one gas through the space occupied by a second gas is another phenomenon related to the speeds at which molecules move. Collisions between molecules limit the rate at which a gas molecule can diffuse.

Departures from ideal behavior increase in magnitude as pressure (36) and as temperature (37). Real gases depart from ideal behavior because the molecules possess (38) volume and because the molecules experience (39) forces for one another. The van der Waals equation is an equation that modifies the (40) law equation to account for molecular volume and intermolecular forces.

18 19 101

17 273 10 76011 76015 atm39 attractive12 barometer25 chaotic4 compressible

37 decreases35 diffusion33 effusion38 finite40 ideal-gas36 increases21 Kelvin6 kelvins

16 kPa5 liters2 low

13 manometer20 dRT/P31 molar mass34 molar mass1 molecular7 moles3 N2 and O2

26 negligible27 no23 XAPtot

22 partial8 Pressure

14 PV = nRT29 speeds19 STP28 temperature30 temperature24 total32 (3RT/MM)½

Phase ChangeSubstances that are gases or liquids at room temperature are usually composed of (1). In gases the

intermolecular attractive forces are (2) compared to the kinetic energies of the molecules; thus, the molecules are widely separated and undergo constant, chaotic motion. In liquids the intermolecular forces

23 122 atmospheric19 condensation16 critical

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 (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 dipole-dipole and dispersion forces depend on the polarity, polarizability, size, and shape of the molecule. Dipole-dipole 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.

4 dipole-dipole5 dispersion

17 evaporate18 evaporation6 hydrogen7 increasing

24 line8 mass

12 melting1 molecules

25 more3 motion9 N, O or F2 negligible

26 normal11 phase15 release10 stronger13 sublimation27 triple21 vapor14 vaporization20 volatile

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.

(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.

12 cations13 electrons8 high

10 high9 Ionic6 low

11 Metallic5 Molecular7 network

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 solvent-solvent 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.

21 billion10 crystallization18 decrease26 density6 endothermic7 exothermic

22 fraction5 hydration9 hydration

17 increase13 ionic8 lattice

20 million14 miscible24 molality25 molar23 Molarity19 percent15 pressure11 saturated16 kPg

12 solubility3 solute1 Solutions4 solvation2 solvent

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 (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 colligative4 higher

12 i5 Kb

6 Kf

14 less2 lowering7 lowering

15 more13 Osmosis3 XPo

11 twice10 two18 water8 Kbm9 Kfm

16 17 MRT

Chemical ReactionsThe study of the quantitative relationships between chemical formulas and chemical equations is known

as (1). One of the important concepts of stoichiometry is the law of conservation of (2), 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 (3) reaction. A (4) chemical equation shows equal numbers of atoms of each element on each side of the equation. Equations are balanced by placing (5) in front of the chemical formulas for the reactants and products of a reaction, not by changing the (6) in chemical formulas.

Among the reaction types described in this unit are (a) combination reactions, in which two reactants combine to form (7) product; (b) decomposition reactions, in which a (8) reactant forms two or more products; (c) combustion reactions in oxygen, in which a hydrocarbon or related compound reacts with O2 to form (9); (d) electron exchange reactions in which an element exchanges (10) with an ion in a compound; (e) ion exchange reactions, in which (11) exchange "partners"; and (f) proton exchange reactions, in which an acid gives its (12) to a hydroxide ion, OH-, to form water. The last three reactions will be discussed in more detail later in the year.

(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 net ionic equation, those ions that go through the reaction unchanged are (15).

The (16) in a balanced equation give the relative numbers of moles of the reactants and products. To calculate the number of grams of a product from the number of grams of a reactant, therefore, first convert grams of reactant to (17) of reactant. We then use the coefficients in the balanced equation to convert the 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 solute in a solution is in terms of molarity. The molarity of a solution is the number of moles of solute per (20) of solution. Molarity makes it possible to interconvert solution (21) and number of moles of solute.

A (22) reactant is completely consumed in a reaction. When it is used up, the reaction stops, thus 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.

24 actual13 Aqueous4 balanced3 chemical9 CO2 and H2O5 coefficients

16 coefficients10 electrons19 grams12 H+

11 ions22 limiting20 liter2 mass

17 moles15 omitted7 one

25 percent18 product8 single

14 spectator1 stoichiometry6 subscripts

23 theoretical21 volume

Gravimetric AnalysisThe (1) formula of any substance can be determined from its (2) 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 (3) of each element in the compound, or if it is a (4) reaction, from the mass of CO2 and H2O produced. If the substance is molecular in nature, its (5) formula can be determined from the empirical formula if the molecular (6) is also known.

4 combustion1 empirical3 mass6 mass5 molecular2 percent

Volumetric AnalysisSolutions of known molarity can be formed either by adding a measured (1) of solute and diluting it to a

known volume or by the dilution of a more (2) 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 (3) of solute in the solution (Mstandard)(Vstandard) = (4).

In the process called (5), we add a solution of known molarity, (6), to a buret and an unknown solution to a flask. The titrant is slowly added to the flask until the (7) changes color; which is called the (8) point, where stoichiometrically equivalent quantities of reactants are brought together (also called the end point). The moles of unknown are calculated from the volume and molarity of titrant (moles = (Mstandard)(Vstandard). By knowing moles of unknown, the (9) of unknown or (10) of unknown solution can be determined.

4 (Mstock)(Vstock)2 concentrated

10 concentration8 equivalence7 indicator1 mass9 molar mass3 moles6 titrant5 titration

Change in Enthalpy (H)Chemical reactions typically involve (1) some bonds between reactant atoms and forming new bonds.

Breaking bonds (2) energy, therefore the chemical system gains bond energy and the surroundings lose

21 (C + mc)T11 +H 19 4.18 J/g•K

energy, typically in the form of heat. In contrast, forming bonds (3) 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 (+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).

2 absorbs23 B.E.r – B.E.p

22 Bond energy1 breaking

16 calorimetry15 coefficient9 cool down7 endothermic

14 enthalpy4 higher6 increases

10 less than8 lose

20 mcT12 products3 releases

18 specific heat13 surroundings17 temperature5 weaker

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.

6 +S1 disorder7 gas2 number5 physical3 spacing4 speed

Thermodynamic DataThe 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, Hf

o = (4). Most compounds have (5) values of Hf

o. 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 Hf

o 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, Hf

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

o is (13) by the number of moles, and involving decomposition, where H = (14). An important use of Hf

o and So is calculate H and S for a wide variety of reactions under laboratory conditions, where H Ho = (15) and S So = (16).

6 strong7 So

9 J/mol•K10 series11 Hess's12 sum13 multiplied1 Hf

o

14 -Hfo

15 Hfo

p – Hfo

r 16 So

p – Sor

5 negative4 0 kJ/mol8 0 kJ/mol•K2 13 298

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 –TS. 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).

6 above7 below3 nonspontaneous2 spontaneous4 temperature1 H – TS5 H/S9 +H and –S8 –H and +S

Reaction RateChemical 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:

16 [A]o

17 [A]t

13 1 – overall order20 1/[A]t

6 coefficients21 half-life15 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/k2 M/s5 negative

11 orders12 overall4 positive

10 rate constant7 rate law1 rates9 reactants3 slows down

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 (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).

7 8.31 J/mol•K4 activated complex3 activation6 -Ea/R2 kinetic5 orientation1 temperature

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) reactants and (2) activation energy. The rate law for each step corresponds exactly to the number of reactant molecules, so that reactant coefficients become (3) in the rate law. An (4) 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 (5) step.

A (6) 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 (7). A (8) catalyst is one that is in the same phase as the reactants. It is consumed in the (9) step and reappears in a later step. As a result, it is not included in the (10) reaction, but is included in the (11). A (12) catalyst has a different phase from the reactants and is written above the reaction arrow.

1 1 or 27 activation energy6 catalyst3 exponents

12 heterogeneous8 homogeneous4 intermediate2 low

10 overall11 rate law5 rate-determining9 slow

1. A student measures the mass of an object to be 75.011 g. The true mass is 77.500 g. What is the percent error?

100(77.500 g – 75.011 g)/77.500 = 3.212 %

2. An empty graduated cylinder (25.044 g) is filled with 50.0 mL of liquid. The total mass of cylinder and liquid is 69.886 g.a, What is the density of the liquid in g/mL?

D = m/V = (69.886 g – 25.044 g)/50.0 mL = 0.897 g/mL

b, What is the density of the liquid in kg/m3?0.897 g x 1 kg x 1 mL x 1 cm 3 = 897 kg/m3

1 mL 103 g 1 cm3 (10-2)3 m3

3. Consider the gas, UF6, which has a density of 15.7 g/L. a. What is the molar mass of UF6?

238 g + 6(19 g) = 352 g

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. What is the volume of 25.0 g of UF6?

25.0 g x 1 L/15.7 g = 1.59 L

4. Write the nuclear symbol that has 26 p and 30 n.56

26Fe5. Calculate the average atomic mass of Pb, which consists

of four stable isotopes whose atomic masses and abundances are as follows:

Isotope 206Pb 207Pb 208PbAtomic Mass 206 u 207 u 208 uAbundance 25.5% 22.1% 52.4%

mav = (206)(0.255) + (207)(0.221) + (208)(0.524) = 207

6. Place the forms of hydrogen (H, H+, H-, H2, NaH) in the table.

Atom Cation Anion Molecule

ionic Compound

H H+ H- H2 NaH7. Predict the product(s) for the following nuclear processes.

alpha emission of Po-218 21884Po 4

2 + 21482Pb

beta emission of Pb-211 21182Pb 0

-1 + 21183Bi

neutron bombardment of S-32 3216S + 1

0n 11H + 32

15P8. Consider a radioactive sample (t½ = 35.5 days).

a. How long will it take for 50 % of the sample to decay?

1 half-life 35.5 days

b. Calculate the rate constant.

k = ln2/t½ = ln2/35.5 days = 0.0195 d-1

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

9. 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/1.06 x 10-19 J = (2.00 x 10-25 J•m)/ = 1.89 x 10-6 m

10. Complete the following chart for magnesium.symbol period group atomic numbermetal/nonmetal

Mg 3 2 12 metal11. 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 Sa/b 3s23p64s2 3s23p1 3s23p4

c [Ar]4s2 [Ne]3s23p1 [Ne]3s23p4

d Ca2+ [Ar] Al3+ [Ar]3d10 S2- [Ne]3s23p4

e/f 3s 3p 4s

3s 3p

3s 3p

(4, 0, 0, -½) (3, 1, -1, +½) (3, 1, 1, +½)g diamagnetic paramagnetic paramagnetic

12. a. Write all the ions that are "isoelectronic" (contain the same electron structure) as Neon.

N3-, O2-, F-, Na+, Mg2+, Al3+

b. List the ions from part (a) from smallest radius to largest.

Al3+, Mg2+, Na+, F-, O2-, N3-

c. 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 largest atomic radius Nlargest ionic radius Nlargest ionization energy Fmost negative electron affinity value Fmost negative ionic charge Ngreatest electronegativity F

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.

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.

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-S4 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 N – O :

••

• ••: N = O :

•• •• ••: O = 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 O5 5 6 5 5 6 5 5 65 4 7 6 4 6 7 4 50 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 NO2

-

N–O Bond strength 3 1 2N–O Bond length 1 3 2

17. Complete the chart for the molecules: SF2, SF4, and SF6.SF2 SF4 SF6

Lewis Structure••

F–S–F••

F••/

F–S–F|F

F F\ /

F–S–F/ \

F FGeom

etry

Electron Domain Tetrahedral Trigonal

bipyramidal Octahedral

Molecular Bent See saw Octahedral

Hybridization sp3 sp3d sp3d2

Bond Angle 109.5o 90o, 120o 90o

Polarity polar polar nonpolar18. Rank bond angles from largest to smallest (NH3, NH2

-, NH4+).

NH4+ > NH3 > NH2

-

19. 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)2NH••

CH3–N–CH3

|H

dimethylamine

CH3COOHCH3-C-OH

|| O

ethanoic acid

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 Clcis 1,2-dichloroethene trans 1,2-

dichloroetheneb. 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 CH3Fnone H-bond H-bond dipole

24. Consider the following heating profile for substance X.

120oC D E

40oC B C A

Heat Added (J)Which phase takes more energy to raise the temperature 1oC? solid liquid

Which process takes more energy melting boiling

per gram substance?Which interval is there a mixture of solid and liquid? B-C D-E

Which temperature is vaporization? 40oC 120oC25. Determine the following using the phase diagram below.

Normal boiling point 68oCDensest phase solidPhase change from 0oC to 40oC at 0.5 atm sublimationPhase change from 1.0 to 0.05 atm at 60oC boiling

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 gb. 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/Ld. 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 Kr29. 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 = nRTP(1.00 L) = (.250+.150+.100)(8.31)(308 K) P = 1280 kPab. 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 gb. How would the following change the amount of water

that evaporates? (increase, decrease, no change)Use a 5.0 L container decreaseFill with humid air decreaseRaise the temperature to 25oC increaseAdd 10 g of water no change

32. Complete the chart for each type of solid.

Metallic Covalent Network Molecular Ionic

Structural Unit ion atom molecule ionBond name metallic covalent molecular ionic

Bond strength variable strong weak strongMelting point variable high low high

Solubility low low variable highConductivity high low low lowMalleability high low variable lowExample copper diamond water salt

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 LatticeBond that breaksBond that formsStronger bondYou 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) = 0.0101c. 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) = 1.05oCTf = 0oC – 1.05oC = -1.05oCc. 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/kgm = molsolute/msolvent(kg)

0.41 mol/kg = molsolute/0.00500 kg molsolute = 0.0021

molMM = 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. _1_C3H9OH(l) + _5_O2(g) _3_CO2(g) + _5_H2O(g)

c. _4_ Al(s) + _3_ O2(g) _2_ Al2O3(s)41. 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 1 L = 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 Li1.00 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 excessc. How many grams of Li3N are formed?0.0480 mol Li3N x 34.8 g Li3N = 1.67 g Li3N 1 mol Li3Nd. If the percent yield is 88.5%, how many grams of Li3N

are produced?

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 KOH0.200 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 molc. 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 OH-

0.0300 mol NiSO4 0.0300 mol Ni2+, 0.0300 mol SO42-

e. 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[SO4

2-]: 0.0300 mol/0.300 L = .100 M44. 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 C1.00 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 C2H5OC2H5O = 2(12) + 5(1) + 1(16) = 45 g/mol90/45 = 2 C4H10O2

45. Consider aspirin, C9H8O4.a. What is the mass percent of carbon?

100(9)(12.0)/[9(12.0) + 8(1.01) + 4(16.0)] = 60.0 %

b. 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 Pb 2+ = 3.25 x 10-4 mol

Pb2+ 1 L 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 Lb. 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 1 L 1 mol NaOH1.000 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 = -mcT 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 CC H–H

414 kJ/mol 347 kJ/mol 820 kJ/mol 436 kJ/molH =BE broken - BE formed

H = BECC + 2 BEC–H + 2 BEH–H – (BEC–C + 6 BEC–H)H = 820 + 2(414) + 2(436) – (347 + 6(414)) kJ = -311 kJ

c. Using the Hfo values.

C2H2 (g) H2 (g) C2H6 (g)

226.7 kJ/mol 0 kJ/mol -84.7 kJ/molH Ho = Ho

f products -Hof reactants

H = -84.7 kJ – (226.7 kJ + 2(0 kJ)) = -311 kJ49. Consider the reaction:

C2H5OH(l) + 3 O2(g) 2 CO2(g) + 3 H2O(l)C2H5OH O2 CO2 H2O(l)

Hfo (kJ/mol) -277.0 0 -393.5 -285.8

So (kJ/mol•K) 0.1607 0.2050 0.2136 0.0699a. Calculate H.

HHo = Hof products – Ho

f reactants

H = 2(-393.5) + 3(-285.8) – (-277.6 + 3(0)) kJ = -1366.8 kJb. Calculate S.

SSo = Soproducts – So

reactants

S = 2(0.2136 ) + 3(0.0699) – (0.1607 + 3(.2050)) kJS = -0.1388 kJ/Kc. Calculate G at 25oC.GHo – TSo

G -1366.8 kJ – (298 K)(-0.1388 kJ/K) -1325.4 kJd. 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 0.0800.500 0.100 0.410.100 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 = 1n = 0 because the rate didn't change when [B2] changedb. 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 = ktln(100/[A]t = (0.80 s-1)(1 s) = 0.80100/[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 temperatureg. 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 + H2ON2O + 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