CLEP CHEMISTRY EXAM FINAL SCRIPT 2026 COMPLETE SOLUTION SET, Exams of Chemical Kinetics

CLEP CHEMISTRY EXAM FINAL SCRIPT 2026 COMPLETE SOLUTION SET

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CLEP CHEMISTRY EXAM FINAL SCRIPT 2026
COMPLETE SOLUTION SET
โ—‰ J.J. Thompson. Answer: Observed deflection of particles in a
cathode ray tube; proposed that atoms are composed of positive and
negative charges; developed the plum pudding model of the atom
โ—‰ Robert Millikan. Answer: Calculated the charge-to-mass ratio of
electrons using oil drops falling in an electric field; surmised the
charge of a single electron
โ—‰ Ernest Rutherford. Answer: Used the deflection of alpha particles
in a cathode ray tube to discover that most of the atom is empty
space, with protons and neutrons centered in the nucleus.
โ—‰ Niels Bohr. Answer: Determined that electrons exist around the
nucleus at a fixed radius; electrons with higher energy exist farther
from the nucleus. Electrons give off electromagnetic radiation when
moving between energy levels.
โ—‰ Max Planck. Answer: Determined that energy is quantized, or
composed of discrete bundles.
โ—‰ 6.63 x 10^-34 J*sec. Answer: Planck's Constant (h)
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CLEP CHEMISTRY EXAM FINAL SCRIPT 2026

COMPLETE SOLUTION SET

โ—‰ J.J. Thompson. Answer: Observed deflection of particles in a cathode ray tube; proposed that atoms are composed of positive and negative charges; developed the plum pudding model of the atom โ—‰ Robert Millikan. Answer: Calculated the charge-to-mass ratio of electrons using oil drops falling in an electric field; surmised the charge of a single electron โ—‰ Ernest Rutherford. Answer: Used the deflection of alpha particles in a cathode ray tube to discover that most of the atom is empty space, with protons and neutrons centered in the nucleus. โ—‰ Niels Bohr. Answer: Determined that electrons exist around the nucleus at a fixed radius; electrons with higher energy exist farther from the nucleus. Electrons give off electromagnetic radiation when moving between energy levels. โ—‰ Max Planck. Answer: Determined that energy is quantized, or composed of discrete bundles. โ—‰ 6.63 x 10^-34 J*sec. Answer: Planck's Constant (h)

โ—‰ 3.00 x 10^8 m/sec. Answer: Speed of Light (c) โ—‰ E = hv. Answer: Energy of a Photon Formula (1) โ—‰ E = hc / wavelength. Answer: Energy of a Photon Formula (2) โ—‰ Louis DeBroglie. Answer: Combined Einstein's relationship between mass and energy and the relationship between velocity and the wavelength of light. All particles with momentum have a corresponding wave nature. โ—‰ Wavelength = h / mv. Answer: Wavelength of Particles Formula โ—‰ Heisenberg's Uncertainty Principle. Answer: It is impossible to simultaneously know the position and momentum of an electron. โ—‰ Erwin Schrodinger. Answer: Attributed a wave function to electrons, describing the probability of where an electron might exist. โ—‰ Orbitals. Answer: Regions of high probability where electrons might exist; broken into four levels: s, p, d, or f

  1. Magnetic (ml)
  2. Magnetic Spin (ms) โ—‰ Principal Quantum Number. Answer: The shell or energy level an electron occupies; values from 1-7. Electrons with higher values are farther from the nucleus. โ—‰ Angular Momentum Quantum Number. Answer: The subshell the electron occupies; describes the shape of an electron's orbital. n = 1: l = 0 (s) n = 2: l = 0 (s), 1 (p) n = 3: l = 0 (s), 1 (p), 2 (d) n = 4: l = 0 (s), 1 (p), 2 (d), 3 (f) โ—‰ Magnetic Quantum Number. Answer: Represents the orbital position. l = 0: ml = 0 (1 possible s orbital) l = 1: ml = - 1, 0, 1 (3 possible p orbitals) l = 2: ml = - 2, - 1, 0, 1, 2 (5 possible d orbitals) l = 3: ml = - 3, - 2, - 1, 0, 1, 2, 3 (7 possible f orbitals)

โ—‰ Magnetic Spin Quantum Number. Answer: Each orbital contains at most 2 electrons: one with a positive spin (+1/2) and one with a negative spin (-1/2) โ—‰ Dimagnetic. Answer: Elements that have paired electrons in each orbital; all subshells are filled. These elements aren't affected by magnetic fields โ—‰ Paramagnetic. Answer: Elements that have an unpaired electron in at least one orbital; creates a magnetic field in the atom that responds to external magnetic fields. โ—‰ Electron Configurations. Answer: Identify the number of electrons in each type of orbital at each energy level โ—‰ Orbital notation. Answer: Identifies where each electron exists in each orbital. Example: Nitrogen (N) = 1s^2:2s^2:2p^ โ—‰ Aufbau Principle. Answer: Electrons exist first at the lowest possible energy level, unless energy has put them into an excited state โ—‰ Hund's Rule. Answer: Electrons enter orbitals of equal energy singly, with the same spin, before becoming paired.

โ—‰ Sigma bond. Answer: The first covalent bond between nonmetals โ—‰ Pi bond. Answer: Each additional covalent bond between non- metals after a sigma bond; much weaker than sigma bonds. โ—‰ Nonpolar covalent bond. Answer: Created when atoms share their electrons equally; usually occurs when two atoms have similar or the same electron affinity. The closer the values of their electron affinity, the stronger the attraction; normally the difference in electronegativity is >0.4. โ—‰ Polar covalent bond. Answer: Occurs when the electrons between atoms are not equally shared. The atom with the higher electronegativity will have a stronger pull for electrons, resulting in the molecule having a slightly positive side and a slightly negative side. Normally the difference in electronegativity is between 0.4 and 1.7. โ—‰ Dipole moment. Answer: The measure of net molecular polarity. The larger the difference in electronegativities of bonded atoms, the larger the moment. โ—‰ Network covalent structure. Answer: A chemical structure in which the atoms are bonded by a group of covalent bonds in a continuous network.

โ—‰ Lewis Structure. Answer: A model of the valence electrons that are involved in covalent bonding โ—‰ Resonance structure. Answer: An attempt to model delocalized electrons โ—‰ Hybridization. Answer: The process by which electrons mix traits of different atomic orbitals to create bonding orbitals; or, the concept of mixing atomic orbitals into new hybrid orbitals (with different energies, shapes, etc., than the component atomic orbitals) suitable for the pairing of electrons to form chemical bonds โ—‰ VSEPR. Answer: Valence shell electron-pair repulsion; electron pairs will repel each other, making each electron pair as far away as possible from every other electron pair. โ—‰ Expanded octets. Answer: Created when highly electronegative atoms bond to large central atoms and there is no space to allow either five or six electron pairs around the central atom. Require d- electrons to participate in hybridization. โ—‰ Isomers. Answer: Molecules that have the same formula but different structure or arrangement of atoms.

โ—‰ T = 0.693 / k. Answer: Half-life formula โ—‰ P1V1 = P2V (The volume of a gas is inversely proportional to its pressure, when temperature is constant). Answer: Boyle's Law โ—‰ V1T2 = V2T (The volume of a gas is directly proportional to temperature, when pressure is constant). Answer: Charles's Law โ—‰ P1T2 = P2T (The pressure of a gas is directly proportional to temperature, when volume is constant). Answer: Law of Gay-Lussac โ—‰ Ptotal = P1 + P2 + ... + Pn (The total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the gases in the mixture.). Answer: Dalton's Law โ—‰ V1n2 = V2n (The volume of a gas is proportional to the number of moles of gas present when temperature is constant). Answer: Avogadro's Law

โ—‰ PV = nRT (Pressure x Volume = number of gas moles x ideal gas constant x absolute temperature). Answer: Ideal gas law formula โ—‰ 0.082 Latm / Kmol. Answer: Ideal gas constant (R) โ—‰ 273 K; 1.0 atm; 1.0 mol gas = 22.4 L gas. Answer: Standard Temperature and Pressure (3 components) โ—‰ P(mm) = dRT (Pressure x molar mass = density x ideal gas constant x absolute temperature). Answer: Ideal gas law formula (in terms of density) โ—‰ 1/2 (ma x va^2) = 1/2 (mb x vb^2) (Two gases at the same temperature and pressure will have the same kinetic energy; v = velocity; m = mass). Answer: Graham's Law โ—‰ ra^2 / rb^2 = Mb / Ma (gas molecules of smaller molar mass move faster than gas molecules of larger molar mass). Answer: Graham's law of effusion

โ—‰ Simple cubic unit cells. Answer: These have one atom at each of the corners of the cube; containing a total of one atom per unit cell. โ—‰ Face-centered crystal. Answer: A simple cubic unit cell with one additional atom shared between two unit cells on each face of the cube; a total of three atoms per unit cell. โ—‰ Body-centered crystal. Answer: A simple cubic unit cell with one additional atom in the center of the cube, for a total of two atoms per unit cell โ—‰ Amorphous solids. Answer: These don't display a specific geometry; example: glass โ—‰ Solvation. Answer: The interaction of solvent molecules with solute molecules to form loosely bonded combinations. โ—‰ Hydration. Answer: The solvation process when water is the solvent. โ—‰ Miscible solutions. Answer: These occur when one substance is soluble in all proportions with another substance.

โ—‰ Saturation. Answer: A solid solute is in equilibrium with dissolved solute. โ—‰ Solubility. Answer: The molar concentration of dissolved solute at saturation โ—‰ Supersaturation. Answer: A solution that contains more solute than required for saturation โ—‰ P = kC (The amount of gas that can dissolve in a liquid is directly proportional to the partial pressure of the gas above the liquid.). Answer: Henry's Law โ—‰ M = moles solute / liters solution. Answer: Molarity formula โ—‰ pH = - log [H+]. Answer: pH formula โ—‰ Molality = Moles solute / kilograms solvent. Answer: Molality formula โ—‰ mole fraction = moles solute / total solution moles. Answer: Mole Fraction formula

โ—‰ Lewis theory. Answer: An acid is an electron-pair acceptor, and a base is an electron-pair donor. โ—‰ Neutralization. Answer: The process where an Arrhenius acid and base are combined to form a salt and water. โ—‰ Amphoteric. Answer: Compounds that can act as either acids or bases โ—‰ Precipitation reactions. Answer: Occur when soluble reactants are mixed together to form an insoluble product (see solubility rules) โ—‰ Net ionic reaction. Answer: A reaction that shows only ions that combine to form the precipitate. โ—‰ Spectator ions. Answer: Ions that remain dissolved in a precipitation reaction. โ—‰ Oxidation reaction. Answer: An atom increases control over an electron; a loss of electrons โ—‰ Reduction reaction. Answer: An atom decreases control over an electron; a gain of electrons

โ—‰ Combustion reaction. Answer: Oxygen combines with another compound to form water and carbon dioxide โ—‰ Synthesis reaction. Answer: Two or more simple compounds combine to form a more complicated one โ—‰ Decomposition reaction. Answer: A complex molecule breaks down to make simpler ones. โ—‰ Single displacement reaction. Answer: One element trades places with another element in a compound. โ—‰ Double displacement reaction. Answer: The anions and cations of two different molecules switch places, forming two entirely different compounds. โ—‰ Voltaic cell. Answer: Converts chemical energy into electrical energy by isolating the oxidation and reduction half-reactions โ—‰ Electromotive force. Answer: The force in a voltaic cell with which the electrons flow through an external wire from the negative electrode to the positive electrode

โ—‰ K(eq) = [C]c [D]d / [A]a [B]b. Answer: Equilibrium constant (Keq); aka law of mass action โ—‰ K(reverse) = 1 / K(forward). Answer: Equilibrium constant for a reverse reaction โ—‰ Q= [C]c [D]d / [A]a [B]b. Answer: Reaction Quotient When Q > K, reaction proceeds left When Q < K, reaction proceeds right When Q = K, reaction is in equilibrium โ—‰ K(p) = K(c)(RT)^โˆ†n (partial pressure equilibrium constant = molar concentration equilibrium constant x ideal gas law x absolute temperature x net moles of gas (products - reactants). Answer: Gas Equilibrium Constant โ—‰ Buffer. Answer: An aqueous combination of a weak acid and its conjugate base, or a weak base and its conjugate acid โ—‰ pH = pK(a) + log ([A-] / [HA]) (= - log of equilibrium constant of acid + log [molar concentration of base / molar concentration of acid]). Answer: Henderson- Hasselbach Equation (pH of solution)

โ—‰ RR = โˆ†concentration / โˆ†time. Answer: Reaction Rate โ—‰ Rate Law. Answer: Describes the rate of the reaction as a function of a rate constant dependent on temperature and the concentrations of reactants Rate = k (rate constant) x [molar concentration of reactants]^reaction order โ—‰ Keq = k forward / k reverse. Answer: Equilibrium Constant โ—‰ ln k = (-Ea / RT) + ln [A] (ln constant = - activation energy / ideal gas constant x temperature

  • ln molar concentration of reactant). Answer: Temperature, rate constant, and activation energy formula โ—‰ Activation energy. Answer: The amount of energy needed to cause a reaction to occur โ—‰ Catalysts. Answer: Increase the rate of a reaction by facilitating a faster pathway without being used up themselves in the reaction. โ—‰ Mechanism. Answer: The combination of steps in a chemical reaction that must add to yield the total reaction