Bonding and intermolecular Forces: Thermodyamics

Which of the following molecules is polar? Question 3 Answer Choices A. C2H6 B. CS2 C. CO D. CCl4

C. Carbon monoxide has two atoms with a slight difference in electronegativity and so it has a dipole moment. Ethane has only C-C and C-H bonds, neither of which are polar. Without polar bonds the molecule cannot be polar (eliminate choice A). While carbon disulfide and carbon tetrachloride contain polar bonds, they are also symmetrical molecules, so the bond dipoles cancel each other out, leaving nonpolar molecules (eliminate choices B and D).

Which of the following molecules has the greatest boiling point? Question 3 Answer Choices A. CH3CH2CH2CH3 B. CH3CH2OH C. CH3CHO D. CH3CO2H

D. CH3CO2H

A substance has a standard heat of formation of 0. This is best explained by which of the following statements? Question 4 Answer Choices A. The substance is an element in its standard state. B. At low temperatures, entropy makes virtually no contribution to changes in Gibbs free energy. C. ΔH is a state function. D. The bond energies within the substance are very negative.

A. A substance has a standard heat of formation of 0. This is best explained by the fact that the substance is an element in its standard state. For elements in their standard state, ΔHf° is equal to zero by definition.

Calculate the standard enthalpy change for the reaction CaCO3(s) CaO(s) + CO2(g) given the standard heats of formation: Question 15 Answer Choices A. 570 kJ/mol B. 178 kJ/mol C. -570 kJ/mol D. -178 kJ/mol

B. 178 kJ/mol is the standard enthalpy change for the reaction CaCO3(s) CaO(s) + CO2(g) given the standard heats of formation: ΔHrxn is the difference between the total ΔHf of the products and the total ΔHf of the reactants. Thus, ΔHrxn = Σ(ΔHf, products) - Σ(ΔHf, reactants) = (-635.5 kJ/mol - 393.5 kJ/mol) - (-1206.9 kJ/mol) = 178 kJ/mol

Rank the average C—O bond length from shortest to longest for CO, CO2, and CO32-. Question 20 Answer Choices A. CO32- < CO2 < CO B. CO < CO2 < CO32- C. CO32- < CO < CO2 D. CO < CO32- < CO2

B. The rank of the average C—O bond length from shortest to longest for CO, CO2, and CO32- is CO < CO2 < CO32-. Bond length can be attributed to the hybrid orbitals which form the bond (sp orbitals form shorter bonds than sp2 orbitals, which form shorter bonds than sp3 orbitals). CO and CO2 have shorter bonds lengths than CO32-, since their carbon atom is sp hybridized, while that of CO32- is sp2 hybridized ("CO32- < CO2 < CO" and "CO32- < CO < CO2" can be eliminated). CO has a shorter bond length than CO2, since its oxygen orbitals are sp hybridized, while they are sp2 hybridized in CO2.

Which of the following compounds will have a dipole moment? A. II only B. III only C. II and III only D. I and II only

Compounds II and III only will have a dipole moment.

What is the formal charge of boron in BH4-? Question 10 Answer Choices A. +1 B. -4 C. +3 D. -1

D. The formal charge of boron in BH4- is -1. Since hydrogen can only have a single bond, the structure must be comprised of boron as the central atom with four hydrogen atoms extending from it. The formula for formal charge is FC = V - (0.5)B - L Where FC = Formal Charge V = # of valence electrons B = # of bonding electrons L = # of lone pair electrons Boron is in the third group, so it has a valence of 3. The four single bonds around it contribute eight bonding electrons. Since the central boron atom already has eight electrons surrounding it, there cannot be any lone electron pairs. Thus, the formal charge of boron would be 3 - (0.5)(8) - 0 = -1. Note that +3 is a trap answer, and is the oxidation state of boron instead of its formal charge.

Rank the indicated bonds in the given molecule from shortest to longest. Question 22 Answer Choices A. b < c < a B. a < b < c C. b < a < c D. a < c < b

D. The rank of the indicated bonds in the given molecule from shortest to longest is a < c < b. CC triple bonds are shorter than C=C double bonds, which are shorter than C—C single bonds. This can be attributed to the increase in percent s character of the hybrid orbitals involved in bond formation (sp, sp2, sp3, respectively).

Which of the following molecules will participate in hydrogen bonding with water? Question 16 Answer Choices A. I only B. II only C. I, II, and III Correct Answer (Blank) D. I and II only

Molecules I, II, and III will participate in hydrogen bonding with water.

Which of the following represents the correct Lewis structure for N2?

A represents the correct Lewis structure for N2. Since each N atom has 5 valence electrons, the Lewis structure for N2 must account for 10 valence electrons.

Given the following bond energies, what is the enthalpy of the combustion of methane? Bond Dissociation Energy H-C 414 kJ/mol H-O 460 kJ/mol O-O 146 kJ/mol O=O 497 kJ/mol C=O 803 kJ/mol Question 1 Answer Choices A. -917 kJ/mol B. -796 kJ/mol C. 796 kJ/mol D. 917 kJ/mol

B. In the combustion of methane (CH4 + 2 O2 → CO2 + 2 H2O), four H-C bonds and two O=O bonds are broken while two C=O bonds and four H-O bonds are formed. To find the enthalpy of the overall reaction, we must use the following equation:

Compared to uncatalyzed reactions, the Ea - ΔHrxn for catalyzed reactions is: Question 9 Answer Choices A. higher because catalysts increase Ea. B. higher because catalysts decrease ΔHrxn. C. lower because catalysts decrease Ea. D. lower because catalysts increase ΔHrxn.

C. Catalysts lower the Ea while having no impact on the ΔHrxn, so choice C is best.

CF2Cl2 is a common freon used in refrigerators. The strongest intermolecular forces holding these molecules together are: Question 1 Answer Choices A. ionic forces. B. hydrogen bonding. C. London dispersion forces. D. dipole-dipole forces.

D. CF2Cl2 is a common freon used in refrigerators. The strongest intermolecular forces holding these molecules together are dipole-dipole forces. CF2Cl2 is not an ion, so it cannot experience ionic forces. In order to experience hydrogen bonding, the molecule must have at least one hydrogen atom bonded to a highly electronegative element (F, O, N), which CF2Cl2 does not. While all molecules have some degree of London dispersion forces, CF2Cl2 is a slightly polar molecule since fluorine is more electronegative than chlorine. Therefore, dipole forces is the best answer since they are stronger than London dispersion forces.

What is the role of the copper(II) cation in the structure of chlorophyllin, shown above? Question 1 Answer Choices A. An electron donor in the coordinate covalent bonds B. A nucleophile C. An oxidizing agent D. A Lewis acid

The role of the copper(II) cation in the structure of chlorophyllin is a Lewis acid. In the coordination complex shown, the organic ring portion is the ligand. The nitrogen atoms have lone pairs of electrons so can act as Lewis bases and donate them to the central metal ion, which is therefore a Lewis acid. The metal can accept the electrons from the ligands into its empty orbitals. This forms the coordinate covalent bond. Thus a Lewis acid is the correct answer. An oxidizing agent is false since coordination of a ligand to a metal is acid base chemistry and not redox chemistry. The remaining choices are effectively identical. Electron pair donors are Lewis bases, or nucleophiles, so both choices must be incorrect.

A reaction occurs that results in a set of products with more stable bonds and more orderly arrangement than were present in the reactants. Which of the following is true of this reaction? Question 16 Answer Choices A. The enthalpy and entropy changes are negative. B. The enthalpy change is positive, and the entropy change is negative. C. The enthalpy change is negative, and the entropy change is positive. D. The enthalpy and entropy changes are positive.

A. A reaction occurs that results in a set of products with more stable bonds and more orderly arrangement than were present in the reactants. The enthalpy and entropy changes are negative is true of this reaction. In order for the products to be more stable, they must have lost energy; thus, ΔH is negative. If the products have a more orderly arrangement than the reactants, then the entropy decreased; thus ΔS is also negative, and the answer is "the enthalpy and entropy changes are negative."

According to the laws of thermodynamics, an endothermic reaction is most likely to be spontaneous if the: Question 2 Answer Choices A. change in entropy is positive and the temperature is high. B. change in entropy is negligible and the temperature is low. C. change in entropy is positive and the temperature is low. D. change in entropy is negative and the temperature is high.

A. According to the laws of thermodynamics, an endothermic reaction is most likely to be spontaneous if the change in entropy is positive and the temperature is high. An endothermic reaction has a positive ΔH. Since a reaction is spontaneous if ΔG = ΔH - TΔS is negative, an endothermic reaction is spontaneous when ΔS is positive and the temperature is high.

How many electrons are found in π bonds in acetylene (C2H2)? Question 7 Answer Choices A. 4 B. 0 C. 6 D. 2

A. Acetylene possesses two π bonds which contain a total of 4 electrons. Acetylene possesses a C≡C which is composed of a σ bond and two π bonds. Each covalent bond is generated by the sharing of two electrons and therefore four total electrons are present in the π bonds of acetylene.

All of the following are examples of polar molecules EXCEPT: Question 11 Answer Choices A. CCl4. B. HF. C. CO. D. H2O.

A. All are examples of polar molecules EXCEPT CCl4. Although each C—Cl bond is very polar, the dipole moments of the four bonds in CCl4 cancel each other out because the molecule is a symmetric tetrahedron.

At room temperature, H2S is a gas, but H2O is a liquid. Which of the following plays an important role in this observation? Question 5 Answer Choices A. The electronegativity of oxygen is greater than that of sulfur. B. Sulfur has a greater ionization energy than oxygen has. C. Sulfur has a smaller atomic radius than oxygen. D. Stronger hydrogen bonds form between molecules of hydrogen sulfide than between molecules of water.

A. At room temperature, H2S is a gas, but H2O is a liquid. The best explanation for this observation is that the electronegativity of oxygen is greater than that of sulfur. H2O molecules are much more polar than H2S molecules because O is more electronegative than S. As a result, the hydrogen bonds between water molecules are much stronger than the dipole-dipole interactions between H2S molecules

Among the following, which molecular geometry CANNOT result in a nonpolar structure? Question 12 Answer Choices A. Bent B. Diatomic covalent C. Square planar D. Trigonal planar

A. Bent is the molecular geometry that CANNOT result in a nonpolar structure. Only molecules with complete symmetry around the central atom can be nonpolar. Bent molecules lack this symmetry, because both noncentral atoms lie on the same side of the central atom.

Each of the following pairs of substances interact through van der Waals forces EXCEPT: Question 3 Answer Choices A. Na+ and H2O B. HCl and Ar C. Cl2 and Br2 D. NH3 and H2S

A. Each of the following pairs of substances interact through van der Waals forces EXCEPT Na+ and H2O. Van der Waals forces are intermolecular forces that occur between neutral molecules or atoms. They include hydrogen bonds and other dipole-dipole forces, dipole-induced dipole forces, and London dispersion forces. "NH3 and H2S" is incorrect because both molecules have permanent dipoles and have a dipole-dipole interaction. "Cl2 and Br2" is incorrect because both molecules are nonpolar and interact through London dispersion forces. "HCl and Ar" is incorrect because it is an example of a dipole-induced dipole interaction. "Na+ and H2O" is an example of an ion-dipole interaction. Since the sodium cation is not neutral, it does not act primarily through van der Waals forces.

Which of the following molecules has the lowest boiling point? Question 7 Answer Choices A. F2 Correct Answer (Blank) B. Cl2 C. H2O D. HCN

A. F2 has the lowest boiling point. Boiling point questions are really an intermolecular forces question in disguise. Because the question is asking for the lowest boiling point, it is really asking which of the choices has the weakest intermolecular forces. Remember that the order from the strongest to weakest intermolecular force is: hydrogen bonding > dipole-dipole forces > London dispersion forces. H2O (water) can hydrogen bond, HCN has dipole-dipole forces, and F2 and Cl2 only have London dispersion forces. Thus, H2O and HCN may immediately be eliminated. To differentiate the two molecules that solely exhibit London dispersion forces, the Cl2 with a larger molecular weight will have the stronger force because it is more polarizable and has a larger surface area of interaction. Thus, F2 will have the lowest boiling point among the choices.

The formation of CO2(g) from the reaction of CO(g) and O2(g) is spontaneous at room temperature. What must also be true for this reaction? Question 6 Answer Choices A. It is exothermic. B. It is endergonic. C. It is endothermic. D. It is isentropic.

A. The formation of CO2(g) from reaction of CO(g) and O2(g) is spontaneous at room temperature. It is also true that it is exothermic. In the reaction , the overall number of moles decreases from reactants to products indicating a decrease in entropy (-ΔS). Entropy is not constant, so it cannot be isentropic. The reaction is spontaneous at room temperature (-ΔG) which means it is exergonic and not endergonic. Since ΔG = ΔH - TΔS, the only way that ΔG can be negative with a negative ΔS is if ΔH is also negative. Therefore, the reaction must be exothermic.

Which of the following lists hydrogen halides in terms of increasing standard heats of formation? Question 8 Answer Choices A. HF < HCl < HBr < HI B. HI < HBr < HCl < HF C. HBr < HF < HCl < HI D. HF < HI < HCl < HBr

A. The hydrogen halides increase in standard heats of formation in the order: HF < HCl < HBr < HI. The general formula for hydrogen halide formation is: . The reaction that forms the most stable HX will be the most exothermic and have the most negative standard heat of formation. Fluorine is the smallest, most electronegative halogen and forms the only hydrogen halide of the group that is not a strong acid. HF is the most stable hydrogen halide and will have the most negative heat of formation. This eliminates choices without HF having the lowest heat of formation. Indeed, the heat of formation of HI is actually endothermic. This eliminates choice HF < HI < HCl < HBr.

Which two atomic orbitals interact to form the D—D bond in D2? Question 18 Answer Choices A. s and s B. p and p C. sp and sp D. sp3 and sp3

A. The two atomic orbitals that interact to form the D—D bond in D2 are s and s. Deuterium (an isotope of hydrogen) only has a 1s orbital, and is therefore not hybridized ("sp and sp" and "sp3 and sp3" can be eliminated). The D—D bond, then, is formed by an s orbital from each deuterium.

Which of the following does not have a closed valence shell? Question 8 Answer Choices A. Cl B. Br- C. Ti4+ D. Kr

A. Chlorine does not have a completely filled p subshell but has an s2p5 electron configuration. The other options have noble gas configurations, giving them the full complement of electrons in their s and p subshells. Br- is isoelectronic with Kr (eliminate choices B and D) and Ti4+ is isoelectronic with Ar (eliminate choice C).

Rank the following elements by increasing electronegativity: Se, Sn, Sb, As. Question 24 Answer Choices A. Sn < Sb < As < Se B. Se < As < Sb < Sn C. Sn < Se < As < Sb D. As < Sn < Sb < Se

A. Electronegativity is the ability of an atom to attract electrons to itself in a covalent bond and is inversely related to atomic radius. Therefore, electronegativity increases as you move from the bottom left to the top right of the periodic table. Selenium is farthest up and right, so choices B and C can be eliminated. Tin is farthest down and left, making choice A best.

Which type of visible light, red or violet, has the highest energy? Question 12 Answer Choices A. Violet light, because it has a lower frequency. B. Violet light, because it has a shorter wavelength. Correct Answer (Blank) C. Red light, because it has a higher frequency. D. Red light, because it has a longer wavelength.

A. High energy is synonymous with short wavelength and high frequency because E = hf and f = c/λ. This eliminates choices A and D. Violet light is adjacent to ultraviolet radiation (waves that are bad to the body) in the electromagnetic spectrum, while red light is adjacent to infrared radiation (waves that are benign to the body, also known as heat). Thus, violet must have higher energy, making choice B correct.

Consider the following reaction: NH3(g) + HCl(g) → NH4Cl(g) ΔG° = -91.1 kJ Which of the following is a true statement? Question 1 Answer Choices A. None of the other answers are correct. B. The entropy change of the reaction is positive. C. The reaction is always spontaneous. D. The reaction is never spontaneous.

A. None of the other answers are correct.

Rank the following elements by increasing atomic radius: N, Ba, F, Te. Question 11 Answer Choices A. F < N < Te < Ba B. F < N < Ba < Te C. Ba < Te < N < F D. N < F < Te < Ba

A. On the periodic table, atomic radius increases from top to bottom due to the increase in the number of electron shells. The element lowest on the periodic table is Ba, so eliminate choices B and C. Atomic radius also decreases from left to right due to the increased pull on the electrons from the increasingly positive nucleus (or greater effective nuclear charge). Therefore, between the remaining choices F should be smaller than N because it has a greater effective nuclear charge, making choice A the best answer.

If an atom's atomic number equals its atomic weight, which of the following would be true? Question 27 Answer Choices A. The number of protons equals the number of electrons. B. The number of neutrons equals the number of protons. C. The number of neutrons equals the number of electrons. D. The number of protons is less than the number of neutrons.

A. The only scenario in which this can occur is if the atom does not contain any neutrons. Thus, for a neutral atom, the number of protons must equal the number of electrons (choice C is correct).

Given the following enthalpies of formation, what is the reaction enthalpy for C3H4(g) + 4 O2(g) → 3 CO2(g) + 2 H2O(l)? ΔH°f for C3H4(g) 185 kJ/mol ΔH°f for CO2(g) -286 kJ/mol ΔH°f for H2O(l) -394 kJ/mol A. -1831 kJ/mol B. -1461 kJ/mol C. -495 kJ/mol D. 865 kJ/mol

A. The reaction enthalpy can be calculated by taking the sum of the enthalpies of formation of the products and subtracting from it the sum of the enthalpies of formation of the reactants after multiplying each by their stoichiometric coefficients. The enthalpy of formation for elements in their standard state is defined as zero (hence we will not need it for our solution). Thus we find that the reaction enthalpy is:

Consider the following reactions: Which reaction proceeds at a faster rate? Reaction 1: S(s) + O2(g) → SO2(g) ΔG° = -300.1 kJ Reaction 2: Cu2S(s) → 2 Cu(s) + SO2(g) ΔG° = -217.3 kJ Question 2 Answer Choices A. The reaction rates cannot be determined with the information provided. B. Reaction 2 C. Reaction 1 D. Reaction 1 and Reaction 2 proceed at the same rate.

A. The reaction rates cannot be determined with the information provided.

Which pair of liquids below will be immiscible? Question 14 Answer Choices A. H2O and CCl4 B. CH3OH and CH3CH2OH C. CH3CH2CH2CH2CH3 and CH3CH2CH2CH2Br D. CCl4 and C6H6

A. The solvent pair that is immiscible will be made of a polar and a nonpolar compound based on the rule of thumb that "like dissolves like." Therefore, compounds that have very different intermolecular forces are least likely to dissolve in each other, and will separate into two layers. Both methanol and ethanol can hydrogen bond, making choice B a miscible pair. Pentane is completely nonpolar, and bromobutane has only one slightly polar C-Br bond. They will both display London dispersion forces (LDFs) (perhaps some weak dipole-dipole forces for the alkyl halide) and be soluble in each other, eliminating choice C. Both CCl4 and C6H6 (benzene) are nonpolar compounds held together by LDFs, so are also soluble, eliminating choice D. Since water is a highly polar, hydrogen-bonding compound while carbon tetrachloride is nonpolar, choice A is the immiscible pair.

Given the reaction below: 2 C2H4 + O2 → 2 C2H4O how many moles of C2H4 remain if 75 moles of C2H4 and 40 moles of O2 were initially present? Question 6 Answer Choices A. 0 Correct Answer (Blank) B. 5 C. 35 D. 40

A. When initial amounts of the two reactants are given, this usually indicates a limiting reactant problem. The first step is determining which of the two reactants is the limiting reactant. If we have 75 moles of methane, we will need half as much (or 32.5 moles) of O2 to obtain a stoichiometric amount because the ratio of O2 to C2H4 is 1:2. However, the problem statement says that we have 40 moles of O2 present, indicating that we have an excess amount. This means that C2H4 is the limiting reactant. Conversely, had we chosen to start with O2, the reasoning is slightly different. If we have 40 moles of oxygen, we will need twice as much (or 80 moles) of C2H4 to obtain a stoichiometric amount, because the ratio of C2H4 to O2 is 2:1. Since the problem statement says that we have 75 moles of C2H4 initially, that reactant has to be limiting because we do not have enough of it. Now that we have concluded that C2H4 is limiting, we still have to solve the problem. By definition, the limiting reactant gets completely consumed, while there will be some moles of the excess reactant (O2 in this case) left over. Thus, there will be 0 moles of C2H4 left over in this reaction (Choice A).

In which of the following structures is the nitrogen sp2 hybridized?

B

Which of the following is the best resonance structure of diazomethane, CH2N2? Question 11 Answer Choices A. Structure A B. Structure C C. Structure D D. Structure B

B Structure C is the best one. Diazomethane is a neutral compound, so Structure D can be eliminated since the formal charges add up to -2. In addition, the central nitrogen violates the octet rule. Good resonance structures minimize formal charge, so Structure B can be eliminated (its C and central N are also electron deficient). For Structures A and C, both have full octets on all atoms and the same amount of formal charges that add to zero. However, only Structure C has the negative formal charge on the more electronegative nitrogen atom, rather than the carbon atom as in Structure A.

How many orbitals make up the 5f subshell? Question 12 Answer Choices A. 10 B. 7 C. 5 D. 14

B. A total of 7 orbitals make up the 5f subshell. There is one s orbital, and there are three p orbitals, five d orbitals, and seven f orbitals. Thus, the answer is 7. (Note: Do not confuse the number of orbitals in a subshell with the number of electrons the subshell can hold. Each orbital can hold two electrons, so the capacity of an nf subshell is 7 × 2 = 14 electrons.)

At 100 K, a certain reaction has the following values for ΔH and ΔS: ΔH = 25 kJ and ?S = 50 J/K. Neglecting any variation in ΔH and ΔS, at what temperature will ΔG = 0? Question 9 Answer Choices A. 100 K B. 500 K C. 1000 K D. 250 K

B. At 100 K, a certain reaction has the following values for ΔH and ΔS: ΔH = 25 kJ and ΔS = 50 J/K. Neglecting any variation in ΔH and ΔS, ΔG = 0 at 500 K.

Which of the following molecules has the greatest dipole moment? Question 2 Answer Choices A. SO3 B. H2O C. SiCl4 D. CO2

B. H2O has the greatest dipole moment. Water's overall molecular shape is bent. Each O—H bond is polar producing an overall dipole moment in the direction of the oxygen atom. Carbon dioxide's overall molecular shape is linear. Each C=O double bond is polar, however they draw electrons with equal strength in opposite directions. Thus, there is no overall dipole moment. Silicon tetrachloride's overall molecular shape is tetrahedral. Each Si—Cl bond is polar, however each of these four bonds has a counterpart that draws electrons with equal strength in the opposite direction. Thus, there is no overall dipole moment. Sulfur trioxide's overall shape is thus trigonal planar. Note that S is capable of exceeding the octet rule. Each S=O double bond is polar, however all three draw electrons with equal strength in opposite directions. Thus, there is no overall dipole moment.

Which of the following pairs of molecules have the same shape? Question 5 Answer Choices A. BF3 and NH3 B. H3O+ and PH3 C. H2O and PH3 D. PH3 and CH4

B. H3O+ and PH3 have the same shape. Use VESPR to understand the shape of each molecule. Accordingly, all the molecules are trigonal pyramid in shape except BF3 (which is trigonal planar), H2O (which is bent), and CH4 (which is tetrahedral).

If a chemical reaction is found to be spontaneous under a given set of conditions, then which of the following must be negative? Question 18 Answer Choices A. Cp B. ΔG C. ΔH D. ΔS

B. If a chemical reaction is found to be spontaneous under a given set of conditions, then ΔG must be negative. ΔG is always negative for spontaneous reactions.

Metal-organic frameworks (MOFs) are systems of metals tethered together by organic linkers forming porous solids with exposed, active metal centers. MOFs are often used to bind and hold gases such as H2, often allowing for a more dense packing of gas than allowed in the pure liquid state. If the binding of gas to metals in the MOF is spontaneous, which must be true? Question 12 Answer Choices A. The enthalpy of binding is positive. B. The enthalpy of binding is negative. C. The change in free energy associated with binding is positive. D. The entropy of reaction is positive.

B. MOFs are often used to bind and hold gases such as H2, often allowing for a more dense packing of gas than allowed in the pure liquid state. If the binding of gas to metals in the MOF is spontaneous, the enthalpy of binding is negative. As we know the reaction is spontaneous (i.e., the change in free energy is negative), and the change in entropy is certainly negative (as gas molecules are transferred to the solid state), the enthalpy of the reaction must be negative to satisfy ΔG = ΔH - TΔS.

What is the formal charge on the central oxygen atom of ozone, O3? Question 2 Answer Choices A. -2 B. +1 C. -1 D. 0

B. The formal charge on the central oxygen atom of ozone, O3 is +1. The Lewis dot structure of ozone is: Using the formal charge formula, formal charge = valence - ½(bonding electrons) - lone electrons, the central oxygen formal charge = 6 - ½(6) - 2 = +1.

How many electrons are shared in the bond between sodium and chlorine in a molecule of NaCl? Question 13 Answer Choices A. 1 B. 0 C. 2 D. 3

B. There are 0 electrons shared in the bond between sodium and chlorine in a molecule of NaCl. NaCl is an ionic compound, and no electrons are shared in an ionic bond.

How many σ bonds and π bonds are there in ethene? Question 19 Answer Choices A. 4 σ and 1 π B. 5 σ and 1 π C. 1 σ and 1 π D. 1 σ and 4 π

B. There are 5 σ and 1 π bonds in ethene. Each C—H bond is a σ bond (there are four). There is also one C—C σ bond and a C—C π bond, for a total of five σ and one π.

Which of the following describes the orbital geometry of an sp2 hybridized atom? Question 14 Answer Choices A. Tetrahedral B. Trigonal planar C. Bent D. Linear

B. Trigonal planar describes the orbital geometry of an sp2 hybridized atom. Remember the following correspondences between the hybridization of the central atom and the molecular geometry: sp - linear, sp2 - trigonal planar, sp3 - tetrahedral. A bent geometry is generally associated with an sp3 hybridized central atom that has two lone pairs of electrons.

As you move from left to right across the periodic table, the atomic radii of elements in the same period: Question 7 Answer Choices A. decrease because the amount of shielding increases. B. decrease because effective nuclear charge increases. C. increase because the amount of shielding decreases. D. increase because of increased electron-electron repulsion.

B. As you move left to right across a period, the elements have more protons, giving the atoms increasing effective nuclear charges after the shielding of the inner core electrons are factored in. This change in effective nuclear charge due to proton number plays a more significant role across a period than changes in shielding (choices A and C are wrong). The increased effective nuclear charge means that the valence electrons feel a stronger pull to the nucleus, thus the radius decreases from left to right. Any electron-electron repulsion is more than counteracted by the increased effective nuclear charge, making B a better answer than D.

Which of the following correctly identifies the hybridization of the C, N-1, and N-2, respectively, in diazomethane above? A. sp 3, sp, sp3 B. sp 2, sp, sp2 C. sp 2, sp2, sp D. sp 3, sp3, sp2

B. Hybridization is determined by the total number of electron groups around a given atom. Each single, double, or triple bond or lone electron pair counts equally as one electron group. The carbon therefore has three electron groups (two single bonds and one double bond), so it needs three hybrid orbitals. Therefore, one s and two p atomic orbitals combine to give sp2 hybrid orbitals (eliminate choices A and D). The central nitrogen has only two electron groups (two double bonds) and should have two sp hybrid orbitals made when one s and one p atomic orbitals combine (eliminate choice C). The terminal nitrogen has three electron groups (one double bond and two lone pairs of electrons), so should have sp2 hybrid orbitals formed from one s and two p atomic orbitals.

Consider the following reaction: CH4 + 2 O2 → CO2 + 2 H2O ΔH° = -860 kJ What is the best explanation for why the reaction is exothermic? Question 4 Answer Choices A. CO2 is less stable than CH2. B. The energy required to break reactant bonds is less than the energy released when product bonds are formed C. More bonds are broken in the reactants than are formed in the products. D. The O=O bond is stronger than the C=O bond, and the C—H bond is stronger than the H—O bond.

B. The energy required to break reactant bonds is less than the energy released when product bonds are formed

Which of the following reactions results in the greatest increase in ΔS? Question 5 Answer Choices A. N2(g) + O2(g) → 2 NO(g) B. CO2(s) → CO2(g) C. 2 C3H8O(l) + 9 O2(g) → 6 CO2(g) + 8 H2O(l) D. NH4NO3(s) → NH4NO3(aq)

B. The entropy of a system can increase with increasing number of moles in a product or a more disordered state in the product. Choice A shows no change in the number of moles of gases so can be eliminated. Choice C decreases from nine to six moles of gas, so is also incorrect. Of the three main phases discussed in the MCAT (solids, liquids, and gases), solids possess the least disorder, while gases possess the most. This means that the sublimation of carbon dioxide (choice B) shows a great increase in disorder. While choice D does show the process of dissolving a solid in water (a clear increase in disorder), the change is not nearly as large as what is observed in the process of sublimation, making choice B better than choice D.

A spontaneous reaction can be made non-spontaneous if: Question 13 Answer Choices A. ΔH > 0, ΔS > 0, and the temperature is raised. B. ΔH < 0, ΔS < 0, and the temperature is raised. C. ΔH > 0, ΔS < 0, and the temperature is lowered. D. ΔH < 0, ΔS > 0, and the temperature is lowered.

B. ΔG is the indicator of spontaneity. If ΔG is negative, the reaction is spontaneous, while if it is positive, the reaction is non-spontaneous. The equation that describes this is: ΔG = ΔH - TΔS A negative ΔH and positive ΔS will always be spontaneous at any temperature, while a positive ΔH and negative ΔS will always be non-spontaneous at any temperature (eliminate choices C and D). A positive ΔH and positive ΔS will be non-spontaneous at low temperatures, and will become spontaneous as temperature increases (i.e. increasing the contribution of the entropy term), so choice A can be eliminated. A negative ΔH and negative ΔS will be spontaneous at low temperatures, and will become non-spontaneous as temperature increases, thus making choice B the best answer.

After studying the results of a freezing point depression experiment, a chemist determined that the molecule Co(Cl4)SO4 dissociates into a complex cation and sulfate when dissolved in water. Based on this information, what is the most likely geometry of the complex cation? Question 16 Answer Choices A. Trigonal bipyramidal B. Trigonal pyramidal C. Tetrahedral D. Octahedral

C. After studying the results of a freezing point depression experiment, a chemist determined that the molecule CoCl4SO4 dissociates into a complex cation and sulfate when dissolved in water. Based on this information, the most likely geometry of the complex cation is tetrahedral. In the complex cation [CoCl4]2+, cobalt possesses four ligands and no lone pairs (four total electron clouds), indicating it should have a tetrahedral geometry.

At 298 K, ?H° = 436 kJ and ?S° = 100 J/K for the reaction H2(g) 2 H(g). Therefore, this reaction: Question 11 Answer Choices A. is at equilibrium at 298 K. B. is spontaneous at 298 K. C. is not spontaneous at 298 K but will be spontaneous at high enough temperatures. Correct Answer (Blank) D. will never be spontaneous, regardless of the temperature.

C. At 298 K, ΔH° = 436 kJ and ΔS° = 100 J/K for the reaction H2(g) 2 H(g). Therefore, this reaction is not spontaneous at 298 K but will be spontaneous at high enough temperatures.

At STP, a certain liquid will spontaneously vaporize, even though the reaction is endothermic. The most likely explanation for this is that: Question 1 Answer Choices A. change in enthalpy has no effect on reaction spontaneity. B. temperature has no effect on reaction spontaneity. C. this reaction results in an increase in entropy. D. endothermic reactions are always spontaneous.

C. At STP, a certain liquid will spontaneously vaporize, even though the reaction is endothermic. The most likely explanation for this is that this reaction results in an increase in entropy. For a reaction to be spontaneous, the Gibbs free energy (ΔG = ΔH - TΔS) must be negative. If the spontaneous reaction is endothermic (i.e, if ΔH is positive), then ΔS must be positive.

In which of the following does hydrogen have a partial negative charge? Question 3 Answer Choices A. H2O B. NH3 C. BH3 D. CH4

C. Hydrogen has a partial negative charge in BH3. The hydrogen has a partial negative charge in BH3 because hydrogen is slightly more electronegative than boron. This can be deduced from periodic trends. The hydrogen in water (H2O) is bonded to oxygen which is the most electronegative of the second period atoms shown in the four choices (B, C, N, O), since electronegativity increases going from left to right across a period. Because oxygen is considerably more electronegative than hydrogen, the hydrogen bears a partial positive charge in H2O, eliminating this choice. Therefore, one can deduce that of the remaining three choices, electronegativity trends would predict that the hydrogens in ammonia (NH3) also bear a partial positive charge, while those in methane (CH4) have essentially no charge as C and H have very similar electronegativities. This leaves BH3 as the only logical answer.

If H denotes enthalpy and S denotes entropy, then a chemical system will always react spontaneously when: Question 10 Answer Choices A. ΔH > 0 and ΔS < 0. B. ΔH < 0 and ΔS < 0. C. ΔH < 0 and ΔS > 0. D. ΔH > 0 and ΔS > 0.

C. If H denotes enthalpy and S denotes entropy, then a chemical system will always react spontaneously when ΔH < 0 and ΔS > 0. Gibbs free energy is ΔG = ΔH - TΔS. A reaction is spontaneous when ΔG is negative, and ΔG is always negative when ΔH is negative and ΔS is positive.

For which of the following materials will hydrogen bonding among its constitutive molecules be important? I. CHCl3 II. CH3CH2OH III. H3C—O—CH3 Question 15 Answer Choices A. III only B. II and III only C. II only D. I and II only

C. In material II only will hydrogen bonding among its constitutive molecules be important. I. CHCl3 II. CH3CH2OH III. H3C—O—CH3 Since Molecules I and III do not possess hydrogen-bond donors (O—H, N—H, or F—H bonds), they will not exhibit hydrogen bonding among themselves. Since Molecule II (ethanol) can act as both a hydrogen-bond donor (using the OH proton) and as a hydrogen-bond acceptor (using the oxygen nonbonding electrons), hydrogen bonding will be important among these molecules.

Which of the following orbitals is NOT found in the carbon-carbon bonds of propene? Question 23 Answer Choices A. Pure p orbitals B. sp3 hybrid orbitals C. Pure s orbitals D. sp2 hybrid orbitals

C. Pure s orbitals are NOT found in the carbon-carbon bonds of propene. There are no pure s orbitals in the carbon atoms of propene:

The BeCl2 molecule is: Question 6 Answer Choices A. linear and polar. B. bent and nonpolar. C. linear and nonpolar. D. bent and polar.

C. The BeCl2 molecule is linear and nonpolar. In this case, beryllium breaks the octet rule. Be has two valence electrons to share. Each of these valance electrons will form a bond with a chloride. From VESPR it is clear that a molecule with only two bonds will form a linear molecule, with the structure Cl-Be-Cl. Given the electronegativity difference, each bond is polar. However, the two polar bonds are exactly opposite in a symmetrical molecule. Thus, they cancel with one another and result in a nonpolar molecule.

Which of the following molecules has a dipole moment?

C. The molecule CHCl3 has a dipole moment. All C—C and C—H bonds are nonpolar, so all hydrocarbons are nonpolar molecules. Chloroform has a net dipole since the C—Cl bonds are polar and are not oriented symmetrically around the central atom.

ow many hybrid orbitals does each fluorine atom in XeF4 have? Question 8 Answer Choices A. Five B. Six C. Four D. Seven

C. There are four hybrid orbitals on each fluorine atom in XeF4. Each fluorine atom has three lone pairs and a single bond, giving a total of four electron groups. These require four orbitals: one s and three p to make four sp3 orbitals.

When a sample of solid NH4NO3 is dissolved in water, the reaction flask becomes cold to the touch. What can be concluded about the given thermodynamic values for the solvation process? Question 3 Answer Choices A. ΔG < 0, ΔH < 0, ΔS > 0 B. ΔG > 0, ΔH < 0, ΔS > 0 C. ΔG < 0, ΔH > 0, ΔS > 0 Correct Answer (Blank) D. ΔG < 0, ΔH > 0, ΔS < 0

C. When a sample of solid N4NO3 is dissolved in water, the reaction flask becomes cold to the touch. In can be concluded that ΔG < 0, ΔH > 0, ΔS > 0 for the solvation process. Since the solvation process is observed to occur, it is spontaneous, so ΔG < 0. If the reaction flask is cold to the touch, then the enthalpy change of solvation is positive (ΔH > 0) as this is an endothermic process. Finally, dissolving a solid in a liquid tremendously increases the disorder of the two components as they mix, so ΔS > 0.

Determine ΔH for the reaction CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) given the following information: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) ΔH = -802 kJ/mol 2 H2O(g) 2 H2O(l) ΔH = -88 kJ/mol Question 13 Answer Choices A. 714 kJ/mol B. -714 kJ/mol C. -890 kJ/mol D. 890 kJ/mol

C. -890 kJ/mol is ΔH for the reaction CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) given the following information: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) ΔH = -802 kJ/mol 2 H2O(g) 2 H2O(l) ΔH = -88 kJ/mol The overall reaction is just the sum of the two individual reactions. Therefore the overall ΔHrxn is (-802 kJ/mol) + (-88 kJ/mol) = -890 kJ/mol.

Why does the number of electrons not play a role in determining the mass number of an atom? Question 2 Answer Choices A. The definition of mass number was created before the discovery of the electron. B. Negative charges interfere with any measurement of mass number. C. Electrons have a significantly lower mass compared to protons and neutrons D. The movement of electrons outside the nucleus interferes with any measurement of mass number.

C. Mass number is defined as the number of protons and neutrons in a nucleus. It is not a measured value, so choices B and D can be eliminated. Electrons are excluded from the determination of mass number since they are much less massive (1/1836 AMU) than a proton or neutron (both approximately 1 AMU), making choice C the best.

Which of the following will release the most energy with the addition of an electron? Question 18 Answer Choices A. Al B. Ga C. Si D. Ge

C. The energy released when an atom gains an electron in its valence shell is its electron affinity. The electron affinity periodic trend becomes more negative as you move from the bottom left to the top right of the periodic table. As the atomic radius of an atom decreases, an electron added to an atom has farther to fall to reach the valence shell closer to the nucleus, thereby giving off more energy. Since silicon is the smallest atom of the options given, it has the most negative electron affinity, indicating the greatest gain in stability.

Which of the following represents the ground state electron configuration of a Co(III) ion? Question 19 Answer Choices A. [Ar] 4s2 3d4 B. [Ar] 4s2 3d10 C. [Ar] 3d6 D. [Ar] 4s1 3d5

C. The ground state electron configuration of a neutral cobalt atom is [Ar] 4s2 3d7. Co(III) is a cation as metals do not gain electrons. Therefore, the electron configuration should have three fewer electrons than a neutral cobalt atom (eliminate choice B). Since the valence electrons of cobalt are in the fourth energy level, these are the first electrons lost (eliminate choice A). While choice D might look appealing since it has a half-filled d subshell, this type of electron configuration is only seen for neutral atoms in the chromium and copper families, so choice D can be eliminated. Both 4s valence electrons will be lost in addition to the then 3d valence electron, making choice C the best answer.

Which of the following pairs will have the strongest intermolecular forces between them? Question 20 Answer Choices A. O2 and N2 B. HCl and CHCl3 C. NaCl and NH3 D. H2O and C8H18

C. The strongest intermolecular forces occur between molecules with a greater magnitude of charge. Na+ and Cl- ions have full +1 and -1 charges, respectively. Polar compounds will have larger partial charges due to differences in electronegativity between the bonded atoms, while nonpolar compounds have little to no partial charge. Therefore, the ions in NaCl and the polar NH3 will form the strongest ion-dipole interactions. All other interactions will be much weaker, including the weakest instantaneous-induced dipole interactions between the nonpolar oxygen and nitrogen (choice A), dipole-dipole forces between the polar HCl and chloroform (choice B), and dipole-induced dipole interactions between the polar water and nonpolar octane (choice D).

What are the formal charges of N-1, N-2, and N-3, respectively, in the azide ion (N3-) based on the structure given? Question 23 Answer Choices A. -2, -1, +2 B. -1, +2, 0 C. -2, +1, 0 D. -1, +1, -1

C. The sum of all formal charges in a molecule is equal to the overall charge on the atom or ion, in this case -1 (eliminate choice B). Formal charge is calculated by subtracting the sum of all electrons in lone pairs plus half the bonding electrons around an atom from the number of valence electrons that element has. Since nitrogen has five valence electrons, the formal charge for each N atom can be determined as follows: N-1: [5 valence - (6 lone pair e-s + 1 bonding e-)] = -2 N-2: [5 valence - 4 bonding e-s] = +1 N-3: [5 valence - (2 lone pair e-s + 3 bonding e-s)] = 0

Given the following reactions, what is the reaction enthalpy for 3 C(graphite) + 4 H2(g) → C3H8(g)? C3H8(g) + 2 O2 → 3 C(graphite) + 4 H2O(l) ΔH° = -1036 kJ/mol H2(g) + ½ O2(g) → H2O(l) ΔH° = -285 kJ/mol Question 17 Answer Choices A. -1321 kJ/mol B. -751 kJ/mol C. -104 kJ/mol D. 75 kJ/mol

C. This question can be solved utilizing Hess's Law, which allows for determination of the reaction enthalpy by summing the enthalpies of a series of other reactions. In order for the given reactions to be added to give the desired equation, we must reverse the first reaction and multiply the second reaction by four. This results in ΔH° values of 1036 kJ/mol and -1140 kJ/mol, respectively. These can then be summed to give the reaction enthalpy of -104 kJ/mol (choice C is correct).

Which of the following compounds would exhibit the strongest intermolecular forces? Question 4 Answer Choices A. CO2 B. CHCl3 C. CH3CO2H D. C6H14

C. This question tests your ability to assign the intermolecular forces for each of these compounds. CH3CO2H (acetic acid) possesses both a hydrogen bound to an oxygen and unbound electrons on another oxygen allowing it to form hydrogen bonds. Since hydrogen bonds are the strongest type of intermolecular force, this is the correct answer. CO2 is capable of only weak London dispersion forces given its linear geometry. C6H14 (hexane) is a nonpolar molecule, making it capable of only London dispersion forces as well (choices A and D are incorrect). CHCl3 can form dipole-dipole interactions which are stronger than dispersion forces, but not as strong as hydrogen bonds (eliminate choice B).

Alkenes can form s bonds with metal centers through donation of p electrons to empty orbitals on the metal, usually drawn as shown below for a generic metal (M) bound to n generic ligands (L). Compared to the C=C bond length in an unbound alkene, the C=C bond length in such a metallated alkene will be: Question 4 Answer Choices A. unaffected. B. longer or shorter depending on the metal. C. shorter. D. longer.

D. Alkenes can form σ bonds with metal centers through donation of π electrons to empty orbitals on the metal, usually drawn as shown below for a generic metal (M) bound to n generic ligands (L). Compared to the C=C bond length in an unbound alkene, the C=C bond length in such a metallated alkene will be longer. The fact that C=C bonds are shorter than C—C bonds is due to the shared electrons in the π bond. If these electrons participate in a bonding interaction with the metal, they are less centralized between the two carbons. Therefore, the bond between the carbons will have less double bond character, and be longer. Any metal which accepts such a bond will cause its lengthening.

Which of the following reactions has the most positive ∆Srxn? Question 5 Answer Choices A. C(diamond) C(graphite) B. 6 H2O(g) 6 H2O(l) C. 6 CO2(g) + 6 H2O(g) C6H12O6(s) D. CO2(s) CO2(g)

D. CO2(s) CO2(g) has the most positive ∆Srxn of all the equations shown. A positive ΔSrxn means that the entropy of the products is greater than that of the reactants, i.e., disorder increases. Keeping in mind that gases typically have much more entropy than liquids or solids, choices with gases as products where reactants are either liquids or solids are best. This eliminates the formation of solid sugar and liquid water. While diamond has less entropy than graphite, both substances are still in the solid phase. As a result, this entropy increase pales in comparison to the process of converting solid carbon dioxide to carbon dioxide gas.

Which of the following molecules contains the most polar covalent bond? Question 4 Answer Choices A. CH4 B. H2O C. NaCl D. HF

D. HF contains the most polar covalent bond. First, "NaCl" can be eliminated since NaCl contains an ionic, not a covalent bond. Bond polarity depends on the electronegativity difference between the two atoms in the bond. Choices "HF," "CH4" and "H2O" all involve a bond to H, which itself has a relatively low electronegativity. Thus, the most polar bond is the bond between H and an atom with a high electronegativity. "HF" is the best choice because F is more electronegative than O or C, since it is higher and further to the right on the periodic table (or remember the mnemonic FONCl BrISCH when going from highest electronegativity to lowest).

Which of the following compounds displays the lowest vapor pressure at 15˚C? Question 10 Answer Choices A. NH3 B. CO2 C. SO2 D. HF

D. HF displays the lowest vapor pressure at 15°C. The compound with the least vapor pressure has the strongest intermolecular forces. Both NH3 and HF can form hydrogen bonds, the strongest type of intermolecular force, so one of these compounds is the most likely correct answer. The polarity of the H?F bond is much greater than that of an H?N bond and HF has a greater molecular weight, both of which result in HF having the lowest vapor pressure of the compounds listed. Carbon dioxide is nonpolar, so it would only have weak London dispersion forces making it a gas at 15°C (CO2 is wrong). Sulfur dioxide is a polar compound and would experience both dipole-dipole interactions and moderate London dispersion forces as it has a relatively large molecular weight. These forces are weaker than the hydrogen bonds in HF, so SO2 is wrong. While this question asks for the lowest vapor pressure rather than to rank all four compounds based on this property, it's good to note that the boiling point of SO2 is higher than the boiling point of NH3 due to the larger molecular weight of SO2.

Rank these three molecules in order of increasing dipole moment: Question 18 Answer Choices A. I < II < III B. II < III < I C. I < III < II D. II < I < III

D. II < I < III is the ranking of these three molecules in order of increasing dipole moment. A dipole moment is the vector sum of the individual bond dipoles in a molecule. The dominating bond dipoles for all of these molecules are the C—O bonds. In Molecule II, the two nearly opposing C—O bonds of the ether nearly cancel each other, resulting in a very small dipole moment. In Molecule I, the C—O and O—H dipoles are also opposed, but of different magnitudes, resulting in a moderate dipole moment. In Molecule III, the C=O double bond, without an opposing bond dipole, dominates the large dipole moment.

N2(g) + 3 H2(g) → 2 NH3(g), ΔH = -22 kcal/mol If ΔG is negative, then the reaction as written is: Question 17 Answer Choices A. spontaneous and endothermic. B. spontaneous with an increase in entropy. C. nonspontaneous due to the decrease in entropy. D. spontaneous and exothermic.

D. If ΔG is negative, then the reaction as written is spontaneous and exothermic. If ΔG is negative, then the reaction is spontaneous. Furthermore, since ΔH is negative, the reaction is exothermic also.

How many of which type of orbitals are used to create the hybrid orbitals on carbon in a molecule of ethylene (C2H4)? Question 6 Answer Choices A. One s and one p orbital B. Two s and two p orbitals C. One s and three p orbitals D. One s and two p orbitals

D. One s and two p orbitals will hybridize to form the bonding structure in ethylene (C2H4). The choice of two s and two p orbitals can be immediately eliminated because there is only ever one s orbital in a given s subshell. The easiest way to answer this question is to look at the areas of electron density surrounding the carbon atoms in ethylene. Since there are three electron groups around each trigonal planar carbon atom, a total of three hybrid orbitals (sp2) are needed. Therefore one s and two p atomic orbitals are required to form these hybrids. Four areas of electron density correspond to a tetrahedral geometry and sp3 hybridization (one s and three p), while two areas of electron density correspond to a linear geometry and sp hybridization (one s and one p).

Perspiration is important in maintaining normal body temperature. Compared to water at its boiling point, which of the following is true about water at normal human body temperature? Question 7 Answer Choices A. Intermolecular forces are weaker B. Average kinetic energy is greater C. More energy is required for gas expansion D. The heat required for vaporization is higher

D. Perspiration is important in maintaining normal body temperature. Compared to water at its boiling point, it is true the heat required for vaporization of water is higher at normal human body temperature. As water on the skin vaporizes it absorbs energy and cools the body. Temperature is a measure of average kinetic energy, so water has less average kinetic energy at 37°C compared to 100°C. As temperature increases, water requires more and more energy to expand the volume of its gaseous phase. Although the elimination of "more energy is required for gas expansion" might not be obvious, "the heat required for vaporization is higher" is a better answer. Water is much more likely to vaporize at its boiling point than at body temperature. It takes additional energy to go from the liquid to gas phase at lower temperatures. Therefore the heat required to vaporize is higher when water is at lower temperatures.

Sodium dodecyl sulfate, also known as SDS, is a common denaturation agent used while separating proteins via gel electrophoresis. It is composed of a highly charged head and a long hydrocarbon tail. Given application of SDS to a protein increases its solubility in water, what forces are responsible for association of SDS with the protein of interest? Question 9 Answer Choices A. Dipole-dipole forces B. Hydrogen bonding C. Ion-dipole forces D. London dispersion forces

D. Sodium dodecyl sulfate, also known as SDS, associates with proteins via London dispersion forces. Given the association of SDS with the protein increases its solubility in water, the highly charged head is likely to interact with water surrounding the protein while the hydrophobic tail interacts with the protein itself. Of the intermolecular forces listed, London dispersion forces predominate in nonpolar molecules and are responsible for the association of SDS with the protein (London dispersion forces is correct). The other types of forces require the attraction of full or partial charges on the associating molecules. If the charged heads of SDS were associated with the protein, the nonpolar tails would be forced to interact with the aqueous environment and would decrease the solubility of the protein instead.

Which of the following must result in a negative free energy change for a reaction? Question 14 Answer Choices A. The enthalpy change is negative, and the entropy change is negative. B. The entropy change is positive. C. The enthalpy change is negative. D. The enthalpy change is negative, and the entropy change is positive.

D. The enthalpy change is negative, and the entropy change is positive must result in a negative free energy change for a reaction. Since ΔG = ΔH — TΔS and T is always positive, a negative ΔH and a positive ΔS always gives a negative ΔG.

The vapor pressure of Liquid I is greater than the vapor pressure of Liquid II. Which of the following is true about Liquid I? Question 13 Answer Choices A. It has stronger intermolecular forces than does Liquid II. B. It has a higher heat of fusion than does Liquid II. C. It has a higher heat of vaporization than does Liquid II. D. It boils at a lower temperature than does Liquid II.

D. The following is true about Liquid I: "it boils at a lower temperature than does Liquid II." Boiling occurs when the vapor pressure of a liquid equals the atmospheric pressure. Since vapor pressure always increases as the temperature increases, Liquid I will boil at a lower temperature than Liquid II, because Liquid I's vapor pressure is closer to the atmospheric pressure. Stronger intermolecular forces means the molecules stick together more, making them less likely to enter the vapor phase. The heat of fusion is irrelevant to this question. We would expect a higher heat of vaporization for a compound that is harder to evaporate, or one with a low vapor pressure.

Identify the hybridizations of the two indicated carbon atoms in the following molecule. A. C-1 = sp2 and C-2 = sp B. C-1 = sp3 and C-2 = sp3 C. C-1 = sp and C-2 = sp D. C-1 = sp and C-2 = sp

D. The hybridizations of the two indicated carbon atoms in the following molecule is C-1 = sp and C-2 = sp2. Carbon 1 has two electron groups (a single bond and a triple bond), and is therefore sp hybridized. Carbon 2 has three electron groups (two single bonds and a double bond), and is therefore sp2 hybridized.

Which mixture of molecules will have the strongest interactions? Question 8 Answer Choices A. F2/Cl2 B. Cl2/NH3 C. CH3I/H2O D. NH3/H2O

D. The mixture of molecules that will have the strongest interactions is NH3/H2O. The strategy for answering this question is to identify the intermolecular forces of each component in the mixture and the intermolecular forces between the components. Remember that the order from the strongest to weakest intermolecular force is: hydrogen bonding > dipole-dipole forces > London dispersion forces. Evaluating each combination, CH3I has dipole-dipole forces while H2O exhibits hydrogen bonding; the two associate via dipole-dipole forces. Cl2 has London dispersion forces while NH3 exhibits hydrogen bonding; the two interact with London dispersion forces. Both F2 and Cl2 exhibit only London dispersion forces (both individually and with each other). However, both NH3/H2O undergo hydrogen bonding (both individually and with each other), and thus would be the mixture with the strongest interaction.

The structure below is shown without complete geometrical detail. What is the correct assignment of the two indicated bond angles? Question 21 Answer Choices A. a = 90°, b = 120° B. a = 120°, b = 109.5° C. a = 120°, b = 120° D. a = 109.5°, b = 120°

D. The structure below is shown without complete geometrical detail. The correct assignment of the two indicated bond angles is a = 109.5°, b = 120°. The nitrogen, having three σ bonds and one nonbonded electron pair, is sp3 hybridized and should therefore display 109.5° between its neighboring atoms, so "a = 109.5°, b = 120°" must be the answer. The benzene carbon atom, having three σ bonds and one π bond, is sp2 hybridized and should display 120° between its neighboring atoms.

Given the same number of moles, which of the following solids will have the lowest melting point? Question 10 Answer Choices A. SiO2 B. Cu C. NaI D. H2Se

D. Hydrogen selenide is a molecular solid. Its intermolecular forces consist of dipole-dipole interactions. Though a relatively strong intermolecular force, they are much weaker than the covalent bonds between the atoms in a network solid (choice A), the covalent bonds between the atoms in a metallic solid (choice B), and the ionic forces between the ions in an ionic solid (choice C). Thus, the weaker intermolecular forces will allow hydrogen selenide to melt more easily than the others.

Which of the following phase changes involves the greatest increase in entropy? Question 5 Answer Choices A. Vaporization B. Freezing C. Deposition D. Sublimation

D. Sublimation

Which of the following species has/have a trigonal planar shape? PF3 CO32- BCl3 Question 15 Answer Choices A. I only B. III only C. I and III only D. II and III only

D. The ?shape of a molecule is determined by the total number of electron groups around the central atom, which determines the angles between the groups, as well as the number of atoms bonded to the central atom. PF3 has four electron groups around the central phosphorus (three single bonds to fluorine and a lone electron pair). It therefore has a trigonal pyramidal shape (eliminate choices A and C). Based on the remaining choices, BCl3 must be trigonal planar; it has three electron groups surrounding the central boron (three single bonds to chlorine). The carbonate ion (CO32-) has three electron groups around the central carbon (two single bonds to oxygen and one double bond to oxygen), making it trigonal planar as well (eliminate choice B).

In electrochemistry, which of the following best demonstrates the first law of thermodynamics? Question 22 Answer Choices A. A galvanic cell proceeding spontaneously B. The overall increase in entropy observed following a reaction C. The chemical reaction proceeding to a more probable state D. Conversion of equal quantities of chemical energy into electrical energy

D. The first law of thermodynamics states that energy is always conserved, thus the conversion of chemical energy into electrical energy is an example of the first law (choice D is correct). A reaction proceeding spontaneously (resulting in an increase in overall entropy) is more a reflection of the second law of thermodynamics (choices A and B are incorrect). Similarly, a reaction always proceeds to a probable outcome, hence why each reaction proceeds in the spontaneous direction (choice C is incorrect).

Two excited state Bohr atoms experience electron transitions, the first atom E3 to E2, and the second atom from E2 to E1. The electron in the first atom loses: Question 25 Answer Choices A. a larger energy, thus emitting a photon with a longer wavelength. B. a larger energy, thus emitting a photon with a shorter wavelength. C. a smaller energy, thus emitting a photon with a shorter wavelength. D. a smaller energy, thus emitting a photon with a longer wavelength.

D. There is an inverse relationship between energy and wavelength. Therefore, a larger energy loss is associated with a photon with a shorter wavelength and vice versa (eliminate choices A and C). The energy differences between energy levels in a Bohr atom become smaller as you increase in energy level. Thus, the E3 to E2 transition is a smaller energy transition compared to the E2 to E1 transition, making choice D correct and choice B incorrect.

What are the hybridizations of Na, Nb, and Nc in roquefortine C, respectively? Question 28 Answer Choices A. sp 2, sp2, sp3 B. sp 2, sp3, sp3 C. sp 3, sp2, sp2 D. sp 2, sp2, sp2

D. To find the hybridization of an atom, count the number of electron groups (single, double, triple bonds or lone pairs of electrons) around the atom. Four e- groups issp3-hybridized, three e- groups is sp2-hybridized, and two e- groups is sp-hybridized. The exception applies to atoms that have lone pairs that are delocalized by resonance. Those atoms are sp2-hybridized even though at first glance they may appear sp3-hybridized. Allof the indicated nitrogens are sp2-hybridized (choice D is correct and A, B, and C are incorrect). Na has three e- groups: lone pair, single bond, double bond. Nb and Nc follow the exception. Each has four e- groups, the lone pair and three single bonds. However, those lone pairs can be delocalized so the atoms are sp2-hybridized.

What are the formal charges of the oxygen atoms in formate? Question 9 Answer Choices A. Oa = 0, Ob = +1 B. Oa = 0, Ob = -1 C. Oa = +1, Ob = 0 D. Oa = -1, Ob = 0

Oa has a formal charge of 0, while Ob has a formal charge of -1 in formate. The formula for formal charge is FC = V - (0.5)B - L Where FC = Formal Charge V = # of valence electrons B = # of bonding electrons L = # of lone pair electrons Oxygen has six valence electrons. Oa has four bonding electrons and two lone pairs of electrons (for a total of 4), so 6 - (0.5)(4) - 4 = 0. Ob has two bonding electrons and three lone pairs of electrons (for a total of 6), so 6 - (0.5)(2) - 6 = -1.