Atomic structure and periodic trends

2 Ag2O 4 Ag + O2 If one mole of Ag is produced in the reaction above, how many grams of oxygen gas are produced? Question 11 Answer Choices A. 16 g B. 2 g C. 4 g D. 8 g

2 Ag2O 4 Ag + O2 If one mole of Ag is produced in the reaction above, 8 g of oxygen gas are produced. According to the given balanced equation, when 4 moles of Ag are produced, 1 mole of O2 is produced. Therefore, if 1 mole of Ag is produced, then 1/4 mole of O2 will be produced. The mass of 1/4 mole of O2 is (1/4)(2)(16 g) = 8 g.

A sample of oxygen contains 2.5 moles of O2 molecules. What is the mass of the sample? Question 6 Answer Choices A. 80 g B. 40 g C. 20 g D. 60 g

A. A sample of oxygen contains 2.5 moles of O2 molecules. The mass of the sample is 80 g. One mole of O2 has a mass of 2(16 g) = 32 g, so 2.5 moles will have a mass of 2.5(32 g) = 80 g.

Compared to carbon, nitrogen has a: Question 12 Answer Choices A. greater ionization energy and more negative electron affinity. B. smaller ionization energy and more positive electron affinity. C. smaller ionization energy and more negative electron affinity. D. greater ionization energy and more positive electron affinity.

A. Compared to carbon, nitrogen has a greater ionization energy and more negative electron affinity. This is a 2 x 2 question and is best approached by evaluating one of the arguments at a time to eliminate answer choices. Carbon and nitrogen are in the same period, and nitrogen is to the right of carbon. Ionization energy increases going left to right in a period due to greater effective nuclear charge, resulting in valence electrons being held more tightly. Thus, nitrogen will have a greater ionization energy, eliminating two answer choices. Evaluating the other argument, electron affinity becomes more negative (i.e., more favorable) going left to right in a period (once again due to the greater effective nuclear charge), making nitrogen more negative compared to carbon. Thus compared to carbon, nitrogen has a greater ionization energy and a more negative electron affinity.

Cu2S 2 Cu + S If two moles of Cu2S are consumed in the reaction above, how many moles of solid copper are produced? Question 10 Answer Choices A. 4 B. 2 C. 8 D. 1

A. Cu2S 2 Cu + S If two moles of Cu2S are consumed in the reaction above, 4 moles of solid copper are produced. If each mole of Cu2S produces 2 moles of Cu, then two moles of Cu2S will produce 4 moles of Cu.

In organic solution, F- deprotonates dissolved HCl. Which of the following explains this observation? Question 10 Answer Choices A. F- has a smaller radius than Cl-. B. F- is a stronger acid than Cl-. C. F- has a greater electronegativity than Cl-. D. F- has a lower ionization energy than Cl-.

A. Since F- has a smaller radius than does Cl-, it is less stable with excess charge, and therefore a better Bronsted base. It outcompetes Cl- for the proton.

What is the molar concentration of Ca2+(aq) in a solution that is prepared by mixing 15 mL of a 0.02 M CaCl2(aq) solution with 10 mL of a 0.04 M CaSO4(aq) solution? Question 16 Answer Choices A. 0.028 M B. 0.035 M C. 0.014 M D. 0.020 M

A. The molar concentration of Ca2+(aq) in a solution that is prepared by mixing 15 mL of a 0.02 M CaCl2(aq) solution with 10 mL of a 0.04 M CaSO4(aq) solution is 0.028 M. For dilution problems, like this one, the concentration (C) times the volume (V) of all the solutions before mixing must equal the concentration times the volume of the final solution: (CV)Soln 1 + (CV)Soln 2 = (CV)Final soln (0.02 M)(15 mL) + (0.04 M)(10 mL) = C(25 mL) 0.028 M = C

The outermost shell of electrons in the noble gases (with the exception of helium) have which of the following electron configurations? Question 18 Answer Choices A. ns2np6 B. ns2np4 C. ns2np2 D. ns2np8

A. The outermost shell of electrons in the noble gases (with the exception of helium) have an ns2np6 electron configuration. Noble-gas configurations are characterized by filled valence s and p subshells. Since the s holds 2 electrons and the p holds 6, the answer is ns2np6.

The reactivity of alkali metals follows the trend Cs > Rb > K > Na > Li. Which of the following is the best explanation for this trend? Question 9 Answer Choices A. The ionization energy increases from Cs to Li. B. The electron affinity increases from Li to Cs. C. The electronegativity decreases from Cs to Li. D. The atomic size decreases from Li to Cs.

A. The reactivity of alkali metals follows the trend Cs > Rb > K > Na > Li. The best explanation for this trend is that the ionization energy increases from Cs to Li. Cesium has the lowest ionization potential of the group and therefore it can give up an electron more easily than the other metals. For this reason, it is the most reactive of the group.

When heated, tungsten(VI) oxide, WO3, reacts with hydrogen gas to produce metallic tungsten and water vapor. Assuming that the reaction goes to completion, how many moles of solid tungsten are produced for every mole of hydrogen gas consumed? Question 5 Answer Choices A. 1/3 B. 2 C. 2/3 D. 3

A. When heated, tungsten(VI) oxide, WO3, reacts with hydrogen gas to produce metallic tungsten and water vapor. Assuming that the reaction goes to completion, 1/3 mole of solid tungsten are produced for every mole of hydrogen gas consumed. Since the balanced equation for the reaction is WO3 + 3 H2 W + 3 H2O, the ratio of W produced to H2 consumed is 1/3.

Which of the following is the correct electron configuration for the iron ions present in rust, Fe2O3? Question 1 Answer Choices A. [Ar] 3d5 B. [Ar] 4s23d3 C. [Ar] 4s23d4 D. [Ar] 3d6

A. [Ar] 3d5 is the correct electron configuration for the iron ions present in rust, Fe2O3. First, determine what the charge on the iron ion should be from the chemical formula for rust. Oxygen will always have a -2 charge in an ionic compound, so in order to make the formula neutral, each of the two iron ions will be Fe3+. The electron configuration for a neutral iron atom is [Ar] 4s23d6, and three electrons must be removed to yield Fe3+. Therefore, choices "[Ar] 4s23d4" and "[Ar] 3d6" can be eliminated since both show only two electrons have been removed. When transition metals lose electrons, they first lose them from the s subshell. The remaining electron that must be removed comes from the 3d subshell, making "[Ar] 3d5" the best answer.

Which of the following atoms/ions has the greatest radius? Question 8 Answer Choices A. Br- B. Kr C. Rb+ D. Sr2+

A. All of these atoms and ions are isoelectronic and therefore have the same amount of electron-electron repulsion. Br- has the fewest number of protons, so its electrons feel the least attractive force, making it largest in size.

The first three electron energy levels of a hydrogen atom are given as: E1 = -13.6 eV E2 = -3.4 eV E2 = -3.4 eV If an electron is in the E2 level, which of the following will occur spontaneously? Question 26 Answer Choices A. The electron transitions to E1, and a photon of 10.2 eV is emitted. B. The electron transitions to E3, and a photon of 1.9 eV is emitted. C. The electron transitions to E1, and a photon of 10.2 eV is absorbed. D. The electron transitions to E3, and a photon of 1.9 eV is absorbed.

A. Hydrogen has only one electron, so if it is in the second energy level it must be in an excited state. The electron will therefore transition from a high energy level to a lower one, in this case E2 to E1, to reach a ground state configuration (eliminate choices B and D). Since the electron is moving closer to the nucleus, a photon will be emitted during the process (eliminate choice C). The difference between the two levels is 10.2 eV, making choice A correct.

Which of the following terms describe the reaction depicted in the diagram above? Question 24 Answer Choices A. Endothermic, because the bonds in the reactants are stronger than the bonds in the products B. Exothermic, because the bonds in the reactants are stronger than the bonds in the products C. Endothermic, because the bonds in the products are stronger than the bonds in the reactants D. Exothermic, because the bonds in the products are stronger than the bonds in the reactants

A. Looking at the diagram, the products are higher in energy than the reactants, indicating that energy was absorbed (i.e., an endothermic reaction). Thus, we can eliminate choices B and D. The ΔH of a reaction is determined by adding how much energy is required to break reactant bonds to the amount of energy released when product bonds are formed. In this situation, the amount needed to break bonds must have been higher, making choice A the best answer.

Given the enthalpies of formation for NO and NO2 are 90 kJ/mol and 33 kJ/mol, respectively, what is the reaction enthalpy for 2 NO(g) + O2(g) ? 2 NO2(g)? Question 6 Answer Choices A. -114 kJ/mol B. -57 kJ/mol C. -24 kJ/mol D. Inadequate information provided to answer the question

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. Thus we find that the reaction enthalpy is (2 × 33 kJ/mol) - (2 × 90 kJ/mol) = -114 kJ/mol (choice A is correct). Note that we define the enthalpy of formation of elements in their standard state as zero.

Which of the following ions will align itself parallel to the poles of a magnetic field? Question 1 Answer Choices A. Ca2+ B. V3+ C. Cr6+ D. Se2-

Any atom or ion that aligns itself parallel to a magnetic field is defined as paramagnetic, and all such species have unpaired electrons in their orbitals. Therefore, eliminate as incorrect any choice that has only completely filled orbital subshells. Ca2+ (Z = 20) has a closed shell noble gas configuration because it is isoelectronic with argon, having 18 e-s (eliminate choice A). V3+ (Z = 23) has 20 e-s, two of which are unpaired in two 3d orbitals since transition metals lose their s electrons first when forming cations (choice B is correct). Both Cr6+ (Z = 24) and Se2- (Z = 34) are also ions with closed shell noble gas configurations with 18 e-s and 36 e-s, respectively. As such, they like Ca2+ will also be diamagnetic with all electrons paired (eliminate choices C and D).

What mass of reactant remains if 27 g of Al is allowed to react with 213 g of Cl2? Question 18 Answer Choices A. 13.5 g of Al B. 107 g of Cl2 C. 6.3 g of Al D. 53.3 g of Cl2

B. 107 g of Cl2 remains if 27 g of Al is allowed to react with 213 g of Cl2, according to the reaction: 2 Al + 3 Cl2 2 AlCl3 27 g of Al is equal to 1 mol Al based on its molar mass from the periodic table. 213 g of Cl2 is equal to: According to the reaction above, 2 moles of Al will react completely with 3 moles of Cl2. As only 1 mole of Al is mixed with 3 moles of Cl2, Al is the limiting reagent. 1 mole of Al will react with only 1.5 moles of Cl2 so: will be left over at the end of the reaction.

4 NH3 + 6 NO 5 N2 + 6 H2O If 85 g of NH3 and 210 g of NO were used in the reaction above, which of the following would be the limiting reagent? Question 17 Answer Choices A. N2 B. NO C. There is no limiting reagent. D. NH3

B. 4 NH3 + 6 NO 5 N2 + 6 H2O If 85 g of NH3 and 210 g of NO were used in the reaction above, NO would be the limiting reagent. 85 g of NH3 is equivalent to: 210 g of NO is equivalent to: According to the reaction, NH3 and NO react in a 4:6 molar ratio. In order for 5 moles of NH3 to react completely, there would have to be: available to react. As there are only 7 moles of NO, NO is the limiting reagent.

Which of the following aqueous mixtures will produce a Cl- concentration of 2 M at 25°C? Question 3 Answer Choices A. 271 g of HgCl2 in 55 mol of H2O B. 53 g of NH4Cl in 22.5 mol of H2O C. 271 g of HgCl2 in 22.5 mol of H2O D. 53 g of NH4Cl in 55 mol of H2O

B. 53 g of NH4Cl in 22.5 mol of H2O has a Cl- concentration of 2 M at 25°C. Pure water has a concentration of 55 M and density of 1 kg/L. 55 moles of water represent 1 L (1 kg) and 22.5 moles of water represent 0.5 L (0.5 kg). HgCl2 demonstrates poor water solubility and will dissociate little. Therefore, 1 mol of HgCl2 placed in either water 1 kg or 0.5 kg of water will not achieve a 2 M Cl- solution, eliminating choices with "HgCl2." NH4Cl is water soluble and has a molecular weight of 53 g/mol. "53 g of NH4Cl in 55 mol of H2O" is a 1 M NH4Cl (1 M Cl-) solution and "53 g of NH4Cl in 22.5 mol of H2O" is a 2 M NH4Cl (2 M Cl-) solution.

Which of the following appropriately ranks the atoms from smallest to largest atomic radius? Question 11 Answer Choices A. Ge < Al < P < C B. C < P < Al < Ge C. C < Al < P < Ge D. Ge < P < Al < C

B. C < P < Al < Ge appropriately ranks the atoms from smallest to largest atomic radius. This is a ranking question, which is best approached by evaluating the extremes. For a given family, atomic radius increases as period increases due to greater shielding, which means that Ge must be larger than C. This eliminates choices Ge < P < Al < C and Ge < Al < P < C. Now, we must evaluate Al vs. P. These two atoms are in the same period. However, atomic radius decreases as you go left to right in the same period due to a greater effective nuclear charge, which means that P is smaller than Al. Thus, the atoms from smallest to largest atomic radius is C < P < Al < Ge.

Why do chlorine atoms form anions more easily than they form cations? Question 13 Answer Choices A. Chlorine has a large positive electron affinity. B. Chlorine can gain one electron in order to complete its outer shell C. Chlorine has a low electronegativity value. D. Chlorine can donate one electron in order to complete its outer shell.

B. Chlorine atoms form anions more easily than they form cations because chlorine can gain one electron in order to complete its outer shell. The other answer choices are all false, and can therefore be eliminated. The answer is "Chlorine can gain one electron in order to complete its outer shell". The quest for closed-shell stability is the primary driving force in chemical processes. When chlorine gains an electron, it completes its octet, so chlorine atoms form anions much more easily than they form cations.

Compared to the electronegativity of iodine, the electronegativity of chlorine is: Question 21 Answer Choices A. less, because chlorine has a smaller nuclear charge which translates into a smaller force exerted on electrons. B. greater, because decreased nuclear shielding allows for a stronger pull on the valence electrons. C. greater, because chlorine has a greater nuclear charge which translates into a greater force exerted on electrons. D. less, because increased nuclear shielding results in a weaker pull on the valence electrons.

B. Compared to the electronegativity of iodine, the electronegativity of chlorine is greater, because decreased nuclear shielding allows for a stronger pull on the valence electrons. The periodic trend for electronegativity predicts that chlorine is more electronegative than iodine, which eliminates "less, because chlorine has a smaller nuclear charge which translates into a smaller force exerted on electrons" and "less, because increased nuclear shielding results in a weaker pull on the valence electrons". Since iodine (Z = 53) has more protons than chlorine (Z = 17), it has a greater nuclear charge, so "greater, because chlorine has a greater nuclear charge which translates into a greater force exerted on electrons" is false.

Consider the unbalanced equation NaHCO3(s) ? Na2CO3(s) + CO2(g) + H2O(g). How many grams of water will be produced by 1 mole of sodium bicarbonate? Question 4 Answer Choices A. 18 B. 9 C. 6 D. 3

B. Consider the unbalanced equation NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g). 9 grams of water will be produced by 1 mole of sodium bicarbonate. The balanced equation is 2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g). Therefore, 1 mole of NaHCO3 will produce ½ mole of H2O, i.e., ½(2 + 16) = 9 grams of H2O.

Which of the following could be the electron configuration of a boron atom in an excited state? Question 15 Answer Choices A. 1s22s22p1 B. 1s22s12p2 C. 1s22s12p1 D. 1s22s22p2

B. The electron configuration of a boron atom in an excited state could be 1s22s12p2. A boron atom contains 5 electrons, so choices "1s22s22p2" and "1s22s12p1" are eliminated immediately, since they account for 6 and 4 electrons, respectively. The configuration in choice "1s22s22p1" is for a boron atom in the ground state, so the answer must be choice "1s22s12p2".

Give the electron configuration of a ground state Zn2+ ion. Question 8 Answer Choices A. [Ar] 4s23d8 B. [Ar] 3d10 C. [Ar] 4s13d9 D. [Ar] 4s23d10

B. The electron configuration of a ground state Zn2+ ion is [Ar] 3d10. First, eliminate choices that involve the wrong number of electrons. Since an atom of zinc contains 30 electrons, a Zn2+ ion contains 28, eliminating [Ar] 4s23d10. Recall that when electrons are removed from a transition metal such as zinc, the electrons always come from the valence s orbital first. Therefore, the electron configuration, [Ar] 3d10 is the correct one.

Rank the following atoms/ions in order of increasing atomic/ionic radius: I. Ne II. Ar III. S2- IV. Cs+ Question 5 Answer Choices A. I < IV < II < III B. I < II < III < IV C. III < II < I < IV D. IV < I < II < III

B. The following atoms/ions in order of increasing atomic/ionic radius is I < II < III < IV. I. Ne II. Ar III. S2- IV. Cs+ For ranking problems, start with the extremes. Cesium ion, despite its positive charge, possesses an additional two electron shells than any of the other answer choices and will be largest (eliminating answer choices "I < IV < II < III" and "IV < I < II < III"). Neon, with its large effective nuclear charge and limited number or shells, is the smallest (choice "I < II < III < IV" is correct). Sulfur and argon possess the same number of electrons, but argon has a greater effective nuclear charge which accounts for its smaller size.

What is the percent mass composition of H2SO4? Question 9 Answer Choices A. 50% oxygen, 48% sulfur, 2% hydrogen B. 65% oxygen, 33% sulfur, 2% hydrogen C. 75% oxygen, 24% sulfur, 1% hydrogen D. 85% oxygen, 14% sulfur, 1% hydrogen

B. The percent mass composition of H2SO4 is 65% oxygen, 33% sulfur, 2% hydrogen. In one mole of sulfuric acid, there are 32.1 g of S and 4(16 g) = 64 g of O. Therefore, the percent mass of O should be twice the percent mass of S, so the answer must be 65% oxygen, 33% sulfur, 2% hydrogen.

What is the total mass of the hydrogen atoms contained in 3 moles of glucose (C6H12O6)? Question 7 Answer Choices A. 12 g B. 36 g C. 3 g D. 48 g

B. The total mass of the hydrogen atoms contained in 3 moles of glucose (C6H12O6) is 36 g. The mass of one mole of hydrogen atoms is 1 g. Glucose contains 12 H atoms, so one mole of glucose contains 12 g of hydrogen. Three moles of glucose would therefore contain 3(12 g) = 36 g of hydrogen.

If a neutral atom loses two electrons, which of the following would be true of the resulting species? Question 18 Answer Choices A. It would be attracted to a positive plate because the resulting species would be negatively charged. B. It would be attracted to a negative plate because the resulting species would be positively charged. C. It would be attracted to neither a positive or negative plate, because the resulting species would be neutral. D. It would be attracted to neither a positive or negative plate, because electrons have a much smaller mass compared to protons.

B. A neutral atom losing two electrons would become positively charged (eliminate choices A and C). Mass has nothing to do with the sign of a charge (eliminate choice D). Opposite charges attract, so a positively charged ion will be attracted to a negative plate, making choice B correct.

Which bond is most polar? Question 2 Answer Choices A. C—Si B. H—Cl C. N—O D. F—F

B. Polarity is determined by the difference in electronegativity between the atoms sharing electrons in a bond. Electronegativity generally increases from the bottom left to the top right corners of the periodic table. Elements that are farther from each other on the periodic table have larger differences in electronegativity, while elements that are closer to each other have smaller differences. Even though fluorine is the most electronegative element, a F—F bond is nonpolar because there is no difference in electronegativity between the two atoms (eliminate choice D). Since the C—Si and N—O pairs of elements are right next to each other on the table, they are only slightly polar (eliminate choices A and C), while hydrogen and chlorine are far from each other, yielding the most uneven sharing of electrons in the bond.

A chemist wishes to determine the energy of a chlorine-chlorine bond. She knows the strength of a H—Cl bond and a H—H bond are 433 kJ/mol and 436 kJ/mol, respectively. Given the following reaction, what is the strength of a chlorine-chlorine bond? H2(g) + Cl2(g) → 2 HCl(g) ΔH° = -187 kJ/mol Question 27 Answer Choices A. 187 kJ/mol B. 243 kJ/mol C. 676 kJ/mol D. 1056 kJ/mol

B. The Cl-Cl bond energy can be determining by utilizing the following equation: Therefore, the dissociation energy of a Cl—Cl bond is 243 kJ/mol.

Which of the following is true about the amount of shielding the highest energy electrons of calcium and arsenic experience? Question 3 Answer Choices A. The electrons of calcium have a greater amount of shielding than the electrons of arsenic. B. The electrons of calcium have a lesser amount of shielding than the electrons of arsenic C. The electrons of calcium have the same amount of shielding as the electrons of arsenic. D. The amount of shielding between the electrons of calcium and of arsenic can only be determined experimentally.

B. The amount of shielding that the highest energy electrons of an atom feels is determined by the number of filled shells in that atom (inner core electrons), as well as subshells within an energy level (eliminate choice D). Because calcium and arsenic are in the same period (row), they have the same number of inner core electrons. However, the highest energy electrons in arsenic are in the p subshell and are shielded by the electrons in the s subshell. Since the highest energy electrons in calcium are in the s subshell, there is more shielding for arsenic, eliminating choices A and C.

A graduate student estimates the chemical energy in a solution before and after a reaction. The chemical potential energy following the reaction is significantly reduced and the student hypothesizes that the unaccounted energy transitioned to thermal energy. Which of the following laws of thermodynamics best explains this observation? Question 11 Answer Choices A. Zeroth law B. First law C. Second law D. Third law

B. The first law of thermodynamics stipulates that energy is always conserved and in this example, it is simply changing form from chemical energy to thermal energy (choice B is correct). The zeroth law states that two systems at equilibrium with a third are in equilibrium with each other (choice A is incorrect). The second law states that all processes proceed toward disorder (choice C is incorrect) and the third law states that a perfect crystal possesses no entropy at absolute zero (choice D is incorrect).

Question 23 Which of the following is the most stable resonance structure for NO2-?

B. The most stable resonance structures occur when charge separation is minimized (eliminate choices C and D). The most stable resonance structures of ions occur when the negative charge is found on the more electronegative atom(s) and the positive charge is found on the more electropositive atom(s). Oxygen is more electronegative than nitrogen, thus oxygen will preferentially have the negative charge (eliminate choice A). Nitrogen is also not able to have an expanded octet, so choice A can be eliminated for this reason as well.

Which of the following compounds is least likely to be soluble in water? Question 10 Answer Choices A. NH3 B. CS2 C. CH3OH D. LiF

B. The rule of thumb for solubility is "like dissolves like," meaning polar solutes are soluble in polar solvents, and likewise for nonpolar things. Since water is a polar molecule that has hydrogen bonds as intermolecular forces, other hydrogen-bonding compounds like NH3 and CH3OH should be very soluble in water (choices A and C are wrong). Since LiF is an ionic compound that dissociates into charged particles in solution, it will need a polar solvent to dissolve (choice D is wrong). CS2, however, is a nonpolar compound and is the least soluble in water, making it the best choice (choice B is correct).

4 CO2(g) + 2 H2O(g) → 2 C2H2(g) + 5 O2(g) Given the following reactions, what is the enthalpy for the reaction above? C2H2(g) + 2 H2(g) → C2H6(g) ΔH° = -94.5 kJ/mol 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g) ΔH° = -566 kJ/mol 2 H2O(g) → 2 H2(g) + O2(g) ΔH° = 142.4 kJ/mol Question 17 Answer Choices A. -519 kJ/mol B. 470 kJ/mol C. 519 kJ/mol D. 1036 kJ/mol

B. 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 all three equations and multiply the first and third equations by two. This results in ΔH° values for the three equations of 189 kJ/mol, 566 kJ/mol, and -284.8 kJ/mol, respectively. These can be summed to give the reaction enthalpy of 470 kJ/mol (choice B is correct).

Rank the following species by increasing boiling point. CH3CHO F2 CH3OH KBr Question 16 Answer Choices A. I < II < III < IV B. II < I < III < IV C. III < I < IV < II D. II < I < IV < III

B. When comparing molecules, the weaker intermolecular forces occur between molecules with a lesser magnitude of charge. The charge (or partial charge) is due to polarity, which occurs due to differences in electronegativity. KBr will have ionic forces, the strongest option, because of the full charges on the K+ and Br- ions (choice B is correct). F2 is a nonpolar molecule and will form instantaneous-induced dipoles (also known as London dispersion forces), the weakest of all intermolecular forces (choices A and C are wrong). CH3OH and CH3CHO are both polar molecules, but methanol is able to hydrogen-bond due to the -OH group, while weaker dipole-dipole forces hold together the CH3CHO.

The combustion of methane is given by the reaction: CH4 + 2 O2 ? CO2 + 2 H2O If 20 moles of methane and 20 moles of oxygen are initially present, how many moles of carbon dioxide are produced? Question 5 Answer Choices A. 1 B. 10 C. 20 D. 30

B. 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. Here, let's choose one (e.g. methane). If we have 20 moles of methane, we will need twice as much (or 40 moles) of oxygen to obtain a stoichiometric amount because the ratio of oxygen to methane is 2:1. However, the problem statement says that we only have 20 moles of oxygen present, indicating that we do not have enough present. This means that oxygen is the limiting reactant. Conversely, had we chosen to start with oxygen, the reasoning is slightly different. If we have 20 moles of oxygen, we will need half as much (or 10 moles) of methane to obtain a stoichiometric amount, because the ratio of methane to oxygen is 1:2. Since the problem statement says that we have 20 moles of methane present, that reactant is in excess, thus making oxygen limiting. Now that we have concluded that oxygen is limiting, we still have to solve the problem. All yield questions should be performed using the limiting reactant. If we have 20 moles of oxygen, we generate half as much (2:1 ratio) of carbon dioxide. Thus, we will generate only 10 moles of carbon dioxide, making choice B the correct answer.

Which one of the following correctly represents the electron configuration of sulfur in an excited state? Question 19 Answer Choices A. 1s22s22p63s23p5 B. 1s22s22p63s23p44s1 C. 1s22s22p63s23p34s1 D. 1s22s22p63s23p4

C. 1s22s22p63s23p34s1 correctly represents the electron configuration of sulfur in an excited state. A sulfur atom contains 16 electrons, so the choices 1s22s22p63s23p5 and 1s22s22p63s23p44s1 are immediately eliminated since they each account for 17 electrons. The configuration in 1s22s22p63s23p4 is that of a ground-state sulfur atom, so the answer must be 1s22s22p63s23p34s1.

A sample of excited hydrogen atoms emits a light spectrum of specific, characteristic wavelengths. The light spectrum is a result of: Question 20 Answer Choices A. energy released as H atoms form H2 molecules. B. the light wavelengths which are not absorbed by valence electrons when white light is passed through the sample. C. excited electrons dropping to lower energy levels. D. particles being emitted as the hydrogen nuclei decay.

C. A sample of excited hydrogen atoms emits a light spectrum of specific, characteristic wavelengths. The light spectrum is a result of excited electrons dropping to lower energy levels. The line spectra of atoms are the result of the photons emitted when excited electrons drop to lower energy levels.

Which of the following elements would be expected to exhibit the greatest ionization energy? Question 22 Answer Choices A. Calcium B. Potassium C. Titanium D. Scandium

C. Of the elements listed, titanium would be expected to exhibit the greatest ionization energy. Ionization energy generally increases as we move to the right across a row in the periodic table. Therefore, of the choices given, titanium should have the highest ionization energy.

Which of the following gives the ground-state electron configuration of a calcium atom once it acquires an electric charge of +2? Question 16 Answer Choices A. [Ar] 4s23d2 B. [Ar] 4s2 C. [Ar] D. [Ar] 4s24p2

C. The ground-state electron configuration of a calcium atom once it acquires an electric charge of +2 is [Ar]. A calcium atom normally contains 20 electrons, so if it acquires a charge of +2, it must have lost 2 electrons, leaving just 18. Only [Ar] accounts for 18 electrons.

Which of the following gives the percent mass composition of acetic acid (CH3COOH)? Question 12 Answer Choices A. 52% carbon, 12% hydrogen, 36% oxygen B. 46% carbon, 10% hydrogen, 44% oxygen C. 40% carbon, 7% hydrogen, 53% oxygen D. 34% carbon, 4% hydrogen, 62% oxygen

C. The percent mass composition of acetic acid (CH3COOH) is 40% carbon, 7% hydrogen, 53% oxygen. In one mole of CH3COOH, the mass of the oxygen atoms is 2(16 g) = 32 g, and the mass of the hydrogen atoms is 4(1 g) = 4 g. Therefore, the mass percent of O must be 8 times greater than that of H. Only the composition 40% carbon, 7% hydrogen, 53% oxygen has a ratio close to this.

What is the reactivity, as indicated by the tendency to lose an electron, of sodium compared to potassium? Question 4 Answer Choices A. This cannot be determined from the information given. B. Potassium has a lower reactivity because potassium has more protons than sodium. C. Potassium has a higher reactivity because the valence electron on potassium is farther from the nucleus. D. Both metals have the same reactivity because the potassium and sodium valence electrons experience the same effective nuclear charge.

C. The reactivity of sodium compared to potassium is that potassium has a higher reactivity because the valence electron on potassium is farther from the nucleus. Reactivity of metallic elements increases down a column of the periodic table. This increase in reactivity corresponds with increased shielding from inner electron shells, increased atomic radius, and decreased ionization energy. Thus, electrons are easier to remove in potassium versus sodium.

Which of the following elements is the most electronegative? Question 19 Answer Choices A. Chlorine B. Bromine C. Oxygen D. Sulfur

C. Electronegativity, which is the ability of an atom to attract electrons to itself in a covalent bond, is inversely related to atomic radius. Therefore, electronegativity increases as you move from the bottom left to the top right of the periodic table. Also remember the relationship F > O > N > Cl > Br > I > S > C ≈ H.

Which of the following statements is true about two neutral isotopes of the same element? Question 22 Answer Choices A. They have a different number of electrons. B. They have a different number of electrons and protons. C. They have a different number of neutrons D. They have a different number of neutrons and protons.

C. Isotopes, by definition, have the same atomic number, but different mass number. Because the number of protons has to be the same, the number of neutrons has to be different making choice C correct. Looking at the other choices, they cannot have a different number of electrons because if they have the same number of protons, they must have the same number of electrons to remain neutral (eliminate choice A). If they had a different number of protons, they cannot be isotopes (eliminate choices B and D).

Which of the following represents the ground state electron configuration for silver? Question 20 Answer Choices A. [Kr] 5s2 5d9 B. [Kr] 5s2 4d9 C. [Kr] 5s1 4d10 D. [Kr] 5s1 3f14 4d10

C. The f subshell appears for the first time in the fourth energy level, so 3f orbitals do not exist (eliminate choice D). The d subshell in any given energy level is one principle quantum number behind the s subshell. Therefore, 5s is followed by 4d, not 5d (eliminate choice A). Since completely filled d subshells are more stable for transition metals, we observe that one s electron is promoted into the d subshell in the copper family, of which silver is a member. This means choice C is better than choice B.

Which of the following correctly describes the molecular shape of AsCl3? Question 25 Answer Choices A. Bent B. Trigonal planar C. Trigonal pyramidal D. Tetrahedral

C. The molecular shape is determined by the total number of electron groups around the central atom (bonds and lone electron pairs) which determine the bond angles, as well as the total number of atoms bonded to the central atom. Since AsCl3 has three atoms attached to it, it cannot have a bent or tetrahedral shape, which require two or four atoms, respectively, bonded to the central atom (eliminate choices A and D). A molecule can only have a trigonal planar shape if it has three bonded atoms with no lone pairs. Since the central arsenic atom has three bonds and a lone pair, its geometry must be three dimensional (eliminate choice B).

A researcher investigates an endothermic reaction found to be non-spontaneous in a temperature-independent manner. Which of the following is the reaction being studied? Question 21 Answer Choices A. 2 H2O(g) → 2 H2(g) + O2(g) B. CO2(s) → CO2(l) C. N2(g) + 2 O2(g) → 2 NO2(g) D. N2O4(g) → 2 NO2(g)

C. The reaction described is non-spontaneous in a temperature-independent manner. As it is endothermic, this indicates that the reaction must have a negative ΔS given ΔG = ΔH - TΔS. The only reaction displaying a negative change in entropy is N2(g) + 2 O2(g) → 2 NO2(g) because the total number of gaseous moles decreases (choice C is correct, and choices A and D are not). In addition, choice B can be eliminated since as a substance changes phase from solid to liquid to gas, the particles are more disordered, so entropy increases.

Of the following valence shell electron configurations, which is associated with the greatest stability for an atom? Question 12 Answer Choices A. ns2np4 B. ns2np5 C. ns2np6 D. ns2np7

C. This is the valence shell electron configuration that correlates with an octet configuration. An octet configuration (eight electrons in the outermost shell) produces great stability.

Which of the molecules has an sp-hybridized atom?

C. 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 is sp3-hybridized, three e- groups is sp2-hybridized, and two e- groups is sp-hybridized. Choice C (DCC) is the only molecule that has an atom with only two electron groups, a carbon with two double bonds, so it is sp-hybridized (choice C is correct and A, B, and D are incorrect).

A compound is found to contain 64% silver, 8% nitrogen, and 28% oxygen by mass. What is the empirical formula of this compound? Question 8 Answer Choices A. Ag3NO B. Ag3NO3 C. AgNO2 D. AgNO3

D. A compound is found to contain 64% silver, 8% nitrogen, and 28% oxygen by mass. The empirical formula of this compound is AgNO3. For purposes of calculation, assume that we have 100 g of the compound. This would mean that we have 64 g of Ag, 8 g of N, and 28 g of O. Therefore, the number of moles of each of these elements is: Ag: (64 g) / (108 g/mol) = 0.6 mol N: (8 g) / (14 g/mol) = 0.6 mol O: (28 g) / (16 g/mol) = 1.8 mol Since the ratio of Ag:N:O is 1:1:3, the empirical formula must be AgNO3.

A researcher experimentally determines the atomic mass of an iron sample to be 55.1 amu. Which of the following most likely explains this result? Question 2 Answer Choices A. The sample has undergone beta decay. B. Part of the sample has been reduced. C. The sample has tin contamination. D. Part of the sample has been oxidized to Fe2O3-.

D. A researcher experimentally determines the atomic mass of an iron sample to be 55.1 amu. The best explanation for this result is that part of the sample has been oxidized to Fe2O3. Atomic mass values listed on the periodic table are weighted averages of the naturally occurring isotopes for each element. 55.1 amu is lower than expected. Fe is metallic and is much more likely to be oxidized than reduced. Beta decay (β-) results in the conversion of a proton to a neutron and an emitted beta particle. Atomic mass does not change significantly in beta decay. Contamination with tin, which is significantly heavier than iron, would yield a higher than expected atomic mass. This leaves the correct answer primarily by process of elimination. When iron is oxidized it forms rust (Fe2O3) that contains oxygen, which has lower atomic mass than iron. Rust contamination would therefore give lower atomic mass than expected.

Does a small atomic radius correspond to a high ionization energy? Question 13 Answer Choices A. No, because the smaller space enhances electron-electron collisions, thus making it easier to remove these electrons. B. No, because the large concentration of negative charge in a smaller space causes increased repulsion, thus making it easier to remove these electrons. C. Yes, because atoms in lower periods exhibit an enhanced strong force on electrons. D. Yes, because the shorter distance between the positive nucleus and the negative electron enhances electrostatic attraction, and thus makes it difficult for these electrons to be removed.

D. A small atomic radius does correspond to a high ionization energy because the shorter distance between the positive nucleus and the negative electron enhances electrostatic attraction, and thus makes it difficult for these electrons to be removed. The strategy to answer this question is to determine whether the question in the stem is true or not, so that we can eliminate half of the choices. Atomic radius increases and ionization energy decreases as period number increases. Thus, it is true that a small atomic radius corresponds to a high ionization energy. This eliminates the choices that begin with "No." Evaluating the final two choices, atoms in lower periods do not exhibit an enhanced strong force on electrons, because the strong force only pertains to holding the protons and neutrons in the nucleus together (and thus works over a very small distance which will never influence electrons). Thus, the choice "Yes, because atoms in lower periods exhibit an enhanced strong force on electrons" may be eliminated. The attractive force exhibited between charged particles such as protons in the nucleus and electrons is the electrostatic force, which increases in strength as the distance between the particles decreases.

Compared to the atomic radius of S, the atomic radius of Al is: Question 3 Answer Choices A. smaller, due to decreased nuclear charge. B. smaller, due to increased nuclear charge. C. larger, due to increased nuclear charge. D. larger, due to decreased nuclear charge.

D. Compared to the atomic radius of S, the atomic radius of Al is larger, due to decreased nuclear charge. Moving left to right across a period, nuclear charge increases due to increasing atomic number (number of protons). As the positive nuclear charge increases, effective nuclear charge increases and electrons are pulled closer to the nucleus, resulting in a decreased atomic radius. Aluminum is therefore larger than sulfur. This problem can also be treated as a two-by-two if you know the periodic trend for atomic radius, which decreases up and to the right. Based on this trend, it can be determined that Al is larger than S, eliminating the choices using "smaller." The distinction between the final two choices can then be determined based on the effect of nuclear charge.

Compared to the atomic radius of calcium, the atomic radius of gallium is: Question 23 Answer Choices A. larger, because increased electron charge requires that the same force be distributed over a greater number of electrons. B. smaller, because gallium gives up more electrons, thereby decreasing its size. C. larger, because its additional electrons increase the atomic volume. D. smaller, because increased nuclear charge causes electrons to be held more tightly.

D. Compared to the atomic radius of calcium, the atomic radius of gallium is smaller, because increased nuclear charge causes electrons to be held more tightly. Atomic radius generally decreases as we move to the right across a row in the periodic table. Therefore, comparing calcium and gallium (both 4th period elements), gallium should have the smaller radius; this eliminates "larger, because increased electron charge requires that the same force be distributed over a greater number of electrons" and "larger, because its additional electrons increase the atomic volume". "Smaller, because gallium gives up more electrons, thereby decreasing its size" can be eliminated since there is no loss of electrons when we are simply comparing atomic size. "Smaller, because increased nuclear charge causes electrons to be held more tightly" is true and provides the correct explanation.

Which of the following is true of ionization energy? Question 14 Answer Choices A. It increases with period, and the second ionization energy is greater than the first. B. It decreases with period, and the first ionization energy is greater than the second. C. It increases with period, and the first ionization energy is greater than the second. D. It decreases with period, and the second ionization energy is greater than the first.

D. Ionization energy decreases with period and the second ionization energy is greater than the first. This is a 2 x 2 question that is best approached by evaluating one of the arguments to eliminate answer choices. As period increases, ionization energy decreases due in part to the greater degree of shielding present. Thus, we can eliminate the two choices that begin with the choice "it increases." Evaluating the other argument, the second ionization energy is always greater than the first because electrons are more tightly bound after one is displaced (a more positively charged atom is less likely to give up an additional electron). Thus, ionization energy decreases with period and the second ionization energy is greater than the first.

Which of the following is true of an electron in an excited state? Question 17 Answer Choices A. It has absorbed a photon, and its energy has decreased. B. It has emitted a photon, and its energy has increased. C. It has emitted a photon, and its energy has decreased. D. It has absorbed a photon, and its energy has increased

D. It is true that an electron in an excited state has absorbed a photon, and its energy has increased. Electrons become excited when they gain energy by absorbing a photon.

Which of the following is the most stable ionization state for Sn? Question 2 Answer Choices A. Sn+1 B. Sn+3 C. Sn+4 D. Sn+2

D. Sn+2 is the most stable ionization state for Sn. When Sn loses two electrons, the resulting configuration has a full and stable 4d10 subshell, making Sn2+ the most stable ionization state. Although losing two more electrons (from the 5s subshell) would also leave a full 4d10, the resulting Sn4+ has a greater reduction potential than Sn2+ due to its higher charge. Therefore, Sn4+ is less stable than Sn2+.

What is the approximate percentage composition by mass of NaNO3? Question 1 Answer Choices A. 33% Sodium, 33% nitrogen, 34% oxygen B. 20% Sodium, 20% nitrogen, 60% oxygen C. 23% Sodium, 28% nitrogen, 49% oxygen D. 27% Sodium, 16% nitrogen, 57% oxygen

D. The approximate percentage composition by mass of NaNO3 is 27% sodium, 16% nitrogen, and 57% oxygen. Since the atomic weights of Na and N are 23 g/mol and 14 g/mol, respectively, the ratio of Na to N in NaNO3 must be 23:14. Two choices are clearly wrong because they give this ratio as 1:1, and the final incorrect choice indicates that the relative mass of N is higher than Na, which is impossible here.

What is the concentration of I- ions in a 0.20 M solution of magnesium iodide? Question 15 Answer Choices A. 0.20 M B. 0.10 M C. 0.05 M D. 0.40 M

D. The concentration of I- ions in a 0.20 M solution of magnesium iodide is 0.40 M. Since the empirical formula for magnesium iodide is MgI2, two moles of dissolved I- result from each mole of dissolved MgI2. Therefore, if [MgI2] = 0.20 M, then [I-] = 2(0.20 M) = 0.40 M.

Which of the following electron transitions (between energy levels, labeled by n) could account for the emission of photons of red, yellow, and blue light from a pure noble gas? Question 6 Answer Choices A. n = 4 to n = 1, n = 4 to n = 2, and n = 4 to n = 3, respectively B. n = 1 to n = 0, n = 2 to n = 1, and n = 3 to n = 2, respectively C. n = 2 to n = 1, n = 3 to n = 2, and n = 4 to n = 3, respectively D. n = 4 to n = 3, n = 3 to n = 2, and n = 2 to n = 1, respectively

D. The electron transition (between energy levels, labeled by n) that could account for the emission of photons of red, yellow, and blue light from a pure noble gas is n = 4 to n = 3, n = 3 to n = 2, and n = 2 to n = 1, respectively. The energy of the photons satisfy red < yellow < blue. Therefore, the energy difference between the energy levels that produce these photons must follow the same order. Since the spacing between energy levels decreases with increasing n, then n = 4 to n = 3, n = 3 to n = 2, and n = 2 to n = 1 is best.

Give the electronic configuration of vanadium, V. Question 7 Answer Choices A. [Ar] 4s13d4 B. [Ar] 3d5 C. [Ar] 4s33d2 D. [Ar] 4s23d3

D. The electronic configuration of vanadium, V is [Ar] 4s23d3. Vanadium, one of the transition metals, contains 23 electrons, so its first 18 electrons have the configuration of the noble gas argon; the question is where to put the remaining 5 electrons. Since the 4s subshell fills before the 3d, the answer must be [Ar] 4s23d3. [Ar] 4s33d2 should have been eliminated right away, since any s subshell can never hold more than 2 electrons.

What is the molar concentration of a solution containing 0.5 mol of solute in 50 cm3 of solution? Question 14 Answer Choices A. 1 M B. 2 M C. 0.1 M D. 10 M

D. The molar concentration of a solution containing 0.5 mol of solute in 50 cm3 of solution is 10 M. By definition, molarity equals moles of solute divided by liters of solution. Since 1 cm3 = 1 mL, the molarity of this solution is (0.5 mol)/(0.05 L) = 10 M.

Which of the following would have the biggest decrease in entropy for a gas in a piston cylinder? Question 14 Answer Choices A. Compression B. Doing work on the gas C. Cooling D. Deposition

D. An increase in entropy means an increase in randomness, or a decrease in order. Therefore, this question is asking for the change that would make the system most ordered. Compression and doing work on the gas are synonymous, so choices A and B can be eliminated. Cooling the piston cylinder would slow the movements of the gas, causing the volume to decrease and bringing molecules closer together. This definitely decreases entropy and could perhaps even increase intermolecular forces to the point that the gas might condense, decreasing entropy even more. Deposition is the process of going from a vapor to a solid, so since this phase change from the least ordered to most ordered phase is made explicit, choice D is a better answer than choice C.

Which of the following would have the highest energy? Question 7 Answer Choices A. Radio waves, because they have longer wavelengths than gamma rays B. Radio waves, because they have higher frequencies than gamma rays C. Gamma rays, because they have longer wavelengths than radio waves D. Gamma rays, because they have higher frequencies than radio waves

D. Gamma rays have the highest energy in the electromagnetic spectrum (eliminate A and B). High energy is synonymous with short wavelength and high frequency because E = hf and f = c/λ, making choice D the correct answer.

Which of the following statements about solids is true? Question 9 Answer Choices A. The intramolecular bonds of a molecular solid are the same as its intermolecular forces. B. Ionic solids have a lower melting point than molecular solids. C. Network solids have greater electrical conductivity than metallic solids. D. Unlike network solids, metallic solids are ductile.

D. Metallic solids are unique in that their covalently-bound lattice of nuclei and inner shell electrons are surrounded by a "sea" of valence electrons. Thus, they are excellent conductors of electricity and heat (eliminate choice C). They are also malleable and ductile. The intramolecular bonds of an ionic solid and of a network solid are the same as their respective intermolecular forces, but molecular solids have covalent intramolecular forces and van der Waals intermolecular forces (hydrogen bonds, dipole-dipole forces, or London dispersion forces), making choice A incorrect. Ionic solids have stronger intermolecular forces than molecular solids, thus they have a higher melting point (eliminate choice B).

What is the hybridation of the central atom in PCl3? Question 15 Answer Choices A. p B. sp C. sp2 D. sp3

D. Phosphorus is the central atom in this compound and will have three bonds to Cl and one lone pair of electrons in the Lewis dot structure. It will therefore need four hybrid orbitals in this compound. As a result, we expect one s and three p atomic orbitals to combine to give four sp3 hybrid orbitals.

An intern in a laboratory is asked to clean up after a new graduate student in the lab who was making a HEPES buffered solution. The intern finds a white solid powder left on the counter from the buffer preparation and is uncertain how to dispose of it. Which of the following is the most likely identity of this white powder? Question 4 Answer Choices A. Na B. H2SO4 C. C3H6O D. NaC3H5O3

D. The intern was disposing of a white powder (solid), thus we are looking for the compound that will have the strongest intermolecular forces (and therefore is most likely to be solid). H2SO4 (sulfuric acid) is capable of hydrogen bonding, C3H6O (acetone) is capable of dipole-dipole interaction, and NaC3H5O3 (sodium lactate) is an ionic compound. Ionic interactions are stronger than any other form of IMF, making NaC3H5O3 the answer (choices B and C are incorrect, D is the correct answer). While sodium is a metal, pure sodium would not appear as a white powder, nor would it be added to any traditional solution given its volatile reactivity when added to water (choice A is incorrect).

Which of the following ions has the largest ionization energy? Question 13 Answer Choices A. Sr+ B. Ca+ C. Mg+ D. Be+

D. The ionization energy periodic trend is opposite the trend for atomic radius. Atoms and ions with valence electrons close to the nucleus require more energy to remove them than atoms/ions with valence electrons farther from the nucleus. Since these ions are in the same group on the periodic table, the smallest one (Be) will have the greatest ionization energy.