Chemistry

At low pressures and temperatures, water can go directly from ice to water vapor. This state change is known as: A. sublimation B. deposition C. vaporization D. gaseous melting

A state change from solid to gas is sublimation. A. sublimation, correct. B. deposition, incorrect, Deposition is gas to solid. C. vaporization, incorrect, vaporization is liquid to gas. D. gaseous melting, incorrect, there is no such thing as gaseous melting.

Which of the following pure compounds form ionic bonds? I. CaCl2 II. HCl III. NaF IV. Mg(OH)2 A. I only B. I and II only C. I, III, and IV D. I, II, III, and IV

An ionic bond is formed between two atoms when one atom loses one or more electrons and another atom gains one or more electrons. This results in a positively charged cation and a negatively charged anion. The electrostatic attraction between the oppositely charged ions creates the bond, which can be measured by the finding the difference in electronegativity of the two atoms. Generally, a metal and nonmetal atom form ionic bonds. Metals prefer to give up electrons to form cations whereas nonmetals prefer their reduced anion form. In the case of CaCl2, NaF, and Mg(OH)2, each atom contains a metal that is bonded to a nonmetal. These all represent ionic bonds. While hydrochloric acid tends to dissociate in water to form H+ and Cl-, a pure solution cannot. Pure HCl contains two nonmetals, which forms a polar covalent bond. Therefore, C is the correct answer.

Sodium nitrate and ammonium sulfate react at 230ºC to yield sodium sulfate, nitrous oxide, and water vapor. If the total pressure of the products is 1.8 atm, what is the partial pressure of the nitrous oxide? The reaction is as follows: 2 NaNO3 + (NH4)2SO4 → Na2SO4 (aq) + 2 N2O (g)+ 4H2O (g) (230—300° С) A. 0.5 atm B. 0.6 atm C. 0.9 atm D. 1.8 atm

B. 0.6 atm, correct, 2 moles nitrous oxide and 4 moles of water vapor in the products indicate that 1/3 of the pressure is from the nitrous oxide A. 0.5 atm, incorrect, this would indicate that 2/7 of the pressure comes from 2 moles of nitrous oxide in the presence of 4 additional moles of water vapor and 1 mole of aqueous sodium sulfate (liquid). C. 0.9 atm, incorrect, this answer incorrectly shows that 1/2 the pressure is from nitrous oxide, but the correct ratio is determined by the molar ratio of gas products D.1.8 atm, incorrect, this would indicate that all pressure is from the nitrous oxide

The scientist adds a small amount of catalyst to a 1L reaction vessel containing 2.03 mol of A2, 6.89 mol of B3C, 14.00 mol of AB2, and 2.50 mol of C2. Which of the following will occur? 3 A2 + 4 B3C → 6 AB2 + 2 C2 Keq = 5.13 x 10-2 A. The reaction will more swiftly proceed to the left. B. The reaction will more swiftly proceed to the right. C. The addition of a catalyst does nothing to a reaction at equilibrium, so the reaction will remain at equilibrium. D. No conclusion can be drawn.

Begin by setting up the reaction quotient equation for this reaction: Q = ([AB2]6[C2]2) / ([A2]3[B3C]4) Then plug in the values given: Q = (1.46 x 2.52) / (2.033 x 6.894) The numbers are tough to deal with, so we'll start by rounding them off: Q = (1.56 x 2.52) / (23 x 74) Notice that we made the numerator bigger by rounding 1.4 up, so we need to make the denominator bigger by rounding up to 7. As much as possible, make your approximations cancel each other out. We'll still want to chip away at this by breaking down the numbers into small pieces: Q = ((1.52)(1.52)(1.52)x 2.52) / (23 x (72)(72)) Q = ((2.25)(2.25)(2.25)x 6.25) / (8 x (49)(49)) Round the 49 up to 50: Q = ((2.25)(2.25)(2.25)x 6.25) / (8 x (50)(50)) Q = ((2.25)(2.25)(2.25)x 6.25) / (8 x (2,500)) Q = ((2.25)(2.25)(2.25)x 6.25) / (20,000) Round the 2.25 down to 2 and the 6.25 up to 7 (remember, make your approximations cancel each other out): Q = (2 x 2 x 2 x 7) / (20,000) Q = (8 x 7) / (20,000) Q = 56 / 20,000 Express in scientific notation: Q = (5.6 x 101) / (2.0 x 104) Q = (5.6 / 2) x (101 / 104) Q = 2.8 x 10-3 Now that we've approximated Q, we can compare it to Keq to see what will happen. Here, the Q is nearly 100,000 times smaller than the Keq. That means our reaction quotient has far too many reactants (bigger denominator means a smaller number). So the reaction will move towards products, to the right. Note that the arithmetic in this question is more difficult than you will likely see on the MCAT, simply due to the number of steps involved. The key lesson from this question, though, is that the MCAT absolutely will expect you to be able to approximate, and will expect you to solve questions that compare Q and Keq. An alternate way to deal with this question would be simply to look at the sizes of the numbers involved. That is, you see a 74 in the denominator. You could look at that and guess "okay this fraction will be less than 1. I don't know how much less, but probably less than 1." Then compare that to the Keq, which is around 500, and arrive at answer (B).

What is the structure of methyl propanoate? A. CH3COCH2CH3 B. CH3COOCH3 C. CH3CH2COOCH3 D. CH3CHOCH2CH3

Break down the compound name: methyl indicates CH3, propane indicates a 3-carbon chain, and "oate" indicates an ester linkage. Only choice C) has the correct groups present. A. CH3COCH2CH3, incorrect, This is a ketone. B. CH3COOCH3, incorrect, This shows an ester but lacks a 3-carbon propane. C. CH3CH2COOCH3, correct. D. CH3CHOCH2CH3, incorrect, This is an ether.

Which of the following characterizes butane? A. high solubility in water B. strong dipole moment C. high molecular weight D. low solubility in water

Butane is a four carbon alkane. As such, it is non-polar and low in molecular weight. Solubility in water is primarily governed by polarity and ability to hydrogen-bond, both of which butane lacks. Choice d is the best choice, low solubility in water.

Which alcohol will most easily react with HCl to form an alkyl halide? A) 1° alcohol B) 2° alcohol C) 3° alcohol D) all alcohols react equally to form alkyl halides

In terms of reactivity, 3° alcohols > 2° alcohols > 1° alcohols in reactions with HCl forming alkyl halides. For 3° alcohols, the reaction proceeds by an SN1 mechanism. For small 1° and 2° alcohols, there is some reactivity by an SN2 mechanism, but the SN1 mechanism with tertiary alcohols is the most reactive. a) 1° alcohol, incorrect. b) 2° alcohol, incorrect. c) 3° alcohol, correct. d) all alcohols react equally to form alkyl halides, incorrect.

Consider the following hypothetical reaction: Q2 (g) + AX4 (g) → 2 QX2 (g) + A (g) ΔH = 23.4 kJ / mol Which of the following is true regarding this reaction? A) Spontaneous at all temperatures B) Nonspontaneous at all temperatures C) Spontaneous at low temperatures D) Spontaneous at high temperatures

Remember that to be spontaneous, a reaction must have a ΔG that is negative. To answer this question, start with the Gibbs Free Energy equation: ΔG = ΔH - TΔS The reaction starts with two moles of gas and ends with three moles of gas, so the ΔS is positive. ΔH is given as a positive number in the question. The equation now looks like this: ΔG = (+) - T(+) To make ΔG negative and the reaction spontaneous, we want a large T so that we're subtracting a large positive number. Thus the answer is (D).

In an SDS-PAGE procedure, the SDS serves as a detergent. Why are the proteins treated with a detergent before being run through the electrophoresis gel? A. To coat the proteins with a large positive charge, since amino acid side chains may have positive, negative, or neutral charges, and a large uniform charge is necessary to get good separation in the gel. B. To allow the electrophoresis to separate the proteins solely on the basis of molecular weight. C. To prevent the protein from denaturing so that the electrophoresis can accurately resolve the proteins on the basis of tertiary structure. D. To break the intramolecular bonds holding the tertiary and primary structure of the protein together, thereby generating linear fragments that may be sorted on size.

SDS is a detergent which denatures the tertiary and secondary structure of a protein. It also coats the protein with a very large negative charge. This electrostatic repulsion pushes the protein in a single long rod shape, allowing the gel to sort various proteins on the basis of molecular weight alone. Thus (B) is the right answer. A: This answer choice is right except is says positive charge. SDS creates a negative charge. C: Detergents cause denaturing, rather than preventing it. D: SDS will break down tertiary and secondary structure, not primary.

According to the abiogenesis theory of the origin of life, the primitive atmosphere consisted primarily of: A) CH4, H2O, O3 B) NH3, O2, H2 C) NH3, CH4, H2 D) O2, CH4, H2O

The abiogenesis theory posits that early in Earth's history, the atmosphere was made up of a "primordial soup" of gases that created a highly reducing environment. Oxygen and ozone were not present, as they would have oxidized any biological macromolecules. Thus, choices (A), (B), and (D) are wrong.

What is the azimuthal quantum number (l) for the orbital to which a fluorine atom gains an electron to form an F- ion? A. 0 B. 1 C. 2 D. 3

The azimuthal quantum number (also known as the orbital quantum number or angular momentum quantum number), l, designates the subshell of an atom and gives the "shape" of an orbital. The number also indicates the number of angular nodes, or regions of zero probability, within the orbital. The orbital shapes commonly referred to as s, p, d, and f, correspond to azimuthal quantum numbers of 0, 1, 2, and 3, respectively. The electron configuration for elemental fluorine is 1s22s22p5. When fluorine gains an electron to become F-, the electron adds to the p orbital. P orbitals are dumbbell shaped with one angular node and correspond to an azimuthal quantum number of 1, making B the correct answer.

The behavior of gas will deviate from ideal behaviors LEAST under which of the following sets of conditions: A. A gas at high pressure and temperature B. A gas at low pressure and temperature C. A gas at high pressure and low temperature D. A gas at low pressure and high temperature

The ideal gas law, PV=nRT, is based on two assumptions: 1) individual molecular volume is negligible, and 2) intermolecular forces do not exist. A real gas at high temperature and low-pressure conditions, best approximates these two assumptions, and thus the ideal gas law. Therefore, answer choice D is the correct answer.

When a packet of light called a photon strikes the electron cloud, the electron may jump to a higher energy shell. Although the electron might stay in the excited state without returning to the ground state, we observe electrons returning to ground rather than staying in excited states. If an additional photon can strike the excited electron to stimulate a return to ground state, how many photons are given off by an electron returning to ground state from a shell at one higher energy level after excitation by a second photon? A. 0 B. 1 C. 2 D. 3

The photoelectric effect describes the phenomenon of electrons jumping to discrete energy levels after excitation by packets of light called photons. Photons can excite electrons to higher energy states in a process called absorption, and photons are given off as electrons return to ground states in a process called emission. We do not observe electrons staying at an excited state and spontaneously returning to ground. In a process called stimulated emission, described by Einstein, excitation from a second photon from incident light again excites the electron, which then returns to ground state. As stated in the question stem, a second photon excites the electron before the electron returns to ground state. This phenomenon of one photon striking an excited electron in order to release two photons in a return to ground state is the theory behind lasers. One in yields two out, but the electron must be in an excited state in order to undergo stimulated emission. A) 0, incorrect, When electrons return to ground state there is an emission of photons. B) 1, incorrect, Although the initial excitation of the electron will yield one photon out in a return to ground state, the question stem states that an additional photon has struck the electron in order to stimulate the emission. C) 2, correct, Both the initial excitatory photon and the second excitatory photon are released. D) 3, incorrect, There is no third photon mentioned in the question.

When an optically inactive reagent and a racemic mixture of an optically active compound are mixed, it can create an optically active product from an initially optically inactive set of reagents. Such a process was described by Donald Cram when formulating Cram's rule when mixing 2-phenylpropionaldehyde and bromobenzene. This process can best be described as: A) Electrophilic substitution B) Nucleophilic substitution C) Racemization D) Enantioinduction

The question presents a lot of information but we can clarify our understanding by focusing on the basic process described: an optically inactive thing became an optically active thing. We have no reason to suppose that this process must (or ever) does involve substitution reactions, so we can eliminate (A) and (B). Next, we know that a racemic mixture is optically inactive. So "racemization" would be the opposite of what's being described in the question. Thus, by process of elimination (D) is the right answer.

A researcher constructs a hypothetical reaction that follows a mechanism that uses the two reactions below: 3 Q2 + 2 AX4 → 2AX3Q + 2XQ + 2Q- 1. AX3+ Q2→ AX3Q + Q- 2. 2 AX4 + Q2 → AX3Q + Q- If the second reaction is experimentally determined to be an order of magnitude slower than the first reaction, the researcher can likely conclude that the rate law followed by the hypothetical reaction would be which of the following? A) R = k[AX4][Q2] B) R = k [AX4]2[Q2] C) R = k [AX4]2[Q2]2 D) R = k [AX4]2[Q2]3

The rate law is determined by the slowest step (the rate-determining step). The problem tells you that the second reaction is the slowest step, so it will determine the rate law. The exponents in the rate law match the stoichiometric coefficients of the reagents in the rate determining step. Here, the AX4 has a coefficient of 2 and the Q2 has a coefficient of 1. That matches the exponents in choice (B).

Consider the following reaction: C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l) ΔH = - 2808 kJ / mol Under what conditions will this reaction proceed spontaneously? A. Low temperature, low pressure B. Low temperature, any pressure C. High temperature, any pressure D. Any temperature, any pressure

Spontaneity is determined by a negative ΔG value in Gibbs free energy equation: ΔG = ΔH - TΔS A reaction with a negative ΔH and a positive ΔS will thus be spontaneous under any conditions. The reaction given has a negative ΔH. It also involves converting 1 mole of solid and 6 moles of gas into 6 moles of liquid and 6 moles of gas. That phase change means entropy is increasing (solid to liquid). Thus, with a negative ΔH and a positive ΔS this reaction is spontaneous under any conditions and (D) is the right answer. A, B: Reactions with a negative ΔH and a negative ΔS are spontaneous at low temperatures (e.g. the freezing of water from liquid to solid). C: Reactions with a positive ΔH and a positive ΔS are spontaneous at high temperatures (e.g. the boiling of liquid water into steam)

A protic acid is being titrated with sodium hydroxide (NaOH) when the pH of the solution stays relatively constant across a significant concentration range of the NaOH. This region of the titration curve includes which of the following: A. the half-equivalence point B. the equivalence point C. the isoelectric point D. Planck's constant

The region of the titration curve at which a protic acid pH remains constant includes the half-equivalence point, corresponding to the pKa of the acid, at which the acid is 50% deprotonated. At the pKa, [HA] is 50% and [A-] is 50%. At the equivalence point, the acid is 100% deprotonated, [A-]=100%. The isoelectric point describes the pH at which a molecule carries equal positive and negative charge, usually used to describe amphoteric molecules such as amino acids. Planck's constant is used in calculations describing quantum phenomenon. Planck's constant and the isoelectric point have nothing to do with titrations. A. the half-equivalence point, correct. B. the equivalence point, incorrect. C. the isoelectric point, incorrect. D. Planck's constant, incorrect.

What is the oxidation state of chromium in the dichromate ion, Cr2O72-? -2 0 +6 +12

The rules for oxidation states are as follows: In a compound, oxidation states of all elements sum to the net charge on the ion or neutral compound. Free elements have an oxidation number of 0. The more electronegative element has a negative oxidation state, fluorine and oxygen being the most and second most electronegative. Group IA elements are +1, Group IIA elements are +2, Group VIIA elements are usually -1 unless bonded to a more electronegative element, oxygen usually -2, hydrogen usually +1, fluorine always -1, chlorine usually -1.

A student observes that mercury forms a convex meniscus in the graduated cylinder but that water forms a concave one. This behavior is best explained by the fact that: A) the two liquids are being kept in graduated cylinders made of different materials. B) the adhesive forces between water and the walls of the graduated cylinder are greater than the adhesive forces between the mercury and the walls of the graduated cylinder. C) the cohesive forces between two mercury atoms are stronger than the cohesive forces between two water molecules. D) the mercury has strong cohesive than adhesive forces, whereas water has strong adhesive than cohesive ones.

The shape of the meniscus lets you know whether the substance has more attractive force for itself (convex meniscus) or for the walls of the cylinder (concave meniscus). Water, for example, experiences more adhesion between the water and the walls of the container, thus "pulling" the water up the sides of the container creating a concave meniscus. The adhesive force is greater than the cohesive force between the water molecules. The opposite is true for mercury, thus making (D) the right answer. A: While this may be true, the student's observation doesn't lead to this conclusion. B, C: The shape of the meniscus doesn't let you compare the forces in water to the forces in mercury directly. It only lets you compare one force (adhesion) to another force (cohesion) for one substance.

In order to determine how much Ag(NO3) can dissolve in aqueous solution, which value must be known? a) Ka b) pKa c) the standard reduction potential, E°, for the silver half-reaction d) Ksp

The solubility product, Ksp, is an equilibrium constant for a solvation reaction. Like any other equilibrium constant, Ksp equals products over reactants raised to the power of their coefficients. Pure solids and pure liquids are left out of the Ksp. For example, Ba(OH)2 (s) dissolved in aqueous solution has Ksp=[Ba2+][OH-]2. a) Ka, incorrect, Ka is a measure of how much an acid dissociates: Ka=[H+][A-]/[HA]. b) pKa, incorrect, pKa is related to the Ka by pKa=-log(Ka) and has nothing to do with solubility. c) the standard reduction potential, E°, for the silver half-reaction, incorrect, The standard reduction potential is useful for calculating the voltage in a galvanic, concentration, or electrolytic cell. d) Ksp, correct.

Given that the [Fe3+]aq= 4 x 10-10 M for dissociated Fe[OH]3, what is the Ksp for Fe[A]3? A. 4 x 10-10 B. 4.8 x 10-19 C. 6.9 x 10-37 D. cannot be determined without knowing the [OH-] concentration

The solubility product, Ksp, is given by the product of the equilibrium concentrations of the dissociated ions [A]a + [B]b in the form Ksp= [A]a[B]b. In this case, FeOH3 dissociates to [Fe3+]aq and 3[OH-]aq. Even though we are not given the concentration of OH-, we know that it must be three times the concentration of Fe3+, since one Fe(OH)3 molecule dissociates into one iron cation and three OH- anions. Therefore, the solubility product Ksp = [Fe3+]1[OH-]3 = [Fe3+]1{3[Fe3+]}3 The math works out to (4 x 10-10) x (27 x 64 x 10-30), and only the correct choice c) is close enough. A. 4 x 10-10, incorrect, This answer disregards the [OH-] concentration. B. 4.8 x 10-19, incorrect, This answer disregards the power of 3 on the [OH-] ion. C. 6.9 x 10-37, correct. D. cannot be determined without knowing the [OH-] concentration, incorrect, The [OH-] concentration is known by the multiplying the given [Fe3+] concentration by 3.

What is the electron configuration of calcium in CaCl 2 ? A) 1s22s22p63s23p5 B) 1s22s22p63s23p6 C) 1s22s22p63s23p64s1 D) 1s22s22p63s23p64s2

This question asks the examinee to determine the electron configuration of the calcium ion, as it exists in calcium chloride. Calcium in its elemental state has a total of 20 electrons. Therefore Ca2+ has a total of 18 electrons. In the answer choices provided, choice B is the only electron configuration that contains 18 electrons making B the correct answer.

The molar heat capacities (J/mole K) for zinc, copper, silver, and gold is 25.2, 24.5, 24.9, and 25.6, respectively. If 1 mole of each substance is heated until the temperature increases by 10K, which metal required the addition of the most heat? A. Zinc B. Copper C. Silver D. Gold

This question asks you to apply information on molar heat capacities to determine the most energetically intensive material. Since the change in temperature and the quantity of each substance is constant, the difference in molar heat capacities dictates the energy requirements for each substance. Molar heat capacity tells you how much energy (J) is necessary to increase one mole of a substance by 1 K. Therefore, the higher the molar heat capacity, the more energy required. In the question above, gold has the highest molar heat capacity and therefore requires the greatest amount of heat. The calculations are given below. Energy = (molar heat capacity) x (# of moles) x (change in temperature) Zinc: (25.2 J/mole K) x (1 mole) x (10 K) = 252 J Copper: (24.5 J/mole K) x (1 mole) x (10 K) = 245 J Silver: (24.9 J/mole K) x (1 mole) x (10 K) = 249 J Gold: (25.6 J/mole K) x (1 mole) x (10 K) = 256 J Multiplying the molar heat capacity by the number of moles and the change in temperature gives you energy required. Gold requires the most heat (256 J) making answer choice D correct.

The combustion of octane is a common reaction in automobile engines. 1) 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g) ΔH1 = -250 kJ/mole 2) 18 H2O (l) → 18 H2O (g) ΔH2 = 88 kJ/mole Net Reaction: 3) 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (l) ΔH3 = ? Given ΔH1 and ΔH2, what is ΔH3 for the net reaction above? A. 163 kJ/mole B. -163 kJ/mole C. 338 kJ/mole D. -338 kJ/mole

This question asks you to determine the change in enthalpy for a net reaction given the enthalpy change for two other reactions. Hess's Law tells us that the sum of two reactions gives us the enthalpy change for the net reaction: ΔH1 + ΔH2 = ΔH3. Since the net reaction has H2O (l) as a product, reaction 2 must be reversed and the sign of the enthalpy change negated (-ΔH2). The sum of the reaction enthalpies for the net reaction is as follows: ΔH3 = ΔH1 + (-ΔH2) ΔH3 = -250 kJ/mole + (-88 kJ/mole) ΔH3 = -338 kJ/mole Therefore, the change in enthalpy (ΔH3) for the net reaction above is -338 kJ/mole, or answer choice D. Furthermore, the combustion of a fuel is a highly exothermic process, making answer choices A and C incorrect.

The production of ATP via cellular respiration is governed by the following balanced reaction: C6H12O6 + 9 O2 > 6 CO2 + 6 H2O + ATP If there are 360 grams of C6H12O6 (MW = 180 g/mole) in the reaction vessel initially and 440 grams of CO2 are produced, what is the limiting reagent? A. C6H12O6 B. O2 C. CO2 D. H2O

This question is asking the examinee to determine the limiting reagent of a chemical reaction. To solve this problem, you must calculate the theoretical amount of CO2 that is produced given 360 grams of C6H12O6. If all 360 grams (2 moles) of glucose initially present react, then 528 g (12 moles) of CO2 should theoretically be produced. 360 g C6H12O6 x (1 mol C6H12O6 / 180 g C6H12O6) x (6 mol CO2 / 1 mol C6H12O6) x (44 g CO2 / 1 mol CO2) = 528 g CO2 Since only 440 grams of carbon dioxide are actually produced, not all of the glucose combusts. Thus, the oxygen must be the limiting reagent in this problem, making answer choice B correct.Choices C and D can be eliminated because products are never limiting reagents. Since there will be leftover C6H12O6, it is not the limiting reagent, eliminating answer choice A.

Thin-layer chromatography (TLC) is a laboratory technique used to separate non-volatile mixtures. Using a silica gel plate and a hexane solvent, which of the following compounds in a mixture containing CH3OH, HCOOH, CH2O, and C6H6 would have the highest Rf value? A. C6H6 B. HCOOH C. CH2O D. CH3OH

This question is testing your familiarity with thin-layer chromatography (TLC), which is a technique used to separate mixtures. Silica gel is composed of a siloxane matrix (-O-Si-O-). The highly polar substance is capable of forming intermolecular bonds, including hydrogen bonding. As a result, polar compounds adhere to the silica gel plate and travel little from the initial starting position. Meanwhile, nonpolar compounds elute away from the starting position with the nonpolar hexane solvent via capillary action. In the mixture above, HCOOH and CH3OH hydrogen bond with the siloxane matrix. Therefore, B and D will have the lowest Rfvalues. The oxygen atom in CH2O makes formaldehyde slightly polar allowing for intermolecular interactions between the gel plates. Benzene is the most nonpolar compound of the choices and therefore travels the farthest from the starting position along with the hexane solvent. Thus, A is the correct answer.

Which of the following aqueous solutions will have the highest boiling point? A. 0.10 m Na3PO4 B. 0.15 m NaCl C. 0.20 m HF D. 0.30 m CH3OH

This question is testing your understanding of colligative properties as they relate to boiling-point elevation. The change in temperature for the boiling point of a solution follows the equation ΔTb = Kbmi, where Kb represents the boiling point elevation constant (for water, Kb = +0.52oC/m), m represents the molality of the solution (moles of solute/kg of solvent), and i represents the van't Hoff factor. Since the boiling point elevation is proportional to the molality multiplied by the van't Hoff factor, you must consider each variable in the overall determination. Upon hydration of each sodium phosphate, three Na+ ions and one PO43- ion are produced, giving a van't Hoff factor of 4. Overall, ΔTb = Kb(0.1)(4) = 0.4Kb. For sodium chloride, one Na+ ion and one Cl- ion are produced, yielding a ΔTb = Kb(0.15)(2) = 0.3Kb. Since hydrogen fluoride is a weak electrolyte (weak acid), i=1 and ΔTb = Kb(0.2)(1) = 0.2Kb. Lastly, methanol is a non-electrolyte and does not dissociate in solution, resulting in a van't Hoff factor of one and ΔTb = Kb(0.3)(1) = 0.3Kb. Thus, sodium phosphate will have the highest boiling point making A the correct answer.

Given the reduction potentials for the following half-reactions, Br2 (l) + 2e- → 2 Br-(aq) Ered= + 1.07 V Au3+ (aq) + 3 e- → Au (s) Ered= + 1.55 V What is the electrochemical potential for the following chemical reaction? 2 Au3+ (aq) + 6 Br-(aq) → 3 Br2 + 2 Au (s) Eocell = ? A. Eo = - 0.48 V B. Eo = - 0.11 V C. Eo = + 0.11 V D. Eo = + 0.48 V

This question is testing your understanding of electrochemical cells and using half-reactions to determine the cell potential. To answer this question, you must first know that the electrochemical potential for a cell is equal to the reduction potential plus the oxidation potential: Eocell = Eored + Eoox In this chemical reaction, Au3+ is reduced to Au making this the reduction half-reaction (Eored= + 1.55 V). Oppositely, Br- is oxidized to Br2 so this would be the oxidation half-reaction. Therefore, you must reverse the reduction potential given for Br2 so that it matches the chemical reaction given, as follows: 2 Br-(aq) → Br2 (l) + 2e- Eoox= - Eored= -(+1.07 V) = -1.07 V Therefore, the Eocell = Eored + Eoox = (+1.55 V) + (-1.07 V) = + 0.48 V making D the correct answer. Note, when calculating the cell potentials, do not multiply the reduction potential or oxidation potential by the coefficients.

The electronegativity of elements increases going from left to right across a row in the periodic table. Which of the following best accompanies this trend in the periodic table? A. A decrease in ionization energy and an increase in atomic radii B. A decrease in atomic radii and a decrease in ionization energy C. An increase in ionization energy and a decrease in atomic radii D. An increase in ionization energy and an increase in atomic radii

This question tests the examinees knowledge of trends within the periodic table. In general, as you move from left to right across a row in the periodic table, electronegativity and ionization energy increases while atomic radii decrease. Therefore, C is the correct answer.

One mole of an unknown compound is determined to have a molecular weight of 86 grams per mole. The mass percent of the compound is as follows: 55.81% carbon, 6.98% hydrogen, and 37.21% oxygen. What is the empirical formula and molecular formula, respectively, of the unknown compound. A. C2H3O and C4H6O2 B. C2H3O and C6H9O3 C. C5H8O and C5H8O D. C5H8O and C10H16O2

To answer this question, you must determine the empirical formula of the unknown compound by treating the mass percent as the number of grams of each element out of a 100g sample. The number of moles of each element is determined by the following calculations: Moles of C = (55.81g C) x (1mol C/12g C) = 4.65 mol C Moles of H = (6.98g H) x (1 mol H/1g H) = 6.98 mol H Moles of O = (37.21g O) x (1 mol O/16g O) = 2.33 mol O Dividing each by the largest common denominator of 2.33 gives the empirical formula of the unknown compound: C2H3O (4.65/2.33) mol C = 2 mol C (6.98/2.33) mol H = 3 mol H (2.33/2.33) mol O = 1 mol O Since the molecular weight of the compound is 86 g/mol, the correct answer must be A. Note that if you were in a hurry on Test Day, or if you forgot how to solve this problem, you could instead use the periodic table to calculate the molar mass of the molecular formula given in each answer choice. If only one answer gave a molar mass of 86 g/mol (as is the case here, with choice A), then you could be certain that this option was correct. This is true of many MCAT questions: there may be multiple ways to solve the same problem!

What is the identity of the reducing agent and oxidizing agent, respectively, in the reaction shown below? 2Cr(OH)3 + 3H2O2 + 4NaOH → 2Na2CrO4 + 8H2O A. Cr(OH)3 and NaOH B. Cr(OH)3 and H2O2 C. H2O2 and Cr(OH)3 D. H2O2 and NaOH

To answer this question, you must identify the atom that is reduced and the atom that is oxidized. In the reaction above, chromium is oxidized from +3 to +6, which makes chromium hydroxide the reducing agent. Alternatively, oxygen in hydrogen peroxide is reduced from -1 to -2, which makes it the oxidizing agent. Thus, the correct answer is B.

What is the product of water and ethyl ethanoate under acidic conditions? A) ethanoic acid and ethanol B) butanol and water C) ethanoic acid, 2 equivalents D) ethanol, 2 equivalents

When mixed with water under acidic or basic conditions, esters undergo hydrolysis. The products are a carboxylic acid and an alcohol. a) ethanoic acid and ethanol, correct. b) butanol and water, incorrect, Ethyl ethanoate does not lose its ester group to become a four-carbon alcohol. c) ethanoic acid, 2 equivalents, incorrect, Hydrolysis of an ester forms carboxylic acid and alcohol. d) ethanol, 2 equivalents, incorrect, Hydrolysis of an ester forms carboxylic acid and alcohol.

How many electrons does gold (Au) possess that have a quantum number l = 2? A) 10 B) 23 C) 30 D) 79

You must know that the quantum number l can take on four different values: 0, 1, 2, and 3. Furthermore, you must know that each number corresponds to a particular orbital: 0=s, 1=p, 2=d, 3=f. So the question is asking us how many d electrons gold possesses. Since gold is in the sixth period, it has the following d orbitals: 3d, 4d, and 5d. The 3d and 4d orbitals are filled with 10 electrons each, and the 5d orbital has 10 electrons, giving a total of 30. Therefore the correct answer is C, 30. (If you're wondering why gold has 10 electrons in its 5d orbital instead of 9, the answer is related to the high stability of a fully-filled d subshell. From the periodic table alone, you would think that gold had 9 5d electrons, but it becomes significantly more stable if one 6s electron moves up to fill the 5d subshell. This explains why the 5d subshell holds 10 electrons!)

What is the most common electron configuration for Chromium (Cr)? A. [Ne] 4s23d4 B. [Ar] 4s23d4 C. [Ar] 4s13d5 D. [Ne] 3d6

[Ne] indicates filled 2s and 2p orbitals, which is incorrect for chromium. [Ar] correctly indicates filled 3s and 3p orbitals. Normally, the 4s would fill with 2 electrons next, followed by 4 electrons in the 3d orbital. However, chromium is an exception which fills 4s1 and 3d5. Copper is a similar exception which fills 4s1 and 3d10 rather than 4s2 and 3d9. These are the two main exceptions you should know for the MCAT. Note that elements in the same periods as Cr and Cu (such as Au, Ag, and Mo) similarly serve as exceptions to the usual rule. A. [Ne] 4s23d4, incorrect, The correct orbital shorthand is [Ar]. B. [Ar] 4s23d4, incorrect, While this would normally be true, Cr and Cu are exceptions which fill their 3d orbitals to 5 and 10 before placing their last electron in 4s1. C. [Ar] 4s13d5, correct. D. [Ne] 3d6, incorrect, [Ar] is the correct shorthand rather than [Ne], and the correct orbital filling is shown in answer c). Note that chromium is an exception to the orbital filling guidelines.

What is the IUPAC name of the following compound: (CH3)2CH-NH-CH-(CH3)2? A. Diisopropylamine B. Diisobutylamine C. Di-tert-propylamine D. 1,1-dimethylethylamine

(CH3)2CH-NH-CH-(CH3)2 is the structure of a secondary amine (NH-R2), with two substituent isopropyl groups, CH(CH3)2. Acceptable names for the compound are Diisopropylamine and N-Isopropylpropan-2-amine. Diisopropylamine is used as an herbicide precursor and in vulcanization processes during the manufacture of rubber. A. Diisopropylamine, correct. B. Diisobutylamine, incorrect, The two substituent groups have 3 carbons each, isobutyl groups have four carbons, -CH2-CH-(CH3)2. C. Di-tert-propylamine, incorrect, There is no such thing as a tert-propyl group; the tert-butyl group requires 4 carbons. D. N,N-dimethylethylamine, incorrect, The IUPAC name is generated by 1) identifying the longest carbon chain (propane), 2) identifying the highest priority functional group (amine on carbon 2 of the parent propane chain, i.e. 2-amine), identify side chains (N-Isopropyl), and put it together: N-Isopropylpropan-2-amine. The structure of N,N-dimethylethylamine is [(CH3)3]N. The ethane parent carbon chain in choice d) reveals this answer to be incorrect; the parent carbon chain in the question stem is a propane chain with three carbons.

What is the effect of adding KNO3 to a saturated aqueous solution of KCl? A. Ksp increases because more K+ ions are present. B. Ksp decreases because NO3- ions displace Cl- ions. C. The concentration of Cl- ions decreases because KNO3 addition drives KCl dissociation towards the solid, undissociated form. D. More KCl dissolves because more aqueous K+ ions require more Cl- ions.

C is correct. Group IA elements, Cl- salts, and nitrate (NO3-) salts are soluble. Ksp for a compound does not change, although the concentration of dissolved species may. When a product concentration is increased in an equilibrium reaction, reactant concentration increases according to Le Chatelier's principle. Here, addition of K+ ions to a saturated aqueous solution of KCl will drive KCl dissociation in the direction of the reactants and cause precipitation of solid KCl. The concentration of Cl- ions in solution will decrease such that the Ksp remains the same for the new product of Ksp=[K+][Cl-]. The only time Ksp changes is when the temperature is changed, which is not the case here. A. This choice states that Ksp increases because more K+ ions are present. Incorrect, Ksp does not change as a result of concentration. B. This answer states that Ksp decreases because NO3- ions displace Cl- ions. Incorrect, Ksp does not change as a result of concentration. C. This option correctly states that the concentration of Cl- ions decreases because KNO3 addition drives KCl dissociation towards the solid, undissociated form. D. In this answer choice, more KCl dissolves because more aqueous K+ ions require more Cl- ions. Incorrect, this contradicts Le Chatelier's principle. In an accurate description of this principle, higher product concentrations should drive the equilibrium towards the reactant, not the product, side. Remember, Le Chatelier's principle states that a reversible process will shift in the direction that relieves stress and re-establishes equilibrium.

Which phase change has the same magnitude of energy to the heat of vaporization for water? A. sublimation B. freezing C. condensation D. deposition

C. condensation, correct, heat of vaporization and condensation both refer to a phase change from gas to liquid and are equal in magnitude and opposite in sense. Explanation- Phase changes require the same energy input or give off the same energy depending on the direction of phase change. Freezing and melting are paired, and the enthalpy change is called the heat of fusion. Evaporation and condensation are paired, and the enthalpy change is the heat of vaporization. Deposition and sublimation are paired, representing solid to gas change and there is no specific term for the enthalpy change. "Heat of sublimation" describes the enthalpy change.

What is the electron configuration of Ca in Ca(OH)2? A) 1s22s22p6 B) 1s22s22p63s2 C) 1s22s22p63s23p6 D) 1s22s22p63s23p64s2

Calcium in Ca(OH)2 exists as an ion Ca2+. The electron configuration of Ca2+ is equivalent to Argon, with orbitals up to 3p filled with electrons. Normally, Ca has the configuration of [Ar]4s2. [Ar] is shorthand for 1s22s22p63s23p6. You may see this shorthand for noble gas electron configurations on the MCAT. A) 1s22s22p6, incorrect, This is the configuration for Neon, Na+, and Mg2+. B) 1s22s22p63s2, incorrect, This is the configuration for Mg. C) 1s22s22p63s23p6, correct, Ca2+has lost its 4s1 and 4s2 electrons. D) 1s22s22p63s23p64s2, incorrect, This is the configuration for Ca, but not Ca2+.

According to Charles's Law, what happens to a gas as the temperature is decreased? A. Volume decreases B. Pressure decreases C. Volume increases D. Number of moles of gas increases

Charles's Law states: V1 / T1 = V2 / T2 It represents the direct relationship between volume and temperature for an ideal gas. Thus, as T decreases, so does V, making (A) the right answer. B: This would by true by Gay-Lussac's Law: P1/T1 = P2/T2 C: The relationship between T and V is direct, not inverse. D: The number of moles of gas figures into Avogadro's Law: V1/n1 = V2/n2

The pressure above a solid compound is slowly reduced. Which of the following is true? A. If the substance is H2O it will transition from solid directly into gas. B. The substance will transition into liquid phase as then gas phase as the pressure is reduced for all substances except H2O. C. The substance will transition from solid directly into gas for all substances except H2O. D. If the substance transitions from solid directly into gas it must be H2O.

For most substances, the solid phase is more dense than the liquid phase, so lowering the pressure will cause the solid phase to transition either to the liquid phase or the gas phase, depending on the temperature. Because this transition (melting vs. sublimation) is temperature-dependent, neither (B) nor (C) is always true. Similarly, both water and other substances are capable of subliming at certain temperature/pressure combinations, so (D) is not always true. You should also remember that although H2O is rare in having a less-dense solid phase, it is not unique. Thus (A) remains as the only option, and it is true. Because reductions in pressure favor the less-dense phase, ice will transition directly to gas as the pressure is reduced.

In comparison with 1,2-dihydroxyethane, ethanol has weaker intermolecular hydrogen bonds resulting in: A. a lower boiling point for ethanol due to weaker intermolecular forces. B. ethanol being a stronger base. C. a higher boiling point because ethanol is a higher molecular weight compound than 1,2-dihydroxyethane. D. weaker intramolecular bonds in ethanol due to overall reduced bonding energy.

Ethane-1,2-diol (ethylene glycol) has one more hydroxyl group than ethanol, resulting in greater intermolecular hydrogen bonding. Due to the higher intermolecular forces, ethane-1,2-diol has a boiling point of 200°C in comparison with 78°C for ethanol. A. a lower boiling point for ethanol due to weaker intermolecular forces. Correct. B. ethanol being a stronger base. Incorrect, The two hydroxyl groups on ethylene glycol confer greater basicity than the one hydroxyl group of ethanol. C. a higher boiling point because ethanol is a higher molecular weight compound than ethane-1,2-diol. Incorrect, ethanol is a lower molecular weight compound than ethylene glycol. D. weaker intramolecular bonds in ethanol due to overall reduced bonding energy. Incorrect, Bond energy does not decrease because of intermolecular hydrogen bonding.

One mole of a compound consists of approximately 72g carbon and 6g hydrogen. What is its empirical formula? A. C6H6 B. CHO C. CH D. C6H12

Explanation, The empirical formula expresses the smallest number of atoms making up a molecule in the correct integer ratio. In this question, converting to moles gives us six moles of carbon and six moles of hydrogen in one mole of the entire compound. This gives us our molecular formula, C6H6. However, remember that empirical formulas must consist of the smallest ratio of all atoms involved! This ratio is CH, which is our answer. A. C6H6, incorrect. This represents the molecular, rather than the empirical, formula. B. CHO, incorrect. There is no oxygen present. C. CH, correct. The empirical formula is the smallest whole-number ratio of the elements making up a molecule. D. C6H12, incorrect. This does not reflect the correct molar ratio of the elements

Consider a very weak base in solution. The pKb of this base would likely be: A) equal to the pOH of the solution. B) higher than the pOH of the solution. C) lower than the pOH of the solution. D) near 7 at 25ºC.

Explanation: The correct answer is B. Normally, to calculate pH or pOH values, we require some numerical information about the concentration of the solution. In this question, this information is not given, so we will have to go off of the concepts alone. The MCAT often requires similar conceptual thinking, and this question is still very doable - so do not panic when you notice that numbers are not provided! Let us write a generic reaction to represent our base in water: B + H2O <—-> HB+ + OH- Kb = [OH-][HB+]/[B] Using an alternate form of the Henderson-Hasselbalch equation, expressed in terms of pOH and pKb: pOH = pKb + log [HB+]/[B] where [HB+] and [B] represent the concentrations of the conjugate acid and original base molecule at equilibrium, respectively. A very weak base will have a very small Kb. This means that equilibrium favors the reactants (including [B]) over the products (including [HB+]). Thus, we can infer that the value of [HB+]/[B] << 1, making the value of log [HB+]/[[B] << 0. This would result in a pOH value that is less than the pKb. Thus, pKb > pOH.

An electrolytic cell consists of two half-reactions: Ag2+(aq) + 2e- → Ag(s) E°= 0.80 Cu2+(aq) + 2e- → Cu(s) E°=0.34 What is the standard state cell potential for the electrolytic cell? A) -1.14 V B) -0.46 V C) 0.46 V D) 1.14 V

For the MCAT, an electrolytic cell will always have a negative standard-state cell potential. A power supply must be used to drive the non-spontaneous reaction. Here, we have two half-reactions: Ag+(aq) + e- → Ag(s) E° = 0.80 Cu2+(aq) + 2e- → Cu(s) E° = 0.34 One of the reactions must run in reverse in order to pair a reduction with an oxidation (which is necessary for a redox reaction). If the copper reaction runs in reverse, the standard reduction potential will be -0.34 + 0.8 = 0.46 V. This cannot be correct as the voltage is positive. Try running the Ag reaction in reverse and the copper forward: the standard-state cell potential is -0.80 + 0.34 = -0.46 V. (Note that we should not multiply the Ag potential by two, even though it only contains one electron while the other reaction contains two. This is important to remember for electrochemistry problems!) This could work for an electrolytic cell being driven by a power source. Both half-reactions cannot run in reverse, as we must always pair an oxidation with a reduction, and both half-reactions cannot run forward for the same reason. a) -1.14 V, incorrect, This answer has both half-reactions running in reverse. b) -0.46 V, correct. This answer has a negative standard-state cell potential, and one reaction runs forward (or reduces) while the other runs in reverse (or oxidizes). c) 0.46 V, incorrect, An electrolytic cell on the MCAT will have a negative standard-state cell potential. d) 1.14 V, incorrect, Both reactions cannot run forward, and the standard-state cell potential must be negative for an electrolytic cell.

What species is oxidized and what is the oxidizing agent in the following spontaneous reaction? 2Au3+(aq) + 3Cu (s) → 2Au (s) +3Cu2+ (aq) A) Au3+, Au B) Cu, Cu2+ C) Au3+, Cu D) Cu, Au3+

For the MCAT, remember: oxidation is loss of electrons (OIL) and reduction is gain of electrons (RIG). The compound containing the element that is reduced is the oxidizing agent, and the compound containing the element that is oxidized is the reducing agent. In this reaction, Cu (s) is oxidized because it loses 2 e-. Au3+ is the oxidizing agent because, although monatomic, it is the element in the compound that is reduced. A) Au3+(aq), Au (s), incorrect, Au3+(aq) is reduced to Au (s), and Au (s) is a product. B) Cu (s), Cu2+ (aq), incorrect, Cu(s) is oxidized, and therefore the compound containing Cu (s) is the reducing agent. Note that although Cu (s) is the reducing agent in this reaction, the entire compound would be the reducing agent if the Cu were in a polyatomic molecule. C) Au3+ (aq), Cu (s), incorrect, Au3+(aq) is reduced, and Cu (s) is oxidized. D) Cu (s), Au3+ (aq), correct.

The standard reduction potentials for Zn and Cu are given below: Zn2+(aq) + 2e- → Zn(s) E°= -0.76 V Cu+(aq) + e- → Cu(s) E°= 0.52 V What is the standard state cell potential for the following reaction? Zn(s) + 2Cu+(aq) → Zn2+(aq) + 2Cu(s) A. -1.8V B. -1.28V C. 0.28V D. 1.28V

Given the standard reduction potentials: Zn2+(aq) + 2e- → Zn(s) E°= -0.76 V Cu+(aq) + e- → Cu(s) E°= 0.52 V The reduction potentials are used as is or multiplied by -1 if the reaction direction is reversed. Stoichiometry does not matter, the reduction potentials are summed for the standard cell potential. Zn(s) + 2Cu+(aq) → Zn2+(aq) + 2Cu(s) 0.76V + 0.52V=1.28V A) -1.8V, incorrect, This uses Zinc in the wrong direction (it should be +0.76V), and copper is also in the wrong direction and multiplied twice. There should be no multiplying for stoichiometric coefficients. B) -1.28V, incorrect, This uses Zinc in the wrong direction and copper in the wrong direction. C) 0.28V, incorrect, This uses Zinc in the right direction and Cu in the wrong direction. D) 1.28V, correct.

Hydrogen bonds form between H2O molecules, but not H2S, H2Se, or H2Te. This explains why: A) H2O is liquid at room temperature while the other compounds are gases. B) H2O is more volatile than the other compounds. C) H2O has the highest molecular weight of all four compounds. D) H2O has a more distinct aroma than the other compounds.

Hydrogen bonds between H2O molecules will keep the molecules closer together than molecules lacking hydrogen bonds. Choice A is the best choice. a) H2O is liquid at room temperature while the other compounds are gases, correct. b) H2O is more volatile than the other compounds, incorrect, Hydrogen bonding will make the H2O less volatile than compounds that do not similarly hydrogen bond. c) H2O has the highest molecular weight of all four compounds, incorrect, Molecular weight has nothing to do with hydrogen bonds. d) H2O has a more distinct aroma than the other compounds, incorrect, Molecular aromas result from some functional groups and conjugated pi systems; aromas do not result from hydrogen bonding. Water is also odorless.

Arrange the following molecules in order of increasing pKa: CH3CH2COOH, CH3CHFCOOH, CH3OCH2COOH, CH3CHBrCOOH A) CH3CHFCOOH, CH3OCH2COOH, CH3CHBrCOOH, CH3CH2COOH B) CH3CHBrCOOH, CH3OCH2COOH, CH3CH2COOH, CH3CHFCOOH C) CH3CHFCOOH, CH3CH2COOH, CH3CHBrCOOH, CH3OCH2COOH D) CH3CH2COOH, CH3OCH2COOH, CH3CHFCOOH, CH3CHBrCOOH

Increasing pKa means decreasing acid strength. A low pKa, just like a low pH, indicates a stronger acid. Here, we need to arrange the molecules from the strongest acid to the weakest. An acid is stronger if the conjugate base is stabilized, and one form of stabilization is inductive stabilization. That is, if an electron withdrawing group pulls electron density off of a carbon, that carbon becomes slightly positive. That slight positive charge can then stabilize the negative charge of the conjugate base. Inductive effects increase with increasing electronegativity. Because fluorine is the most electronegative atom, it has the biggest inductive effect and will most help stabilize the conjugate acid. Thus CH3CHFCOOH is the strongest acid and must be first in our list. That narrows us down to (A) and (C). Propanoic acid with no inductive stabilization would be the weakest acid here, with the highest pKa. So it must be the last item on the list. That tells us choice (A) is the right answer

In order to extract isobutyric acid from a solution of diethly ether, one should wash the solution with: A) 1 30mL wash of water B) 3 10mL washes of water C) 1 30mL wash of hexane D) 3 10mL washes of hexane

Isobutyric acid has a carboxylic acid group, is polar, and can readily hydrogen bond. It can be extracted from the organic solvent, diethly ether, by washing with water. Multiple smaller volume washer extract better than one large volume wash. Use of a second organic solvent, hexane, will not extract the isobutyric acid from the diethyl ether. a) 1 30mL wash of water, incorrect, Multiple smaller volume washes will yield higher extraction results. b) 3 10mL washes of water, correct, Isobutyric acid will enter the aqueous layer and wash off of the organic layer of diethyl ether. c) 1 30mL wash of hexane, incorrect, Hexane is another organic solvent like diethyl ether. d) 3 10mL washes of hexane, incorrect, See c).

The human body takes biological monomers in the form of amino acids and sugars and builds highly complex physiological structures from them. As such, the human body's relationship with the second law of thermodynamics can best be described by which of the following: A) The human body is an example of how biological systems are not part of the domain of the second law of thermodynamics, since they decrease entropy through anabolic processes. B) There is an equilibrium between anabolism and catabolism in the human body, thus it falls under the second law of thermodynamics by not increasing entropy. C) The human body follows the second law of thermodynamics by creating decreasing entropy within the system by increasing the entropy of the environment. D) It is impossible to relate thermodynamic laws to complex physiological systems.

Living things don't get a free pass when it comes to thermodynamics. The human body gets its energy by taking highly organized molecules (imagine the rings of a sugar molecule) and turning them into disorganized things (breathing out CO2 as waste). Thus, the decrease in entropy that is seen in anabolic processes in the human body comes at the expense of increasing disorder in the rest of the universe, making (C) the right answer. A, D: These are both false. Living systems are governed by the same rules of chemistry as every other chemical system. B: While (B) may occasionally be true, it need not be. Most notably in a growing child or adolescent, there will be more anabolism than catabolism as the body grows

Which of the following pairs of molecules have bond angles that are the most similar? A. CH4, CO2 B. BF3, NH3 C. H2O, SO2 D. CH4, H2O

Similar bond angles generally - but not always - imply that the molecules have similar VSEPR geometries. However, we also need to consider factors including the presence of lone pairs and the overall Lewis structures. Let's look at the shapes of each pair individually to find the right answer. A. tetrahedral, linear. CH4, like a typical tetrahedral molecule, has approximate bond angles of 109.5 degrees. In contrast, a linear arrangement implies that bond angles are 180 degrees. These are too distant to be correct. B. trigonal planar, trigonal pyramidal. These may seem similar initially! However, trigonal planar bond angles are 120 degrees, while a trigonal pyramidal arrangement resembles a tetrahedral structure with one substituent replaced by a lone pair. As a result, the associated angles are slightly smaller than 109.5 (107.8, if you want to be accurate). C. bent, bent. This, too, it tempting! But H2O possesses two lone pairs on its central atom. As a result, its bond angles are even smaller than 107.8. In fact, they're around 104.5 degrees. This differs significantly from the 120-degree angles expected in SO2. D. tetrahedral, bent. Finally, here's our correct choice! Tetrahedral bond angles are typically 109.5 degrees, while that between the O-H bonds of water is 104.5. This is only a 5-degree difference (smaller, if you consult some textbooks, though we don't really care about the exact details). The key is that, although their shapes differ, the presence of different numbers of lone pairs cause these angles to be nearly identical.

Similarities in chemical behavior can be observed between elements within the same family in the periodic table. For instance, sodium and potassium react with oxygen to produce a metal oxide. Which of the following gives the balanced reaction for the oxidation of an alkali metal (M)? A. 2M(s) + O2(g) → 2MO(s) B. 4M(s) + O2(g) → 2M2O(s) C. 3M(s) + O2(g) → M3O2(s) D. M(s) + O2(g) → MO2(s)

Sodium and potassium are congeners found in the first column of the periodic table. They are included in the alkali metal family and exhibit similar chemical reactivity. In the presence of oxygen, alkali metals are oxidized. To form a complete octet, alkali metals lose one valence electron and take on their preferred oxidation state of +1. Oxygen, on the other hand, completes its octet by gaining two electrons to form an anion with a charge of -2. Thus, in the reaction above, two metal (M) atoms must each lose one electron in order for each oxygen atom to gain two electrons. Of the reactions above, B is the only choice where the metal (M) has a charge of +1 and oxygen had a charge of -2 with a 2:1 coefficient ratio of metal to oxygen. Thus, B is the correct answer for the oxidation of alkali metals.