MCAT Chemistry Review

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Courtenay
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231597
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MCAT Chemistry Review
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2013-09-24 17:13:29
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MCAT Chemistry
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Atoms, elements, and the periodic table, bonding and molecular formation, chemical reactions: fundamental phenomena, equilibrium dynamics, and thermodynamics, solutions and solubility, gases, phase changes, acid-base chemistry, and electrochemistry
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  1. What is the charge on an aluminum atom that contains 10 total electrons?
    +3
  2. A scientist is attempting to determine the accurate atomic mass of a sample of naturally occurring chlorine. Which of the following would introduce the most error into her measurements?
    A. Neglecting the mass of the electrons
    B. Neglecting the mass of the protons
    C. Neglecting the mass of the chlorine-36 isotope
    D. Neglecting the mass of the neutrons
    D. Neutrons are slightly heavier than protons, which are much heavier than electrons.
    (this multiple choice question has been scrambled)
  3. What is the principal quantum number?
    It is the shell. A shell is equivalent to a row in the periodic table. It is designated n and can take the value of any positive number.
  4. What is the second quantum number (the Azimuthal quantum number)?
    The subshell and can take any integral value from 0 to n-1. It has the designation of l. Each subshell has a letter designation: 0=s, 1=p, 2=d, 3=f
  5. What is the third quantum number (the magnetic quantum number)?
    It is the orbital, or the region surrounding an atom's nucleus where an electron is most likely to be. It can be any integral value from -l to l including zero. There are 2l+1 orbitals in the l subshell.
  6. What is the fourth quantum number (spin quantum number?
    It is the electron spin state. It is experimentally derive and it can either be +1/2 or -1/2 (they either spin one direction or the other direction).
  7. What is a quick way to determine the electron configuration of a particular atom?
    • Using the Aufbau principle (follow the arrows)
    •        ↙
    • 2   1s ↙  ↙
    • 2   2s 2p ↙  ↙
    • 8   3s 3p 3d ↙  ↙
    • 8   4s 4p 4d 4f ↙
    • 18 5s 5p 5d 5f
    • 18 6s 6p 6d

    Add the numbers in the left column for number of electrons (at most) at each stage.
  8. Which of the following represents a possible electron configuration of a neutral magnesium atom?
    A. 1s22s22p8
    B. 1s22s22p6
    C. 2s22p63d4
    D. 1s22s12p6
    C.
    (this multiple choice question has been scrambled)
  9. Looking at the periodic table, electronegativity increases in what directions?
    It increases as you move to the right and upward of the periodic table. So does ionization energy and electron affinity.
  10. What does the first column of the periodic table list?
    Alkali metals
  11. What does the second column of the periodic table list?
    Alkaline earth metals
  12. What does group 17 (or VIIA) of the periodic table represent?
    Halogens
  13. What does group 18 (or VIIIA) of the periodic table represent?
    Noble gases
  14. What is the "s block" of the periodic table?
    Columns 1 and 2 (including Helium)
  15. What is the "p block" of the periodic table?
    Groups 13-18 (IIIA-VIIIA)
  16. What is the "d block" of the periodic table?
    Groups 3-12 (IB-VIIIB) - the transition metals
  17. Every element in the first period of transition elements except scandium (AN=21) forms a +2 cation. Scandium most likely does not form this cation because:
    A. the third ionization energy of scandium is low.
    B. the first ionization energy of scandium is low, the second ionization energy of scandium is high.
    C. the two electrons removed are 4s electrons.
    D. removing two electrons from scandium would make it isoelectronic with potassium.
    A. Choices A and B represent true statements, however, these statements do not answer the question. Choice C is true, since the second ionization energy of any atom is higher than the first ionization energy. However, it fails to address why a +2 cation is not formed, since it would imply that a +1 cation is the most common ion for scandium. Choice D is the correct answer, since the most common cation formed from scandium is the +3 ion. Therefore, the +2 ion is not formed because if it was formed ,it would be immediately transformed into the +3 ion, which corresponds to a noble gas configuration for scandium.
    (this multiple choice question has been scrambled)
  18. The first ionization energy of potassium (K) is 0.4189MJ/mol. The first ionization energy of barium (Ba) is 0.5029 MJ/mol. Therefore:
    A. barium is more metallic than potassium.
    B. potassium is more metallic than barium.
    C. barium has a greater electron affinity than potassium.
    D. potassium has a greater electron affinity than barium.
    B. Metallic character has to do with the tendency of an atom to give up electrons. A low first ionization energy indicates a valence electron that can be easily lost. Electron affinity is the willingness of an atom to gain an electron. While it may seem straightforward to assume that an atom that is less willing to give up an electron must be more willing to take an electron, it is not quite so clear. By knowing the first ionization energy of an atom, we can say nothing about the electron affinity of that atom.
    (this multiple choice question has been scrambled)
  19. Would methane or ethane be expected to have the lower ionization energy?
    A. Ethane, because a primary cation is more stable than a methyl cation.
    B. Methane, because a methyl cation is more stable than a primary cation.
    C. Methane, because a methyl anion is more stable than a primary anion.
    D. Ethane, because a primary anion is more stable than a methyl anion.
    A. This requires integration of organic chemistry knowledge with inorganic chemistry knowledge. The MCAT will frequently require this. First, to eliminate choices B and D, we need only to consider the fact that in removing electrons, cations are made, not anions. From organic chemistry, remember that a tertiary cation is more stable than a secondary cation, a secondary cation is more stable than a primary cation, and a primary cation is more stable than a methyl cation.
    (this multiple choice question has been scrambled)
  20. The most common ions of iron are Fe2+ and Fe3+. Are these ions diamagnetic or paramagnetic?
    A. Fe3+ is diamagnetic, Fe2+ is paramagnetic.
    B. Both ions are paramagnetic.
    C. Fe2+ is diamagnetic, Fe3+ is paramagnetic.
    D. Both ions are diamagnetic.
    B. Iron has an atomic number of 26. Therefore, Fe2+ has 24 electrons and Fe3+ has 23 electrons. An atom, or ion, with an odd number of electrons cannot be diamagnetic (all electrons paired). For this reason, choices B and C are eliminated. Since iron is filling a d subshell, it has five orbitals to fill before it will electron pair. Fe2+ has six electrons past the nearest noble gas configuration. It is irrelevant whether two of these electrons are paired in the 4s orbital, or one elctron is in the 4s orbital, or no electrons are in the 4s orbital. In any case, the remaining 4, 5, 6 electrons, respectively, will not all pair in the 3d orbitals. Therefore, Fe2+ is paramagnetic, and the answer must be D.
    (this multiple choice question has been scrambled)
  21. An atom has the electron configuration 1s22s22p63s23p64s2. This configuration could represent any of the following species EXCEPT:
    A. Mn:5+
    B. Ca:
    C. Ti:2+
    D. K
    D. The electron configuration given is the ground state electron configuration for a calcium atom. This electron configuration also represents possible electron configurations for the dipositive titanium ion and the pentapositive manganese ion. In its ground state, the potassium atom (K) has only one electron in the 4s subshell; therefore, its electron configuration is 1s22s22p63s23p64s1.
    (this multiple choice question has been scrambled)
  22. Which of the following molecules is expected to have the largest individual bond diapole moment?
    A. H2O
    B. NaH
    C. N2
    D. CH4
    B. The nitrogen molecule will not have an overall dipole moment, nor will it have a bond dipole moment. The two nitrogens are equivalent, and they share a pure covalent bond. Methane (CH4) will not have an overall dipole moment due to the symmetry of the molecule; however, it will have an individual bond dipole in each C-H bonds. These dipoles result from the difference in electronegativity of carbon and hydrogen. It is important to recognize that carbon is slightly more electronegative than hydrogen, so each dipole points in the direction of the carbon atom. Sodium has an electronegativity of 0.92 on the Pauling scale (relative to fluorine, which has an electronegativity of 4.0). The electronegativity of hydrogen is 2.20. The difference in electronegativity between these two atoms is therefore 1.27. Recalling that an electronegativity difference of greater than 1.7 is needed for an ionic bond, it can be stated that this is a polar covalent bond.  The electronegativity difference between hydrogen and oxygen is 1.24. This is close to the difference between sodium and hydrogen, but not quite as large. (You do not need to know the values of electronegativity for the MCAT, but must recognize that electronegativity is a periodic trend that increases from bottom to top and left to right. Also, imagine hydrogen as falling between boron and carbon in the periodic table. Using this scenario, without resorting to electronegativity values, it can be seen that Na-H is the most polar bond in the above question.
    (this multiple choice question has been scrambled)
  23. The sodium cation, Na+, is more soluble in water than the silver cation, Ag2+. The best explanation for this fact is:
    A. silver has a greater electronegativity than sodium.
    B. sodium has a greater electronegativity than silver.
    C. the silver cation is larger than the sodium cation.
    D. the sodium cation experiences an ion-dipole interactions with the water.
    A. Because silver is more electronegative than sodium, we expect sodium to maintain itself as a 'naked' cation (not accepting electron density from the water). This leads to a large ion-dipole interaction between the sodium and the water. It is true that the sodium cation experiences ion-dipole interactions with the water; however, it does not provide the complete explanation of the problem at hand, since the silver cation also experiences ion-dipole interactions with the water. What is a polar molecule; as such, it will dissolve polar entities (the more polar, the better). The extremely low electronegativity of sodium makes the sodium cation a better "positive charge" than the silver cation.
    (this multiple choice question has been scrambled)
  24. Give the type of interaction occurring with the following examples:
    1) Na+/Cl-
    2) Na+/H2O
    3) H2O/H2O
    4) Na+/CCl4
    5) H3OH/CCl4
    6) CCl4/CCl4
    • 1) ion-ion
    • 2) ion-dipole
    • 3) dipole-dipole
    • 4) ion-induced dipole
    • 5) dipole-induced dipole
    • 6) induced dipole-induced dipole
    • Note: even molecules that do not possess a permanent dipole have interatomic or intermolecular forces that are created by induced dipole-induced dipole interactions. These forces, also known as London dispersion forces, are the weakest intermolecular forces.
  25. What is a hydrogen bond?
    • A hydrogen bond consists of two components: a hydrogen atom attached to an electronegative atom X and a second electronegative atom Y that contains a lone electron pair. Since the hydrogen that is attached to X is participating in a net dipole with X, the hydrogen posses a net partial positive charge. The electrons that form the bond between the hydrogen and the electronegative atom spend a majority of time associated with the electronegative atom
    • Y: →H-X ⇒ Y---H−X
    • Note: On the MCAT, the term hydrogen bond is generally restricted to situations in which hydrogen interacts, as shown, with nitrogen (N), fluorine (F) or oxygen (O). Hydrogen bonding, and the degree of hydrogen bonding that occurs within a given substance, tends to prevent its molecules from undergoing physical separation, and this tends to increase its boiling point.
  26. The boiling point of water is 373 K. The boiling point of ethanol (C2H5OH) is 351.5 K. The difference in boiling points observed is most likely accounted for by the fact that:
    A. water is involved in hydrogen bonding, ethanol is not.
    B. ethanol has a higher molecular weight than water and therefore is expected to have a lower boiling point.
    C. both are involved in hydrogen bonding, but the hydrogen bonds in water are stronger.
    D. ethanol has a higher molecular weight than water and therefore is expected to have a higher boiling point.
    C. is correct. Both molecules can form hydrogen bonds with themselves, since they both contain a hydrogen attached to an electronegative oxygen atom, and they both contain an electronegative oxygen with lone pairs to donate to the hydrogen. The hydrogen bonds I water are much stronger than the hydrogen bonds in ethanol, which makes the boiling point of water higher than the boiling point of ethanol. Ethanol does have a higher molecular weight than water and is expected to have a higher boiling point but does not solve the problem that has been presented.
    (this multiple choice question has been scrambled)
  27. Which of the following gases is expected to have the strongest intramolecular bond(s)?
    A. H2
    B. N2
    C. O2
    D. F2
    • B. is correct. If we consider the Lewis do structures of each of these molecules, we see:
    •            ..   ..    ..   ..     .. ..            
    •    H-H  N=N    O=O    :F-F:
    •                       ..   ..     .. ..
    • Both hydrogen gas and fluorine gas have single bonds. Oxygen gas has a double bond. Nitrogen gas has a triple bond.
  28. Give the formulas of some important polyatomic ions:
    ammonium
    nitrate
    carbonate
    sulfate
    phosphate
    • ammonium - NH4+
    • nitrate - NO3-
    • carbonate - CO32-
    • sulfate - SO42-
    • phosphate - PO43-
  29. How many resonance structures exit for the phosphate ion (PO43-)?
    A. 2
    B. 1
    C. 3
    D. 4
    D. is correct. Envision a structure with a central phosphorus atom and four attached oxygen atoms. Since the phosphate ion has a net charge of -3, only one of these oxygens will have a double bond to phosphorus in any resonance structure, and each of the other three oxygens will form a single bond with phosphorus. Since there are four different oxygen atoms that could form the double bond, there are four resonance structures.
    (this multiple choice question has been scrambled)
  30. The chemical formula for aluminum sulfate is:
    A. Al3SO4.
    B. Al3(SO4)2.
    C. Al2(SO4)3.
    D. AlSO4.
    C. is correct. Since aluminum is a group 13 element, it has a valence state of 3+. Sulfate ion should be memorized as SO42- so it has a charge of 2-. From this it can be seen that two aluminum ions (3+ x 2 = 6+) are needed for every three sulfate ions (2- x 3 = 6-).
    (this multiple choice question has been scrambled)
  31. What is the molecular geometry of CO2?
    • Linear
    • _       _
    • O=C=O
  32. What is the molecular geometry of BF3?
    • Trigonal Planar
    •                  _
    •                 |F|
    •                  |
    •                  B
    •                /  
    •             |F|   |F|
  33. What is the molecular geometry of NO2-?
    • Bent (Trigonal Bent)
    •         ..
    •         N
    •      //  
    •    |O|  |O|
  34. What is the molecular geometry of CH4?
    • Tetrahedral
    •         H
    •         |
    •         C
    •       / | 
    •     H  H   H
    • The middle H on the bottom should be coming out of the C toward the reader.
  35. What is the molecular geometry of NH3?
    • Trigonal Pyramidal
    •          ..
    •          N
    •       / |  
    •     H  H   H
    • The middle H on the bottom should be coming out of the N toward the reader.
  36. What is the molecular geometry of H2O?
    • Bent (Tetrahedral, Bent)
    •      ..
    •      O:
    •    / /
    •   H H
    • The second H should be coming out of the O toward the reader.
  37. Give the molecular geometry of each of the following:
    1) PCl5
    2) SF4
    3) ClF3
    4) XeF2
    • 1) Trigonal Bipyramidal
    •              Cl
    •               |
    •         Cl - P - Cl
    •             /  |
    •          Cl   Cl
    • 2) Seesaw
    •               F
    •               |
    •             : S - F
    •             /  |
    •           F    F
    • 3) T-shaped
    •               F
    •               |
    •         F - Cl - F
    •           ..    ..
    • 4) Linear
    •              ..
    •         F - Xe - F
    •            ..    ..
  38. Give the molecular geometry of each of the following:
    1) SF6
    2) BrF5
    3) XeF4
    • 1) Octahedral
    •            F    F
    •            |   /
    •       F - S - F
    •         /  |
    •       F    F
    • 2) Square Pyramidal
    •            F    F
    •            |   /
    •       F - Br - F
    •         /  ..
    •       F
    • 3) Square Planar
    •                   F
    •             ..  /
    •       F - Xe - F
    •          /  ..
    •         F
  39. The predicted molecular shape of MnBr3 is:
    A. T-shaped.
    B. seesaw.
    C. linear.
    D. trigonal planar.
    A. is correct. Mn - 7 valance electrons; since three are allocated to bonding with the three bromine atoms, there are four nonbonding electrons, or two lone pairs.
    (this multiple choice question has been scrambled)
  40. What are the five types of chemical reactions?
    • Synthesis: A + B ->AB
    • Decomposition (analysis): AB -> A + B
    • Single replacement: AB -> AX + B
    • Double replacement: AX + BY -> AY + BX
    • Oxidation-reduction: may take the form of any of the four other classes, and is when the oxidation number of an atom changes during a reaction; a redox reaction.
  41. All of the following are examples of synthesis reactions EXCEPT:
    A. H2O(l) + O2(g) → H2O2(aq)
    B. 4FeO(s) +O2(g) → 2Fe2O3(s)
    C. FeCl3(s) + 3Na(s) →Fe(s) +3NaCl(s)
    D. 2Na(s) + Cl2(g) → 2NaCl(s)
    C. is correct. This reaction is an example of single replacement reaction.
    (this multiple choice question has been scrambled)
  42. All of the following are examples of unbalanced decomposition reactions EXCEPT:
    A. HIO4(aq) → HIO3(aq) + O2(g)
    B. FeO(s) → Fe(s) + O2(g)
    C. HIO3(aq) + O2(g) → HIO4(aq)
    D. Fe2O3(s) → FeO(s) + O2(g)
    C. is correct. This reaction is an example of a synthesis reaction.
    (this multiple choice question has been scrambled)
  43. All of the following are examples of single replacement reactions EXCEPT:
    A. 2Mn(s) + 10HCl(aq) → 2MnCl5(s) + 5H2(g)
    B. 16NaCl(s) + S8(s) → 8Na2S(aq) + 8Cl2(g)
    C. Na(s) + KCl(s) → NaCl(s) + K(s)
    D. 3NaOH(aq) + FeBr3(aq) → 3NaBr(aq) + Fe(OH)3(s)
    D. is correct. This reaction is an example of a double replacement reaction.
    (this multiple choice question has been scrambled)
  44. Iron (III) hydroxide may be formed via a reaction between sodium hydroxide and iron (III) bromide. In the balanced equation, what is the ratio of Fe(OH)3 to NaBr?
    A. 1:1
    B. 1:2
    C. 1:3
    D. 3:1
    • C. is correct. Balance the equation
    •      FeBr3 + NaOH → Fe(OH)3 + NaBr
    •      FeBr3 + NaOH → Fe(OH)3 + 3NaBr
    •      FeBr3 + 3NaOH → Fe(OH)3 + 3NaBr
    • Now that the equation is balanced, it can be seen that the ratio of Fe(OH)3 to NaBr is 1:3.
  45. What is a limiting reagent?
    It is the reagent that is present in the smallest quantity (based on equivalents). It limits the reaction.
  46. If 28 grams of nitrogen gas are allowed to react with 4 grams of hydrogen gas to produce ammonia, which is the limiting reagent? How much ammonia will be produced?
    • First, convert gram masses into moles:
    •   28gN2/28g/mol = 1 mole N2
    •   4gH2/2g/mol = 2 mole H2
    • Next, divide each molar quantity by the appropriate stoichiometric coefficient to arrive at a number of "equivalents" (the amount of each substance that is needed to balance the reaction):
    •   N2 + 3H2 → 2NH3
    •       1mol N2/1mol/equiv = 1 equivalent N2
    •       2mol H2/3mol/equiv = 0.667 equivalent H2
    • Since there are fewer equivalents of hydrogen gas than of nitrogen, hydrogen is the limiting reagent.
    •    To find ammonia produced, multiply the number of equivalents of the limiting reagent by the stoichiometric coefficient of ammonia.
    •   0.667equiv x 2mol NH3/equiv = 1.333mol NH3
    •   1.333mol x 17g/mol = 22.667g NH3
  47. To form calcium hydroxide and sodium carbonate, 100 grams of calcium carbonate are allowed to react with 100 grams of sodium hydroxide in 100 grams of water. In this reaction, the limiting reagent is:
    A. calcium carbonate.
    B. sodium hydroxide.
    C. water.
    D. sodium.
    • A. is correct.
    • First, write out a balanced equation for the reaction:
    • CaCO3 + 2NaOH → Ca(OH)2 + Na2CO3
    • Next, calculated equivalents for each reactant:
    • 100g CaCO3 x 1mol/100g = 1 mol CaCO3
    • 1mol CaCO3 x 1 eq/1mol = 1 equiv CaCO3
    • 100g NaOH x 1mol/40g = 2.5 mol NaOH
    • 2.5mol NaOH x 1 eq/2mol = 1.25 equiv NaOH
    • Since CaCO3 is available in the least number of equivalents, it is the limiting reagent.
  48. Oxidation numbers indicate the number of electrons an atom would either gain or lose in the formation of the molecule of which it is a part. Fill in the following:
    All group IA elements have an oxidation number of _____.
    All group IIA elements have an oxidation number of _____.
    All group IIIB elements have an oxidation number of _____.
    Fluorine always has an oxidation number of _____.
    Oxygen almost always has an oxidation number of ______ (oxygen has a ____ oxidation state in peroxides, e.g., H2O2).
    Hydrogen has an oxidation number of either ____ or ____ (when bound to group IA).
    • +1
    • +2
    • +3
    • -1
    • -2, -1
    • +1, -1
  49. Give the representative information (formula, charge) of each common anion:
    1) Fluoride
    2) Chloride
    3) Bromide
    4) Iodide
    5) Oxide
    • 1) F-
    • 2) Cl-
    • 3) Br-
    • 4) I-
    • 5) O2-
  50. Give the representative information (formula, charge) of each common anion
    1) Sulfide
    2) Nitride
    3) Hydroxide
    4) Nitrate
    5) Perchlorate
    • 1) S2-
    • 2) N3-
    • 3) OH-
    • 4) NO3-
    • 5) ClO4-
  51. Give the representative information (formula, charge) of each common anion
    1) Carbonate
    2) Sulfate
    3) Phosphate
    4) Acetate
    • 1) CO32-
    • 2) SO42-
    • 3) PO43-
    • 4) CH3CO2-
  52. As a rule, highly charged species have a (lesser/greater) force of attraction. What does this mean for its solubility?
    Highly charged species (ex: PO43- →  ALPO4) have a greater force of attraction thus are much less soluble in water than those with little charge (NaCl)
  53. Give the representative information (formula, charge) of each common cation
    1) Sodium
    2) Lithium
    3) Potassium
    4) Ammonium
    5) Hydronium
    • 1) Na+
    • 2) Li+
    • 3) K+
    • 4) NH4+
    • 5) H3O+
  54. Give the representative information (formula, charge) of each common cation
    1) Hydrogen
    2) Calcium
    3) Magnesium
    4) Iron (II)
    5) Iron (III)
    • 1) H+
    • 2) Ca2+
    • 3) Mg2+
    • 4) Fe2+
    • 5) Fe3+
  55. Consider K2Cr2O7. Give the oxidation state or chromium.
    • Oxygen has an oxidation number of -2.
    • Potassium (being a group I) has an oxidation number of +1
    •   K  2 x +1 = +2
    •   O  7 x -2 = -14
    •              = -12 total
    • since the sum must be zero, Cr must balance
    •   Cr 2 x __ = +12; so Cr must be +6
  56. What is the oxidation number of iodine in periodic acid (HIO4)?
    A. -5
    B. -1
    C. +5
    D. +7
    • D. is correct
    • H = +1 x 1 = +1
    • O = -2 x 4 = -8
    •              = -7
    • So, I = +7 to balance
  57. Consider the oxidation-reduction reaction between potassium dichromate and hydrogen iodide:
    K2Cr2O7 + HI → KI + CrI3 + I2 + H2O
    Balance the equation.
    • First, determine which atoms have been oxidized and which have been reduced.
    • K2  Cr2  O7 + H  I  →  K  I + Cr  I3 + I2 + H2  O
    • +1  +6  -2  +1  -1    +1 -1  +3  -1   0    +1  -2
    • Note that chromium is reduced and iodine oxidized 
    •     Cr6+ + 3e- → Cr3+      and
    •     2I1- → I2 + 2e-
    • These two reactions are not balanced in terms of electrons:
    •   2 x (Cr6+ + 3e- → Cr3+)
    •   3 x (2I1- → I2 +2e-)
    • giving
    •   2Cr6+ + 6I1- → 2Cr3+ + 3I2
    • Therefore
    •   K2Cr2O7 + 6HI → KI +2CrI3 + 3I2 + H2O
    • Now balance for potassium, oxygen, and hydrogen:
    •   K2Cr2O7 + 14HI → 2KI + 2CrI3 + 3I2 + 7H2O
  58. What is the ratio of copper to nitric acid (HNO3) in the redox reaction between these substances to form copper (II) nitrate, water, and nitric oxide?
    A. 1:4
    B. 3:8
    C. 8:3
    D. 4:1
    • B. is correct.
    • Cu + HNO3 → Cu(NO3)2 + H2O + NO
    • (0) (+1)(+5)(-2) →(+2)(+5)(-2) (+1)(-2) (+2)(-2)
    • Look at the redox:
    • 3(Cu → Cu2+ + 2e-) - oxidation
    • ?(N5+ + 3e- → N2+) - reduction
    • However, not all of the N5+ was reduced (which is why the ? is used in the above equation).
    • 3Cu + 8HNO3 → 3Cu(NO3)2 + 4H2O + 2NO
  59. What is activation energy? How does it relate to the rate-determining step?
    For each elementary process, or step, there is an energy barrier that must be overcome in order for that step to proceed. This energy barrier is termed the activation energy for that step. The higher the activation energy of a step, the less likely that step is to occur and, therefore, the slower that step will be. For any mechanism, the step with the highest activation energy is the slowest step. The slowest step in a series of steps will limit the overall progression from reactants to products, so this step is termed the rate-determining step.
  60. What is microscopic reversibility?
    An important feature of elementary processes; while an overall chemical reaction may not be considered to be reversible, given the order and combination of mechanical steps, ALL elementary processes are reversible. It is this reversibility of elementary processes that is termed the principle of microscopic reversibility. It says that for every forward process, the reverse process also occurs in exactly the reverse manner.
  61. How is an intermediate different from a transition state?
    A transition state is a high-energy species found at the peak of the reaction curve. Intermediates are found in the troughs of reaction curves.
  62. All of the following affect reaction kinetics EXCEPT:
    A. catalysts.
    B. stoichiometry.
    C. microscopic reversibility.
    D. temperature.
    B. is correct. Kinetics is not concerned with the stoichiometry of the reaction under investigation. Kinetics is affected by catalysts and other reaction conditions, such as temperature and pressure. Kinetics is concerned with the mechanism of reactions. Mechanisms of reactions are comprised of steps, or elementary processes, which are all reversible.
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  63. The rate of the reaction 2NO(g) + O2(g) → 2NO2(g) is given by:
    A. Δ[N2]/Δt
    B. Δ[NO2]/2Δt
    C. Δ[NO]/2Δt
    D. Δ[O2]/Δt
    B. is correct. The rate of reaction is given by either the rate of disappearance of reactants in a given time or the rate of formation of products in a given time. Choices Δ[NO]/2Δt and Δ[O2]/Δt would be rate of reaction expressions for the reverse reaction. The last choice is nonsense.
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  64. What is rate law? What is the overall order of a reaction?
    • Rate = k[A]x[B]y
    • where [A] and [B] are the concentrations of A and B, respectively The value of x is the order of the reaction with respect to A and the value of y is the order of the reaction with respect to B. The values of x and y must be experimentally determined and bear no relationship to the stoichiometric numbers a and b.
    • The overall order of this reaction is given by the sum of x and y.
  65. Give some important points to know in regards to the specific rate constant (or rate constant).
    • The specific rate constant is denoted by the letter k in the following equation
    •  rate = k[A]x[B]y and
    • The value of k is unique to each reaction at a given temperature.
    • The value of k will change if the temperature is changed.
    • The value of k does not change with time.
    • The value of k is not dependent on the concentrations of either reactants or products.
    • The value of k must be determined experimentally.
    • The units of k depend on the overall order of the reaction.
    • ** The magnitude of the change in the rate constant for different temperatures is DIRECTLY PROPORTIONAL to the activation energy of the reaction -- the larger the activation energy, the greater the change in rate constant for the same temperature change.
  66. All of the following factors do not affect the specific rate constant, k, EXCEPT:
    A. initial concentration of reactants.
    B. time.
    C. temperature.
    D. volume of reaction system
    C. is correct. Be careful of the double negative questions on the MCAT. Try rephrasing the question to:"Which of the following affects k?"
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  67. As temperature is increased:
    A. k increases for all reactions.
    B. k increases for some reactions.
    C. k decreases for some reactions.
    D. k decreases for all reactions.
    • A. is correct. Given the equation
    • Because temperature is measured in Kelvin, the temperature term must always be positive when temperature increases (T2>T1). By definition, the activation energy, Ea, is always positive, as is the gas constant, R. Therefore, for increasing temperature, the right side of the Arrhenius equation is always positive. The equation can now be reduced to
    •                log K=P
    • where K=k2/k1 and P is meant to designate the positive value of the right side of the equation.
    • Solving for K: K=10P and
    • since 10P is always greater than I, K must always be greater than I, implying that the numerator (k2) must always be greater than the denominator (k1).
  68. A catalyst does NOT:
    A. affect the elementary processes of a reaction.
    B. affect the activation energy of a reaction.
    C. affect the concentration of products at equilibrium.
    D. affect the mechanism of a reaction.
    C. is correct. Catalysts operate by altering the activation energy of a reaction. The alteration in activation energy is achieved by providing an alternative mechanism -- via altered elementary processes. Catalysts do not affect equilibrium.
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  69. Acid catalysts such as p-toluensulfonic acid are often used to dehydrate alcohols. The role of the acid catalyst is to:
    A. increase ΔG° and increase the activation energy for the dehydration reaction.
    B. increase ΔG° and lower the activation energy for the dehydration reaction.
    C. maintain ΔG° at the same time and lower the activation energy for the dehydration reaction.
    D. lower ΔG° and increase the activation energy for the dehydration reaction.
    C. is correct. This question tests your understanding of a catalyst: they speed up the rate of a reaction (kinetics), they decrease the activation energy, they do not affect Keq, they are not used up in a reaction, and finally, they do not affect thermodynamics, ΔG°
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  70. The reaction between a secondary alkyl halide and a base would tend to yield a higher percentage of elimination product by treatment with:
    A. a very weak base which is also a very hindered electrophile.
    B. a very weak base which is also a very hindered nucleophile.
    C. a very strong base which is also a very hindered electrophile.
    D. a very strong base which is also a very hindered nucleophile.
    • D. is correct. This question will be fully explored below using LDA as an example.
    • This question presents two important points: (i) it describes a compound called LDA as hindered (= bulky) and (ii) its conjugate acid has a pKa of about 40. Of course, pKequals -log Ka and Ka is the acid dissociation constant. A Ka which is extremely low (approx. 10-40), indicating that the conjugate acid is verrrrry weak, would give a pKa of about 40. All to say, weak conjugate acid = strong base!
    • Translation of this question: what does a strong bulky base like to do? Before we answer, what does a strong small base like to do in organic chemistry? They tend to engage in nucleophilic substitution reactions (i.e. OH-, CN-). A big bulky base usually has difficulty accessing carbon for a nucleophilic reaction. Instead, it takes the easy way out: it plucks protons off a molecule like oranges off a tree! Of course some H+'s are easier to remove than others. This depends on the stability of the suggested product. For example, removing a H+ from the methoxy substituent would create a primary carbanion (verrrry unstable) with no real stable options to get rid of the negative charge. On the other hand, examine the carbon in the ring which is attached to the methyl group. If you remove the only hydrogen attached to that 'ringed' carbon (= a-hydrogen which is happy to be plucked, you get a tertiary carbanion. Furthermore, you get a logical ensuing reaction: the negative charge is quickly attracted to the carbonyl carbon which is delta positive thus forming a double bond. Simultaneously, the pi electrons from the carbonyl group are kicked up to the oxygen creating an enolate anion (almost identical to keto-enol tautomerism,).
  71. 2SO2(g)+O2<-->2SO3(g) ΔH=-197kJ/mol
    During the reaction, before equilibrium was reached, the mole fractions of SO2 (g) and SO3 (g) were 1/2 and 1/6 respectively. What was the partial pressure of O2 (g)? 
    A. 0.16 atm
    B. 0.50 atm
    C. 0.66 atm
    D. 0.33 atm
    D. is correct. The sum of the mole fractions of each species present must be unity (= 1). Therefore, the partial pressure of O2 is 1 - (1/6 + 1/2) = 1/3. The partial pressure of a species is given by the product of its mole fraction and the total pressure of the system. Therefore, the answer is given by 1/3 x 1 atm = 0.33 atm, approximately.
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  72. If the temperature was decreased in the following reaction,
    2SO2(g)+O2<-->2SO3(g) ΔH=-197kJ/mol
    which of the following would experience an increase in its partial pressure when the same system reaches equilibrium?
    A. There will be no change in the partial pressure of any of the reactants.
    B. SO3 (g)
    C. SO2 (g)
    D. O2 (g) and SO2 (g)
    • B. is correct. Reaction I is exothermic (ΔH is negative ∴ heat is released). The reaction could be written as:2SO2(g) + O2(g) ↔ 2SO3(g) + Heat
    • Adding heat (increasing the temperature) adds to the right hand side of the equilibrium forcing a shift to the left in order to compensate. The reverse occurs by decreasing the temperature thus creating a shift to the right which would produce more SO3(g) and more heat is released.
  73. Reaction I: 2SO2(g)+O2<-->2SO3(g) ΔH=-197kJ/mol
    If the SO2 is very carefully dissolved in water, sulfurous acid is obtained. The first proton from this acid ionizes as if from a strong acid while the second ionizes as if from a weak acid. Reaction II:
    H2SO3 + H2O --> H3O+ + HSO3-
    Reaction III: H2SO3 + H2O-->H3O++SO32-
    Ka=5.0x10-6
    What is the pH of 0.01 M H2SO3?
    A. 1.0
    B. 2.0
    C. 3.0
    D. 4.0
    • B. is correct. Since we are told to consider the first ionization of H2SO3 as if from a strong acid (Reaction II), we assume that the first proton completely dissociates. However, the second proton ionizes as if from a very weak acid (Reaction III). After all, a Ka ≈ 10-6 means that the product of the reactants is about 1 000 000 (one million!) times greater than the product of the products (which includes the second proton; for Ka see CHM 6.1). The preceding fact combined with the imprecision of the available multiple choice answers means that our answer can be estimated by assuming that H2SO3 acts as a strong monoprotic acid like HCl.
    • Therefore, one proton completely dissociates while the second proton's concentration is relatively negligible; thus [H+] = 0.01 mol dm-3 (1 dm-3= 1 L-1, see App. B.2) The pH is equal to the negative logarithm of [H+] = -log (0.01) = -log (10-2) = 2.
  74. The refractive index of a transparent material is related to a number of the physical properties of light. In terms of velocity, the refractive index is equal to the ratio of the velocities of light in a vacuum and in the medium. From this ratio, it can be seen that light is retarded when it passes through most types of matter.
    A ship went out on a search for a sunken treasure chest. In order to located the chest, they shone a beam of light down into the water using a high intensity white light source. The refractive index for sea water is 1.33 while that of air is 1.00.
    From the information above, how would you expect the speed of light in air to compare with the speed of light in a vacuum (which is given by "c")? 
    A. It would be approximately the same as c.
    B. It would be greater than c.
    C. It would be less than c.
    D. This cannot be determined from the information given.
    • A. is correct. The passage states that the refractive index is equal to the ratio of the velocities of light in a vacuum and in the medium.
    • nmedium = c / vmedium
    • Since the value for nmedium is 1.00 when the medium is air, the velocity of light in air (vmedium) must be approximately that of light in a vacuum (that is, "c").
  75. Group II Carbonate || Solubility (mol L-1 H2O)
                MgCO3          1.3x10-3
                CaCO3           0.13x10-3
                SrCO3           0.07x10-3
                BaCO3           0.09x10-3
    The solubility product for MgCO3 is: 
    A. 1.3 x 10-4
    B. 2.6 x 10-4
    C. 1.7 x 10-6
    D. 6.7 x 10-8.
    • C. is correct.  Since MgCO3 ↔ Mg2+ + CO32-,
    • Ksp = [Mg2+][CO32-] = s x s = s2 where s is the solubility.
    • From the table provided, s = 1.30 x 10-3 mol L-1. Thus Ksp = [Mg2+][CO32-] = s2 = (1.30 x 10-3)2 = 1.69 x 10-6 ≈ 1.7 x 10-6. {Mathfax: (13)2= 169 ∴ (1.3)2 = 1.69; also, to increase speed you should at least be able to recognize the squares of numbers between 1-15, i.e. (12)2 = 144; rules of exponents.
  76. Group II Carbonate || Solubility (mol L-1 H2O)
                MgCO3          1.3x10-3
                CaCO3           0.13x10-3
                SrCO3           0.07x10-3
                BaCO3           0.09x10-3
    Ca(OH)2 has approximately the same Ksp as CaSO4. Which of them has the greater solubility in terms of mol L-1
    A. They both have the same solubility.
    B. Ca(OH)2
    C. CaSO4
    D. It depends on the temperature at the time.
    • B. is correct. More moles.
    • Consider these two equations:
    • (i) CaSO4 ↔ Ca2+ + SO42-
    • (ii) Ca(OH)2 ↔ Ca2+ + 2OH-
    • On the Surface: for the same Ksp, the more ions produced, means the more soluble
    • (Ca(OH)2 produces 3 ions as opposed to just 2). Though temperature is indeed related to solubility, it would affect both compounds in a similar way so having 3 ions will still mean that it is more soluble than just 2 ions.
    • Going Deeper: If [Ca2+] = s = solubility of salt, then:
    • For (i): s2 = Ksp, therefore s = square root of Ksp
    • For (ii): (2s)2 x s = Ksp, therefore s = cube root of (1/4 x Ksp)Since Ksp is always less than 1 for sparingly soluble salts (i.e. see the table provided in the problem), the value for s obtained in (ii) necessarily is greater than that obtained in (i). {For fun(!), work out values for 's' in (i) and (ii) using any value for Ksp less than 1}
  77. There is one exception to the general rules of solubility. If the cation of the salt is approximately the same size as the anion, the arrangement of ions in the crystal lattice is more uniform and hence the lattice is more stable and ΔHlatt is more negative.
    Group II Carbonate || Solubility (mol L-1 H2O)
                MgCO3          1.3x10-3
                CaCO3           0.13x10-3
                SrCO3           0.07x10-3
                BaCO3           0.09x10-3
    The CO32- anion is approximately the same size as: 
    A. Ba2+.
    B. Ca2+.
    C. Mg2+.
    D. Sr2+.
    D. is correct. SrCO3 is actually less soluble than the salt below it (see the table in the question) despite the fact that Ba is situated lower down Group II in the periodic table. Thus the explanation given in the question text has to be considered to explain the stability or extremely low solubility of SrCO3.
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  78. A solution of SrCO3 in water boils at a higher temperature than pure water. Why is this? 
    A. SrCO3 has a low solubility in water.
    B. SrCO3 increases the density of water.
    C. SrCO3 decreases the vapor pressure of the water.
    D. SrCO3 decreases the surface tension of the water.
    C. is correct. A non-volatile solute causes a lowering of the vapor pressure of the solvent and hence an elevation of the boiling point. The preceding is a colligative property.
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  79. Which of the following is a correct representation of the phase diagram for carbon dioxide?

    A. Phase diagram A
    B. Phase diagram B.
    C. Phase diagram C.
    D. Phase diagram D.
    B is correct. This answer is a typical phase diagram for a substance. The exceptions to this rule include substances which exhibit larger intermolecular forces than usual, for example hydrogen bonding. These include water and ammonia, which would yield a phase diagram such as that in answer phase diagram A.
  80. Graphite possesses what is commonly known as a layer structure: carbon atoms form three covalent bonds with each other to yield layers of carbon assemblies parallel with each other. These layers are held together via weak Van der Waals' forces which permit some movement of the layers relative to one another.
    The properties of the layer-like structure of solid graphite stated in the passage would lend it to which of the following industrial uses?
    A. Corrosive
    B. Structural
    C. Lubricant
    D. Insulator
    C. is correct. The text says that the layers in graphite can move relative to one another. Hence, it can be used as a lubricant. No evidence is provided for any of the other properties as they relate to the layered structure of graphite.
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  81. 20 mL of 0.05 M Mg2+ in solution is desired.
    It is attempted to achieve this by adding
    5 mL of 0.005 M MgCl2 and 15 mL of Mg3(PO4)2.
    What is the concentration of Mg3(PO4)2?
    A. 0.065 M
    B. 0.022 M
    C. 0.150 M
    D. 0.100 M
    • B. is correct. On the Surface: Don’t convert mL to litres because it will cancel out.
    • x = molarity of Mg3(PO4)2 --> Solve for x
    • (.005M Mg2+)(5mL)+(3x)(15mL) =
    • (20mL)(.05M Mg2+
    • Note the 3 in front of x because there are 3 potential Mg2+ generated from each Mg3(PO4)2 in aqueous solution.
    • .025 + 45x = 1 
    • 45x = 0.975 thus x = 0.022
    • Going Deeper: a detailed calculation . . . Let's begin with the total number of moles of Mg2+ present in the final solution: 0.05 moles/L x 0.02 L = 0.001 moles of Mg2+. Next, let's look at the number of moles of Mg2+ obtained from MgCl2: 0.005 moles/L x 0.005 L = 0.000025 moles of Mg2+. Now we know the number of moles of Mg2+ we need supplied from Mg3(PO4)2: (0.001 - 0.000025) moles = 0.000975 moles. Thus from the 15 mL of Mg3(PO4)2 we need 0.000975 moles of Mg2+. But each mole of Mg3(PO4)2 contains 3 moles of Mg2+ Therefore, the concentration of Mg3(PO4)2 = [(0.000975 moles)/(0.015 L)] x 1/3 = 0.022 mol/L= 2.2 x 10-2 M.{Note: the math is much faster if you notice that the answer to the second step is negligible thus you can skip the third step and the final calculation is easier}
  82. Which of the following is the strongest reducing agent?
    Electrochemical Reaction         || Eo value(V)
    MnO2+4H++2e-<->Mn2+2H2O ||+1.23
                     Fe3++e- <-> Fe2+  ||+0.771
              N2+5H++4e-<->N2H5+  ||-0.230
                       Cr3++e- <-> C2+  ||-0.410
    A. Cr3+
    B. Cr2+
    C. Mn2+
    D. MnO2
    • B. is correct. The reduced species of the electrochemical equilibrium with the most negative Eo value is the strongest reducing agent. Memory aside (!), it is of value to note that a reducing agent reduces the other substance, thus a reducing agent is oxidized. Note that only answer choices Cr2+ and Mn2+ are oxidized (= lose electrons). When you write the two relevant equations as oxidations, instead of reductions like the table provided, you will note that only answer Cr2+ has a positive Eo value indicating the spontaneous nature of the reaction. The table provided demonstrates half-reactions written as reduction potentials. In order to write the oxidation, simply reverse the reaction and change the sign of Eo:
    • Oxidation: Cr2+ ↔ Cr3+ + e- Eo = 0.410
  83. As the atomic number increases as one moves across the periodic table, the numerical value for electron affinity generally:
    A. becomes more positive because of the increasing atomic radius.
    B. becomes more negative because of the increasing effective nuclear charge.
    C. becomes more positive because of the decreasing effective nuclear charge.
    D. becomes more negative because of the increasing atomic radius.
    B. is correct. As one moves across the periodic table, the atomic radius decreases as a result of the increasing effective nuclear charge (without an increase in the number of atomic orbitals). In other words, the nucleus becomes more and more positive from left to right on the periodic table resulting in the drawing of negatively charged orbital electrons nearer and nearer to the nucleus. As a result, atoms will accept electrons more readily as we go across the periodic table and the electron affinity (EA) becomes more negative (less positive). For example, halogens have very negative EA values because of their strong tendencies to form anions. Alkaline earths have positive EA values.
    (this multiple choice question has been scrambled)

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