UWorld Chem and Ochem

effective nuclear charge (Zeff)

amount of the positive charge from the nucleus that is experienced by the valence electrons. - increase when atomic number increase - Zeff=Z-S (S=shielding constant=number of core e-)

electron affinity

assesses the tendency of an atom to accept an additional electron by measuring the energy change when an electron is added to an atom.

What approximate volume of the oxytocin solution with the 10 mM Zn2+ additive was analyzed if 2.2 × 10−6 moles of oxytocin acetate (MW = 1067 g/mol) were recovered from the sample after 4 weeks at 50 °C? A.9 μL B.900 μL C.2 mL D.10 mL

at 10mM Zn2+, the oxytocin solution recover rate is 80% to find the original moles in solution: 2.2*10^-6 mol / 0.8 = 2.75*10^-6 mol 2.75*10^-6 mol * 1067 g/mol * mL/0.3mg = 9.5 mL correct answer is D

Compared to ΔG° for the dissociation of acetic acid at a temperature yielding a pH of 2.6, ΔG° for the dissociation of acetic acid at another temperature yielding a pH of 2.0 would be: A.lower.[47%] B.the same.[5%] C.higher.[37%] D.unrelated to pH.[10%]

according to the table in the passage, pH decreases when temperature and Ka increase. by using dG=-RTlnK, while both T and K increase, dG is lower at pH=2.0 correct answer is A

equilibrium

achieved when two opposing chemical reactions occur simultaneously at the same rate such that the concentrations of the chemical species become constant. Equilibrium reaction rates are equal, but the equilibrium concentrations of chemical species may be unequal.

Formal charge and oxidation state are two different ways of accounting for electrons between atoms. Based on the Lewis structure of the ClO− anion, which of the following pairs of values corresponds to the formal charge of the chlorine atom and the oxidation state of the oxygen atom, respectively? A.+1, −1[8%] B.0, −1[40%] C.0, −2[33%] D.−1, −2[17%]

Formal charge assesses the allocation of charge to the atoms in a Lewis structure based on the bonding configuration of the atoms. - FC = group valence - non-bonding e - 1/2* bonding e -> FC_Cl = 7-6-1/2(2)=0 Oxidation state assesses the gain or loss of electrons by an atom (relative to the elemental valence configuration) due to bond formation. - OS = group valence - non-bonding e - bonding e correct answer is C - OS_O = 6-6-2 = -2

ionization energy

The amount of energy required to remove an electron from an atom

Suppose that the (p-n) nuclear reaction described in the passage was attempted but resulted in the 18O nucleus absorbing a proton without ejecting a neutron. The resulting nucleus formed would be an isotope of: A.fluorine with different chemical properties than 18F.[47%] B.oxygen with the same chemical properties as 18O.[7%] C.fluorine with the same chemical properties as 18F.[40%] D.an element other than O and F with different chemical properties.

The chemical behavior of an atom is determined primarily by its electron configuration, not by the number of neutrons in the nucleus. As a result, isotopes of the same element have nearly identical chemical properties (such as bonding and reactivity) but differ in their physical properties (such as density and mass). - 18F and 19F are isotopes of the same element with the same electron configuration, the chemical properties of these nuclei would be nearly identical. correct answer is C

According to trends in the first ionization energy, which of the following alkali metals is the LEAST reactive? A.Li B.Na C.K D.Cs

The first ionization energy is the energy required to remove the first, most loosely bound valence electron from a neutral atom. - The first ionization energy tends to increase with increasing atomic number moving across a period and decrease moving down a group on the periodic table. - Elements with a lower first ionization energy are easier to ionize in reactions than elements with a higher ionization energy. in summary, less energy is required to remove the electron from the valence shell of larger alkali metals, which makes them more reactive toward an ionizing reaction. correct answer is A

Lewis Acid vs Lewis Base

lewis acid: electron pair acceptor lewis base: electron pair acceptor

Electronegativity

the ability of an atom to attract electrons when the atom is in a compound

rate determining step

the slowest step in a reaction mechanism -> step with the largest energy barrier

pressure

1 atm = 760 mmHg = 760 torr = 101,325 Pa = 101.325 kPa

beta decay

1. beta- decay: electron emission -> atomic number increase -> converts a neutron into a proton and emits an electron 2. beca+ decay: positron emission -> atomic number decrease -> convert a proton into a neutron 3. electron capture -> atomic number decrease -> convert a proton into a neutron

Tosylate

A compound containing the functional group -SO3C6H4CH3, derived from toluenesulfonic acid

electronegativity and bonding

A large difference in electronegativity promotes ionic bond formation, but a small difference in electronegativity promotes covalent bond formation.

Brønsted-Lowry acid vs base

B-L acid: an acid is a molecule that donates a proton (H+ ion) B-L base: a base is a molecule that accepts a proton.

The chemical structures of sulfur compounds H2S, SF6, S2Cl2, and S4N4 each contain only single (sigma) bonds. The compound that contains the longest bond between sulfur and an atom of another element is: A.H2S[43%] B.SF6[16%] C.S2Cl2[27%] D.S4N4[13%]

Because a σ bond involves end-to-end overlap of two orbitals, the length of a σ bond can be estimated as the sum of the atomic radii of the bonded atoms. Bonding between atoms with larger atomic radii positions the atomic nuclei farther apart and results in a longer σ bond (and vice versa). -> atomic radius decreases across the periodic table d/t increase effective nuclear charge, and increase down the column d/t additional electron shell -> larger the atomic radius, longer the bond length => atomic radius (large->small): Cl>N>F>H

Shiny nickel metal granules added to orange liquid bromine mixed in alcohol produced a combination reaction yielding a blue-green solution of nickel(II) bromide. In the reaction between Ni and Br2: A.bromine forms a chelate with nickel. B.nickel undergoes an oxidation-reduction reaction with bromine. C.the reaction forms a precipitate of nickel(II) bromide D.nickel and bromine participate in a Brønsted-Lowry acid-base neutralization

D is wrong because bromine or nickel is not bronsted-lowry acid and base, and there is no formation of water correct answer is B

Suppose that citric acid (H3C6H5O7) is titrated with 0.1 M NaOH to form a citrate buffer solution with a pH of 4.5. What is the pH at the first equivalence point? (Note: pKa1 = 3.13, pKa2 = 4.76, pKa3 = 6.40) A.Less than 3.13 B.Between 3.13 and 4.76 C.Equal to 3.13 D.Greater than 4.76

During the titration of a weak polyprotic acid with a strong base, the first stoichiometric equivalence point occurs when the moles of added base are equal to the moles of the initial acid species. Prior to the equivalence point, a buffering region forms in which both the weak acid and its conjugate base are present and the pH is determined by the pKa and the ratio of the acid and base species concentrations. When the concentrations of the acid and base species are equal, pH = pKa. the pH at the first equivalence point results from pKa1 and pKa2 and must be between them. correct answer is B

When comparing atoms across the same row of the periodic table, which of the following groups will contain the atom with the lowest second ionization energy? A.Group 1 B.Group 2 C.Group 14 D.Group 16

Group 14 and 16 are nonmental and their ionization energy is higher than metal, so eliminate choice C and D though ionization energy increase across row and decrease down the column, there is only 1 valence electron in Group 1. to remove an additional electron require loss of core electron that will take more energy than removing valence electron -> eliminate choice A correct answer is B

The above chromatogram shows the separation of a mixture containing an ester, a carboxylic acid, an alcohol, and an alkane. If normal-phase high-performance liquid chromatography is used to separate the mixture components, which peak corresponds to the alcohol? ** graph in world doc A.I B.II C.III D.IV

HPLC is used to separate small compounds based on polarity, and NP-HPLC consists of polar stationary phase and nonpolar mobile phase. if the mixture is separated by NP-HPLC, nonpolar compounds will be eluded before polar compounds - more retention time for polar compounds elude order: alkane -> ester -> alcohol -> carboxylic acid so peak 3 is alcohol correct answer is C

The overall reaction for the generation of an NO2+ electrophile: H2SO4+HNO3 ⇄ HSO−4+H2O+NO+2 can be stated in two separate, reversible steps as a protonation followed by a decomposition (Reactions 1 and 2). H2SO4+HNO3 ⇄ HSO−4+H2NO+3 Reaction 1 H2NO+3 ⇄ NO+2+H2O Reaction 2 Which molecule acts as a Lewis acid, but NOT as a Brønsted-Lowry acid, in these reactions? A.H2SO4[15%] B.HNO3[24%] C.H2NO3+[25%] D.NO2+[33%]

In Reactions 1 and 2, all chemical species except H2O and NO2+ follow the Brønsted-Lowry definition of acid-base behavior. Notably, NO2+ lacks H atoms and cannot be classified by the Brønsted-Lowry definition. However, in the reverse reaction of Reaction 2, the NO2+ electrophile accepts a pair of electrons donated by H2O. Therefore, applying the Lewis definitions of acids and bases, H2O is acting as a Lewis base (electron pair donor) and NO2+ is acting as a Lewis acid (electron pair acceptor). - H2SO4 functions as a Brønsted-Lowry acid by donating H+ ions. - HNO3 acts as a Brønsted-Lowry base (H+ acceptor) and also as a Lewis base (electron pair donor) in Reaction 1 - H2NO3+ is an H+ donor (Brønsted-Lowry acid) in the reverse of Reaction 1 and decomposes in Reaction 2. correct answer is D

The equilibrium HC2H3O2 + H2O ⇄ H3O^+ + C2H3O−2 exists in a 1 M aqueous solution of HC2H3O2 with an equilibrium constant Keq = [H3O+][C2H3O2−]/[HC2H3O2]. Which of the following expressions explains why Keq does not include a term for [H2O]? A.[H2O] < [C2H3O2−] B.[H2O] ≈ 55.5 ≈ constant C.[H3O+] > [H2O] D.[H2O] ≈ 0

In this reaction, the concentration of water is omitted from the Keq expression because, although water is a reactant, it is also the solvent. As the solvent, the number of water molecules is significantly greater than the number of all other species in the solution; the HC2H3O2 is dilute and dissociates only to a small extent. Therefore, in relative terms, the [H2O] (55.5 M) is essentially constant and is omitted D is wrong b/c water is solvent and can't have concentration equal 0 correct answer is B

Moving down the column, the alkaline-earth metals are observed to give increasingly vigorous reactions when forming ionic bonds with nonmetals (Reactions 1-3). Based on atomic properties, this trend in reactivity is best explained by comparing: A.the energy required to remove an electron from each atom. B.the tendency of each atom to attract electrons within a bond. C.the extent to which the electron cloud of each atom can be distorted by an external charge. D.the energy released when an electron is added to each atom

Ionization energy is the energy required to remove an electron from an atom. Because of the associated ionization and electron transfer involved in forming ionic bonds, the reactivity of atoms forming ionic compounds increases as the ionization energy decreases. -> moving down the column, ionization energy decrease => this makes removing an electron more favorable and increase reactivity Choice B is wrong -> Although the reactivity of the alkaline-earth metals does increase as electronegativity decreases, this correlation does not explain the cause of the reactivity trend. choice C is wrong -> Reactions 1-3 depend on the transfer of electrons (ionic bond formation) rather than on the distortion of the electron cloud. correct answer is A

An unknown metal M forms an ionic hydroxide with the formula M(OH)2 that exhibits the equilibrium: M(OH)2(s)⇄M2+(aq)+2 OH−(aq) in a saturated aqueous solution. If the solution pH is 10, the solubility product constant Ksp of the compound is: A.5.0 × 10−31 B.5.0 × 10−13. C.1.0 × 10−12. D.1.0 × 10−8.

Ksp: solubility product constant [M2+] = 0.5*[OH-] Ksp = [OH-][M2+] = (1*10^-4)^2 * (5*10^-5) = 5 * 10^-13 correct answer is B

Compared to the effective nuclear charge of 18O, the effective nuclear charge of 18F is: A.higher, because 18F and 18O have the same number of core electrons but 18F has fewer neutrons.[8%] B.lower, because 18F has more valence electrons than 18O.[20%] C.higher, because 18F and 18O have the same number of core electrons but 18F has more protons.[67%] D.lower, because 18F has an equal number of protons and electrons.[3%]

O and F have the same number of core electrons, but b/c F has more protons, its effective nuclear charge will be greater as its Z (nuclear charge) is 9 and O's nuclear charge is 8. Zeff_F = 9 - 2 = 7 Zeff_O = 8 - 2 = 6 However, When comparing atoms within the same period (row) of the periodic table, this approximation shows that Zeff increases as the atomic number increases, but this approximation does not distinguish between different rows of the periodic table and only shows that Zeff increases as the group (column) number increases. correct answer is C

Equilibrium is reached in a reversible reaction when the rate of the forward reaction equals the rate of the reverse reaction. The graph shows the changes in concentration during the formation of HI(g) from a mixture of H2(g) and I2(g). If the measurement is taken at equilibrium, how many moles of HI(g) are present in a 750 mL sample of the reaction mixture? ** #1 on word doc A.0.8 mol B.1.9 mol C.4.5 mol D.6.0 mol

Once equilibrium is achieved, the forward reaction generates products as fast as the reverse reaction converts those products back into the original reactants, and this causes the concentrations of the reactants and the products to become constant. -> concentrations are not equal though. only the rate is equal the equilibrium concentration of Hl is 6.0M 750ml = 0.75L 0.75L * 6mol/L = 4.5 mol choice B is wrong -> the intersection pt does not indicate equilibrium correct answer is C

Four alkyl halides: bromomethane, 2-bromobutane, 2-bromo-2-methylpropane, and 1-bromobutane, each react with water through solvolysis by an SN1 reaction in which the solvent acts as the nucleophile. In what order will these alkyl halides react to form alcohols, from fastest to slowest? A.bromomethane → 1-bromobutane → 2-bromobutane → 2-bromo-2-methylpropane B.bromomethane → 1-bromobutane → 2-bromo-2-methylpropane → 2-bromobutane C.2-bromo-2-methylpropane → 2-bromobutane → 1-bromobutane → bromomethane D.2-bromobutane → 2-bromo-2-methylpropane → 1-bromobutane → bromomethane

Sn1 rxn: most stable carbocation (= most substituded compound) will have the fastest rxn tertiary > secondary > primary > methyl alkyl halide correct answer is C

Suppose an experiment is to be performed utilizing a reaction that reduces ClO4− to ClO3−. Which experimental approach can be expected to make the reduction of ClO4− more favorable during the reaction? *** graph #1 on word doc A.Perform the reaction under acidic conditions[62%] B.Perform the reaction under neutral conditions[3%] C.Perform the reaction under basic conditions[27%] D.The pH of the reaction mixture does not make a difference[6%]

The passage states that on a Frost diagram, the slope of a line segment joining two species is equal to E° for the couple. As a result, when comparing line segments, the segment with the greatest positive slope is the more favorable reaction. -> the slope for acidic conditions (pH 0) is greater than the slope for basic conditions (pH 14). answer is A

chelate formation reactions

a metal cation and a ligand react to form one or more rings via a pincer-like coordinate bonding arrangement. - will form ring structure

high-performance liquid chromatography (HPLC)

a purification technique ideal for small sample sizes, separates compounds based on polarity 1. normal phase (NP-HPLC) - polar stationary phase and nonpolar mobile phase - more retention time for polar molecules 2. reverse phase (RR-HPLC) - nonpolar stationary phase and polar mobile phase - more retention time for nonpolar molecules

disproportionation reaction

a reaction in which an element in one oxidation state is both oxidized and reduced - to find this rx, need to compare oxidation number is reactants and in product ex: 3 K2MnO4 + 2 CO2 → 2 KMnO4 + 2 K2CO3 + MnO2 Mn in reactant +6, but in product is +7 and +4

What is the approximate molar concentration of Na+ ions in a solution prepared by dissolving 8.2 g of Na3PO4 (MM = 163.94 g/mol) into 250 mL of water? A.6.0 × 10−4 M[7%] B.3.3 × 10−2 M[15%] C.2.0 × 10−1 M[35%] D.6.0 × 10−1 M[40%]

be careful what question is asking correct answer is D

If a channel of a particular cell preferentially transmits smaller ionic species across the membrane, given the following biologically relevant, isoelectronic ions, the channel is LEAST likely to transmit: A.S2− B.Cl− C.K+ D.Ca2+

compare to natural element, its cation will be smaller but its anion will be bigger - losing electron to form cation will cause the remaining electrons to experience a greater effective nuclear charge (Zeff_, pulling the electrons closer to the nucleus -> the size is reduced - gaining electron to form anion produces greater electronic repulsion and nuclear shielding (lesser Zeff), which pushes electrons farther from the nucleus. -> size increases isoelectronic mean anions, cations, and natural atom have the same number of electrons. but the number of protons is different in each ion, the electrons in each ion experience a different Zeff. - in an isoelectronic series, ionic radii decrease as the atomic number increases - Because sulfur has the lowest atomic number (fewest number of protons) in this isoelectronic series, Zeff is smallest in S2−, making it the largest ion. - in isoelectronic series, cation are smaller than anions and neutral atom, so eliminate C and D correct answer is A

If radium-226 were to undergo radioactive decay by electron capture (a type of beta decay) instead of by alpha emission, the resulting nucleus would be: A.Rn B.Fr C.Ra D.Ac

electron capture will decrease atomic number by 1. Ra has 88 protons, so the result atom should have 87 protons, that give us Fr correct answer is B

Ozone (O3) in the atmosphere protects against harmful UV radiation. Its formation proceeds in two steps and is initiated when molecular oxygen (O2) splits into two oxygen atoms upon absorption of UV light. Which of the following would be the second step if ozone formation goes to completion? A.O + O2 → O3 B.2O + 2O2 → 2O3 C.3O2 → 2O3 D.O + 2O2 → O3

first rxn: O2 → 2O second rxn: 2O + 2O2 → 2O3 net rxn: 3O2 → 2O3 choice A is wrong -> if this were the second step it would not allow the reaction to go to completion because the resulting net reaction would be 2O2 → O3 + O. correct answer is B

The overall reaction for the electrolysis of molten sodium chloride, 2 NaCl(l) −→−−−−−−electric current 2 Na(l)+Cl2(g) can be expressed as two separate, simultaneous redox processes (Reactions 1 and 2): 2 Cl− → Cl2(g)+2 e− Reaction 1 2 Na+ + 2 e− → 2 Na(l) Reaction 2 What is the amount of electric charge required to produce 0.80 mol of Na(l) during the electrolysis? A.0.8 faradays B.1.6 faradays C.2.4 faradays D.3.2 faradays

from rxn 2 , 0.8 mol of NaCl must undergo electrolysis since 2 mol of e- is required for every 2 mol of NaCl that undergo electrolysis, 0.8 mol NaCl will require 0.8 mol of electrons The charge used during electrolysis can be measured in a unit called the faraday, where 1 faraday equals the electric charge present in 1 mole of electrons: 0.8 mol of electron = 0.80 faradays of electric charge are required correct answer is A

Suppose that Experiment 2 is repeated using chloride salts of the Group 1 metals. Compared with the concentration of Cl− in the 10 mM Mg2+ solution, the concentration of Cl− in the 5 mM Na+ solution would be: A.higher. B.the same C.2 times lower D.4 times lower

moles of Cl- in MgCl2 solution: 1 L * 0.01M MgCl2 * 2 moles Cl- = 0.02 mole Cl- moles of Cl- in NaCl solution: 1L * 0.005 M NaCl * 1 mole Cl- = 0.005 moles Cl- 0.02/0.005 = 4 correct answer is D

SN1 reaction

nucleophilic substitutions that occur in two steps - step 1: the bond between the leaving group and carbon is broken and a carbocation intermediate is formed. - step 2: a nucleophile attacks the carbocation , forming a bond between the nucleophile and the carbocation to yield the final product. - Reactivity of the alkyl halide substrate is dependent on its level of substitution. -> tertiary compound is the most stable carbocation and undergo SN1 rxn fastest

Based on the reaction in Figure 1 and the energy profile in Figure 2, which statement correctly describes both Compound 2 and Compound 3? ** #2 on word doc A.Compound 2 and Compound 3 are both the kinetic products of the reaction. B.Compound 2 is the kinetic product, and Compound 3 is the thermodynamic product C.Compound 2 is the thermodynamic product, and Compound 3 is the kinetic product. D.Compound 2 and Compound 3 are both the thermodynamic products of the reaction.

product with the lowest energy form fast -> making it kinetic product product with the lowest energy level indicate its stable -> making it thermodynamic product => but it form slowly b/c it has higher energy barrier correct answer is B

Polarizability

the extent to which an electron cloud of an atom can be distorted by an external charge or by an applied electric field to produce a dipole