Chemistry & Orgo

Consider the cyclic form of the sugar from which Compound 1 is derived. Can this sugar be classified as a pyranose and a hemiacetal? A) Yes; pyranoses are six-membered rings and hemiacetals are formed from aldoses B) Yes; pyranoses are five-membered rings and hemiacetals are formed from ketoses C) No; pyranoses are five-membered rings and hemiacetals are formed from aldoses D) No; pyranoses are six-membered rings and hemiacetals are formed from ketoses

A - A cyclic sugar can be classified as a furanose or a pyranose based on its structural similarity to the heterocycles furan or pyran, respectively. Sugars that are five-membered rings (four carbon atoms and one oxygen atom) are furanoses, and sugars that are six-membered rings (five carbon atoms and one oxygen atom) are pyranoses. Compound 1 is a six-membered ring containing one oxygen and five carbon atoms; therefore, Compound 1 is derived from a pyranose. The structure from which Compound 1 is derived from is a hemiacetal because C-1 is bonded to two oxygen atoms (a hydroxyl group and an -OR group), a hydrogen atom, and an alkyl group that formed from an aldose.

In a hyperbaric chamber, the volume of intravascular air bubbles decreases as pressure increases. Which of the following gas laws best explains this behavior? A) Boyle's law B) Charles' law C) Avogadro's law D) Gay-Lussac's law

A - According to the kinetic molecular theory of gases, pressure results from the number of collisions that gas particles make with the walls of a container. Decreasing volume while maintaining the same temperature (ie, speed of gas molecules) and the same number of gas particles increases the number of collisions with the walls of the container and increases the pressure. Consequently, pressure and volume are inversely proportional, and every gas has a particular pressure-volume constant at a given temperature. Boyle's law describes this inverse relationship between volume V and pressure P for a fixed number of molecules of an ideal gas at constant temperature according to the following equations: PV = k where k is a constant. These equations show that as pressure increases, volume decreases (and vice versa). Therefore, Boyle's law best explains the decrease in the volume of intervascular air bubbles as the pressure in the hyperbaric chamber increases. (Choice B) Charles' law describes the directly proportional relationship between the volume and the absolute temperature of a fixed number of molecules of an ideal gas at a constant pressure, and V/T is a constant. (Choice C) Avogadro's law describes the directly proportional relationship between the number of moles and the volume of an ideal gas at constant temperature and pressure, and V/n is a constant. (Choice D) Gay-Lussac's law describes the directly proportional relationship between the pressure and the absolute temperature of a fixed number of molecules of an ideal gas kept at a constant volume, where P∝TP∝T and P/T is a constant.

Zinc, titanium, iron, and aluminum are all ingredients found in inorganic sunscreens. Which of the following choices correctly lists the atoms in order of increasing atomic radius? A) Al < Zn < Fe < Ti B) Zn < Fe < Ti < Al C) Ti < Fe < Zn < Al D) Al < Ti < Fe < Zn

A - Atomic radius is defined as the distance from the nucleus of the atom to the outermost electron. The periodic table has the following trends for the atomic radius of atoms: Within a group (column), the atomic radius increases from top to bottom because each successive atom in the group has an additional electron shell. The outer electrons are shielded from the pull of the protons by the inner electrons. The electrons experience a decreased attractive force toward the nucleus (effective nuclear charge, Zeff), so they orbit farther from the nucleus. Within each period (row), the atomic radius gradually increases from right to left. With each step to the right, an electron and a proton are added. The charge in the nucleus increases with each additional proton, causing a gradual drawing in of the electron cloud. Because the number of electron shells is constant within a period, each new electron is added to the same shell and cannot effectively shield other electrons from the nucleus. Because the addition of shells has a strong effect on shielding, changes in atomic radius down a group are more pronounced than for those across a period. Aluminum has the smallest atomic radius because it is located in the highest row and farthest to the right in the periodic table. Zinc, iron, and titanium are all in the same period, so the trend from right to left is Zn < Fe < Ti.

Which gas would occupy more volume at a constant temperature and pressure, 1.5 g of N2 gas or 1.5 g of O2 gas? A) N2 because there are more moles of N2 B) O2 because its molecules are larger C) Neither gas occupies any volume D) Both gases occupy equal volumes

A - Because the masses of both N2 and O2 are equal (1.5 g each), the gas with the lower molar mass has more molecules. Using the periodic table, the molar mass of N2 is calculated to be 28 g/mol, and the molar mass of O2 is 32 g/mol. Therefore, the N2 gas has more moles and occupies a larger volume.

High school chemistry student is given a flask containing n-octane, boiling point equals 125°C. If she must react this compound in such a way as to raise boiling point above 145°C, what should she do? A) Use free radical halogenation to add a chlorine atom to one of the compounds terminal carbons. B) Utilize a Gringnard reaction to further alkylate the compound, thus increasing its molecular weight. C) Hydrolyze the compound into two smaller alkanes. D) Isomerize the compound to make it more highly branched without changing its molecular formula.

A - Boiling point is mainly determined by intermolecular forces in molecular weight. Since unsubstituted alkanes are able to exert only London dispersion forces, they typically display low boiling points. However, if the student were able to halogenate n-octane, she would be giving it a polar bond between the chlorine atom in one carbon. This would allow the product to experience the dipole-dipole forces, raising it's boiling temperature. B: Grignard processes are typically used to alkylate compounds that contain carbon all functionalities. C: Higher, not lower, molecular weights make compounds more difficult to boil. D: Branching, which decreases the surface area available for intermolecular attractions, tends to lower the boiling point

If you wish to promote an E2 reaction, you should use a substrate with: A) An anti-periplanar hydrogen atom, a strong base, heat, and a polar aprotic solvent. B) A substrate with an anti-periplanar hydrogen atom, mild base, heat, and a polar aprotic solvent. C) A tertiary substrate, a strong base, heat, and a polar aprotic solvent. D) A tertiary substrate, mild base, heat and polar product solvent.

A - E2 reactions are by molecular eliminations. These reactions are analogous to us and to mechanisms, as both our second order and involve a one step mechanism. Unlike us into reactions, however, E2 processes must involve a hydrogen that is anti-to the living group. Additionally, use of a strong base tends to push a reaction towards E2, while E1 reactions can proceed with weaker basic species, finally, like as in two processes, E2 reactions prefer aprotic solvents.

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

A - Elements with loosely bound valence electrons have lower first ionization energies (i.e., are easier to ionize) than elements with tightly bound valence electrons. On the periodic table, the first ionization energy tends to increase with increasing atomic number moving across a period (row), but it tends to decrease moving down a group (column) with some intermittent exceptions. Alkali metals occupy Group 1 and have the lowest first ionization energies of all the element groups. As the atoms in the alkali metal group increase in size moving down the column, the valence electron occupies an increasingly higher energy level, and therefore is less tightly bound. As a result, 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. As the smallest atom in the group with a valence electron in a lower energy level, lithium is the least reactive of the alkali metals in Reactions 1-3.

What can be inferred about the HPLC column used in the passage if tryptophan has a shorter retention time than Compound 2? A) A reverse-phase HPLC column was used with a nonpolar stationary phase and polar mobile phase B) A reverse-phase HPLC column was used with a polar stationary phase and nonpolar mobile phase C) A normal-phase HPLC column was used with a polar stationary phase and nonpolar mobile phase D) A normal-phase HPLC column was used with a nonpolar stationary phase and polar mobile phase

A - High-performance liquid chromatography (HPLC) is a purification technique used for small sample sizes. The instrumentation consists of a sample injector, a column (stationary phase), solvent under pressure (mobile phase), a detector, and a computer for data acquisition. Two types of columns—normal-phase (NP) or reverse-phase (RP)—can be used, depending on the polarity of the compounds being separated. NP-HPLC is used to separate relatively nonpolar compounds and consists of a polar stationary phase (typically silica) and a nonpolar mobile phase. RP-HPLC is used to separate polar compounds and has a nonpolar stationary phase(typically C-18 alkyl hydrocarbon) and a polar mobile phase. Nonpolar compounds in a mixture will interact more with the stationary phase than polar compounds, causing nonpolar compounds to have a longer retention time. The only structural difference between Compound 2 and tryptophan is the Fmoc protecting group on Compound 2, which contains a polycyclic aromatic hydrocarbon and an ester. The large nonpolar hydrocarbon group decreases the polarity of Compound 2 relative to tryptophan. Because tryptophan has a shorter retention time and is more polar than Compound 2, tryptophan interacted less with the stationary phase than Compound 2. Therefore, the mobile phase is polar and the stationary phase is nonpolar. These conditions indicate that an RP-HPLC column was used.

Which of the following molecules is unable to form hydrogen bonds with water? A) C2H4Br2 B) C3H9N C) C9H18O2 D) C2F4

A - Hydrogen bonding is a strong intermolecular force that occurs between a hydrogen atom on one molecule and an electronegative atom with a lone electron pair on another molecule. The hydrogen atom is the bond donor and must be covalently attached to a small electronegative atom that withdraws electrons from hydrogen, giving it a partial positive charge. The partial positive charge is then attracted to the partial negative charge of the lone pair of electrons in the bond acceptor. These interactions can only occur over short distances, so they require sufficiently small electronegative atomsin both the donor and acceptor molecules, and are essentially limited to fluorine, nitrogen, and oxygen. Carbon is a small atom but not sufficiently electronegative to participate in hydrogen bonding. On the other hand, bromine is electronegative but too large to participate in hydrogen bonding. Therefore, C2H4Br2 is unable to form hydrogen bonds with water.

Which of the following does NOT describe why a nucleophile will react more quickly with acetic anhydride than with N,N-diisopropylisobutyramide? A) N,N-diisopropylisobutyramide has a greater inductive effect than acetic anhydride B) Acetic anhydride is a stronger electrophile than N,N-diisopropylisobutyramide C) Acetic anhydride is less sterically hindered than N,N-diisopropylisobutyramide D) N,N-diisopropylisobutyramide has a less stable leaving group than acetic anhydride

A - Induction is an electronic property in which electrons are donated through sigma bonds. Electronegative atoms or electron withdrawing substituents pull electrons away from an adjacent atom toward themselves, creating a dipole with a partial negative charge on the electronegative atom and a partial positive charge on the adjacent atom. The closer the electron withdrawing group to an atom, the greater the inductive effect that atom experiences. Electron withdrawing groups also make good leaving groups because they can stabilize the negative charge acquired after being eliminated. Acetic anhydride and N,N-diisopropylisobutyramide both have two electronegative atoms bonded to each carbonyl carbon. The anhydride carbonyl carbons are bound to two oxygen atoms whereas the amide carbonyl carbon is bound to an oxygen and a nitrogen. Because oxygen is more electronegative than nitrogen, it draws more electron density away from the carbonyl carbon than nitrogen does, creating a larger dipole on the anhydride than on the amide, as well as a more electrophilic carbonyl carbon on the anhydride (Choice B). Therefore, an anhydride, such as acetic anhydride, has a greater inductive effect than an amide, such as N,N-diisopropylisobutyramide.

After isolation of compound 1, an infrared spectrum was obtained. The spectrum gives all of the following information EXCEPT: A) The spin-spin splitting of atoms in a compound B) The signals corresponding to stretching vibrations and rotations C) The amount of light absorbed at a certain frequency D) The relative amount of energy needed to stretch a bond

A - Infrared spectroscopy is a technique used to determine the functional groups present in a sample; the data collected are plotted as percent transmittance vs. wavenumber. The signals in the spectrum correspond to bond-stretching vibrations and rotations at a certain frequency, and the signal intensity is dependent on the amount of energy absorbed.

What types of functional groups react to form the imine intermediate in the Strecker synthesis of amino acids? A) An aldehyde and ammonia B) An aldehyde and a secondary amine C) A ketone and a tertiary amine D) A ketone and a quaternary amine

A - Ketones and aldehydes react with ammonia (NH3) or primary (1°) amines to form imines, which are ketone and aldehyde analogues that contain a carbon-nitrogen double bond. The reaction begins with an acid-catalyzed addition of the amine to the carbonyl, which involves protonation of the carbonyl, nucleophilic attack of the carbonyl by the amine, and deprotonation of the amine. This forms an intermediate known as an α-aminoalcohol. The second part of the reaction is an acid-catalyzed dehydration involving protonation of the -OH group, loss of H2O, and deprotonation. The Strecker synthesis is used to make α-amino acids from aldehydes using NH3 and potassium cyanide (KCN). The first step of the reaction proceeds with protonation of the carbonyl oxygen by H3O+, followed by nucleophilic attack of the carbonyl carbon by NH3, resulting in dehydration and imine formation. Therefore, an aldehyde and NH3 are used to form the imine intermediate in the Strecker synthesis.

What functional group forms in an acid-catalyzed reaction between a secondary amine and a ketone? A) Enamine B) Nitrile C) Imine D) Alkene

A - Ketones and aldehydes react with primary (1°) and secondary (2°) amines to form a carbon-nitrogen bond. The reaction begins with an acid-catalyzedaddition of the amine to the carbonyl, which involves protonation of the carbonyl, nucleophilic attack of the carbonyl by the amine, and deprotonation of the amine. This forms an intermediate known as an α-aminoalcohol. The second part of the reaction is an acid-catalyzed dehydration involving protonation of the -OH group, loss of H2O, and deprotonation. The product formed is dependent on the type of amine used (1° or 2°). If a 1° amine is used, an imine (nitrogen analogue to ketones and aldehydes) is formed because the nitrogen has a proton available for removal in the last step, permitting C=N bond formation. If a 2° amine is used, an enamine (nitrogen analogue to an enol) is formed because the nitrogen does not have a proton available to be removed in the last step. Therefore, a proton is removed from the α-carbon, forming a C=C bond.

Nickel metal hydride (NiMH) batteries are rechargeable cells used in many hybrid at all electric vehicles. And NiMH battery in the process of recharging is similar to a: A) An electrolytic cell. B) A galvanic cell. C) A concentration cell. D) A half cell.

A - Rechargeable batteries work via reversible redox processes. One such as cell is discharging, it is releasing stored energy. This is similar to the spontaneous processing and galvanic cells. However, this question asks about recharging. Here, we must use an external power source to move electrons in a way that would be naturally non-spontaneous, just as an electrolytic cell does. C: A concentration cell is a type of galvanic cell. It is that's more closely related to discharge than to recharge. D: A half cell includes only 1/2 have a redox reaction, either reduction oxidation. As no redox process can occur without 2/2 reactions, this choice does not make sense.

The strongest attractive force for linkage, of those below, is: A) A nonpolar covalent bond. B) A hydrogen bond. C) A dipole dipole force. D) London dispersion force.

A - Remember, intramolecular bonds, which attach atoms of the same molecule, always stronger than intermolecular attractions. Non-polar covalent bonds are intramolecular. B: This is the strongest of the listed intermolecular forces.

Considering the type of element represented by each of the following, which of the elements below is most likely to have the highest electrical conductivity? A) Ca B) Si C) Se D) I

A - Silicon is a metalloid. Metalloids tend to be more conductive than nonmetals but less conductive than metals. (Choices C and D) Selenium and iodine are both nonmetals. Nonmetals tend to be the least conductive when compared to metalloids and metals.

The C=S double bond in carbon disulfide (CS2) consists of a combination of one σ bond and one π bond. Compared to the bond dissociation energy (bond strength) of the π bond, the bond dissociation energy of the σ bond portion of the C=S bond is: A) Greater B) Less C) The same D) Proportional

A - Stronger bonds require more energy to break (dissociate) than weaker bonds. The overall strength of a bond results from the sum of all σ and π bonding contributors. As a result, when comparing bonds involving the same two types of atoms, a triple bond is stronger than a double bond, and a double bond is stronger than a single bond because double and triple bonds are composed of both σ and π bonds. However, if the strengths of the σ bond and π bond contributors are considered separately, π bonds are weaker than σ bonds. The end-to-end orbital overlap in σ bonds is more efficient than the side-to-side orbital overlap in π bonds. This causes σ bonds to exist in a more stable, lower energy state. As a result, breaking a σ bond requires more added energy than breaking a π bond (ie, a σ bond has a greater dissociation energy).

If researchers used the Gabriel synthesis to make the dansylalanine backbone, which of the following statements is true? A) Potassium phthalimide is a starting material B) Diethyl malonate is a starting material C) Potassium cyanide is a starting material D) An aldehyde is a starting material

A - The Gabriel synthesis, also known as the malonic ester synthesis, is a method used to make primary amines, including α-amino acids, without overalkylation of the amine. The amine is generated from potassium phthalimide, a "protected" form of ammonia that prevents multiple alkylations due to the steric hindrance of the phthalimide group. The synthesis begins with an SN2 reaction, where potassium phthalimide is the nucleophile that attacks an alkyl halide. The amine is then "deprotected" by removal of the phthalimide group. For the synthesis of amino acids, potassium phthalimide is the nucleophile that attacks the electrophilic carbon of diethyl bromomalonate, and bromide is the leaving group. The α-carbon is deprotonated with a base and attacks an alkyl halide to add the amino acid side chain via another SN2 reaction. The phthalimide is removed through a basic hydrolysis reaction, yielding a dicarboxylic acid. Lastly, the diacid is decarboxylated with acid and heat. The resulting loss of CO2yields the final amino acid. The starting reagents for this synthesis are planar and do not have chiral centers; therefore, the amino acid product will be a mixture of L- and D-amino acids.

Suppose that Experiment 1 was done in diethyl ether instead of acetone, and NaI was replaced with another nucleophile. The conjugate base of which of the following compounds would be the most nucleophilic? A) CH3(CH2)2CH3, pKa = 50 B) (CH3)2NH, pKa = 40 C) CH3CH2OH, pKa = 15.9 D) HF, pKa = 3.2

A - The conjugate bases of CH3(CH2)2CH3, NH(CH3)2, HOCH2CH3, and HF are −:CH2(CH2)2CH3, −:N(CH3)2, −OCH2CH3, and F−, respectively. To compare the relative nucleophilicity of each, the electronegativity of the negatively charged atoms must be considered. The charged carbon atom of butane is less electronegative than the nitrogen atom of the amine, the oxygen atom of the alcohol, and the fluorine atom of the hydrogen halide. Therefore, electrons on a negatively charged carbon atom are less stabilized and more easily shared with an electrophile, giving butane (a very weak acid with a high pKa) the strongest conjugate base and making −:CH2(CH2)2CH3 the best nucleophile.

The acid dissociation constant Ka for HCN is approximately 1 × 10−9. What is the value of the pKb? A) 5 B) 9 C) 14 D) −5

A - The pKb may be found from the pKa using the relationship: pKb = pKw − pKa pKb = 14 − 9 = 5 (Choice B) 9 is the pKa rather than the pKb. (Choice C) 14 is the pKw rather than the pKb. (Choice D) −5 results from a calculation error. The negative sign is incorrect and would result from either dividing Ka by Kw or from subtracting pKw from pKa.

In an aqueous solution of 1 × 10−3 M HCN (Ka ≈ 1 × 10−9), HCN can best be described as a: A) Weak acid, because [H+] < 1 × 10−3 M B) Strong acid, because [H+] = 1 × 10−6 M C) Weak base, because [H+] < 1 × 10−7 M D) Strong base, because [H+] ≈ 1 × 10−3 M

A - The small Ka value for HCN indicates that the non-ionized form is favored, and it is a weak acid. Evaluation of the [H+] gives [H+] = 1 × 10−6 M. This [H+] is much less than the HCN concentration (1 × 10−3 M) and confirms that only a small percentage of the dissolved HCN molecules in solution are ionized (consistent with a weak acid). (Choice B) Strong acids ionize essentially 100% into H+ and A−. If HCN were a strong acid, the [H+] would be very close to 1 × 10−3 M (the initial HCN concentration). (Choice C) Although the nitrogen atom of the cyanide group does have an available lone pair of electrons that could act as a weak proton acceptor (weak base), HCN is donating protons (acting as a weak acid). The calculated value of [H+] = 1 × 10−6 M (pH 6) is greater (more acidic) than a neutral solution ([H+] = 1 × 10−7 M, pH 7) and confirms that HCN acts overall as a weak acid rather than as a weak base. (Choice D) A strong base would fully dissociate and provide a large [OH−]. This would result in a very small [H+] and a solution pH much higher than 7. An [H+] ≈ 1 × 10−3 M would indicate acidic conditions.

Would the concentration of H2SO4 remain constant as a lead storage battery is discharged? A) No, because water is one of the products produced when the battery is discharged B) Yes, because PbSO4 increases as the battery is discharged C) No, because the SO42- ions are oxidized in the reaction D) Yes, because the H+ ions are not oxidized or reduced in the reaction

A - When a lead storage battery is discharged, lead oxide (PbO2) gains electrons, or is reduced. Based on the oxidation-reduction reactions for the lead storage battery, water is one of the products from the reduction of PbO2. This oxidation-reduction reaction occurs in an acidic medium. Before discharging, the concentration of H2SO4 is 4 M, but after the battery is completely discharged, the acid is diluted by the additional water, lowering the concentration of H2SO4. Because the concentration of H2SO4 changes when the battery is discharged, it is often used as an indicator of when the battery needs to be charged.

A certain electrochemical cell contains a positive (+) anode. What type of cell must it be? A) An electrolytic cell. B) A galvanic cell. C) Either an electrolytic or a galvanic cell. D) Neither in electric nor a galvanic cell.

A - While the anode is a site of oxidation in all cells, it's sign is not always the same. Specifically, remember that electrolytic cells are non-spontaneous. Since electrons always flow from anode to cathode, the anode of such a cell must be positive. The resulting positives and negative electron flow exemplifies a non-spontaneous process. Galvanic cells have negative anode. This way, electrons can flow from a negative to a positive region, a process that is spontaneous in nature.

Unlike SN1 reaction, which require at least two steps, S and two reactions can proceed there a single transition state. This high energy species is best described as: A) Pentavalent, because a nucleophile approaches from the side opposite the living group. B) Pentavalent, because a nucleophile approach is from the same side as a leaving group. C) Tetrahedral, because a nuclear file approaches the vacant orbital from the same side as a leaving group. D) Tetrahedral, because a nucleophile can approach the vacant orbit off from either side of the species being attacked.

A - as a result of sterile constraints imposed upon the geometry of the transition state, the nucleophile always approaches the electro file from the side opposite the living group, that's why I sent two reactions are described as backside attack. Since none of the orbitals are empty, all are occupied by Adams or lone pairs, the transition state reaction center is transiently coordinated by five different substituents.

Which of the following statements regarding chemical kinetics are always true? I. According to collision theory, reaction rate increases with increasing concentration. II. Raising the temperature will always increase the total amount of products formed. III. The activation energy for a forward reaction is equal to that of the reverse reaction. IV. Catalyst decreases the delta G of a reaction. A) I only B) I and II only C) I, II, and IV D) I, II, III, and IV

A - collision theory states that reaction rate is dependent on the number of molecules that collide in a favorable way. As concentration increases, so will the total number of favorable collisions. II. virtually all reactions to increasing rate of temperature is increased, however, this is separate from the change that occurred to the equilibria. exothermic reactions, which release heat, looks great experience a shift toward the reactants and the temperature is raised. for this reason, the relative amount of products compared to reactants will drop. III. Activation energy is a difference in energy between the starting species as a high energy transition rate. Since reactants and products typically have different potential energy, the energetic distance between them and the transition state will differ. This will yield different activation energy for the forward and reverse reactions. IV. Catalyst do lower the activation energy, which corresponds to the energy of the transition state however, they do not impact delta G, which is the free energy difference between reactants and products.

Grignard Regent

A Grignard reagent is an organometallic compound, for example CH3MgBr, that can be used to add alklyl groups to aldehyde or ketone's. Such reagents are generally produced by adding the associated alcohol halide, CH3Br, to a vessel containing diethyl ether and small pieces of solid Mg.

Extraction

Acylation reactions between anhydrides and amines generate amides and carboxylic acids. Amides and carboxylic acids are polar, water-soluble molecules because they form hydrogen bonds in an aqueous solution. However, long alkyl side chains (R groups) of amides and carboxylic acids increase the hydrophobic (nonpolar) nature of the molecules and make them insoluble in aqueous solvents. Long-chain amides and carboxylic acids are the products of the reaction between long-chain anhydrides and the amine group of phenylethylamine. To separate long-chain amides from carboxylic acids in the organic layer, a base must be added. Bases deprotonate the carboxylic acid and generate carboxylate anions that are more soluble in the aqueous layer than in the organic layer. Dilute LiOH is strong enough to deprotonate a carboxylic acid but not to deprotonate an amide N-H. Therefore, addition of the base LiOH will produce carboxylate anions that enter the aqueous layer and allow long-chain amides to remain in the organic layer.

Alkali metals

Alkali metals are reactive especially with water due to the low ionization energies, and they become more reactive as you move down a group.

Salt bridge

Anions and cations traveled through a salt bridge in opposite directions. Since electrons move from right to left, we might previously have assumed that S04 2- ions will travel the other way to neutralize the resulting charge gradient. However, now we realize an Na+ ions are present instead. These cat ions will move from right to left in a 1:1 ratio with electrons, preventing charge buildup and allowing the reaction to continue as long as possible. Sulfate certainly did not participate in the redox reaction described, but it still could have function is an important species due to his relationship with a salt bridge. Na+ has a highly negative reduction potential and is this incredibly unlikely to reduce.

Anomers vs Epimers

Anomers are a subtype of epimer that differ only at the anomeric carbon. Epimer's are molecules that differ a single stereogenic center.

Atomic radius

Atomic radius increases as you move down and to the left along the periodic table.

In the alveoli, hypoventilation (decreased gas exchange) and hyperventilation (increased gas exchange) can lead to abnormalities in acid-base homeostasis. Which of the following is true of the immediate effects of abnormal ventilation? A) Hyperventilation decreases blood pH and decreases HCO3− levels. B) Hyperventilation increases blood pH and decreases HCO3− levels C) Hypoventilation increases blood pH and increases HCO3− levels D) Hypoventilation decreases blood pH and decreases HCO3− levels

B - According to Le Châtelier principle, chemical reactions are in a state of dynamic equilibrium to maintain optimal ratios of reactants and products. At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. If there are changes in conditions (eg, changes in concentrations or changes in temperature or pressure), the equilibrium will shift to counteract these changes. A decrease in the concentration of a chemical species shifts the equilibrium toward that species whereas an increase in concentration shifts the equilibrium away from that species. During ventilation, CO2 is expelled from the lungs. An increase in ventilation (hyperventilation) decreases levels of CO2in the blood. Removal of CO2 then shifts the reaction to the left to maintain equilibrium. This shift results in a decrease in carbonic acid levels, which translates to a decrease in HCO3− and H+ (increase in pH) in pulmonary circulation. Clinically, this is referred to as respiratory alkalosis.

What happens to the pOH of the gastric acid entering the duodenum (pH 6.0) as a result of the introduction of bicarbonate ions from incoming pancreatic secretions? A) The pOH increases as [OH−] increases B) The pOH decreases as [OH−] increases C) The pOH increases as [OH−] decreases D) The pOH decreases as [OH−] decreases

B - Accordingly, lower pOH values indicate more alkaline solutions with higher [OH−]. This is opposite to the pH scale, in which lower pH values indicate more acidic solutions. Consequently, pH and pOH are inversely correlated. As pH rises, pOH falls (and vice versa) while maintaining the relationship pH + pOH = 14. The bicarbonate ion (HCO3−) is a base (proton acceptor) capable of neutralizing gastric acid through the formation of carbon dioxide and water: H+ + HCO3− → CO2 + H2O The [H+] decreases as bicarbonate consumes the available H+ from the gastric acid, and the relative [OH−] increases. As a result, the duodenum (pH 6.0) is less acidic than the upper stomach (pH = 1.0). Therefore, going from the highly acidic stomach to the less acidic duodenum gives a higher pH and a lower pOH, meaning [H+] decreases and [OH−] increases.

The bonds of four salts (MgBr2, NaCl, MgCl2, and NaBr) are evaluated for ionic character based on electronegativity differences between atoms. What is the expected order of these four salts if listed according to bond character from the least ionic character to the greatest ionic character? A) NaBr < NaCl < MgBr2 < MgCl2 B) MgBr2 < MgCl2 < NaBr < NaCl C) NaCl < NaBr < MgCl2 < MgBr2 D) MgBr2 < MgCl2 < NaCl < NaBr

B - An ionic bond between two atoms requires the transfer of valence electrons from one atom to the other, creating a cation and an anion. The ionic character increases as the difference in electronegativity between two atoms increases. Electronegativity tends to increase when moving across a period (row) from left to right and when ascending a group (column) on the periodic table, with some exceptions in the transition metals. Comparing NaBr and MgBr2, NaBr has a higher electronegativity difference (and higher ionic character) because Na is further to the left of Br on the periodic table than Mg. Similarly, NaCl has a higher ionic character than MgCl2. Of the two higher-ranking sodium salts, NaCl has the greatest electronegativity difference (ionic character) as Cl is above Br in the halogens column. Therefore, the expected order of these four salts if listed from the least ionic character to the greatest ionic character is MgBr2 < MgCl2 < NaBr < NaCl.

Barbiturates are known to inhibit complex I but do not affect complexes II-IV. Based on this fact, how will barbiturates affect the proton motive force? A) The proton motive force increases B) The proton motive force decreases C) The proton motive force is unchanged D) The proton motive force becomes zero

B - As stated in the passage, the proton motive force (pmf) arises as hydrogen ions (protons) are pumped from the mitochondrial matrix to the intermembrane space by complexes I, III, and IV. It is the driving force behind ATP synthesis in the mitochondria. When complex I is inhibited, the ETC can still operate from complex II onward as long as FADH2can continue to provide complex II with electrons. Complexes III and IV continue to pump protons, and the pmf remains above zero (Choice D). Because complex I normally participates in proton pumping, its inhibition will result in fewer protons being pumped across the mitochondrial membrane. Therefore, the H+ concentration gradient is reduced and the pmf decreases, resulting in decreased ATP synthesis.

Ethyl acetate could be better separated from ethanol by doing which of the following? A) Replacing helium with nitrogen as the carrier gas B) Running the mixture through a longer column C) Starting the chromatograph at a higher temperature D) Increasing the carrier gas flow rate

B - Chromatography methods separate molecules based on their relative affinities for a stationary phase versus a mobile phase. For compounds to separate efficiently on any column, they need sufficient time to interact with the stationary phase. Given enough time, even subtle differences in affinity for the stationary phase can be amplified, allowing separation of compounds with similar properties. Increasing the length of the stationary phase through which the compounds must travel can provide the necessary time. Ethanol and ethyl acetate have similar, but not identical, boiling points and can be separated on a gas chromatograph by running them through a longer column.

Which of these choices best exemplifies a concentration cell? A) A galvanic cell with to half cells that contain 2M Sn(NO3)2 and 0.05 M MgCl2, respectively. B) A galvanic cell with to half cells that contain 2 M Sn(NO3)2 and 0.10 M SnCl2, respectively. C) An electrolytic cell with 2 half cells that contain 2 M Sn(NO3)2 and 0.0 5 M MgCl2, respectively. D) An electrolytic cell with 2 half cells that contain 2 M Sn(NO3)2 and 0.10 M SnCl2, respectively.

B - Concentration cells are a form of galvanic, spontaneous, cell. Specifically, they include the same species in both half cells, and that function sonly if the two sides container different concentrate is. Here, for example one and a half cell is highly concentrate it an Sn 2+, making that side very positive. The other side of significantly more dilute. Thus, electron will move to the concentrated side until an equilibrium is established. A: This choice properly identifies the cell as galvanic however, it's half cells contain entirely separate species. C, D: Concentration cells are not electrolytic. In fact, they function only because different concentrations of the same ion can produce a very small positive potential.

In metal complexes such as copper(II) glycinate, the coordination number refers to: A) The number of ligands that form the complex. B) The number of coordinate bonds formed. C) The number of electrons involved in coordinate bonding. D) The oxidation number of the metal ion.

B - Coordinate covalent bonds, unlike covalent bonds, are formed between two atoms when both of the shared electrons are donated by the same atom. Such coordinate bonds are often formed between electron-poor metal ions and molecules called ligands that contain one or more electron-rich atoms with available lone-pair electrons. The coordinately bonded metal and its ligands are called a complex. As originally defined by Alfred Werner, the coordination number of a metal complex refers to the number of coordinate bonds formed between the central metal ion and its nearest neighboring atoms. When all of these nearest neighboring atoms are from separate molecules or ions, the number of ligands will equal the coordination number. However, if two or more of these nearest neighboring atoms are joined to the same coordinating ligand unit, then the number ligands will not equal the coordination number. In both cases, the number of nearest neighboring atoms and coordinate bonds is unchanged, but the number of ligand units is different. For example, if bonding interactions from other ligands such as water are excluded (as shown in Figure 1), both isomers of the copper(II) glycinate complex have a coordination number of 4. The Cu2+ atom has two oxygen atoms and two nitrogen atoms as its nearest neighbors, forming four coordinate bonds. However, each pair of N and O atoms is joined to the same glycine unit, and the four neighboring atoms come from only two glycine ligands.

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

B - For example, reactions producing an insoluble solid (precipitate) crashing from the solution are said to be precipitate formation reactions whereas those involving Brønsted-Lowry acids and bases that react to produce an ionic salt and water are classified as acid-base neutralization reactions. Two other important reaction classifications are chelate formation and redox reactions. In chelate formation reactions, a metal cation and a ligand react to form one or more rings via a pincer-like coordinate bonding arrangement. In an oxidation-reduction (redox) reaction, the oxidation states of some atoms change during the conversion of the reactants to the products. The given reaction between Ni and Br2 is a combination reaction in which two reactants form a single ionic product, NiBr2. The assigned oxidation states for the elements in this reaction show that nickel goes from an oxidation state of zero in its elemental metal form to a state of +2 in the product (an oxidation), and bromine goes from an oxidation state of zero to a state of −1 (a reduction). Therefore, in addition to being a combination reaction, the reaction between Ni and Br2 is also an oxidation-reduction reaction.

Which of the following characterize gas-liquid chromatography? I. Gas mobile phase II. Liquid stationary phase III. Separation based on polarity IV. Room-temperature column conditions A) I only B) I and II only C) I and III only D) I, II, and IV only

B - Gas-liquid chromatography is a technique used to separate molecules in a mixture based on their boiling points. A gas chromatograph consists of an injection port, mobile and stationary phases, a column in a heated oven, a detector, and a computer for data analysis. The mobile phase is an inert gas such as helium or nitrogen (Number I), and the stationary phase is a liquid that coats a solid support on the inside of the column (Number II). A small amount of a liquid mixture is injected into the gas chromatograph, and the compounds in the mixture are vaporized by heating. The vapors then travel through the column in a heated oven to the detector. The most volatile molecules (low boiling points) spend more time in the gas phase than they spend interacting with the stationary phase of the column, so they migrate to the detector rapidly. However, molecules with higher boiling points condense more readily and spend more time interacting with the liquid stationary phase. These molecules make their way through the column slowly as the temperature in the oven increases. (Number III) Components of a mixture are separated by boiling point rather than polarity. Although polarity impacts a compound's boiling point, it is not the only determining factor. A sufficiently large nonpolar molecule can have a higher boiling point than a polar molecule. For example, a nonpolar 20-carbon alkane chain has a higher boiling point than a polar 4-carbon carboxylic acid (343 °C vs 164 °C) and would move through the column more slowly despite being non-polar. (Number IV) The column must be placed in a heated environment to allow volatile molecules to remain in the gas phase. Columns kept at room temperature would not separate the molecules.

Which of the following are naturally occurring R amino acids? I. Glycine. II. Cysteine. III. Aspartic acid. IV. Proline. A) I only B) II only C) III and IV only D) II, III, and IV

B - Glycine is the only amino acid that is achiral. All other common residues have S configurations, with the exception of Cysteine, the only common amino acid designated as R.

Which of the following changes could be made to this experiment to increase the signal intensity of peaks that correspond to smaller mass-to-charge (m/z) values relative to larger m/z values? A) Increase the pH of the sample solvent B) Decrease the pH of the sample solvent C) Decrease the concentration of the sample D) Increase the concentration of the sample

B - In MS analysis, signal intensity, or the height of a data peak relative to other peaks, corresponds to the relative quantity of ions at each mass-to-charge ratio m/z. During electrospray ionization (ESI), particles are charged via protonation within droplets. Ions that have multiple charges have a smaller m/z ratio because the denominator (charge) has increased. In spectrometry data, this translates into greater signal intensities for these ions. Lowering the pH of the solvent increases the concentration of protons and results in a greater number of multiply charged molecules (z > 1). An increase in multiply charged molecules would therefore result in greater signal intensity for smaller m/z values.

What are the bond angles of the water molecule formed during Reaction 1, the methylated carbon atom of the product from Reaction 1, and the alkene formed in Reaction 2, respectively? A) 104.5°, 104.5°, and 120° B) 104.5°, 109.5°, and 120° C) 109.5°, 109.5°, and 180° D) 180°, 109.5°, and 120°

B - In general, sp3-, sp2-, and sp-hybridized atoms have bond angles of 109.5°, 120°, and 180°, respectively. Lone pairs and electrons involved in pi bonding, especially in molecules with larger dipoles, may further distort predicted bond angles. For example, the water molecule (H2O) displays a bond angle of 104.5° due to increased electron repulsion of its lone pairs despite its sp3-hybridized orbitals, which would normally suggest a bond angle of 109.5°. The bond angles of water, the methylated carbon atom on vitamin K 2,3-epoxide, and the newly formed alkene in vitamin K quinone are 104.5° (sp3), 109.5° (sp3), and 120° (sp2), respectively.

At higher concentrations, alkali metal ions close to the size of a potassium ion can displace potassium ions in biological systems and promote low potassium levels (hypokalemia). Which cation is most likely to displace K+ ions? A) Li+ B) Rb+ C) Cs+ D) Fr+

B - Ionic radii tend to decrease in size across a period (row) of the periodic table (left to right) and increase moving down a group (column). This trend occurs for metal cations, and then resets and repeats for anions beginning near the division between metals and nonmetals, past which anions tend to preferentially form. Losing electrons to form a cation causes the remaining electrons to experience a greater effective nuclear charge (Zeff), pulling the electrons closer to the nucleus. Conversely, gaining electrons to form an anion produces greater electronic repulsion and nuclear shielding (lesser Zeff), which pushes electrons farther from the nucleus. Alkali metal ions that are more similar in size to K+ ions are the most likely to displace a K+ ion. Because the size of the alkali metal ions increases moving down the column, the elements adjacent to potassium will form ions that are closer to the size of the K+ ion. As a result, out of the options given, a Rb+ ion is closer to the size of a K+ ion and will be more likely to displace K+ ions than Li+, Cs+, or Fr+ ions. (Choice A) Because alkali metal ions decrease in size moving up the column, Li+ is much smaller than K+.

A wax is made up of a long-chain fatty acid and a long-chain alcohol joined by: A) n amide linkage B) An ester linkage C) An anhydride linkage D) A glycosidic linkage

B - Lipids are hydrophobic molecules that are classified based on characteristics such as length, linkage type, and solubility. These molecules are broadly classified as hydrolyzable (lipid contains hydrolyzable ester or amide linkages) or nonhydrolyzable (lipid does not contain hydrolyzable ester or amide linkages). The hydrolyzable lipids, such as phospholipids, glycolipids, and waxes, are further classified based on the structure of their backbone (glycerophospholipids, glycosphingolipids, etc.). Waxes are an example of a hydrolyzable lipid that serves as a form of protection in plants and animals. Waxes are made up of a long-chain fatty acid and a long-chain alcohol. The wax forms when the acid and the alcohol combine through dehydration to form an ester bond.

An element that exists in its standard state as a diatomic gas and contains a double covalent bond is: A) Nitrogen B) Oxygen C) Chlorine D) Krypton

B - Of the common representative elements, seven exist as diatomic molecules in their standard elemental state. These seven elements are hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine. To achieve a more stable electron configuration with a full valence shell, atoms of these elements form covalently bonded pairs by sharing valence electrons and occur naturally as H2, N2, O2, F2, Cl2, Br2, and I2, respectively. These elements exist as gases at room temperature with the exception of bromine (a liquid) and iodine (a solid). Due to differences in orbitals and the number of valence electrons, H2, F2, Cl2, Br2, and I2 can achieve a full valence shell by forming a pair with only a single covalent bond, but O2 must form a double bond and N2 must form a triple bond. In the second period (row) on the periodic table, nitrogen, oxygen, fluorine, and neon are all elemental gases, but only oxygen exists as a diatomic molecule with a double covalent bond in its elemental state. (Choice A) Nitrogen is an elemental gas in the second period, but it forms a diatomic molecule with a triple covalent bond in its elemental state. (Choice C) Chlorine is an elemental gas in the third period. Also, chlorine forms a diatomic molecule with only a singlecovalent bond in its elemental state. (Choice D) Krypton is an elemental gas in the fourth period. As a noble gas, it does not form a diatomic molecule in its elemental state.

For biological processes, which of the following is true of oxidation-reduction reactions? I. Reduction often involves the formation of bonds to hydrogen. II. Oxidation often involves the formation of bonds to oxygen. III. Reducing agents undergo a change to a lower oxidation state. A) I only B) I and II only C) II and III only D) I, II, and III

B - Oxidation-reduction (redox) reactions are reactions in which electrons are transferred from one molecule to another. One way to remember the flow of electrons in an oxidation-reduction reaction is with the acronym "OIL RIG": oxidation is loss of electrons, reduction is gain of electrons. However, different compounds transfer electrons in different ways. Unlike metals, which transfer electrons directly and have an oxidation state equal to their net charge, organic molecules generally transfer electrons via hydrogen or oxygen. When hydrogen forms a bond to carbon, the more electronegative carbon atom takes the electron from hydrogen and is reduced (Number I). Loss of a bond to hydrogen or formation of a bond to electronegative oxygen results in carbon losing electrons, which is oxidation (Number II). A simple way to determine the oxidation state of an organic molecule is to count the number of C-O and C-H bonds. For example, an aldehyde is in a higher oxidation state than an alcohol because it has two C-O bonds in the form of a double bond compared to the single C-O bond in alcohols. Alkanes are more reduced than alkenes because they have more bonds to hydrogen. (Number III) Reducing agents cause the reduction of other molecules and become oxidized in the process. They change toward a higher oxidation state.

The radioisotope gallium-68 is used in nuclear medicine as a traceable radioactive emitter for diagnostic imaging procedures. The 68Ga isotope is generated from 68Ge by: A) Adding a proton to the nucleus B) Converting a proton in the nucleus to a neutron by electron capture C) Converting a neutron in the nucleus to a proton by β− decay D) Removing three electrons from the atom

B - Radioactive beta decay can occur in three forms: β− decay (electron emission), β+ decay (positron emission), and electron capture. In all three forms of beta decay, the mass number remains unchanged whereas the atomic number either increases (β− decay) or decreases (β+ decay and electron capture) by 1. As the atomic number changes, the identity of the element changes accordingly. During electron capture, a nuclear proton captures an inner electron near the nucleus and then converts to a neutron. This decreases the atomic number by 1 but leaves the mass number the same. In the conversion of 68Ge into 68Ga, the identity of the element changes; therefore, the number of protons (atomic number) must decrease by 1. Because the mass number of 68 is unchanged, the loss of 1 proton must then be accompanied by the gain of 1 neutron. Therefore, 68Ga is generated from 68Ge by converting a proton to a neutron (as occurs during electron capture).

During the synthesis of a tosylate, which of the following pairs of atoms form a covalent bond? A) C and Cl B) S and O C) S and Cl D) C and O

B - Substitution reactions require strong nucleophiles and electrophiles, as well as good leaving groups. The nucleophile attacks the electrophile and displaces the leaving group. A bond is formed between the nucleophile and electrophile, whereas the bond between the electrophile and the leaving group is broken. Alcohols are weak electrophiles and poor leaving groups; therefore, a nucleophile is unlikely to displace the hydroxyl group (-OH) of an alcohol. To improve its leaving group ability, an alcohol can be converted into a tosylate (-SO3C6H4CH3) or a mesylate (-SO3CH3). Tosylates are produced by the reaction of an alcohol with p-toluenesulfonyl chloride (TsCl) and a base, such as pyridine. The hydroxyl O acts as the nucleophile to attack the S of TsCl, and Cl acts as the leaving group. Then pyridine deprotonates the O atom, giving the tosylate product. Therefore, a bond between S and O is formed during the synthesis of a tosylate.

In the ultraviolet (UV) range, pyrimidine has maximum absorptions at 240 nm and 280 nm, and the absorption at 280 nm is due to nonbonding electron excitation. The nonbonding electrons on nitrogen in pyrimidine undergo which of the following transitions when absorbing UV light? A) π → π* B) n → π* C) π* → π D) π* → n

B - The electromagnetic spectrum consists of waves of varying wavelengths, energy, and frequency. The spectrum spans from radio waves (long wavelength and low frequency and energy) to gamma rays (short wavelength and high frequency and energy). Ultraviolet (UV) light, which corresponds to wavelengths between 200 nm and 400 nm, is used in UV spectroscopy. In UV spectroscopy, a compound in solution is irradiated with UV light; the amount of UV light absorbed by the compound is measured at each wavelength and a spectrum is generated by plotting absorbance as a function of wavelength. Absorption of UV light causes an electron transition, or excitation, from the ground state to a higher energy level. In the ground state, π electrons from double bonds are in the π bonding molecular orbital, and nonbonding electrons are in the n nonbonding molecular orbital. Upon interactions with UV light of sufficient energy, these electrons are excited to the Lowest Unoccupied Molecular Orbital (LUMO), called the π* antibonding orbital. The lone electron pairs on nitrogen in pyrimidine are nonbonding (n) electrons. Therefore, when absorbing UV light, these electrons are excited to the antibonding (π*) orbital and undergo a n → π* transition.

If the electromotive force of the battery in an AED is found to be −2.0 V while it is charging, the battery is functioning as a: A) Galvanic cell B) Electrolytic cell C) Concentration cell D) Fuel cell

B - The electromotive force Eocell for a cell is the difference between the standard reduction potential of the reaction at the cathode and the anode. The standard reduction potential of a particular metal or molecule is the potential (in volts) required to reduce the compound. The more negative the value of Eo for any given compound, the less likely it is to be reduced. The equation below is used to determine the standard potential for an oxidation-reduction pair for a particular electrochemical cell: E∘cell=E∘cathode−E∘anode For an electrolytic cell, this value is negative, indicating that the oxidation-reduction reaction is not spontaneous. During the charging phase of a rechargeable battery, an external potential is applied to force the oxidation-reduction reaction to proceed in a nonspontaneous direction. Because Eocell for this battery is a negative value (−2.0 V), the cell is functioning as an electrolytic cell. (Choice A) A galvanic cell is an electrochemical cell that converts chemical energy into electrical energy via an oxidation-reduction reaction. This reaction occurs spontaneously, and Eocell is a positive value. (Choice C) A concentration cell is a type of galvanic cell in which ions diffuse across a membrane, creating a potential. Because concentration is the driving force behind electron transfer, the standard potential of a concentration cell is positive whereas the concentration is different between the two compartments of the cell until it reaches equilibrium. (Choice D) A fuel cell is a type of galvanic cell whose reactants are continuously supplied at the anode and cathode, and whose products are continuously removed from the system. Although outside material is continuously supplied to the fuel cell, its reaction is still spontaneous, meaning that the Eocell is positive.

Which of the following affects the freezing point of an ideal solution? I. The strength of intermolecular forces II. The strength of intramolecular forces III. The reactivity of the solute A) I only B) II only C) I and III only D) II and III only

B - The freezing point is the temperature at which solid and liquid phases of a substance are in equilibrium. Freezing occurs when the kinetic energy of a molecule can no longer overcome the intermolecular forces binding it to nearby molecules. If a molecule experiences strong intermolecular forces with surrounding molecules, the kinetic energy required to overcome those forces is greater, and the freezing point temperature is correspondingly higher. Therefore, the freezing point of water is impacted by the strength of intermolecular forces (Number I). (Number II) Intramolecular forces are the forces within a molecule, and these forces do not impact interactions between molecules. Therefore, they do not affect the freezing point of any substance. (Number III) Raoult law states that for an ideal solution, the addition of a solute lowers the freezing point of a liquid according to the amount of solute added. As solute is added, the solvent's concentration decreases, leading to fewer solvent-solvent intermolecular interactions. However, the freezing point (a physical property) is not affected by the reactivity (a chemical property) of the solute. Properties that are affected by the amount but not the identity of the solute are called colligative properties.

Phospholipids contain all of the following structural features EXCEPT: A) A phosphate B) Phenol C) A glycerol backbone D) Fatty acid chains

B - The general structure of a phospholipid includes a polar head group and nonpolar tail joined together by a molecule referred to as the backbone. The hydrophilic head contains a substituted phosphate, and the hydrophobic tail is made of two fatty acid chains (Choices A and D). The hydrophilic head and the hydrophobic tail are linked together by a glycerol backbone (Choice C). Two adjacent carbons of glycerol are bonded to the fatty acid chains through an ester bond, and the third glycerol carbon is bonded to phosphate through a phosphodiester bond. A variety of phospholipids can be made with different head groups on phosphate, a different backbone, and different fatty acids (saturated or unsaturated). The fatty acid double bond configuration causes the hydrophobic tail to take on different shapes and affects the properties of the phospholipid, such as membrane fluidity.

Which of the following values gives the number of parts per million (ppm) of NaCl in a 0.90% saline solution? A) 900 B) 9,000 C) 90,000 D) 900,000

B - The relationship between percent and ppm is a matter of scale because the basis for ppm is 10,000 times larger than the basis for percent (1,000,000 ÷ 100 = 10,000). As a result 1% = 10,000 ppm. Using this equality as a conversion factor, the equivalent ppm concentration for the 0.90% aqueous NaCl solution can be determined by multiplying the percentage by the conversion factor: 0.90% × 10,000 ppm/percent = 9,000 ppm.

What is true of diastereomers? I. they are impossible to separate via fractional distillation or extraction. II. They most likely have at least somewhat different solubility as an aqueous solution. III. Their physical properties are identical. A) I only B) II only C) III only D) I, II, and III

B) Diasteromers, unlike enantiomers, exhibit different physical properties. This quality allows them to be separated fairly easily. In other words, while their solubilities, boiling points, and melting points are likely similar, they are not identical. They are easier to separate than enantiomers, not diasteromers, have indistinguishable chemical and physical properties.

A student needs to separate a mixture of chloroform (bp 61°C) and benzene (bp 80°C). What type(s) of distillation would be expected to give the best separation of the two compounds? I. Simple distillation II. Fractional distillation III. Vacuum distillation A) I only B) II only C) I and III only D) II and III only

B. - There are two variations of simple distillation: Fractional distillation works best for compounds that have boiling points that are <25°C apart, such as chloroform and benzene (19°C difference in boiling points). The setup is the same as a simple distillation, with the addition of a fractionating column between the flask and the condenser to provide a larger surface area on which the vapors of the distillate condense before distilling into the receiving flask. The increased area allows for better separation of compounds with similar boiling points (Number II). Vacuum distillation is appropriate for compounds that have boiling points >150°C, which may decompose when heated beyond this temperature. To avoid decomposition, the distillation apparatus is connected to a vacuum, decreasing the pressure of the system and consequently lowering the boiling points of the mixture components. Neither chloroform nor benzene has a boiling point >150°C (Number III).

How is the activation energy for the decomposition of H2O2 affected by the addition of NaI to the coffee cup? A) Increased, because adding NaI changes the concentration of the reactants in the reaction vessel B) Increased, because NaI lowers the bond dissociation energy of H2O2 C) Decreased, because NaI stabilizes the transition state of the reaction D) Decreased, because NaI increases the amount of heat produced from the reaction

C - A catalyst is a compound that increases the rate of a reaction without being consumed by the reaction. It does this by lowering the reaction's activation energy (the difference between the energy of the transition state and the energy of the reactants). The transition state is an unstable, transient formation (activated complex) that is much higher in energy than the reactants. A catalyst stabilizes the transition state for a reaction (creating a less energetic transition state). After the products are formed, the catalyst is recovered. The iodide ion (I−) from NaI acts as a catalyst in the decomposition of H2O2 by stabilizing the transition state formed when the bonds in H2O2 break to form H2O and O2. This decreases the activation energy and increases the rate of the decomposition of H2O2.

Which of the following is a combustion reaction? A) H2CO3 → CO2 + H2O B) C + H2O → CO + H2 C) CH4 + 2 O2 → CO2 + 2 H2O D) C + CO2 → 2 CO

C - Chemical reactions can be categorized broadly (but not exclusively) into five main types: combination, decomposition, single replacement, double replacement, and combustion. Combustion reactions involve the burning of a fuel (often a hydrocarbon, an alcohol, or another flammable organic compound) in the presence of oxygen, often from the air. When the supply of oxygen is plentiful, complete combustion occurs to form carbon dioxide (CO2) and water. The reaction of methane (CH4), a hydrocarbon fuel, with oxygen to form CO2 and water is an example of a complete combustion reaction. Variations of combustion reactions also exist. For example, if the supply of oxygen is limited, carbon monoxide (CO) may form. Additionally, if the fuel contains atoms of nitrogen, sulfur, or other elements, then oxides of these atoms can also be produced as products.

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 B) SF6 C) S2Cl2 D) S4N4

C - Covalent bonds between atoms are formed by sharing electrons, which is made possible through the overlap of the atomic orbitals from each atom. Covalent bonds made by the end-to-end overlap of atomic orbitals are called sigma (σ) bonds, whereas covalent bonds made by the side-to-side overlap of p orbitals are called pi (π) bonds. A single covalent bond consists only of a σ bond, a double bond consists of one σ bond and one π bond, and a triple bond consists of one σ bond and two π bonds. 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). On the periodic table, atomic radii tend to decrease across a row(due to an increasing effective nuclear charge) and increase down a column (due to having an additional electron shell). In this question, the given compounds all have σ bonds to a sulfur atom. As a result, the longest bond between sulfur and another element will be formed with the element that has the largest atomic radius. Because Cl has a larger atomic radius than H, N, or F, the S-Cl bond in S2Cl2 will the longest bond with sulfur among the given compounds.

Researchers purify two samples of the same mixture, one by simple distillation and the other by vacuum distillation. Which of the following expressions correctly compares the relationship between the mixture components' boiling points (bp) during each distillation? A) Simple distillation bp = vacuum distillation bp B) Simple distillation bp < vacuum distillation bp C) Simple distillation bp > vacuum distillation bp D) Simple distillation bp = 2(vacuum distillation bp)

C - Distillation is a technique used to separate liquid molecules based on their boiling point, the temperature at which the vapor pressure equals the pressure of the system. There are three common types of distillation: simple, fractional, and vacuum. The type of distillation suitable for a mixture is dependent on the boiling points of the mixture components. Simple distillation is used for compounds with boiling points less than 150 °C, and that are more than 25 °C apart from each other. Vacuum distillation is used for compounds that decompose at their boiling point (typically boiling points greater than 150 °C), causing the boiling point to decrease and thereby preventing degradation. Simple and vacuum distillations are set up in the same way, except that simple distillations are done at atmospheric pressure, whereas vacuum distillations are connected to a vacuum pump and performed at reduced pressure. Because vacuum distillations are performed under reduced pressure, the boiling point of a compound under vacuum will decrease relative to its boiling point at atmospheric pressure. Therefore, the simple distillation boiling point of a molecule is greater than the boiling point of that molecule under vacuum (ie, simple distillation bp > vacuum distillation bp).

An optically active molecule that rotates plane polarized light clockwise reacts with methanol in solution via an sn1 mechanism. In which direction will the product solution rotate plane polarized light? A) Clockwise. B) Counter clockwise. C) No rotation will be observed. D) Optical rotation can only be determined experimentally.

C - During an SN1 reaction, the stereochemistry of the original substrate is lost when the leaving group detaches in the carbocation intermediate forms. The nucleophile will attack this ion without a stereo chemical preference, thus, a racemic mixture will form, containing 50% of each enantiome. Racemic mixture is do not rotate light overall.

Researchers separated fatty acid sodium salts from glycerol by adding HCl and then extracting with hexanes. Which statement correctly describes the extraction process? A) HCl protonates the carboxylate, and the fatty acid is found in the aqueous layer B) The carboxyl group is deprotonated, and the fatty acid is dissolved in the polar solvent C) HCl protonates the carboxylate, and the fatty acid is found in the organic layer D) The carboxyl group is deprotonated, and the fatty acid is dissolved in the nonpolar solvent

C - Fatty acid sodium salts are soluble in aqueous solutions because the charged carboxyl group and sodium ion interact with water through ion-dipole interactions. Protonation of fatty acid sodium salts renders the molecule insoluble in water because the fatty acid is no longer a charged species, and therefore cannot interact as readily with water. In addition, the long-chain hydrocarbons on fatty acids greatly increase the hydrophobicity of these molecules, causing fatty acids to be insoluble in water and soluble in an organic solvent. Separation of glycerol from fatty acid sodium salts can be accomplished by protonation of the negatively charged carboxyl group, which can be achieved by the addition of a strong acid, such as HCl. Hexanes are nonpolar organic molecules that can readily interact with the protonated fatty acids. Therefore, protonated fatty acids will enter the organic layer after extraction with the organic solvent hexanes.

Which separation technique is optimal for purification of small sample sizes and employs a stationary phase, a solvent mobile phase (under pressure), and a detector to separate compound mixtures based on polarity? A) Gas chromatography B) Extraction C) High-performance liquid chromatography D) Thin-layer chromatography

C - High-performance liquid chromatography (HPLC) is a purification technique comprised of a mobile phase, stationary phase, detector, and computer for data acquisition. The mobile phase is a solvent or mixture of solvents pumped through the system under pressure. The stationary phase is a column made of either a polar (normal-phase) or nonpolar (reversed-phase) material, depending on the nature of the compounds to be separated. HPLC is optimal for small sample sizes and separates compounds based on their polarity, giving each compound a different affinity for the mobile and stationary phases and causing the compounds to pass through the column at different rates. Compounds are detected as they come off the column and go into the waste or are collected separately. Data is transmitted to the computer, creating a chromatograph.

The regulation of mineral ions in the cellular fluids of biological systems is performed by ion pumps that selectively transport ions of specific size and charge across cellular membranes. Which of the following isoelectronic species is the smallest ion? A) Na+ B) F− C) Mg2+ D) O2−

C - Ionic radii tend to decrease in size across a period (row) of the periodic table (left to right) and increase moving down a group (column). This trend occurs for metal cations, and then resets and repeats for anions beginning near the division between metals and nonmetals, past which anions tend to preferentially form. Compared to the neutral atom of a given element, its cation will be smaller but its anion will be larger. Losing electrons to form a cation causes the remaining electrons to experience a greater effective nuclear charge (Zeff), pulling the electrons closer to the nucleus. Conversely, gaining electrons to form an anion produces greater electronic repulsion and nuclear shielding (lesser Zeff), which pushes electrons farther from the nucleus. Na+, F−, Mg2+, and O2− ions are isoelectronic (have the same number of electrons), but because the number of protons is different in each ion, the electrons in each ion experience a different Zeff. Therefore, in an isoelectronic series, ionic radii decrease with increasing atomic number. Because magnesium has the highest atomic number (greatest number of protons) in the isoelectronic series, Zeff is greatest in this ion, making it the smallest within the given series.

Litmus paper will turn blue when immersed in all of the following EXCEPT: A) Ca(OH)2 B) NH3 C) HNO3 D) NaCN

C - Litmus paper is used as a pH indicator to determine whether a solution is acidic or basic. The paper is coated with litmus dye that changes color depending on the pH of the solution in which it is immersed. An acidic (pH < 7) solution will turn litmus paper red whereas a basic (pH > 7) solution will turn litmus paper blue. The Brønsted-Lowry definitions of acids and bases can be used to identify acidic and basic molecules. Brønsted-Lowry acids donate a proton (H+ ion), and Brønsted-Lowry bases accept a proton. The negatively charged ions OH− and CN− in Ca(OH)2 and NaCN, respectively, can accept a proton (Choices A and D). The N atom in NH3 (Choice B) has a lone pair of electrons, which can accept an H+ ion. Because these molecules are bases (proton acceptors), they will turn litmus paper blue. HNO3 can donate a proton, making it an acid. Therefore, when litmus paper is immersed in HNO3, it will turn red rather than blue.

The experimental data obtained when a chiral molecule interacts with plane-polarized light is called: A) The polarimeter B) The R or S configuration C) The specific rotation D) The infrared spectrum

C - Normal light is made of light waves that oscillate in all planes perpendicular to the direction of propagation. When light encounters a polarizer, only waves that are aligned with the polarizer can pass. Therefore, all the light that passes through the polarizer oscillates in one direction. This resulting uniform light is known as plane-polarized (linear) light. Chiral molecules rotate plane-polarized light; the angle of rotation, known as the specific rotation, is unique for each chiral molecule and must be determined experimentally. A molecule's specific rotation contains direction (+ or −) and magnitude. Rotations that are clockwise are (+) whereas those that are counterclockwise are (−). Enantiomers have specific rotations of equal magnitude but in opposite directions.

Which of the following compounds is a strong acid? A) Nitrous acid. B) Hypochlorous acid. C) Perchloric acid. D) Sulfurous acid.

C - Perchloric acid (HClO4) is one of the strong acids you should know. A: Nitrous acid (HNO2) is weak. Make sure you do not confuse it with nitric acid (HNO3). B: Hypochlorous acid (HClO) is weak. D: Sulfurous acid (H2S03) is weak. Do not confuse it with sulfuric acid (H2SO4)

In a galvanic cell at standard conditions: Zn 2+ = -0.76 Cu 2+ = 0.34 A) Zn 2+ will reduce at the cathode and Cu (s) will oxidize at the anode. B) Zn(s) will oxidize at the cathode and Cu 2+ will reduce at the anode. C) Zn(s) will oxidize at the anode and Cu 2+ will reduce at the cathode. D) Zn 2+ will reduce at the anode and Cu(s) will oxidize at the cathode.

C - Reactions in galvanic cells occur spontaneously. Since Cu 2+ has a more positive reduction potential than Zn 2+, it is Cu 2+ that will reduce and Zn(s) that will oxidize. Oxidation and reduction always occur according to the mnemonic redcat, reduction happens at the cathode and oxidation takes place at the anode. Regardless of the type of cell, oxidation never occurs at the cathode in reduction never occurs at the anode.

Which term can be used to classify the relationship between two isomers that have the same connectivity but specific rotations of +40° and −25°, respectively? A) Enantiomers B) Racemic mixture C) Diastereomers D) Conformational isomers

C - Specific rotation, the degree to which chiral molecules rotate plane-polarized light, is unique to each chiral molecule. The value of specific rotation consists of direction (+ or −) and magnitude (number of degrees). Clockwise rotations are designated as (+), whereas counterclockwise rotations are (−). Isomers are molecules with the same molecular formula but have either different connectivity (constitutional isomers) or spatial orientation (stereoisomers). Diastereomers differ at one or more stereocenters but have the orientation of at least one stereocenter in common. The specific rotations of diastereomers differ in magnitude and may differ in direction. The same is true for constitutional isomers. Because the question states that the two molecules have the same connectivity, they cannot be constitutional isomers. Because they have different specific rotations (direction and magnitude), it can be concluded that the two molecules are diastereomers.

If an alcohol were to undergo a substitution reaction, which of the following modifications would improve the alcohol's leaving group ability? A) Protection of the alcohol B) Deprotonation of the alcohol C) Conversion of the alcohol to a mesylate D) Oxidation of the alcohol

C - Substitution reactions require strong nucleophiles and electrophiles and good leaving groups. When a nucleophile attacks an electrophile, a bond is formed between the nucleophile and electrophile whereas the bond between the electrophile and the leaving group is broken. A leaving group containing an electron withdrawing group can enhance the partial positive charge of the electrophile. Leaving groups are weak bases and are able to accept electrons; they often can be stabilized by the delocalization of electrons through resonance and an inductive effect. Alcohols are weak electrophiles and poor leaving groups; therefore, nucleophiles are unlikely to displace the hydroxyl group (-OH). The alcohol can be converted into a mesylate (-SO3CH3) to improve its leaving group ability. Mesylates are produced by the reaction of an alcohol with methanesulfonyl chloride and a base, such as triethylamine, and are good leaving groups because the mesylate anion can stabilize the negative charge through resonance. Similar to mesylates, tosylates (-SO3C6H4CH3) can also be used to make an alcohol a better leaving group.

Which of the following statements accurately describes the stereochemical outcome of the Strecker synthesis of alanine? A) The product is L-alanine B) The product is D-alanine C) The product is a mixture of L- and D-alanine D) Alanine does not contain a stereocenter

C - The Strecker synthesis is used to make α-amino acids from an aldehyde using ammonium chloride (NH4Cl) and potassium cyanide (KCN). The first step of the reaction proceeds with protonation of the carbonyl oxygen by ammonium (NH4+), followed by nucleophilic attack of the carbonyl carbon by ammonia (NH3), resulting in dehydration and imine formation. In the second step, a cyanide anion is added to the imine to form an aminonitrile. Finally, in the third step, the nitrile (R-CN) nitrogen is protonated, and two water molecules add to the nitrile carbon in succession, eliminating ammonia from the nitrile and forming the carboxylic acid of the α-amino acid. Alanine can be synthesized via the Strecker synthesis from acetaldehyde, NH4Cl, and KCN. Acetaldehyde does not have a stereocenter, and the imine intermediate also does not have a stereocenter and is a planar molecule, allowing nucleophilic attack to occur from either above or below the plane. Therefore, the reaction is not stereospecific and the product will be a mixture of L- and D-alanine. The same outcome is obtained for amino acids produced by the Gabriel synthesis because the final step (decarboxylation) is not stereospecific.

A linear five-carbon ketone has a boiling point most similar to: A) Five-carbon alkane B) A five-carbon carboxylic acid C) A branched aldehyde with the same molecular weight D) An alcohol with a higher molecular weight

C - The boiling point of a substance depends on the intermolecular forces holding the molecules together. Strong intermolecular forces require more energy to break, causing higher boiling points. Molecules with functional groups that participate in strong bonds experience stronger intermolecular forces and have higher boiling points than molecules with similar molecular weights (MWs) that participate only in weak bonds. Molecules with a greater surface area (higher MW or less branching) experience more intermolecular forces and have higher boiling points than similar molecules with smaller surface areas. The carbonyls of ketones and aldehydes have a partial positive charge on the carbon atom and a partial negative charge on the oxygen atom, forming a dipole moment and allowing dipole-dipole interactions to occur. Branching causes a relatively small boiling point difference in compounds with similar functional groups and MWs. Therefore, a five-carbon ketone and a branched aldehyde of the same MW have similar boiling points.

If the hydroxide concentration in common duct bile is 1 × 10−6 M, what is the pH of the bile? A) 6 B) 7 C) 8 D) 14

C - The pH of a solution relates directly to the hydrogen ion (or hydronium) concentration [H3O+] by the relationship pH = −log[H3O+]. If the hydroxide concentration [OH−] is known instead, it is related to the [H3O+] by the self-ionization constant Kw of water as given by: Kw = [H3O+][OH−] = 1.0 × 10−14 M2 Accordingly, dividing Kw by the known [OH−] permits the calculation of [H3O+] as follows: [H3O+] = Kw /[OH−] = (1.0 × 10−14 M2)/(1.0 × 10−6 M) [H3O+] = 1.0 × 10−8 M The pH of the solution can then be determined: pH = −log[H3O+] = −log(1.0 × 10−8) = 8 Alternatively, the pOH can be used to find the pH, given that pOH = −log[OH−] and that pH + pOH = 14 (as derived from Kw). By this method, calculation of the pOH is given by: pOH = −log[OH−] = −log(1.0 × 10−6) = 6 and the subsequent evaluation of the pH is given by: pH = (14 − pOH) = (14 − 6) = 8

Ketoses such as fructose are expected to give a positive Tollens test because: A) Ketoses cannot mutarotate B) Ketoses are not reducing sugars C) Ketoses tautomerize to aldoses D) Ketoses are not hemiacetals

C - Tollens test is used to identify the presence of aldehydes and hydroxy ketones, including reducing sugars, which have a free anomeric carbon. It utilizes the oxidizing agent Ag(NH3)2+ to oxidize aldehydes to carboxylic acids. A positive Tollens test results in the formation of metallic silver that looks like a silver mirror deposited on the glassware. Ketoses such as fructose are hydroxy ketones. Ketones cannot be directly oxidized to carboxylic acids; however, hydroxy ketones can undergo tautomerization via an enediol intermediate to produce an aldehyde under Tollens test basic conditions. Because of this tautomerization, ketoses give a positive result to Tollens test.

Study the equilibrium of an unknown reaction when he determines that the initial product and reactant concentration is too small. In this concentration profile: A) Q is equal to K. B) Q is greater than K. C) Q is less than K. D) None of the above.

C - if the relative amount of products is too small, the reaction is not a equilibrium, specifically, it must travel to the right to reach is equilibrium value. The ratio in question is Q, the reaction quotient, which year is less than K.

Carboxylic acids

Carboxylic acids, like asters, or relatively less prone to reduction that he tones and aldehydes. Reduction of a carboxylic acid requires a strong reducing agent, such as lithium aluminum hydride, LAH.

Which statements correctly describe electrolysis? During electrolysis: I. An electrochemical decomposition reaction occurs. II. Electric current is supplied as the energy to perform an endergonic reaction. III. The electrochemical cell operates with a negative ΔG. IV. electric current flows in a nonspontaneous direction within an electrochemical cell. A) I and IV only B) I, II, and III only C) II and IV only D) I, II, and IV only

D

The (R)-enantiomer of the antiasthma drug albuterol is the active isomer. If a researcher wants to separate a racemic mixture of albuterol, which of the following methods will most likely separate the enantiomers? A) Extraction with dilute base B) Thin-layer chromatography C) Fractional distillation D) Addition of a resolving agent

D - A 50:50 mixture of enantiomers is known as a racemic mixture. Enantiomers have many of the same chemical and physical properties, including melting and boiling points, solubility, and polarity, although they differ in the way that they interact with plane-polarized (linear) light. Therefore, separation of enantiomers relies on changing the physical properties of the molecules. The separation of enantiomers, such as those in the racemic mixture of albuterol, requires the addition of a resolving agent (a chiral molecule). When a resolving agent is added to a racemic mixture, it reacts with each enantiomer, forming a covalent bond or an ionic salt. Because the resolving agent is chiral, it incorporates a new chiral center into each enantiomer, creating a pair of diastereomers. Diastereomers can be separated from each other because, unlike enantiomers, they have different physical properties. Once the diastereomers are separated, the resolving agent is removed, yielding the original molecules as single enantiomers.

Carboxylic acid derivatives include: A) Aldehydes and ketones. B) Amides and ethers. C) Esthers and nitro groups. d) Thioesters and acyl halides.

D - A carboxylic acid derivative can be synthesized from a carboxylic acid via nucleophilic acyl substitution. Alternatively, such a compound is one that contains a carbonyl group immediately adjacent to a heteroatom, generally one that is electronegative. Esther includes a carbonyl next to a sulfur atom, while in acyl halide contains a carbonyl next to a halogen

Fructose for the synthesis of the Shi catalyst can be obtained by cleaving sucrose. What type of reaction occurs when sucrose is cleaved? A) Condensation of a glycosidic bond B) Condensation of a peptide bond C) Hydrolysis of a peptide bond D) Hydrolysis of a glycosidic bond

D - A glycosidic bond is the linkage between two sugar molecules in a disaccharide or polysaccharide, or between a sugar molecule and a hydroxyl group of another molecule. This bond is made up of the hemiacetal or hemiketal of one sugar and the hydroxyl (-OH) group of another and are connected via an α- or β-linkage. A glycosidic bond is broken via a hydrolysis reaction, frequently with the aid of an enzyme. In this reaction, water acts as a nucleophile and breaks the glycosidic bond, with one of the sugars acting as a leaving group. Therefore, the sugar is broken into two smaller components. The disaccharide sucrose is made up of fructose and glucose joined together by a glycosidic bond. To obtain fructose from sucrose, the glycosidic bond must be hydrolyzed.

Which of the following statements correctly describes the peaks displayed in a mass spectrum? A) The tallest peak is the molecular ion B) The peaks are uncharged molecules C) The height of each peak is dependent on molecular weight D) The peaks represent ionized fragments of the sample

D - An electric field accelerates the ions toward a magnet, which deflects them according to mass. The strength of the magnetic field is gradually changed during the experiment, and each field strength causes ions of a specific mass to reach the detector while all others are deflected into the walls of the tube. The ions are detected, and a mass spectrum is generated. The y-axis of the mass spectrum represents the ion mass abundance, and the x-axis represents the mass-to-charge ratio (m/z). The mass spectrum can be used to identify the mass of a molecule's fragments by taking the m/zdifference between peaks. Samples analyzed by mass spectrometry are ionized and fragmented before detection. Therefore, the peaks observed in the mass spectrum represent ionized fragments of the sample.

Which of the following statements describe a characteristic of an ideal gas? I. There are no attractive or repulsive forces between the gas molecules. II. The size (molecular volume) of the individual gas molecules is negligible. III. Collisions between the gas molecules are completely elastic. A) I and II only B) I and III only C) II and III only D) I, II, and III

D - An ideal gas is a hypothetical gas with characteristics that are consistent with certain simplifying assumptions made for systems involving real gases. An ideal gas has four characteristics: - An ideal gas has no attractive or repulsive forces between the gas molecules (Number I). - The size (molecular volume) of the individual gas molecules of an ideal gas is negligible (taken to be zero) compared to the volume (space) of the container the gas occupies (Number II). - Collisions between the molecules of an ideal gas are completely elastic (no energy is lost by interactions or friction) (Number III). - Ideal gas molecules have an average kinetic energy (energy of motion) that is directly proportional to the gas temperature.

If a student adds bromothymol blue as an indicator to determine the amount of titratable acid in 25 mL of urine according to Folin's method, what effect will the bromothymol blue have on the pH of the urine sample? A) It will increase the pH of the sample B) It will decrease the pH of the sample C) It will neutralize the sample D) It will have no effect on the pH of the sample

D - An indicator is used in acid-base titrations to determine the endpoint of a titration. A good indicator should change color close to the equivalence point of a titration. The equivalence point is the stoichiometric amount of titrant needed to react with all of the solution being titrated. In the case of a buffer solution, a titration is halfway to the equivalence point when a pH is achieved that is near the pKa value of the buffer's acid. Bromothymol blue is an indicator that changes from pale yellow to dark blue, from pH 6 to 8, but it has no effect on the pH of the solution. (Choices A and B) An indicator does not affect the pH of the system because it does not participate in the reaction. The pH will neither increase nor decrease. (Choice C) An indicator only responds to changes in pH; it does not aid in neutralizing the solution.

Propane is exposed to H2 and a solution containing a chunk of solid platinum. If it reacts successfully, the original alkyne will be: A) Be partially oxidized form propene. B) Be partially reduced to form propene. C) Be fully oxidized form propane. D) Be fully reduced to form propane.

D - Catalytic hydrogenation is a form of reduction. Here an alkaline is reduced to form the corresponding alkane, or single bonded carbon structure. B: While this partial reduction can occur, it requires the presence of Lindlars catalyst. Simple reaction with H2 in a Platinum or Pallidium catalyst will fully reduce an alkyne to an alkane.

Which of these combinations of solutions could be used to create an effective buffer? I. 500 mL of 1 M HCN in 500 mL of 1M KCN. II. 250 mL of 1.5 M HC2H302 and 250 mL of 1.5 M NH3. III. 500 mL of 1 M HCN 350 mL of 1 M KCN. IV. 500 mL of 1.5 M HC2H302 at 500 mL of 0.7 5 M NaOH. A) I only B) I and II only C) I, II, and IV D) I, III, and IV

D - Choice I is a classic buffer, with equal amounts of a weak acid and its conjugate base. Choice III contains the same conjugate pair but we're with more moles of acid. The concentrations of acid and conjugate base do not have to be equal to produce a buffer, just a produce one with pH equals PKA. Choice IV appears to contain a strong base, which should not appear in a buffer solution. However, note that the number of moles of NaOH is exactly half that of HC2H302, meaning that the base will be protonate 1/2 of the acidic acid molecules. This will leave a solution with equal amounts of HC2H302 and NAC 2H302, a perfect buffer. II: while this involves a weak acid and a weak base, they are not a conjugate pair.

Which of the following alkali metals is LEAST likely to accept an additional electron? A) Na B) K C) Rb D) Cs

D - Electron affinity is defined as the change in energy resulting from adding an electron (e−) to a neutral atom of an element (X) in the gas state to form an anion with a −1 charge: X + e− → X− When a stable anion is formed, energy is released as a product and a negative value results for the change in energy (energy of the products minus energy of the reactants) from the process. Therefore, electron affinity values with a more negative magnitude indicate elements that more readily accept the addition of an electron. In general, an additional electron is often more easily accepted by nonmetals than by metals. Periodic trends show that apart from some intermittent, regional exceptions caused by atomic size and orbital-filling effects, electron affinity tends to become stronger (more negative) moving left to right across a period and tends to become weaker (less negative) moving down a group on the periodic table. Consistent with these general trends, the electron affinity of the alkali metals in Group 1 weakens moving down the column. Therefore, cesium (Cs) has the weakest electron affinity and is the least likely (least favorable) to accept an additional electron. (Choices A, B, and C) Sodium (Na), potassium (K), and rubidium (Rb) are all alkali metals, but each has a stronger electron affinity than cesium (Cs). Adding an additional electron to Na, K, or Rb forms a more stable anion and releases more energy than is released from Cs.

Which statement correctly describes both the 3s and 3p energy levels in the lowest-energy electron configuration of the phosphide anion P3− ? A) Both the 3s and 3p levels are half-filled B) The 3s level is filled and the 3p level is half-filled C) The 3s level is half-filled and the 3p level is filled D) Both the 3s and 3p levels are filled

D - Elemental atoms contain an equal number of protons and electrons, resulting in no net charge. If an elemental atom gains or loses electrons and an ion is formed, the charge of the ion is determined by how many electrons are gained or lost. If electrons are lost (cation formation), electrons in the highest-energy filled shell/subshell level are removed first. Likewise, if electrons are gained (anion formation), the lowest-energy unfilled shell/subshell level is filled first. An elemental atom of phosphorus has a total of 15 electrons and 15 protons (no net charge) with an electron configuration of 1s22s22p63s23p3. Therefore, a phosphide anion is the result of an elemental phosphorus atom gaining 3 additional electrons within the half-filled 3p subshell. As a result, the phosphide anion has a total of 18 electrons with a configuration of 1s22s22p63s23p6. This gives the phosphide anion a −3 charge and filled 3s and 3p energy levels.

The solutes in an aqueous reaction mixture are separated by extraction. The function of the nonpolar organic solvent is to: A) Transfer hydrophilic molecules to the organic layer B) Combine the components of a reaction mixture C) Increase the volume of the solvent D) Remove hydrophobic molecules from the aqueous layer

D - Extraction is a technique used to separate molecules based on their solubility. It relies on the principle that "like dissolves like", meaning that molecules will dissolve in a solvent of the same polarity (ie, nonpolar molecules will dissolve in nonpolar solvents and polar molecules will dissolve in polar solvents). If the initial reaction mixture contains hydrophobic and hydrophilic molecules in an aqueous (polar) solution, a nonpolar organic solvent is added to the aqueous solution in a separatory funnel for an extraction. The separatory funnel is shaken, and the funnel is vented to release pressure. The organic and aqueous layers are immiscible (meaning they do not mix) so after shaking, the layers separate based on their densities. Hydrophilic ("water-loving") molecules are polar and will remain in the aqueous layer. Hydrophobic ("water-fearing") molecules are nonpolar and will dissolve better in the organic layer. Therefore, a nonpolar organic solvent is used in an extraction to remove the hydrophobic molecules from the polar aqueous layer while the hydrophilic molecules remain in the aqueous layer.

If dichloramine formation is spontaneous under standard conditions, the equilibrium constant must be: A) < 0 B) > 0 and < 1 C) = 1 D) > 1.

D - If ln(Keq) is positive, ΔG° is negative and the reaction is spontaneous. The natural logarithm (ln) of any number greater than 1 is positive, so Keq must be greater than 1 for the reaction to be spontaneous under standard state conditions. If ln(Keq) is equal to 0, the entire right side of the equation is equal to 0 and the system is at equilibrium in the standard state. This occurs when Keq = 1 because ln(1) = 0 (Choice C). If ln(Keq) is negative, then a negative number (−RT) is multiplied by another negative number (ln[Keq]), making ΔG° positive and the reaction nonspontaneous. The natural logarithm of any number between 0 and 1 is a negative number (Choice B).

The boiling point of CH3CH2Br (bromoethane) is 38.4 °C whereas the boiling point of CH3CH2I (iodoethane) is 71.5 °C. Which of the following best explains the difference between the boiling points of the two alkyl halides? A) The C-Br bond in bromoethane requires less energy to break than the C-I bond in iodoethane B) The bromine atom in bromoethane is more polarizable than the iodine atom in iodoethane C) The C-Br bond in bromoethane is more polar than the C-I bond in iodoethane D) Iodoethane has greater London dispersion forces than bromoethane.

D - London dispersion forces are intermolecular attractions between instantaneous dipoles induced by momentary distortions in the distribution of the electron "cloud" around atoms. As such, London dispersion forces are greater in larger molecules, which have a larger electron cloud and thus are more polarizable. London dispersion forces are among the weakest intermolecular forces (as an individual interaction). However, the combined effect of numerous dispersion force interactions often make London dispersion forces a very strong net influence on physical properties, especially in larger molecules. Comparing bromoethane and iodoethane, both molecules have an equivalent, nonpolar CH3CH2 chain, but the iodine substituent of iodoethane is larger and more polarizable than the bromine substituent of bromoethane. As such, iodoethane experiences greater London dispersion forces and has a higher boiling point than bromoethane.

Which of the following reactions will convert hexanoic acid and the carboxylic acid derivative described in the passage to the same product? A) Base-catalyzed aldol condensation B) Decarboxylation C) Oxidation with NaBH4 D) Reduction with LiAlH4

D - Reducing agents decrease an atom's oxidation state. The reduction of a carbon atom decreases the number of bonds it has to electronegative atoms (including oxygen, nitrogen, and halogens), and increases the number of bonds to less electronegative atoms (typically hydrogen). Hydride reagents, such as LiAlH4 and NaBH4, are common reducing agents and are made of a metal surrounded by one or more hydride ions. The hydride ion acts as a nucleophile and attacks an electrophilic atom. LiAlH4 is a strong reducing agent and will reduce carbonyl compounds to alcohols. Aldehydes, esters, and carboxylic acids will be reduced to primary alcohols, and ketones will be reduced to secondary alcohols. The reaction mixture contains the ester ethyl hexanoate and unreacted hexanoic acid. If these two compounds are reacted with LiAlH4, they would both be reduced to the six-carbon primary alcohol hexanol. Ethyl hexanoate will also produce ethanol as a side product when reduced with LiAlH4.

How could one best increase the rate of reaction of 1-Bromopropane with sodium cyanide? A) By doubling the concentration of 1-Bromopropane. B) By doubling the concentration of sodium cyanide. C) By acidifying the reaction medium. D) Both choices A and B would increase the reaction rate.

D - Since 1-Bromopropane is primary, this reaction must proceed according to an S and two pathway. The rate of this reaction depends entirely on the probability that the substrate, one Bromo propane, will collide favorably with a nucleophile, cyanide and I am common solution. Doubling the concentration of either reactant will increase the likelihood of these favorable interactions.

Which additional extraction steps would cause phosphatidylethanolamine and 2,6-dimethoxyphenol to enter the aqueous layer consecutively? A) Add 0.05 M NaOH(aq) to the organic layer followed by 0.01 M H2SO4(aq) B) Add 0.01 M NaHCO3(aq) to the organic layer followed by 0.05 M HCl(aq) C) Add 0.01 M H2CO3(aq) to the organic layer followed by 0.05 M NaHCO3(aq) D) Add 0.05 M H2SO4(aq) to the organic layer followed by 0.01 M NaOH(aq

D - The acid-base properties of organic compounds can be used to increase their separation during extraction. Compounds with acidic or basic functional groups can enter the aqueous layer as ionic salts, which are formed by deprotonation or protonation by a base or acid of varying strengths, respectively. Strong organic bases form ionic salts easily when weak acids are added to a mixture. On the other hand, weaker bases such as amines are only protonated by strong acids. Likewise, carboxylic acids, which are considered relatively strong organic acids, can be deprotonated by strong and weak bases. However, phenols are much weaker acids that require strong bases to be deprotonated. For phosphatidylethanolamine and 2,6-dimethoxyphenol to enter the aqueous layer consecutively, their ionic salts must be formed in sequence. Phosphatidylethanolamine contains an amine group that can be protonated to form a water-soluble ammonium salt; therefore, a strong acid such as H2SO4 or HCl should be added to protonate the amino group. In contrast, 2,6-dimethoxyphenol has a weakly acidic hydroxyl group on the benzene ring that must be deprotonated to form an ionic salt. Only strong bases such as NaOH or KOH are able to remove protons from phenols. Consequently, the correct sequence of extraction steps would involve the addition of a strong acid followed by a strong base.

Acetic acid most likely eluted from the gas chromatograph at what time point? A) 1 min B) 2 min C) 5 min D) 15 min

D - The chromatogram shown in Figure 1 has four peaks. To determine which one corresponds to acetic acid, the boiling point of acetic acid relative to the other compounds must be determined. Boiling point is a function of intermolecular forces, with fewer forces corresponding to lower boiling points. Organic molecules with the same number of carbon atoms follow a pattern of increasing intermolecular forces and boiling points: Alkanes have the weakest intermolecular forces and the lowest boiling points, followed by aldehydes and ketones, then alcohols, and carboxylic acids, which have the highest boiling points. Acetic acid is a carboxylic acid and has the highest boiling point of the compounds produced; therefore, it has the longest retention time and must elute at 15 min.

Which of the following reagents CANNOT be used to make benzoic acid from benzaldehyde? A) CrO3 B) KMnO4 C) H2CrO4 D) PCC

D - The oxidation of an aldehyde to a carboxylic acid requires an increase in the number of C-O bonds and a decrease in the number of C-H bonds. An increase in C-O bonds results in the loss of electron density around carbon and a partial positive charge, while the oxidizing agent gains electrons and becomes reduced. Aldehydes are readily oxidized to carboxylic acids by numerous oxidizing agents, including chromium reagents, in an aqueous environment. The aldehyde is believed to go through a hydrate intermediate, also known as a geminal diol, before being oxidized to a carboxylic acid. However, pyridinium chlorochromate (PCC) can only oxidize primary and secondary alcohols to aldehydes and ketones, respectively. PCC cannot oxidize aldehydes further because it is an anhydrous oxidizing reagent and does not contain the water necessary to convert the aldehyde to a hydrate, a necessary intermediate to be oxidized to the carboxylic acid. (Choices A, B, and C) CrO3, KMnO4, and H2CrO4 can all oxidize aldehydes to carboxylic acids because these reagents are all in aqueous environments, meaning they contain the water necessary to convert the aldehyde to the hydrate intermediate that is oxidized to the carboxylic acid.

The acid strength of hydrogen halides is ranked from strongest to weakest as follows: H−I > H−Br > H−Cl >> H−F Which of the following best explains why H−F is not a strong acid? A) Fluorine has the largest atomic radius, making the H−F bond weak B) Fluorine has the largest atomic radius, making the H−F bond strong C) Fluorine has the smallest atomic radius, making the H−F bond weak D) Fluorine has the smallest atomic radius, making the H−F bond strong

D - The size of the halogen atom impacts the bond length and strength of hydrogen halide acids (H−X). Halogens are located in Group 7A (Group 17) on the right side of the periodic table and increase in size moving down the group. Therefore, halogens with a smaller atomic number have a smalleratomic radius and form shorter, stronger H−X bondsthat are more difficult to ionize. Smaller halogens are more electronegative and more readily attract electrons in the H−X bond toward itself, creating a polar covalent bond. The stability of the conjugate base of the hydrogen halides (X-) also affects acidity. The conjugate base formed from a smaller halogen is less stable because it has a smaller atomic radius (less surface area) and cannot spread out (stabilize) the negative charge as well as halogens with larger atomic radii (greater surface area). Fluorine is the most electronegative halogen and has the smallest atomic radius of the halogens. Therefore, H−F forms the shortest, strongest bond of all the hydrogen halides. A short, strong H−X bond that is difficult to ionize in aqueous solution indicates a weak acid.

Given that acetic acid (CH3COOH) has pKa ≈ 5, the estimated pH of an aqueous 0.001 M sodium acetate (CH3COONa) salt solution is: A) < 4 B) > 4 and < 7 C) = 7 D) > 7

D - The tendency of the salt ions to act as an acid or a base depends on the strengths of the acid and the base involved in forming the salt. Strong acids form weak conjugate bases, and strong bases form weak conjugate acids. Therefore, examining the strengths of the ionic species permits a qualitative assessment of the salt solution pH: - Neutral salts formed from a strong acid reacting with a strong base give neutral solutions (pH = 7). - Acidic salts formed from a strong acid reacting with a weak base give acidic solutions (pH < 7). - Basic salts formed from a weak acid reacting with a strong base give basic solutions (pH > 7). - Salts formed from a weak acid reacting with a weak base are more difficult to broadly classify, and the solution pH of these salts often must be determined experimentally or by calculation. Although the pH of a 0.001 M CH3COONa salt solution can be calculated, an exact solution pH is not requested; the nature of the solution can be assessed qualitatively. Considering the ions of CH3COONa in detail, CH3COONa is a basic salt produced by the reaction of CH3COOH (a weak acid) with NaOH (a strong base). Therefore, CH3COONa will undergo hydrolysis to produce a basic solution (pH > 7).

Platinum most likely increases the reaction rate by stabilizing: A) The reactants B) The products C) The intermediate step D) The transition state

D - The transition state of a reaction is the state of the molecules where new bonds are forming and old bonds are being broken. It is an unstable state and normally requires a large amount of energy to form, which is called the activation energy. The activation energy provides an energy barrier to a reaction's progress. A catalyst is a substance that increases the rate of a reaction without being consumed by the reaction. It increases the rate by stabilizing the transition state and thereby lowering the activation energy of the reaction. The passage states that platinum increases the reaction rate of hydrogen combustion without being consumed; therefore, platinum must be acting as a catalyst, stabilizing the transition state of the combustion reaction. (Choices A and B) Stabilization of reactants or products would not help a reaction proceed more quickly because the energy barrier between them (the transition state) would still be present. (Choice C) An intermediate step is a relatively stable step in a reaction that exists between two transition states. Its stabilization would not help increase the rate of a reaction.

Which choice is an accurate statement about electron affinity? A) It reflects an endothermic process for both bromine and potassium. B) It reflects an exothermic process for both bromine and potassium. C) The process that it reflects is endothermic for bromine by exothermic for potassium. D) The process that it reflects is exothermic for bromine but endothermic for potassium.

D - To establish a complete octet, bromine needs only one additional electron. Therefore, when bromine adds an electron, it will become more stable and release energy. Potassium, and contrast refers to lose an electron to have a full octet. For this element, adding another electron is unfavorable and will require energy in an endothermic process.

Which of the following does not explain water's ability to act as a solvent? A) Water has the geometry of a bent molecule B) Hydrogen is less electronegative than oxygen C) Water is a relatively small molecule D) Water has a relatively high surface tension

D - Water readily dissolves many polar and ionic compounds because it is able to interact well with them through hydrogen bonding and other electrostatic interactions. Several important characteristics contribute to water's ability to participate in these interactions, including the following: Because oxygen is more electronegative than hydrogen, the bonding electron pairs are more tightly drawn to oxygen. This property results in the oxygen atom having a partial negative charge whereas the hydrogen atoms have a partial positive charge, allowing interactions with other charged molecules (Choice B). The bent geometry of water contributes to its polarity by grouping positive charges at one end of the molecule and negative charges at the other. The charged regions of a water molecule are attracted to opposite charges on other polar compounds (Choice A). Water's small size allows efficient formation of a hydration shell around solutes. This process acts to evenly distribute and isolate solute particles (Choice C). On the other hand, surface tension is a force induced at the interface between a liquid and a gas. The molecules in the liquid interact with each other more strongly than they interact with molecules in the air, causing the surface of the liquid to behave as a thin film. Water has a high surface tension, but this characteristic does not aid in water's ability to act as a solvent.

Gaseous hydrogen as a diatomic gas. Which intermolecular forces can H2 exert? A) Ionic bonding. B) Dipole dipole forces. C) Hydrogen bonding. D) London forces.

D - gaseous hydrogen is composed of two atoms of identical electronegativities. Therefore, it lacks a typo entirely can only participate in the week as attractions, London forces.

Alkali metals when placed in water, are highly reactive and sometimes capable of causing large explosions. Which of the following properties of the these metals serves as a cause of this reactivity? A) Alkali metals have very low electronegativity's. B) Alkali metals have very high electron affinity. C) Alkali metals are toddlers in combustion reactions. D) Alkali metals have very low ionization energy

D - members of the alkali metals have one valance electron. When this electron is lost, these elements gain a noble gas configuration, making them very stable. This reason, alkali metals a very low ionization energies and lose an electron extremely easily. In this reaction, hydrogen gas is produced in the metal forms its corresponding oxide. While alkali metals do you have low electronegativity's, this relates more to their ability to form ionic bonds, Not the reactivity in water. Alkali metals are non-catalyst for combustion reaction, nor with this explain the reactivity.

Transition metals

Do you have the most well-known chemical hallmarks of transition metals are their ability to possess multiple oxidation numbers of the tendency to form brightly colored compounds. This second characteristics stems from the arrangement of their d electrons. Somehow it is a brightly colored in their atomic states. They do not, however, have multiple stable oxidation states,

Le Chatelier principle

Due to Le Chatelier's principle, basic compounds dissolve best in acidic solution. Specifically, the presence of protons neutralize some of the free hydroxide ions, pushing the equilibrium towards disassociation. 2M of a strong acid would certainly accomplish the students goal of raising the number of calcium ions in solution. Bases dissolve especially poorly in basic solution due to the common ion effect.

Photoelectron

Due to conservation of energy, a photo electrons energy cannot be greater than that of the incident photon. This concept is due to the existence of the work function, the quality of energy required for an electron to be released at all. Since some of the incident photon energies devoted to overcoming the special. Not all of it can be converted to kinetic energy for the ejected electron. It is the energy of the incident photon, and whether it surpasses the special, that determines whether ejection occurs which is the threshold frequency. The intensity of the incident ray determines the number, not the individual energies, reject a photo electrons.

Nuclear Decay

During alpha (α) decay, an unstable nucleus ejects an α particle consisting of 2 protons and 2 neutrons (a helium-4 nucleus without its electrons). Accordingly, the resulting nucleus formed following an α decay will have a mass number that is 4 units less and an atomic number that is 2 units less than the nucleus that underwent the α decay. Beta decay can occur in three forms: β− decay (electron emission), β+ decay (positron emission), and electron capture. In all three forms of beta decay, the mass number is unchanged but the atomic number either increases (β− decay) or decreases (β+ decay) by 1 unit.

Newman projection

Eclipsed confirmation has a substituents on the front carbon arranged as closest possible to those on the back Harbin additionally, the two largest groups are directly adjacent to one another, this energetically unfavorable arrangement is known as fully eclipsed Anti implies that the largest substituents are 180° from each other. A staggered confirmation exists when substituents on the front and back carbons are 60° for each other. The most stable potential conformer, it's staggered and anti

Effective nuclear charge

Effective nuclear charge to be found by subtracting the total number of electrons in all shells preceding the one of the questions from the nuclear charge, the number of protons. Selenium has an atomic number of 34. In other words, the diagram includes 32 electrons. Starting from the first round the number of electrons in each shell is respectively, 2, 8, 18, and 32. The effective nuclear charge can be found by subtracting the number of electrons in all shells proceeding the one question from the nuclear charge.

Nitrogen exception to the rule

Electron affinity relates to an elements desire to gain an electron. For the majority of elements, this quality increases as you move upwards into the right on the periodic table. However, the nitrogen containing group is an exception to this rule. Carbon has an electron configuration of [He] 2S^2 2P^2, so the addition of another electron would give carbon a half filled P orbital and is this highly favorable. Nitrogen on the other hand, has an electron configuration of [He] 2S^2 2P^3. Addition of another electronic alter nitrogens half filled Valecen state and produce a partially filled order. Therefore, nitrogen will have a lower electron affinity than carbon.

IR

Functional groups show absorption in the infrared spectrum at different frequencies depending on the bond type present in the particular functional group. Characteristic functional group absorptions include 3650-3200 cm−1 (alcohol and phenol O−H stretch); 3550-3060 cm−1 (amide N-H stretch); 3100 cm−1 (sp2 C-H stretch); 3000-2875 cm−1 (sp3 C-H stretch); 2260-2,100 cm−1 (triple bonds); and 1850-1650 cm−1 (C=O stretch).

Deposition

Gas to solid

Ideal gases

Gases behave least ideally at high pressure and low temperature. In other words, they deviate most from ideal behavior under conditions that favor formation of a liquid.

Bond order

I had ordered notes whether a bond a single, double, or triple in nature. The higher the bond order, the shorter the associated length.

Nerst Equationn E = Eknot + ET/nF ln(Q)

If Q > 3, ln(Q) must be larger than 1. For this reason, the quantity RT/nF ln(Q) gives rise to a positive value that must be added to Eknot. Thus, E, the cell potential, will be greater than E knot, the standard states cell potential. Q cannot possibly be equal to zero, as ln(0) represents a quantity that does not exist. When Q equals one, LMQ equals zero. In this case, the value of tea does not matter, the product of zero in any other number is always zero.

Nuclear Decay

In alpha decay, an unstable nucleus ejects an alpha particle (represented as α24α24 or He24He24) consisting of 2 protons and 2 neutrons (a helium-4 nucleus without its electrons), as indicated by the nuclear notation of the particle. Beta decay can occur in more than one form. During beta minus (β−) decay, an electron (represented as e−, e−10e-10, or β−) and an antineutrino (represented as V⎯⎯⎯eV¯e or υ⎯⎯00υ¯00) are emitted from the nucleus via the conversion of a neutron to a proton. Alternatively, during beta plus (β+) decay, a positron (represented as e+, e+10e+10, or β+) and a neutrino (represented as νe or υ00υ00) are emitted from the nucleus via the conversion of a proton to a neutron through direct transmutation or through electron capture. Although present, the neutrino and antineutrino may not always be explicitly mentioned when discussing beta decay in a chemistry context. During gamma emission, an unstable nucleus in an excited state releases excess energy by emitting a gamma ray (a high-energy photon, which is an electromagnetic wave packet (represented as γγ or γ00γ00), but no particle is emitted and the number of proton and neutrons in the nucleus remains unchanged. As a result, the isotopic and elemental identity of the nucleus are unchanged following gamma emission.

Visible Spectrum

In highly conjugated systems, the difference in energy between the ground and excited states of the electrons is equal to the energy of a particular wavelength of visible light. Photons of this wavelength are absorbed by the molecule, causing an electron to enter the excited state. The remaining wavelengths are either reflected or transmitted and are ultimately perceived by the eye as various colors. In general, the perceived color of a substance is complementary to the color of the wavelength that is maximally absorbed by that substance. For example, a molecule that absorbs blue light will appear orange and vice versa. When the basic colors of the visible spectrum (red, orange, yellow, green, blue, and violet, or ROYGBV) are arranged in a wheel with red and violet next to each other, a color's complement is the color directly across from it on the wheel. According to the given information, Compound 1 absorbs violet light (448 nm), so it appears yellow (complement of violet). After Cu2+ is added, the absorption maximum changes to 623 nm and the compound absorbs orange light, causing it to appear blue (complement of orange). Therefore, upon the addition of Cu2+, Compound 1 must undergo a change in electronic structure that causes the solution to change from yellow to blue.

Absorption and Emission

In the Bohr model of the atom, several assumptions are made regarding electrons and their locations around the nucleus. The Bohr model asserts that: - Electrons move around the nucleus in fixed circular orbits, which are only allowed at particular intervals from the nucleus. - Electrons in orbits farther from the nucleus have higher energy than electrons in orbits closer to the nucleus. - Energy is absorbed by an electron moving from a lower orbit to a higher orbit, but energy is emitted by an electron returning from a higher orbit back to a lower orbit. - The energy that is absorbed or emitted by an electron equals the energy difference between two orbits.

Data Interpretation

Isomers are molecules with the same molecular formula but different structural arrangements. They can be generally classified as constitutional isomers and stereoisomers. Constitutional isomers differ in atom connectivity whereas stereoisomers have the same atom connectivity but differ in spatial arrangement. Stereoisomers that arise from disubstituted double bonds are called geometric isomers. They are classified as cis or transbased on the relative positions of the nonhydrogen substituents. Geometric isomers can be separated by gas chromatography because they have slightly different boiling points. Docosenoic acid is an unsaturated fatty acid, meaning it contains a disubstituted double bond, and can be in the cis or trans form. Both isomers have the same molecular weight and m/z ratio, but will come off the GC column at different times, yielding two distinct peaks. Therefore, it can be inferred that Peaks A and B are cis/trans isomers of docosenoic acid.

KMnO4

KMnO4, like a number of other oxygen containing regions, is an oxidizing agent. The original molecule contains a primary and a secondary alcohol, which will be oxidized into a carboxylic acid and ketone, respectively. It also contains a tertiary alcohol, as well as a phenol ring, neither which can be oxidized without breaking the carbon carbon bond. These substituents will thus remain in the form of alcohols. Note that KMno4 is a strong oxidant, which is why can oxidize a primary alcohol directly to a carboxylic acid without first forming an aldehyde. An aldehyde form only if the students were reacting to primary alcohol with a weak accident, such as PCC.

Auto ionization of water

KW is only equal to approximately 10^ -14 at 25°C. Since pH decreases with increasing temperature, more disassociation must take place when more heat energy is available. Thus, this reaction must be endothermic. PH equals POH for pure water at all temperatures. While pH does drop when the temperature is high, POH changes in the same manner. A solution is only acidic and more protons are present than hydroxide ions.

Keq

Keq who is not affected by pressure changes, although it does vary with temperature.

Vapor pressure

Low vapor pressure always corresponds to high boiling point, property that directly relates to the strength of intermolecular attractions.

Diamagnetic materials

No unpaired electrons, which possess full orbitals.

Primary alcohols

Only primary alcohols, which are positioned at the end of carbon chains, can be oxidized formaldehydes. Additionally, this can only be accomplished using week accidents, such as PCC. A stronger oxidizing agent would cause a primary alcohol to gain two bonds of oxygen form of carboxylic acid.

Pure water

Pure water always has an equal number of hydroxide and hydrogen ions. While pH does decrease with increasing temperature, POH decreases as well.

Galvanic cell

Reactions occurring in galvanic cells are always spontaneous, meaning that they have positive Ecell values. Positive cell potentials correlate to the negative delta G values, since both refer to spontaneous reactions. Similarly, these reactions must have K values that are greater than one, because they favor the products, not the reactants.

SN1 reactions

SN1 reaction require the loss of a leaving group before such a charge species can form. In this question, the living group is a halogen atom. For this reason, we are looking for an answer in which the halogen is located on a highly substituted carbon.

Purity

Several techniques can be used to evaluate the purity of a compound, including thin-layer chromatography (TLC), which separates compounds in a mixture based on polarity. In TLC, the tested sample is dissolved in an organic solvent and a small amount is spotted onto the stationary phase (commonly a polar silica [SiO2]) coated on a TLC plate. After spotting, the TLC plate is placed in a solvent chamber, and the mobile phase (organic solvent) travels up the plate and carries the spotted compounds along at different rates, which separates the mixture components. Nonpolar compounds have less affinity for a polar stationary phase and will travel further up the plate than polar compounds. The conversion of theobromine to caffeine results in N-methylation (conversion of N-H to N-CH3). The N-H in theobromine can hydrogen bond to the polar stationary phase on the TLC plate whereas the N-CH3 in caffeine cannot hydrogen bond. Because theobromine can hydrogen bond, it is more polar and has a smaller Rf value than caffeine. Therefore, TLC can be used to assess product purity by comparing a standard of pure theobromine to the isolated product. If a single spot with a larger Rf than that of theobromine is visible on the product sample track on the TLC plate, then the isolated product is pure.

Tautomerization

Tautomerization is a conversion between specific, easily interchangeable isomers. The most common form of this reaction, known as keto enol tautomerization, involves conversion between a ketone or aldehyde and its corresponding Enol, a structure with an OH group immediately adjacent to a C=C double bond. However a similar reaction can occur with mines and enamines, nitrogen containing functional groups analogous to the keto and email forms, respectively

Stereo isomers

The number of stereoisomers of a molecule is equal to 2^nth, where n denotes number of stereo centers. Since this compound has three Chiro centers, eight stereo isomers exist. However, 2 are enantiomers, leaving six diastereoisomers remaining.

Ka

The product of the Ka of an acid and the Kb of its conjugate base is always equal to the Kw. HS04- and S04 2- are conjugates, so this statement is true. Ka multiplied by Kb is always equal to Kw, but the value of Kw is only 10 to the -14 at 25°C.

Cis versus trans

The sis isomer of this alkene should melt at a lower temperature than the trans isomer. In general, the trend is true Alkeenes, as a trans form is better able to stack on top of like molecules. Better stock and equates to more surface area over which intermolecular forces may be exerted, making a substance more difficult to melt.

Stability of an anion

The two factors that contribute most of the stability of an anion our residence and the presence of electron withdrawing groups, also known as inductive effect, on the molecule. Fluorine atoms and nitro groups are classic examples of electron withdrawing substituents.

Paper Chromatography

Thin-layer chromatography is a technique used to separate compounds based on polarity. In normal-phase thin-layer chromatography, the stationary phase is made up of a polar, adsorbent material, typically silica (SiO2), and the mobile phase is an organic solvent that travels up the stationary phase via capillary action. The rate at which a compound travels up the plate is a function of the relative polarities of the compound and the solvent. If a mixture remains near the plate's origin, a more polar solvent is needed to increase the compound's affinity for the mobile phase relative to its affinity for the stationary phase. Conversely, if a mixture travels with the solvent front, a more nonpolar solvent is needed to decrease the compounds' affinity for the mobile phase relative to its affinity for the stationary phase. The 1:1 hexanes/ethyl acetate mixture is more polar than hexanes and less polar than ethyl acetate and achieved the compound separation essential for column chromatography. This solvent mixture decreased the compounds' affinity for the mobile phase relative to ethyl acetate.

Determine a final principal quantum number

To determine this, we must be familiar with a different species inspector. The Lyman series, ultraviolet rays, involves any emotion in which the ground state of the electron is N = 1. The Balmer series, visible rays, involves any emissions in which the final state is N = 2. Finally, the Paschen series, infrared, contains any emission with the final state of an N = 3.

R/S

To establish the R/S configuration of the chiral amino acids, the priority of the groups bonded to the α-carbon should be determined as follows: -NH2 -CO2H -R (side chain) -H

UV-vis

UV-vis spectroscopy, is mainly use analyze conjugated systems. The larger the system, the more shifted the UV absorption peak, implying that anthracene speak would likely appear distinct enough to notice its presence. Simple distillation separates compounds on the basis of boiling point. As a 12 carbon structure with an alcohol and a carboxylic acid group, five hydroxy double-decker know if acid is probably similar enough to antecedents of oil at close temperature. Additionally, just dilation is a separation method and cannot be directly use for detection. Extraction, another separation technique, utilizes differences in solubility, particularly acid-base properties. I'll be in Alkeenes or two somewhere to be isolated in this manner. Finally, though TLC could theoretically be used as an exact detection method, separate compounds on the basis of polarity. Two of the potential contaminants or nonpolar, just like anthracene.

Markovnikov Reactions: the anti product

When an alkene is reacted with hydrobromic acid, the Markovnikov product typically forms. Since this product includes a halogen bound to the more substituted end of the bond, it tends to be especially stable. Radical reactions, particularly those initiated by peroxide, for one exception to this rule. In these processes, hydrogen adds to the less substituted end, while the bromine ion adds to the other position.