3/20 Questions

. Most of the amino acid side chains are uncharged at physiological pH because the side chain functional groups do not have acidic or basic protons. The uncharged amino acid side chains may be hydrophobic or hydrophilic. Tryptophan (Trp, W) and phenylalanine (Phe, F) are uncharged hydrophobic amino acids (Choices B and C). Threonine (Thr, T) is a polar amino acid that is uncharged at physiological pH (Choice D). . Acidic side chains are deprotonated at physiological pH and have a −1 charge, and the basic side chains have a fully protonated nitrogen atom and a +1 charge. Therefore, the overall charge of these amino acids is not zero. Aspartate (Asp, D) has an acidic side chain and therefore has a negative charge at physiological pH and cannot be a zwitterion at physiological pH.

A researcher needs to use an amino acid that is a zwitterion at physiological pH. Which of the following amino acids could NOT be used in the experiment? . A. D B. W C. F D. T . . . . . .

The cardiac output CO represents the volumetric flow rate of blood exiting the left ventricle and entering the aorta. The CO is the product of the cross-sectional area A of the aorta and the velocity v of the blood: CO=A·v . This equation can be rearranged to solve for v: . v=CO/A . In this question, A is given as 5 cm2 and the exercise output of the runner is given as 250 W. From Figure 1, an exercise output of 250 W equates to a CO value of approximately 7.5 L/min. Converting the units of CO from L/min to cm3/s yields: . CO=7.5 L/min*1,000 cm3/L*1/60 min/s= 125 cm3/s . Substituting the values of CO and A into the equation for v gives: . v=125 cm3s5 cm2=25 cms=0.25 ms .

A runner's exercise output is 250 W. If the cross-sectional area of their aorta is 5 cm2, what is the average velocity of the blood flowing through their aorta? . . A. 0.25 m/s B. 1.5 m/s C. 2.5 m/s D. 15 m/s

Power is the amount of work done per unit time. If the power generated during exercise is known, the time required to do a specific amount of work can be calculated. p = w/t w = f.d or kgJ t = w/p (W = work) P is for power, t is for time. Solve for t: t = 150Kj/1000W, Unive for (Watts) is J/s

How long does a person need to exercise to do 150 kJ of work when they consume oxygen at 2.3 L/min? . A. 6.7 s B. 65 s C. 150 s D. 345 s

In this question, Compounds 1 and 2 are stereoisomers, where the configuration of the stereocenter in Compound 1 is R and the configuration of the stereocenter in Compound 2 is S. . . Consequently, Compounds 1 and 2 are enantiomers, and the specific rotations of Compounds 1 and 2 are equal in magnitude and opposite in direction. The question states that the specific rotation of Compound 1 is −5.2 (°·mL)/(g·dm); therefore, the specific rotation of Compound 2 must be +5.2 (°·mL)/(g·dm).

If Compound 1 has a specific rotation of −5.2 (°·mL)/(g·dm), the specific rotation of Compound 2 must be: . . A. −5.2 (°·mL)/(g·dm) B. 0 (°·mL)/(g·dm) C. +5.2 (°·mL)/(g·dm) D. +10.4 (°·mL)/(g·dm)

The conjugate of an acid (or base) is the species that the acid (or base) becomes after losing (or gaining) one proton (ie, an H+ ion). A base is converted into its conjugate acid when the base accepts a proton. Likewise, an acid is converted into its conjugate base when the acid donates a proton. Therefore, an acid (or base) and its conjugate have chemical formulas that differ by only one proton. . In the forward direction of Reaction 2, the [M(H2O)6]n+(aq) complex acts as an acid by donating a proton to a H2O(l) molecule. Therefore, [M(H2O)5OH](n − 1)+(aq) is the conjugate base of the acid in the forward reaction (ie, the species that the [M(H2O)6]n+(aq) acid becomes after losing one proton).

In Reaction 2, which species is the conjugate base in the forward reaction? . A. [M(H2O)6]n+(aq) B. H2O(l) C. [M(H2O)5OH](n − 1)+(aq) D. H3O+(aq)

The kinematic equation that describes the distance traveled by a uniformly accelerated object is d=v0t+1/2at^2, where d is distance, v0 is the initial velocity, t is time, and a is acceleration. This equation can be rearranged to solve for an object's acceleration.

In one experimental trial, the projectile was uniformly accelerated from rest to a distance of 2 m in 0.1 s. What was the acceleration of the projectile? . A. 20 m/s2 B. 40 m/s2 C. 200 m/s2 D. 400 m/s2 . . . . . . .

At physiological pH (7.4), amino acid backbones have a positively charged amino group (+1 charge) and a negatively charged carboxyl group (−1 charge), and the charges cancel one another out, yielding a net charge of 0. These compounds are known as zwitterions. Acidic and basic amino acid side chains are charged at pH 7.4, so the overall charge of the molecule is not zero and these amino acids are not zwitterions at physiological pH.

Physiological pH (7.4) is above the carboxyl group's pKa of 2, causing the carboxyl group to become deprotonated and gain a −1 charge. However, the pH is below the amine group's pKa of 9, so the amine is fully protonated and has a +1 charge. This doubly charged species is known as a zwitterion, or a dipolar ion, in which the charges of the amine and carboxyl groups cancel one another out.

In this question, the head accelerated from a stationary position to a velocity of 10 m/s (Δv = 10 m/s), and the collision lasted for 10 ms (t = 10 ms). The value of t is converted from milliseconds (ms) to seconds (s) using the conversion 1 s = 1,000 ms: . t=(10 ms)(1 s/1,000 ms)=0.01 s . Using these values, the average acceleration of the head was: . a=∆v/t=10 m/s/0.01 s=1,000 m/s2 . In Figure 1, the point that represents an average acceleration of 1,000 m/s2 for a duration of 10 ms lies above the threshold for a closed TBI but below the threshold for an open TBI (Choices A and C). Therefore, a closed TBI is expected, which is described in the passage as a brain injury that occurs without a skull fracture (Choice D).

Suppose the physical characteristics of the crash test dummy accurately represent those of a human. What is the expected injury from an impact lasting 10 ms that accelerated the head to a velocity of 10 m/s? . . . "Open TBI = skull fracture, close TBI = no skull fracture. TBI traumatic brain injury" . A. No injury B. TBI and no skull fracture C. Skull fracture and no TBI D. TBI and skull fracture . . . . . According to the passage, the type of injury can be determined from the collision duration and the average acceleration of the head during the collision. The average acceleration (a) of an object is its change in velocity (Δv) over the time (t) the object is accelerated: . a=∆v/t

Each stereocenter can adopt two different conformations, so the expression 2n provides the maximum number of stereoisomers possible for the particular compound, where n is the number of stereocenters in the molecule. For example, labetalol has two stereocenters, and therefore 22 = 4 possible stereoisomers. A compound with three stereocenters has 23 = 8 possible configurations, and so on. . (Choices A and C) Enantiomers and diastereomers are types of stereoisomers, and are included in the number of total stereoisomers that can exist. . (Choice D) Epimers are a type of diastereomer that differs in configuration at only one stereocenter. The number of epimers is not factored into the calculation of the number of stereoisomers possible for a particular molecule.

The number of stereoisomers possible for a given molecule is 2n. The variable n denotes the number of: . A. enantiomers. B. stereocenters. C. diastereomers. D. epimers. . . . . Stereoisomers are molecules that differ in spatial arrangement but have the same connectivity and molecular formula. The number of stereoisomers for a particular compound corresponds to the number of configurations possible for that compound and is directly related to the number of stereocenters in the molecule.

The relative speed between two objects depends on their directions of motion. The minimum relative speed occurs when two objects move in the same direction, and the maximum relative speed occurs when two objects move in opposite directions. . Correct answer is 10m/s

A baseball traveling at 20 m/s strikes a player running at 5 m/s. The relative speed of the ball with respect to the player depends on the direction of travel of the ball relative to the player's direction of travel. What is the difference between the minimum and maximum possible speeds of the ball relative to the player? . A. 10 m/s B. 15 m/s C. 20 m/s D. 25 m/s

In this question, changes in the KE of a bicycle rider are considered when the duration of exercise, the velocity, the distance traveled, or the gravitational acceleration are changed. From the equation above, the KE depends only on m and v. Therefore, only changes in m or v will affect the KE. . (Choices A, C, and D) The duration of exercise, distance traveled, or gravitational acceleration do not necessarily influence the KE of the bicycle rider. For example, all these parameters could change but the KE of the rider will remain unchanged if traveling along a horizontal road at constant v. . Educational objective:The kinetic energy of an object depends only the object's mass and velocity. Other parameters, such as the duration of motion, distance traveled, or gravitational acceleration, do not affect the kinetic energy of an object.

Changing which of the following will always change the kinetic energy of a bicycle rider? . A. Duration of exercise B. Velocity C. Distance traveled D. Gravitational acceleration

Elements can be broadly classified on the periodic table as metals, metalloids, or nonmetals. Due to many shared characteristics within each of these categories, the general traits of each type of element are often useful for qualitative comparisons between elements.

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

Therefore, the power of the collision shown in Figure 2 can be determined from its ΔKE, which depends on the head's mass and its initial and final velocities during the collision. The head is stated to be stationary initially, and the final velocity of the head can be determined from the information given in Figure 2 (Choice D). Therefore, the mass of the head is needed in addition to the given information to determine the power of the collision. . (Choice B) The power of the collision is determined from the duration of the collision. The time until the head stops moving may occur after the collision has ended. . (Choice C) Tension is the force experienced by a string (or similar object) due to opposing forces at each of its ends. The forces at the neck do not need to be known to estimate the power of the collision.

What additional information do the researchers need from the crash test dummy to estimate the power of the collision in the experiment shown in Figure 2? . . A. Mass of the head B. Time until the head stops moving C. Tension of the neck during the collision D. Velocity of the head at the end of the collisioned . . . Power (P) is the rate of work (W) per unit time (t), and work is the energy transferred by a force through a distance. From the work-energy theorem, work is equal to the change in the kinetic energy (KE) of an object. Therefore, power can be calculated from ΔKE and t: . P=W/t=∆KE/t . An object's kinetic energy is the energy associated with its motion, and it is calculated from its mass m and velocity v: . KE=1/2mv^2

Small, highly charged metal cations can form hydrated complexes that are weakly acidic. The acid dissociation constant Ka (or pKa) indicates the extent to which a weak acid ionizes in solution. The pH of the solution is determined by the equilibrium concentration of the hydronium ion produced by the ionization of the acid.

What is the pH of the Zn2+(aq) solution in Experiment 1? . "A sample of a zinc salt was dissolved in water to produce a 0.10 M Zn2+(aq) solution. The solution was checked with a pH probe, and the measured pH confirmed that the solution was weakly acidic. [Zn(H2O)6]2+(aq) was found to have a pKa of 9.0, as reported in the scientific literature. . A. 4 B. 5 C. 7 D. 9 . . . . . . . . .

The carbon at position 3 is sp3 hybridized and bonded to two carbon atoms, an oxygen atom, and a hydrogen atom. One of the carbon substituents is bonded to oxygen and the other is bonded to nitrogen, so the two carbon substituents are different and the central carbon is bonded to four unique groups. Therefore, the carbon at position 3 is a stereocenter. . A stereocenter is any atom where when two substituents switch positions, a stereoisomer is formed. An sp3 carbon is a stereocenter if it is bonded to four unique groups. The sp2 carbons in alkenes are stereocenters if the alkene can be classified as E/Z or cis/trans.

What position(s) on the analog of labetalol shown above can be classified as a stereocenter? . A. Position 1 only B. Positions 2 and 3 only C. Position 3 only D. Positions 3 and 4 only

(Number II) Enantiomers are stereoisomers that are nonsuperimposable mirror images. These molecules contain one or more stereocenters, and each stereocenter is the opposite configuration from the corresponding stereocenter in the other molecule. For example, the enantiomer of (S,R)-labetalol is (R,S)-labetalol. . (Number III) Conformational isomers are different forms of the same molecule that are generated as atoms rotate about their bonds. Unlike other stereoisomers, conformational isomers can rapidly interconvert by rotation without the need to break any bonds. . Educational objective: Diastereomers are stereoisomers that are not mirror images. They contain at least two stereocenters in which one or more (but not all) are in opposite configurations.

(S,R)- and (R,R)-labetalol, the active forms of the drug, can be described as which of the following? . Diastereomers Enantiomers Conformational isomers . A. I only B. I and II only C. II and III only D. I, II, and III . . . . . Stereoisomers are molecules that contain the same atom connectivity but differ in the spatial arrangements of the atoms. Diastereomers are stereoisomers that are not mirror images. . These molecules contain at least two stereocenters in which one or more (but not all) of the stereocenters on corresponding carbon atoms are in opposite configurations. . (S,R)- and (R,R)-labetalol are stereoisomers that are not mirror images, and each compound contains two stereocenters. The two compounds have different configurations at the stereocenter in position 1 (one S and one R) and have the same configuration at the stereocenter in position 2 (both R). Therefore, (S,R)- and (R,R)-labetalol are diastereomers (Number I).

. In the given reaction, the As atom is oxidized (ie, gains bonds to oxygen), causing the oxidation number of As to change from +3 in H3AsO3(aq) to +5 in H3AsO4(aq). The electrons lost by the As atom cause I2(aq) to be reduced, which changes the oxidation number of I from 0 in I2(aq) (ie, the elemental state) to −1 in the I−(aq) ion. Therefore, H3AsO3(aq) causes reduction and is the reducing agent in the reaction. . (Choice B) I2(aq) is the oxidizing agent in the reaction. . (Choice C) The oxidation numbers of the H and O atoms do not change during the reaction. As such, the atoms are neither oxidized nor reduced, and H2O(l) cannot be the reducing agent. . (Choice D) H3AsO4(aq) is a product of the reaction and cannot be the reducing agent.

Arsenous acid reacts with iodine according to the equation below. . H3AsO3(aq) + I2(aq) + H2O(l) → H3AsO4(aq) + 2 I−(aq) + 2 H+(aq) . Which chemical species acts as the reducing agent in the reaction? . A. H3AsO3(aq) B. I2(aq) C. H2O(l) D. H3AsO4(aq) . . . The reducing agent in a reaction is the chemical species that causes reduction in another atom by giving up electrons (ie, the reducing agent itself gets oxidized in the process). Conversely, the oxidizing agent is the chemical species that causes oxidation by taking electrons from (oxidizing) another atom and getting reduced in the process. .

Small, highly charged metal cations can form hydrated complexes that are weakly acidic. The acid dissociation constant Ka (or pKa) indicates the extent to which a weak acid ionizes in solution. The pH of the solution is determined by the equilibrium concentration of the hydronium ion produced by the ionization of the acid.

Based on the results from Experiment 2, the pKa of [Al(H2O)6]3+(aq) is: . "A sample of an aluminum salt was dissolved in water to produce a 0.10 M Al3+(aq) solution. The solution was observed to turn the blue litmus paper red and had a measured pH of 3.0. The pKa of [Al(H2O)6]3+(aq) was not found in the scientific literature." . A. 3 B. 5 C. 7 D. 8 . . . . . . . . .

. (Choice B) Increasing blood flow to the skin does not increase blood flow to the skeletal muscles. . (Choice C) Increasing blood flow to the skin may have the detrimental effect of decreasing blood flow to the skeletal muscles. However, the need to maintain normal body temperature is more important than maximizing muscle blood flow. . (Choice D) Increasing blood flow to the skin does not decrease the cardiac output. . During exercise, blood flow to the skin increases to maintain normal body temperature. Excess body heat generated during exercise can be transferred to the surface through convection and dissipated into the environment through radiation.

Does blood flow to the skin increase during exercise? . A. Yes, because regulation of body temperature is improved B. Yes, because blood supply to skeletal muscles is improved C. No, because blood supply to skeletal muscles is reduced D. No, because cardiac output is reduced . . . . . Once blood flow transports excess heat to the surface of the body, the heat can be exchanged with the environment more effectively. This transfer of heat occurs through radiation in the form of electromagnetic waves in the infrared region of the spectrum emitted from the skin into the environment. In this question, blood flow to the skin increases during exercise to increase the transfer of excess body heat to the environment and maintain normal body temperature.

In Figure 2, the acceleration of the dummy's head (y-axis) is given as a function of time (x-axis). Acceleration (a) is the rate at which velocity changes, and it is calculated as the change in velocity (Δv) divided by the duration (t) of the acceleration: . a=∆v/t . Therefore, the change in velocity of the head can be determined as the product of acceleration and its duration: . ∆v=at . On a graph, the product of the axes' quantities is represented by the area under a curve. Therefore, the area under the curve between two points in time in Figure 2 represents the Δv of the head during that time interval. Because the velocity of the head is 0 m/s at the start of the graph, the area under the curve from 0 to 10 ms represents the velocity of the head at 10 ms.

In the range from 0 to 10 ms in Figure 2, which of the following represents the velocity of the head at 10 ms? . A. The length of the curve B. The area under the curve C. The maximum height of the curve D. The maximum slope of the curve . . . . Acceleration is the rate at which velocity changes over time, and therefore the product of acceleration and time is the change in velocity. On a graph, the product of the axes' quantities is represented by the area under a curve. Therefore, the area under the curve on an acceleration vs. time graph represents the change in velocity.

Double bond 1 contains a trisubstituted double bond. On the left side of the double bond, the ring is higher priority because carbon has a greater atomic number than hydrogen. The right side of the double bond is attached to two carbon atoms: . a carboxyl carbon (COOH), which has three bonds to oxygen (ie, the double bond to oxygen counts as two bonds) . a methylene group (-CH2), which has two bonds to hydrogen and one bond to a carbon in a ring. . Comparing the atom with the highest molecular weight bonded to the carboxyl and methylene carbons (ie, oxygen and carbon), oxygen has a greater atomic weight than carbon, and thus the carboxyl group has a higher priority. The high-priority groups are on opposite sides of the double bond, so it is classified as E. . Double bond 2 is a disubstituted double bond in which both hydrogen atoms (or both high-priority groups) are on the same side; therefore, it is cis.

The structure of eprosartan mesylate is shown below. Double bonds 1 and 2, respectively, can be classified as: . A. E for double bond 1 and cis for double bond 2 B. E for double bond 1 and trans for double bond 2 C. Z for double bond 1 and cis for double bond 2 D. Z for double bond 1 and trans for double bond 2 . . . E and Z designations distinguish stereoisomers about double bonds, in which Z signifies high-priority groups on the same side of the bond and E indicates the high-priority groups on opposite sides of the bond. In disubstituted double bonds, the Z conformation is also called cis and the E conformation is called trans.

. Therefore, in an aqueous solution at pH 8.0, the majority of the L-phosphotyrosine molecules will have two negative charges from the phosphate group, one negative charge from the carboxyl group, and one positive charge from the amino group which sum to a net charge of −2. . . (Choice A) Zero net charge would require that no ions formed in the structure (or that the charges formed cancelled out). The acid-base behavior of thephosphate, carboxyl, and amino groups excludes this possibility. . (Choice B) A net charge of −1 would occur if only the amino and the phosphate groups formed ions. However, the acidic nature of carboxylic acids ensures that the carboxyl group will ionize also. . (Choice D) A net charge of −3 would be possible if only the acidic phosphate and carboxyl groups formed ions. The basic nature of amines, however, causes the amino group to be protonated and form a +1 charge.

The structure of the molecules in a sample of solid L-phosphotyrosine is shown below. . What will be the net charge of the majority of L-phosphotyrosine molecules when placed in an aqueous solution at pH 8.0? (Note: The pKa values of the phosphate group are 2.2 and 7.2.) . A. 0 B. −1 C. −2 D. −3 . . . . Groups attached to amino acids can also act as an acid or a base. A phosphate group behaves like an attached molecule of phosphoric acid, and at pH 8.0, it also ionizes to acquire two negative charges because its pKa values are lower than the pH.

Triacylglycerols are lipids made up of three fatty acids attached to a glycerol molecule through ester linkages. All three fatty acids may be identical, two of the three may be identical to each other, or all three may be distinct. . If at least one fatty acid is different from the other two, then the middle carbon may be a stereocenter. . Both Triacylglycerol A and Triacylglycerol B have two identical fatty acids and one distinct one, but Triacylglycerol A is arranged symmetrically, so it is not chiral and will not rotate plane-polarized light (Choice C). Triacylglycerol B is not arranged symmetrically, so it is chiral and will rotate plane-polarized light. . (Choices A and B) Only Triacylglycerol B has a stereocenter; therefore only it, and not Triacylglycerol A, will rotate plane-polarized light.

Triacylglycerols A and B are shown below. . Which of the triacylglycerols would rotate plane-polarized light? . A. Both triacylglycerols B. Neither triacylglycerol C. Triacylglycerol A only D. Triacylglycerol B only

By the Brønsted-Lowry definitions, an acid is a species that donates a proton (ie, an H+ ion) and a base is a species that accepts a proton. Although the Brønsted-Lowry definitions are sufficient to identify acids and bases in many cases, these definitions are inadequate for species that lack acidic protons in their structure yet exhibit acid-base behavior. To overcome this limitation, the Lewis definitions were developed. Accordingly, a Lewis acid is defined as an electron pair acceptor, and a Lewis base is defined as an electron pair donor. . In Reaction 1, Mn+(aq) acts as a Lewis acid by accepting an electron pair from each of the six H2O molecules. In Reaction 2, the [M(H2O)6]n+ complex acts as a Brønsted-Lowry acid by donating an H+ ion to a H2O molecule in solution.

What roles do Mn+ and [M(H2O)6]n+ have in Reactions 1 and 2? . A. Both Mn+(aq) and [M(H2O)6]n+(aq) act as Lewis acids. . B. Mn+(aq) acts as a Lewis acid, whereas [M(H2O)6]n+(aq) acts as a Brønsted-Lowry acid. . C. Mn+(aq) acts as a Brønsted-Lowry acid, whereas [M(H2O)6]n+(aq) acts as a Lewis acid. . D. Both Mn+(aq) and [M(H2O)6]n+(aq) act as Brønsted-Lowry acids.

. When a stationary head is hit by a moving object, the skull and the brain are both accelerated in the same direction. Because the brain can move within the skull, the brain will continue to move due to its inertia even after the skull had slowed significantly or stopped. An object's inertia is proportional to its mass, and it is the tendency of an object to resist changes to its speed; an object in motion will stay in motion, and an object at rest will stay at rest (Newton's first law). Therefore, the inertia of the brain can lead to a brain-skull collision on the opposite side of the head from the initial impact (contrecoup injury). . An object's potential energy is its "stored" energy due to its height above the ground. The potential energy of the projectile does not cause the continued movement of the brain relative to the skull.

When a stationary head is hit by a moving projectile, a contrecoup injury is likely to occur due to: . . A. the inertia of the brain. B. the weight of the skull. C. the center of mass of the head. D. the potential energy of the projectile. . . . A contrecoup injury is described in the passage as a brain injury that occurs on the side of the head opposite an impact. This form of brain injury involves the movement of the brain relative to the skull due to the rapid acceleration and deceleration of the head. . An object's inertia is its resistance to changes in its velocity; objects tend to stay in motion or stay at rest. Due to the brain's inertia, the brain can continue to move independently of the skull and result in a contrecoup injury. . . . . . .

. (Choice A) The side chain for leucine is a four-carbon isobutyl group. In this question, the side chain precursor portion of Compound A is a three-carbon isopropyl group. . (Choice B) The Gabriel synthesis of α-amino acids utilizes a phthalimide-protected nitrogen component. These features are absent in the synthesis shown in this question. . (Choice C) In the Strecker synthesis, the planar iminium ion intermediate can be attacked by the cyanide ion from either face and will ultimately result in a racemic (D,L) amino acid product. This reaction cannot preferentially result in the formation of only the L-isomer. . α-Amino acids can be prepared through the Strecker synthesis from appropriate aldehyde precursors. The products of a Strecker synthesis will be racemic (D,L).

Which statement most accurately describes the reaction sequence shown below? . A. Strecker synthesis of D,L-leucine B. Gabriel synthesis of L-valine C. Strecker synthesis of L-valine D. Strecker synthesis of D,L-valine . . . . The Strecker synthesis is a man-made (synthetic) approach to prepare α-amino acids from aldehyde precursors. The carbon of the aldehyde functional group ultimately becomes the α-carbon (Cα) in the final amino acid, and the aldehyde's alkyl chain becomes the side chain (R). Since the side chain in this question is an isopropyl group, the target α-amino acid is valine. . Treatment with ammonia (NH3) generates a transient iminium ion, which then undergoes a nucleophilic addition reaction with a cyanide ion (-CN) to form an aminonitrile. The nucleophilic cyanide ion adds to either the top or bottom face of the planar (achiral) iminium ion and leads to a racemic outcome for the newly generated chiral carbon atom and a D,L-amino acid product. . Treatment of the aminonitrile intermediate with aqueous acid promotes hydrolysis of the nitrile functional group and its conversion to a carboxylic acid. Combining the features of this synthesis, this reaction sequence depicts the Strecker synthesis of D,L-valine.

Standard temperature and pressure (STP) refers to the situation where T = 273 K and P = 1 atm. Under STP, an ideal gas has the same ratio of V and n, known as the molar volume Vm. Rearranging the ideal gas law to solve for Vm yields: . Vm=V/n=RT/P=0.082L·atmK·mol *273 K / 1 atm≈ 22 L/mol . According to the passage, metabolizing carbohydrates generates 1 liter of carbon dioxide for each liter of oxygen consumed. However, the metabolism of fats generates 0.7 liters of carbon dioxide per liter of oxygen consumed. In this question, the person's cardiac output is 15 L/min, which corresponds to an oxygen consumption VO2 of 2 L/min based on Figure 1. At STP, the VO2 in mol/min is: VO2= (2 L/min)(1 mol/ 22 L )= 0.091 mol/min Therefore, the VO2 and carbon dioxide production VCO2 are equal, indicating that this person is metabolizing only carbohydrates: VO2= VCO2 = 0.091 mol/min

While exercising on a treadmill, a person has a cardiac output of 15 L/min and produces carbon dioxide at 0.091 mol/min. . What is their ratio of metabolizing carbohydrates to fats during the exercise? (Assume all gasses are ideal and are at STP.) . A. 1:2 B. 1:1 C. 1:0 D. 2:1 . . . . An ideal gas follows the equation that the product of pressure P and volume V equals the product of the number of moles n, the gas constant R, and the temperature T: PV=nRT

(Choice A) (R,R)- and (S,S)-labetalol are enantiomers, not diastereomers. No correlation exists between diastereomers and the magnitude or direction of plane-polarized light rotation. . (Choice B) If two samples rotated plane-polarized light by the same magnitude and direction, they would likely be the same compound. Although enantiomers (R,R)- and (S,S)-labetalol rotate plane-polarized light by the same magnitude, they cause the light to rotate in opposite directions. . (Choice D) If a sample were to have a rotation of zero, it would mean that the rotations have canceled each other out. This occurs when the sample is a 50:50 mixture of enantiomers (ie, 50% R and 50% S) known as a racemic mixture. .

Would (R,R)- and (S,S)-labetalol be expected to rotate plane-polarized light in the same direction? . A. Yes, because they are diastereomers. B. Yes, because the rotations would have the same magnitude. C. No, because they are enantiomers. D. No, because the rotations would cancel out each other. . . . . Normal light is made up of waves that vibrate in all planes perpendicular to the direction of travel. When the light meets a polarizer, only waves that are aligned with the polarizer can pass through, so all the light that passes through vibrates in one direction. This resulting light is known as plane-polarized (linear) light. . When chiral molecules interact with polarized light, the interactions cause the angle of oscillation to rotate. . An R or S configuration is not predictive of the direction of rotation. Every molecule is different, and the concentration and length of the path through which the light travels affect the angle of rotation; the direction and magnitude of rotation can only be determined experimentally. . Considering experimental parameters such as concentration and path length, the observed rotation can be standardized with the calculation of specific rotation. Enantiomers always have specific rotations of equal magnitude but opposite directions. (R,R)-labetalol and (S,S)-labetalol are enantiomers, and therefore rotate plane-polarized light by the same amount but in opposite directions. . Enantiomers rotate plane-polarized light by the same magnitude but in opposite directions. No correlation exists between diastereomers and the magnitude or direction of rotation. Racemic mixtures have a rotation of zero because they consist of equal amounts of each enantiomer.

Structural constitutional . The other available choices possess the same molecular formula as valsartan. The only differences are that each compound has one or more functional groups in a different location than in the original molecule, making each a structural isomer of valsartan.

_______________ isomers, also known as _________________ isomers, have the same molecular formula and thus the same molecular weight but differ in the order of atom connectivity in at least one position. . Because a methylene group (-CH2-) is missing from the carbonyl-containing (C=O) alkyl chain in Choice C, this structure does not have the same molecular formula as the original compound, and therefore is not a structural isomer of valsartan. . . Below is the structure of valsartan. All the following are structural isomers of valsartan EXCEPT: . A structural isomer, or constitutional isomer, is a molecule that contains the same molecular formula as another molecule. The atoms in each molecule differ in their connectivity.

Specific rotation Isomers Enantiomers

_________________ is the degree to which chiral molecules rotate plane-polarized light, and it is unique to each chiral molecule. . Clockwise rotations are designated as (+) whereas counterclockwise rotations are (−). . The specific rotation is a constant and describes the direction (+ or −) and magnitude (number of degrees) of a solution of normalized concentration (1 g/mL) and pathlength (1 dm). . . _______________ are molecules that have the same molecular formula but have either different connectivity (constitutional isomers) or spatial orientation (stereoisomers). . _________________ are nonsuperimposable mirror images where each corresponding stereocenter is opposite in configuration. . The specific rotations of enantiomers are equal in magnitude but opposite in direction (sign).