Part I of IVChemical and Physical Foundations of Biological Systems59 Questions (44 Passage Based Questions + 15 Discrete Questions)95 Minutes

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Passage 1A hospital is considering revamping its rehabilitation center. In an effort to facilitate movement for handicapped individuals, administrators commissioned a project to reconstruct a specific ramp. Preliminary investigations involved three volunteer patients wheeling their way up the ramp individually and providing feedback on the difficulty of the exercise.

Figure 1. A diagram of the ramp under consideration for reconstruction.

The hospital is further interested in obtaining better wheelchairs. The hospital staff is pursuing qualities such as improved balance, weight, torque, and wheel grip. To consolidate this entire project, they requested the patient volunteers to each use a different wheelchair under consideration for widespread hospital use during their trial.

Wheelchair A has a frame composed of aluminum and has thin rubber wheels. On the other hand, wheelchair B is made of plastic and has thicker rubber wheels. Finally, wheelchair C has increased rigidity due to its titanium build and large leather wheels.

Information involving each wheelchair’s mass, type, and coefficient of friction is compiled in Table 1.

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Table 1. Compilation of wheelchair data.

Mass Wheel material μκμκ μκ

Wheelchair A 20kg Thin rubber 0.5 0.2

Wheelchair B 20kg Thick rubber 0.6 0.3

Wheelchair C 30kg Leather 0.7 0.4

The masses of volunteers A, B, and C are 60, 50, and 70kg respectively.

Table 2. Trigonometric function table of angles 30 and 45 degrees.

Angle sinθ cosθ tanθ

30∘ 0.5 0.87 0.6

45∘ 0.7 0.7 1

1. Volunteer A realizes she is moving too quickly up the ramp, and engages her brakes. This causes her to begin skidding up the ramp. Which of the following regarding forces on the wheelchair is true?Please choose from one of the following options.

1 The force of the brakes and the force of friction are pointing in the same direction

2 The force of friction and horizontal force of gravity are pointing in opposite directions

3 The force of friction and horizontal force of gravity are pointing in the same direction

4 The force of the brakes and the force of friction are pointing in opposite directions

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2.Volunteer B makes her way up the ramp on wheelchair B. She pauses and engages her brakes to catch her breath. What is the force of friction exerted on the wheelchair while at rest on the ramp? (Use 10 m/s2 for the acceleration due to gravity)Please choose from one of the following options.

1 260N2 350N3 210N4 150N

3. Volunteer C is moving up the ramp with increasing velocity. How many forces are acting on the wheelchair?Please choose from one of the following options.

1 12 23 34 4

4. Volunteer B is moving at constant velocity up the ramp. What must be true about the forces on the wheelchair?Please choose from one of the following options.

1 The force of kinetic friction equals the normal force2 The force of static friction equals the horizontal component of the

normal force3 The force of kinetic friction equals the horizontal component of the

normal force4 The net force on the volunteer is equal to zero

5. Volunteer A is moving at a constant acceleration of a up the ramp. She is pushing herself with a force of Fp. What is the force of friction on the wheelchair?Please choose from one of the following options.

1 Ff = ma - Fp + (.87)mg2 Ff = -ma + Fp - (.5)mg3 Ff = -ma + Fp - (.87)mg4 Ff = ma - Fp + (.5)mg

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Passage 2

Fick's first law of diffusion states that the flux J of a substance (units:mol x m-2 x s-1) is proportional to the gradient for the substance: J = (−D)( ΔC /Δx)

where ΔC is the difference in initial and final concentration of the substance (units: mol × m−3), Δx is the distance of diffusion (in meters), and D is the diffusion constant, which depends on environmental conditions and increases with temperature. Fick's second law of diffusion states that the time it takes a particle to diffuse a given distance is proportional to the square of that distance:

t = x2 / 2D

Fick's laws can be used to model the rate of oxygen diffusion across biological membranes (MO2) using the following equation

MO2 = K x A/L X Δ PO2

where K is Krogh's diffusion coefficient, A is the surface area of the membrane, L is the thickness of the membrane, and ΔPO2 is the oxygen partial pressure difference across the membrane.

The physical restrictions modeled by these equations have lead to interesting evolutionary adaptations in the structures that animals use to absorb oxygen. For instance, Atlantic salmon hatchlings absorb the oxygen needed for cellular respiration cutaneously (through the skin). However, over the course of juvenile development, these animals experience a rapid increase in body size, and absorption of oxygen shifts almost totally to the branchial tissue of the gills. Researchers studied the measurable consequences of this developmental process by measuring oxygen absorption (MO2) in branchial and cutaneous tissue and compared the results to measurements of surface area and anatomical diffusion factor (ADF), two different measures of the oxygen-absorbing capacity of tissues (figure 2). ADF is defined as the ratio of mass-specific surface area to

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harmonic mean diffusion distance. The harmonic mean of a data set is a measure of the average value of the data set. Data for measurements of harmonic mean diffusion distance are given in figure 2.

Figure 1. Relation between body mass and change in cutaneous and branchial oxygen absorption (MO2), surface area (area), and anatomical diffusion factor (ADF) as a proportion of total body MO2, area, and ADF over the course of juvenile development of atlantic salmon hatchlings.

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Figure 2. Measurement of harmonic mean diffusion distance for filaments, skin, and lamellae for atlantic salmon (filaments and lamellae are types of branchial tissue).

Wells, P., & Pinder, A. (1996). The respiratory development of Atlantic salmon. I. Morphometry of gills, yolk sac and body surface. J Exp Biol, 199 (Pt 12), 2725-2736.

Wells, P., & Pinder, A. (1996). The respiratory development of Atlantic salmon. II. Partitioning of oxygen uptake among gills, yolk sac and body surfaces. J Exp Biol, 199 (Pt 12), 2737-2744.

6. What are the SI units for D, the diffusion coefficient?Please choose from one of the following options.

1 s−1 × m2 × mol2 s−1 × m2

3 s2 × m-2 × mol4 s × m-2

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7. According to figure one, cutaneous surface area as a percentage of total surface area decreased ~20% over the course of Atlantic salmon juvenile development, while cutaneous ADF as a percentage of total ADF decreased ~ 80% over the the same time frame. What is the best explanation for this divergence?Please choose from one of the following options.

1 Harmonic mean diffusion distance for filaments increased.2 Harmonic mean diffusion distance for filaments decreased.3 Harmonic mean diffusion distance for skin increased.4 Harmonic mean diffusion distance for skin decreased.

8. Why did researchers chose to measure ADF in addition to surface area?Please choose from one of the following options.

1 ADF takes the dependence of oxygen absorption on temperature into account.

2 ADF takes the dependence of oxygen absorption on distance traveled into account.

3 ADF takes the dependence of oxygen absorption on membrane type into account.

4 ADF takes the dependence of oxygen absorption on partial pressure into account.

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9. If researchers measured O2 flux as a function of gradient under different temperatures, which of the following graphs would most likely represent the results?Please choose from one of the following options.

1 Graph A.2 Graph B.3 Graph C.4 Graph D.

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Passage 3

Caffeine is most valued for its ability to temporarily ward off drowsiness and restore alertness. It belongs to a large class of compounds known as alkaloids, compounds that contain mostly basic nitrogen atoms. These are of plant origin and can be extracted from tea bags, for example, through a series of purification steps.

Figure 1. Caffeine Molecule

Extraction Purification

After brewing tea bags in water, the liquid extract is filtered to remove any insoluble material. The remaining supernatant can then be separated through liquid/liquid extraction. The two solvents used in this extraction step are water and dichloromethane.The two solutions then separate into two layers; a brown aqueous solution and an organic layer with dichloromethane. Evaporation of the organic solvent yields a caffeine compound with impurities such as green chlorophyll.

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Sublimation Purification

Further purification of caffeine is possible through sublimation if impurities are nonvolatile. Crude material is placed on the bottom of a sublimation chamber below a cool surface test tube. Impure caffeine extract is then heated to sublimation. A vacuum is applied to the airtight environment by sucking all of the air out of the chamber through the vacuum/gas line as shown in Figure 2.

(1)Cooling water goes in test tube.

(2)Cooling water goes out of test tube, generating a cold surface.

(3)Vacuum/gas line creates a near zero pressure environment. It constantly removes gas it is exposed to from the system.

(4)Crude caffeine material is placed inside the base of the sublimation apparatus.

(5)When heated, sublimed caffeine gas rises from the base of the apparatus, forming solid crystals when it makes contact with the cold surface test tube.

(6)Solid non-volatile impurities are left at the base of the apparatus after sublimation.

(7)External heating drives the sublimation process. The gaseous caffeine will be separated from the less volatile impurities and then forms crystal caffeine deposits along the cool surface. The caffeine deposits are then collected for the next step in processing.

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Figure 2. Sublimation apparatus

This vacuum is key to the process of sublimation. For a substance to change phase into a gas, its molecules must reach a vapor pressure equal to that of its external air pressure. Because a vacuum lowers the air pressure, the caffeine can reach this vapor pressure at a lower temperature and bypass the liquid phase.

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Figure 3. Caffeine Phase diagram with the arrow indicating the path of the caffeine sublimation.

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10. How would the caffeine sublimation phase diagram be affected if the vacuum in the apparatus allowed more air pressure in the chamber?Please choose from one of the following options.

1

2

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3

4

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11. Which of the following types of hybridizations describe carbon atoms on the caffeine molecule?I.spII.sp2

III.sp3

Please choose from one of the following options.1 II and III2 I and II only3 I and III only4 I, II, and III

12. Why was dichloromethane (CH2Cl2) used in the above extraction?Please choose from one of the following options.

1 It has a binding affinity with caffeine2 It is immiscible with water3 It has a binding affinity with water4 It is immiscible with caffeine

13. If the sublimation purification were to be performed with the cool surface tube situated higher than the original setup as shown in the left image of Figure 2, how would the results be affected?Please choose from one of the following options.

1 There would be more actual yield as gaseous caffeine will take more time to reach ambient temperature before reaching the cool surface where it will be crystallized

2 The sublimation arrow in the phase diagram would move right due to the warmer temperature experienced by the gaseous caffeine

3 There would be less actual yield as much of the gaseous caffeine will exit the sublimation chamber without making contact with the cool surface

4 The sublimation arrow in the phase diagram would move left due to the cooler temperatures experienced by the gaseous caffeine

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14. Despite heating the apparatus, the scientist detects no temperature change for a short period during the sublimation step. Which of the following explains this observationPlease choose from one of the following options.

1 The energy added was used to break caffeine's intermolecular bonds2 The energy added was stored as chemical potential energy3 The energy added had a negligible effect on temperature, because

caffeine has a high specific heat capacity4 The energy added was used to increase the pressure inside the

apparatus

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Questions 15. 16. 17. 18. 19. are NOT based on Passages.

15. An ambulance is currently traveling at 15m/s, and is accelerating with a constant acceleration of 5 m/s2 . The ambulance is attempting to pass a car which is moving at a constant velocity of 30m/s. How far must the ambulance travel until it matches the car’s velocity?Please choose from one of the following options.

1 67.5 m2 45 m3 90 m4 67.5 km

16.Which of the following statements describes what must be true in the context of Newton’s First Law?Please choose from one of the following options.

1 The object is difficult to bring to a complete stop due to its high initial speed.

2 The tendency for drivers to keep moving linearly while the car makes a sharp turn on the road is an example of the concept of inertia.

3 Mass is a measure of an object’s ability to resist motion or movement of any kind.

4 An object with zero acceleration and an object traveling at a constant acceleration are considered similar states.

17. A 100 kg block of wood is traveling with a constant velocity on ice. What is its normal force?Please choose from one of the following options.

1 0 N2 10 N3 100 N4 1000 N

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18. 3 kg of wet clothes are hung on the middle of a clothesline with posts 6ft apart. The clothesline sags down by 3 feet. What is the total tension upon the clothesline?Please choose from one of the following options.

1 15√(2)/2 N2 30√(2) N3 60√(2) N4 15√(2) N

19. Which of the following statements is true regarding a rider in a roller coaster cart moving with a constant speed through a loop?Please choose from one of the following options.

1 Gravity is the only force doing work on the rider.2 The sum of all the forces acting on the rider is zero.3 The rider is accelerating.4 There are two forces acting on the rider, but neither does any work on

the rider.

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Passage 4

Noble gases have numerous medicinal functions due to their relative non-reactivity. Helium is beneficial in respiratory treatments due to its low density, low solubility, and high thermal conductivity. Argon is useful for cryotherapy, in which tissue is exposed to sub-zero temperatures for surgical purposes. The anesthetic application of Xenon has been demonstrated to be more effective than previous techniques, although the significant financial expense of the gas makes its use less common. Finally, Krypton finds use in the field of cardiology in its use of detecting abnormal heart openings.

A research assistant in a lab is investigating the properties of Helium, Argon, Krypton, and Xenon for further use in the medical field. In the lab, there are four sealed containers, each of which contains a certain amount of a respective gas. The canisters are labeled with the type of gas contained, the pressure of the gas contained, the volume of the canister, and the # of moles of the gas.

Table 1. Each canister and the element's corresponding pressure, volume, and number of moles.

Pressure of gas Volume of canister

# of moles of gas

Helium 100,000 Pa 0.02m3 4

Argon 200,000 Pa 0.04m3 8

Krypton 300,000 Pa 0.06m3 1

Xenon 400,000 Pa 0.08m3 2Sources:Berganza, C. Zhang, J. (2013). The role of helium in medicine. Medical Gas Research, 3, 18. Loma Linda, CA.Jordan, B. Wright, E. (2010). Xenon as an Anasthetic Agent. American Association of Nurse Anesthetists Journal, 78.

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Chemical Properties. (2015). Krypton. Chemical Properties of Krypton. Lenntech BV. Rotterdamseweg, NED.

20.What would be the correct ranking of the average kinetic energy of the gas molecules of the gases based on how they are labeled?Please choose from one of the following options.

1 KEXenon > KEArgon > KE Krypton > KE Helium2 KEXenon > KE Krypton > KE Argon > KE Helium3 KE Krypton > KE Xenon > KE Argon > KE Helium4 KEXenon > KE Krypton > KE Helium > KE Argon

21. What would happen to the temperature of the Helium container if the number of moles was kept constant but the pressure was changed to 400,000Pa and the volume of the canister was changed to 0.01 m3 ?Please choose from one of the following options.

1 The temperature would increase by a factor of 82 The temperature would quadruple 3 The temperature would be cut in half4 The temperature would double

22. Based on how the canisters are labeled, how much greater is the average kinetic energy of the molecules of Xenon compared to Argon?Please choose from one of the following options.

1 8 times larger2 16 times larger3 32 times larger4 The average kinetic energy is the same for each gas

23. Based on how the canisters are labeled, how would you rank the heat capacities of the gases at constant volume Cv?

1 Cv Helium > CvArgon > Cv Krypton > Cv Xenon2 Cv Argon > Cv Helium > Cv Xenon > Cv Krypton3 Cv Krypton > Cv Xenon > Cv Helium > Cv Argon4 Cv Krypton > Cv Xenon > Cv Argon > Cv Helium

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24.You decide to add the same amount of heat to each canister, in their labeled states, and allow the canisters to expand so the pressure of the gas remains constant while the heat is added. During this process which gas would change its temperature by the greatest amount?Please choose from one of the following options.

1 Helium2 Argon3 Krypton4 Xenon

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Passage 5

Sickle cell disease is an autosomal recessive disease commonly found in African populations. The disease gets its name from the fact that patients’ red blood cells become sickle-shaped when passing through the capillaries of metabolically active tissues. These red blood cells become frail and can rupture long before their normal lifespan. The sickled red blood cells block capillaries and inhibit red blood cell function, causing severe anemia in sufferers.

The most common hemoglobin of adults is hemoglobin A (HbA). However patients suffering from sickle cell disease are homozygous for the allele coding for the abnormal variant of hemoglobin (HbS) due to a missense mutation in the gene encoding the β subunit of hemoglobin. This missense mutation replaces glutamic acid with valine at the sixth position of the β-globin chain.

The absence of a polar amino acid at this position promotes the non-covalent aggregation of hemoglobin in a low-oxygen environment which distorts red blood cells into a sickle shape and decreases their elasticity. Biochemically, the low oxygen environment causes the beta chain of neighboring hemoglobin molecules to hook together, becoming rigid and polymerized. These cells fail to return to their normal shape when oxygen is restored and thus fail to deform as they pass through narrow vessels, leading to blockage in the capillaries.

In vitro studies of deoxygenation and reoxygenation of sickle-cell hemoglobin indicates that the process is reversible. The hemoglobin molecules polymerize and form crystals as oxygen concentration is lowered. But as the oxygen concentration is increased again, hemoglobin molecules can depolymerize and return to their soluble state. This can be written as:

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Figure 1. The equilibrium reaction equation relates non-sickled hemoglobin with sickle cell hemoglobin polymers.

The discovery of the sickle cell mutation came from isoelectric focusing, a type of electrophoresis that separates complex mixtures of large molecules by applying an electric current. Hemoglobin taken from patients suffering from sickle cell anemia, individuals who were heterozygous for the abnormal variant, and those that did not have the sickle cell allele were subjected to isoelectric focusing, the following results were obtained:

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Figure 2. Distribution of hemoglobin protein position along gel after isoelectric focusing.

Data adapted from: Natural Sciences Learning Center, Washington University. (2003). The Molecular Biology of Sickle Cell Anemia.

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25. Which of the following amino acids, if present at the sixth position of the β-globin chain, would yield a non-covalent aggregation of hemoglobin in a low-oxygen environment?Please choose from one of the following options.

1 Aspartic Acid2 Serine3 Threonine4 Isoleucine

26.How can the amount of soluble non-sickled hemoglobin be increased in an in vitro study?Please choose from one of the following options.

1 Introduce additional sickle cell hemoglobin polymers2 Reduce the oxygen concentration level3 Reduce the amount of sickle cell hemoglobin polymers4 Reduce the amount of soluble non-sickled hemoglobin

27. How does the beta chain of neighboring HbS molecules cause hemoglobin to aggregate?Please choose from one of the following options.

1 Valine causes the hemoglobin molecules to exhibit increased van der Waal forces due to hemoglobin’s increase in size.

2 The interaction between glutamic acid and other hydrophilic amino acids creates a hydrophilic pocket leading to hemoglobin aggregation.

3 Glutamic acid hydrogen bonds to other amino acids creating a shell of polymerized amino acids.

4 The interaction between valine and other hydrophobic amino acids creates a hydrophobic pocket leading to hemoglobin aggregation.

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28. What does the distribution in the isoelectric focusing experiment indicate about heterozygous individuals?Please choose from one of the following options.

1 They exhibit two different types of hemoglobin proteins, separated from each other in the electrophoresis by their difference in length.

2 They exhibit multiple different types of hemoglobin proteins, separated from one another in the electrophoresis by their difference in mass.

3 They exhibit two different types of hemoglobin proteins, separated from each other in the electrophoresis by their difference in charge.

4 They exhibit multiple different types of hemoglobin proteins, separated from one another in the electrophoresis by their difference in hydrophobicity.

29. How would the distribution of hemoglobin proteins in the 50:50 mixture be different if a higher voltage difference were to be applied between the electrodes?Please choose from one of the following options.

1 The bimodal distribution would become less pronounced as the electric field would be less effective in imparting a charge to the proteins.

2 The bimodal distribution would become more pronounced as the electric field would be more effective in separating the hemoglobin proteins from each other.

3 The bimodal distribution would become more pronounced as the electric field would be more effective in imparting a charge to the proteins.

4 The bimodal distribution would become less pronounced as the electric field would be less effective in separating the hemoglobin proteins from each other.

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Passage 6

An incompressible fluid is flowing without turbulence through a tube that has different diameter sections seen in the diagram. The absolute pressure at point 1 is 220,000 Pascals and the speed of the fluid at point 1 is 8.0m/s. The radius of the tube at point 1 is 0.50m. The radius of the tube at point 3 is 0.25m. The radius of the tube at point 4 is 1.0m.

Figure 1. A diagram of the fluid passing through a shaped tube.

30. What would be the correct ranking of the speeds of the fluids at points 1,2,3,4?Please choose from one of the following options.

1 v1 > v2 > v3 > v42 v2 > v3 > v1 > v43 v4 > v1 > v2 > v34 v3 > v2 > v1 > v4

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31. How does the pressure at point 2 compare to the pressure at point 4?Please choose from one of the following options.

1 The pressure at point 2 is the same as the pressure at point 4 since the fluid is incompressible

2 The pressure at point 2 is greater since the water is moving faster there

3 The pressure at point 2 is smaller since the water is moving faster there

4 The pressure at point 2 is greater since the water is more compressed together

32. What will be the speed of the fluid at point 3 and point 4?Please choose from one of the following options.

1 v3=16m/s, v4=4m/s2 v3=4m/s, v4=16m/s3 v3=32m/s, v4=2m/s4 v3=2m/s, v4=32m/s

33. What would be a possible pressure of the fluid at point 3 based on the information given?Please choose from one of the following options.

1 280,000 Pa2 250,000 Pa3 220,000 Pa4 200,000 Pa

34. What is the gauge pressure at point 1?Please choose from one of the following options.

1 329,000 Pa2 219,000 Pa3 119,000 Pa4 19,000 Pa

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Questions 35. 36. 37. 38. 39 are NOT based on passages.

35. According to your understanding of Bernoulli’s principle and continuity equation, which of the following statements accurately describes the phenomenon of vascular flutter which occurs when an artery becomes constricted due to the accumulated plaque in its inner walls?Please choose from one of the following options.

1 Constriction in the blood vessel causes the pressure to build up right at the narrowing.

2 The constriction will cause the blood to travel faster and the extra blood will create pressure after the narrowing.

3 The constriction causes the pressure to drop and the vessel to collapse.

4 Constriction in the blood vessel causes backup and the pressure to build before the narrowing.

36. All of the following are properties of ideal gases except:Please choose from one of the following options.

1 Volume occupied by molecules is negligible compared to the volume occupied by the gas

2 Small amounts of energy are lost during collisions between gas molecules

3 Gas molecules do not interact with each other except during collisions

4 Collisions between gas molecules are completely elastic

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37. Suppose two parallel plates are inserted into a solution, and the current that passes between them for a known potential difference is used to measure the electrolytic resistivity of the solution. How would this resistivity change if the area of both of the two plates was doubled, at a fixed separation distance?Please choose from one of the following options.

1 It would remain the same2 It would be halved3 It would quadruple4 It would double

38. Early researchers found that an electric or magnetic field could split radioactive emissions into three types of beams as illustrated below. Which of the following statements accurately describes these characteristic radioactive decay products?

Please choose from one of the following options.1 Particle B represents a neutron since there is no deflection in its path

due to its larger mass.2 Particle A represents an alpha particle because of the upward and

smaller deflection than that of particle C. 3 Particle C represents a positron because it experiences the most

deflection from passing through the magnetic field.4 Particle A represents a proton because a particle with a positive

charge would have such a characteristic path.

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39. Which of the following statements accurately describes the Nernst equation shown below?

Ecell = Eocell − RT lnQ nF

Please choose from one of the following options.1 At standard conditions, Ecell = E°cell because all the concentrations

are 1 M at an ambient temperature of 273 K.2 Increasing the concentration of the reactants will increase the value

of Ecell.3 This equation is used to figure out the standard cell potential at any

concentration, partial pressure, or temperature.4 At equilibrium, Ecell calculates to a value of 0 and the battery is

considered dead because Q becomes Keq and K eq has a value of 0.

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Passage 7

A variety of chronic disorders may develop from the accumulation of non-essential trace elements. Humans have been exposed to such elements only recently due to extensive mining. Because humans did not encounter these elements during the course of evolution, they have not developed homeostatic mechanisms to accommodate or eliminate these elements from their bodies. For example, cadmium has been linked to the development to itai-itai disease which has symptoms such as renal failure, weak bones, and joint and spine pain. Studies indicate that elevated cadmium levels, due to inefficient removal from the body by the kidneys ,leads to mitochondrial damage and eventually renal failure.

Patients all originating from one manufacturing town and its environs have been reporting at various hospitals with wide-ranging symptoms, including headache, vomiting, muscle pain, memory loss, constipation, and high blood pressure. To determine what may be responsible, an investigator can use qualitative analysis to identify the ions present in a sample by systematically conducting different chemical tests. The most common test is selective precipitation, which separates cations into groups based on their solubility under specific conditions. The reagents are added sequentially to the same solution throughout testing. After separation, the individual ions can be identified by confirmatory tests, which can include observing their color, odor, appearance, and solubility.

A systematic qualitative analysis is conducted on a water sample at a local lab, and the results are in the table below:

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Table 1. Qualitative Analysis of Water Sample by Precipitating Cation Groups with Reagents

Cation Category

Anion Precipitated

Reagent Added (in sequential

order)

Precipitate Color

1 Cl- dilute HCl White

2 S2-

(acid insoluble)H2S in

concentrated HNO3

No precipitate

3 OH- NH4Cl/NH3 buffer (solid

NH4CL and 16M NH3

Brown

4 S2-

(base insoluble)H2S in NH4Cl/

NH3 bufferWhite

5 PO43- (NH4)3PO4 and concentrated NH3

White

6 None None No precipitate

Many of the tests for identifying cations have been superseded by techniques like mass spectrometry and absorption spectrophotometry due to the range of chemicals the instruments can detect. However, traditional qualitative analysis can be still useful in field work or other situations where access is not expedient. A neighboring facility conducted a spectroscopic

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analysis of a sample with another suspected pollutant, and the results are below:

Figure 1. Absorption Spectrogram of Water Sample from Adjacent Town

40. The addition of nitric acid to the water sample produces a colorless and odorless gas. What is the likely reaction when nitric acid is added to the sample?1. MX + HNO3⇋ M+(aq) + NO2(g) + NO(g)2. MX + HNO3⇋ M+(aq) + NO3-(g) + SO2(g) + H20(l)3. MX + HNO3⇋ M+(aq) + NO3-(g) + CO2(g) + H20(l)4. MX + HNO3⇋ M+(aq) + Cl2(g) + NO3-(g)

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41. Which of the following can be best detected by UV-Vis absorption spectroscopy?Please choose from one of the following options.

12 Cr (s)

3

4

"

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42. The Category 6 cations consists of the Group IA elements and ammonium, which cannot be further partitioned by selective precipitation. Based on Table 1, what is the identity of the Category 5 precipitate?Please choose from one of the following options.1. Zn3(PO4)2

2. CrPO4

3. Sr3(PO4)2

4. Pb3 (PO4)2

43. Which of the following statements correctly explains the role of the reagents in the qualitative analysis?Please choose from one of the following options.

1 The buffer solution increases the OH− ion available to precipitate the Category 3 cations.

2 Concentrated NH3 decreases the PO43−ion available to precipitate the Category 5 cations.

3 H2S is placed in HNO3 to increase the S2− ion available to precipitate Category 2 cations.

4 The buffer solution increases the S2− ion available to precipitate the Category 4 cations when NH3 reacts with H3O+.

44. Based on the absorption spectrum in Figure 1, what was the color of the original sample?Please choose from one of the following options.

1 Red2 Orange3 Blue-Green4 Violet

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Passage 8

A tank of water is filled to the top and has a small hole through which the water can flow out. The top of the tank is open to the atmosphere. The tank is full so the water level is a height D above the hole, and the hole is a height H above the table.

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45. As the water flows out of the tube, and the water level drops, what will be seen to happen to the water coming out of the hole?Please choose from one of the following options.

1 The water will flow out with less speed since the density of the water will be smaller as the water level reaches the hole

2 The water will flow out with greater speed since more potential energy is being turned into kinetic energy

3 The water will flow out with less speed since the pressure exerted by the water above the level of the hole will get smaller

4 The water will flow out with more speed since the water level will be closer to the hole

46. The water flowing out of the hole is given by which expression?Please choose from one of the following options.

1 v = (2gH)1/2

2 v = (2gD+2gH))1/2

3 v = (2gH-2gD)1/2

4 v = (2gD)1/2

47. If while the water was flowing someone were to place an airtight cover over the top of the container, what would they likely see and why?Please choose from one of the following options.

1 The water would at first flow the same but soon stop flowing out of the tank since the dropping water level would create a partial vacuum above the water in the tank

2 The water would immediately not flow at all out of the tank since there would be no air above the water in the tank

3 The water would flow more quickly and continue to flow more quickly since the air pressure would be trapped inside the top of the tank

4 The water would flow the same as always and at a consistent constant rate since the air pressure will remain constant above the water

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48. If the open top tank experiment were repeated, but a liquid of a larger density than water (but the same viscosity of water) were used, what would be seen to happen?Please choose from one of the following options.

1 The higher density fluid would flow with smaller speed since it has a greater inertia

2 The higher density fluid would flow with greater speed since the pressure at the location of the hole would be greater

3 The higher density fluid would flow with greater speed since it has a greater weight

4 The higher density fluid would flow with the same speed since it has both a greater weight and a larger inertia

49. If someone were to take the tank off of the table while the water is flowing out, and drop the entire tank allowing it to free fall downwards, what would be seen to happen to the water flowing out of the hole?Please choose from one of the following options.

1 The water would continue to flow out the same as before dropping it since the water at the top and at the hole are effected by free fall equally

2 The water would flow out faster since the water above the hole is moving faster while free falling

3 The water would flow out of the hole only a little less slowly since the density of the water would decrease slightly while free falling

4 The water would not flow out, since the water above the hole would be effectively weightless by free falling

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Passage 9

The HSAB (Hard Soft Acid Base) theory categorizes chemical species as acids or bases and as “hard”, “soft”, or “borderline”. It explains that soft acids or bases tend to be large and very polarizable, while hard acids or bases are small and non-polarizable. Since these categories are not absolute, there are species that are considered borderline, which lie in between hard and soft. Generally, we could characterize hard acids and bases as having:

1 Small atomic radii2 Low polarizability3 High positive charge (3+ or higher)4 High electronegativity (base)5 High charge density6 High oxidation state

A hard acid or base may have some or all of these characteristics, but the hard-soft distinction is most directly linked to polarizability. Polarizability is the relative tendency of the electron cloud to be distorted from its normal shape.

Table 1. Table of representative hard, borderline, and soft acids and bases

Hard Borderline Soft

Acids K+, Al3+, Sr2+, Cr3+ Ni2+, Rh3+, Ir3+ Cu+, Pd2+, Tl+

Bases OH-, NH3, SO2-4, F-

Br-, NO2-, N2 SCN-, RS-, I-

The theory elaborates that hard acids prefer to bond with hard bases, and the resulting adduct tend to have more ionic character in its bonding. Correspondingly, soft acids prefer to bond with soft bases, and their adducts are more covalent in nature.

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The concept has been used in chemistry to explain the stability of various compounds, pathways, or reaction mechanisms. In transition metal chemistry, it offers a qualitative approach to understanding metal-ligand interactions in complexation reactions. Here are the Keq values for two such ligand exchange reactions:

[CH3Hg(H2O)]+ HF ⇋ CH3HgF + H3O+ Keq = 4.5 x 10-2

[CH3Hg(H2O)]+ HBr ⇋ CH3HgBr + H3O+ Keq = 4.2 x 1015

In toxicology, the HSAB theory has recently proved useful in predicting toxicant-target interactions, whereby most electrophiles are the toxicants and most nucleophiles are biological macromolecules, in particular, thiols and thiolates in cysteine and amines in lysine. Cytotoxicity occurs when toxicants like acrolein and vinyl chloride form adducts with biomacromolecules of similar hardness or softness, thus impair cellular function. One of the compounds under investigation is curcumin, a phenol found in turmeric, and its structure is below:

Figure 1. Structure of Curcumin

Source: Adapted from a paper by R. LoPachin, et al. Copyright 2011 by American Chemical Society.

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50. The HSAB theory can be used to predict solubility trends. Based on the determination of the hardness of the ions, what happens whenAgF and LiI are placed together into solution?Please choose from one of the following options.

1 No precipitate is formed.2 Only LiF precipitates.3 Both LiF and AgI precipitate.4 Only AgI precipitates initially, but dissolves back into solution when

Ag+ forms a complex ion with F− and H2O.

51. Which of the following trends has a negative correlation with hardness?Please choose from one of the following options.

1 Charge density2 Polarizability3 Oxidation state4 Electronegativity

52. Recent clinical trials have suggested that 1,3-dicarbonyl enolates offer cytoprotective effects against toxicants of a similar softness or hardness. Your research team has identified methyl vinyl ketone (butenone) as a soft electrophile and 2,5-hexanedione as a hard electrophile. Against which of the following, if any, will curcumin offer the best protection?Please choose from one of the following options.

1 2,5-hexanedione2 Both3 Methyl vinyl ketone4 None

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53. Acrolein or propenal is considered a soft electrophile and extremely toxic. Which of the following targets would acrolein have the greatest affinity for?Please choose from one of the following options.

1 Secondary amine of a histidine residue2 Thiol group of a cysteine residue3 Guanidinium group (RHNC(NH2)2)

of an arginine residue4 Ring oxygen of deoxyribofuranose moiety

54. How does the HSAB theory account for the difference inKeq values for the two ligand exchange reactions given in the passage?Please choose from one of the following options.

1 Hg+ and F− are soft ions and cannot form a covalent bond in solution.

2 Hg+ and Br− do not match in terms of their hardness.3 Hg+ and Br− are soft ions and prefer to bind forming a bond

covalent in nature.4 Hg+ and F− are hard ions and favorably bind together to form an

ionic compound.

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Questions 55. 56. 57. 58. 59 are NOT based on passages.

55. The speed of sound traveling through a gaseous sample of mixed gases can be affected by:I. TemperatureII. DensityIII. PressureIV. Composition of the gaseous samplePlease choose from one of the following options.

1 One of the above2 Two of the above3 Three of the above4 All of the above

56. Which of the following is not a property of an electromagnetic (EM) wave?Please choose from one of the following options.

1 Momentum2 Heat energy3 Energy4 Pressure

57. Which of the following statements accurately describes the molecular vibrations characteristic of IR spectroscopy?Please choose from one of the following options.

1 Stretching frequencies are lower than corresponding bending frequencies.

2 Triple bonds have lower stretching frequencies than corresponding double bonds, which in turn have lower frequencies than single bonds.

3 Bonds to hydrogen have higher stretching frequencies than those to heavier atoms.

4 Stretching frequencies appear mostly in the fingerprint region.

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58. During an experiment, a thin-lens system in its initial state produces a real, inverted, and enlarged image and in its final state produces a virtual, upright, and enlarged image. Which of the following statements best elucidates on what kind of change occurred for the thin-lens system?Please choose from one of the following options.

1 An additional converging lens was placed outside the focal point of the original lens.

2 The object is moved from a point outside the radius of curvature to a point within the focal length of a concave lens.

3 The object is moved from a point outside the focal length to a point within the focal length of a convex lens.

4 The converging lens is swapped for a diverging lens to create the switch in orientation.

59. Which of the following is true of reflected light rays from a surface?I.The angle of incidence is equal to the angle of reflection for diffuse reflections.II. The angle measured between the surface and the ray is the same magnitude as the angle of reflectionIII. The angle of incidence is found by subtracting the angle made with the surface from the normalPlease choose from one of the following options.

1 I + II2 II only3 I + III4 I only

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