FL 7 - Chem/Phys

Question 1

Suppose that a moderately hard surface is found to decelerate a person making a feet-first landing to a stop in an average of 0.09 seconds. What is the person’s deceleration in terms of g (where 1 g = 9.81 m/s2) when falling from a height of two meters?

Answer 1

Let’s start by figuring out exactly what the question is asking. A person is falling from a 2-meter height, then hitting the ground. From the time of impact (where he hits at some unknown velocity) to the end of the collision (where v = 0 m/s), it takes 0.09 s.

To find the magnitude of this acceleration, we need to know the velocity at impact. For this, we can use v_f² = v_i² + 2ad. Initial velocity is 0 m/s, distance is 2 m, and acceleration is g, or 9.81 m/s². Again, since answer choices are far apart, you’ll likely experience no problems if you estimate g as 10 m/s².

v_f² = (0 m/s)² + 2(10 m/s²)(2 m)

v_f² = 40 m²/s²

We know that 6² = 36 and 7² = 49, so ?40 must fall between 6 and 7. Let’s call it 6.4. Now, acceleration = (final velocity - initial velocity) / time.

a = (0 m/s - 6.4 m/s) / 0.09 s

We get a negative value in the numerator, but all of our answer choices are positive, and the question asks for the person’s deceleration. Since the negative sign simply denotes that he is decelerating, we can ignore it going forward. We can also round 0.09 s to 0.1 s.

a = 6.4 m/s / 0.1 s

a = 64 m/s²

Finally, note that our answer must be written in terms of g. Estimating g as 10 m/s², 64 m/s² is equal to 6.4g. However, the value of g is actually lower, so our real answer should be slightly higher than 6.4g. Choice C, 7g, is closest.

Question 2

While playing, a 40-kg child is accidentally pushed directly forward off a ledge, causing him to fall and hit the ground at a 60° angle. If the child has 720 J of kinetic energy at the instant before impact, at what velocity was he pushed? Assume negligible air resistance.

Answer 2

Let’s begin with the information that we have, most notably the kinetic energy value. Remember, KE = ½ mv², so we can use this value to calculate the child’s total velocity immediately before impact.

720 J = ½ (40 kg)(v²)

720 J = 20 kg (v²)

36 = v²

v = 6 m/s

Note that this is the child’s total final velocity, so it includes both a horizontal and a vertical component. Since air resistance is negligible and gravity acts only in the vertical direction, the child’s final horizontal velocity is the same as his horizontal velocity at the moment of the push. This value can be found using the cosine of the given angle (if in doubt, draw out the situation).

v_horizontal = v_total cos(60°)

v_horizontal = 0.5v_total = 0.5 (6 m/s) = 3 m/s

Question 3

A student ran a thin-layer chromatography (TLC) plate and obtained a lengthy streak for one of the samples. What are some likely factors to have caused the TLC plate to streak?

Answer 3

In TLC, streaking can be caused by overloading the spot with sample.

In TLC, streaking can be caused by using a sample that is too concentrated.

If a TLC sample is impure, its multiple components will travel different distances along the plate, which could lead to a streaked appearance.

Question 4

How much does the rate of effusion of dichloromethane at 200 K change when the temperature is changed to 800 K? Assume all other conditions are identical.

Answer 4

It increases by 100%

Rate of effusion depends on root-mean-square speed, which is calculated using the formula v_rms = √(3RT)/M. Here, T is temperature and M is molar mass. Since the identity of the gas is constant, the only factor changing is T. Therefore, for two samples of the same gas with identical pressures but different temperatures, the following relationship holds:

rate₁/rate₂ = √(T₁/T₂)

rate₂/rate₁ = √(800/200) = 2

This means that the new rate is twice as large, or 100% larger than the original rate.

Question 5

What is the function of a ribonuclease in E. coli?

Answer 5

It cleaves RNA.

We can assess the function of this enzyme by analyzing its name. The ending “-ase” is typical for enzymes, and often (though not always) implies that some larger molecule is being broken down. Specifically, nucleases break down nucleic acids. The two major types of nucleic acids are deoxyribonucleic acids (DNA) and ribonucleic acids (RNA). Given the name, we can figure out that a ribonuclease breaks down RNA.

Question 6

On which amino acid residue(s) does phosphorylation usually occur in eukaryotic proteins?

Answer 6

Phosphorylation usually occurs on serine (S), threonine (T), tyrosine (Y), and histidine (H) residues in eukaryotic proteins.

Question 7

We see d²sp³ hybridization in the ___________and sp³d² hybridization in the _______.

Answer 7

We see d²sp³ hybridization in the transition metals and sp³d² hybridization in the nonmetals.

Question 8

Prior to the collagen protein analysis used in scaffold design, protein samples must undergo proteolysis. What kind of reaction is required to accomplish this task?

Answer 8

The question stem indicates that the preparation requires proteolysis. Proteolysis is the breakdown of proteins into smaller polypeptides or individual amino acids via hydrolysis reactions, as shown below:

Question 9

Refractive myopia is corrected by the use of a diverging lens of appropriate optical power. When compared to a healthy eye, a myopic, but otherwise normal, eye’s near point is most likely:

Answer 9

smaller than that of a healthy eye.

A diverging lens decreases the optical power of an instrument by increasing its effective focal length. Such a lens is corrective for the myopic eye, where its optical power exceeds that required for the axial length of the eye. As a result of this increased optical power, the myopic near point should be smaller than that found in a healthy eye. This is due to the myopic eye’s capacity to focus light of great divergence on the retina.

Question 10

To create a virtual image with a converging lens, the object distance must be _______ than the focal distance of the lens in question. Here, this means that the object must fall _______ the focal length of the eyepiece lens.

Answer 10

To create a virtual image with a converging lens, the object distance must be smaller than the focal distance of the lens in question. Here, this means that the object must fall within the focal length of the eyepiece lens.

Question 11

The compound 2,3-dihydroxybutanedioic acid (tartaric acid) has how many stereoisomers?

Answer 11

3

The IUPAC name for tartaric acid is 2,3-dihydroxybutanedioic acid. As seen in Figure 1, this compound has two chiral carbons. Based on the 2ⁿ rule, it should have 2² = 4 possible stereoisomers, but two of these possibilities are meso compounds, so the best answer is 3, as shown below:

Question 12

Which of the following are bases?

Potassium Hydroxide, Calcium hydroxide, Sodium carbonate, or Methyl acetate.

Answer 12

Methyl acetate is an organic ester, not a base.

Question 13

In an isolated system, entropy is maximized when:

I. the system is at equilibrium.

II. the system is far from equilibrium.

III. the system is unable to perform work.

Answer 13

I and III only

Recall that at equilibrium, ∆G = 0. With no free energy change, the system is unable to perform work (III). At equilibrium, there are no energy gradients within the isolated system, so energy is maximally dispersed, resulting in maximal entropy (I).

Question 13

In an isolated system, entropy is maximized when:

I. the system is at equilibrium.

II. the system is far from equilibrium.

III. the system is unable to perform work.

Answer 13

I and III only

Recall that at equilibrium, ∆G = 0. With no free energy change, the system is unable to perform work (III). At equilibrium, there are no energy gradients within the isolated system, so energy is maximally dispersed, resulting in maximal entropy (I).

Question 14

A student is conducting an experiment in which he collects gaseous CO₂ produced during a particular reaction. Under which conditions would the student’s CO₂ samples behave most ideally?

A. T = 285 K and P = 0.05 atm

B. T = 10 K and P = 50 atm

C. T = 273°C and P = 50 Pa

D. T = 273°C and P = 50 kPa

Answer 14

Gases behave most ideally under high temperatures and low pressures. The tricky part of this question is noticing that not all answer choices are given in the same units. Of the temperatures listed, the highest is 273°C (approximately 546 K). For the pressure values, remember that 1 atm is roughly equal to 101,500 Pa. 50 Pa (1 atm / 1.01 x 10⁵ Pa) yields a value on the order of 10^-4 atmospheres, so 50 Pa is the lowest pressure given as part of an answer choice as well.

Question 15

What is the electronic configuration of Fe³⁺?

A. [Ar] 4s²3d⁶

B. [Ar] 3d⁵

C. [Ar] 4s²3d³

D. [Ar] 4s²3d⁹

Answer 15

B. [Ar] 3d⁵

Elemental iron, Fe, has 8 more electrons than argon. The electronic configuration of Fe is [Ar] 4s²3d⁶. However, we are asked about Fe³⁺, which has 3 fewer electrons than Fe. But which electrons should we remove first? When writing the electronic configuration of a cation, begin by removing the electrons with the highest principal quantum number. Here, the 4s electrons are less tightly held than the 3d electrons, so they will be lost first. The loss of one additional electron from the d subshell is favored because it results in stable, half-filled d orbitals.

Question 16

An organic synthesist seeks to identify an efficient stereoselective reagent with which to produce the enantiomer of the biologically active sphingosine molecule pictured in the passage. Which of the following would be the most logical choice to try?

A.Tert-butoxide

B. DMSO

C. Diethyl tartrate

D. Propanol

Answer 16

C. Diethyl tartrate

This question relies upon prior knowledge that you should master prior to taking the MCAT; the only stereoselective reagent listed is diethyl tartrate.

To be a stereoselective reagent, the molecule must be chiral

Question 17

Addition of aluminium hydride to sphingosine would most likely result in which of the following products?

Figure 1 Sphingosine

A. (2S,3R)-2-aminooctadec-4-ene-3-ol

B. (2S)-2-aminooctadec-4-ene-1-ol

C. (3R)-octadec-4-ene-1,3-diol

D. (2S,3R)-2-aminooctadecan-1,3-diol

Answer 17

D. (2S,3R)-2-aminooctadecan-1,3-diol

Aluminium hydride is a powerful hydrogen donor that is commonly used as a reducing agent; the hydrogens donated are in a negatively-charged state, and therefore are extremely unlikely to behave as Brønsted-Lowry acids. The most highly oxidized area of sphingosine is the alkene bond, since all of the electronegative atoms (O and N) have their electron affinity partially offset by sigma bonds to other elements. Choice D correctly recognizes that hydrogen donation by aluminium hydride would result in the loss of the alkene bond, making it the best choice for this question.

Question 18

The conversion of dihydrofolate to THF occurs via the enzymatic addition of a hydride to dihydrofolate. The enzyme that catalyzes this reaction is an example of a:

Answer 18

reductase.

Hydride additions represent electrochemical reduction. Biological reductions are catalyzed by reductases, as is the case with the reduction of DHF to THF. This reaction is catalyzed by dihydrofolate reductase, which transfers a hydride from the cofactor NADPH to DHF:

Question 19

An object floats in water with 4/5 of its volume submerged. If the object is then placed in a type of oil with a density half that of water, which of the following is true about the object placed in the oil?

A. The object will float just as it did before.

B. The object will float lower than it did in the water.

C.The object will float higher than it did in the water.

D. The object will sink to the bottom.

Answer 19

The object will sink to the bottom.

When dealing with the principles of floating, buoyancy, and density (as in the case of a object floating with a certain portion of it submerged), we can use a shortcut equation:

V_fluid / V_obj = ρ _obj / ρ _fluid , where V _fluid is the volume of displaced fluid (here, it’s water).

V_fluid / V_obj = 4/5, so ρ _obj / ρ_water = 4/5.

The question tells us that ρ_water = 2ρ_oil. This means that 4/5 = ρ_obj / 2ρ_oil .

Therefore, ρ_obj / ρ_oil = 8/5. Since the ratio of densities is larger than 1, we know that the object is more dense than oil, and it will sink to the bottom. The diagram shown below presents the scenario in the question:

Question 20

A student performs an experiment and determines that the aqueous decomposition of chloromethyl methyl ether at 650°C is first order with a half-life of 118 seconds.

If the student began the experiment with 5 moles of chloromethyl methyl ether, how much remained after 354 seconds?

A. 125 mg

B. 625 mg

C. 5 ✕ 10⁴ mg

D. 5 ✕ 10⁵ mg

Answer 20

C. 5 ✕ 10⁴ mg

Don’t let all of the extra information (i.e. the reaction schematic) distract you! This question provides a half-life and then requires us to find out how much of the original molecule is left after (354/118) = 3 half-lives. The molar mass of chloromethyl methyl ether (C₂H₅ClO) is 80.51 g, which we can round to 80 g. A compound loses 50% of its current amount after each half-life, so 3 half-lives means (½)³ = 1/8 will be left. 5 moles of chloromethyl methyl ether (80 g/mol) = 400 g. After 3 half-lives, 400 g (1/8) = 50 g = 50 g • 10³ mg/g = 5 ✕ 10⁴ mg.

Question 21

A strong wind lifts a 2-g leaf and propels it 3 m along a spiral path to reach a height of 2 m above the ground. What is the work done by gravity throughout this process?

A. 0 J

B. -0.04 J

C. -0.06 J

D. -0.10 J

Answer 21

B. -0.04 J

In this question, the work done by gravity can be calculated as the change in potential energy of the leaf, noting that the work done by gravity will be a negative value since the force of gravity acts opposite the displacement of the leaf. The initial potential energy of the leaf is zero. The final potential energy of the leaf at a height of 2 m can be calculated using the potential energy formula:

U = mgh = (0.002 kg)(10 m/s²)(2 m) = 0.04 J

Gravity is a conservative force, so the work done by gravity is path-independent. Therefore, the potential energy depends on only the final and initial position of the leaf, as shown in the not-to-scale diagram below:

Note that this problem could also be solved by calculating work as force times distance, as shown in the diagram.

Question 22

Radiation oncologists use machines called gamma knives to perform radiosurgery. Lead vests are used to protect non-cancerous areas and medical personnel from radiation exposure. A typical gamma knife uses a photon beam with a frequency of 1.6 x 10¹⁹ Hz. Suppose a stray photon hits a lead atom and ejects an electron and a lower-frequency photon of 1.599 x 10¹⁹ Hz. What is the maximum kinetic energy of the ejected electron? (Lead has a work function of 4.14 eV, Planck’s constant is 6.626 x 10^-34 J∙s, and 1 eV = 1.6 x 10^-19 J.)

A. 6.0 x 10^-18 J

B. 6.6 x 10^-18 J

C. 1.0 x 10^-14 J

D. 0 J

Answer 22

A. 6.0 x 10^-18 J

The photons used for radiation therapy are so high-energy that they cannot be absorbed by an atom all at once. Instead, they display a scattering behavior in which a collision produces a second, lower-energy photon that moves off at an angle. Because of the law of conservation of energy, the energy of the initial photon must equal the sum of the energies of the emitted photon and the electron. However, we are given only frequencies. Since frequency and energy are directly proportional, we can use E = hf to find the energies required to solve. First, though, we must find the frequency of the electron, which is equal to the difference between the two photon frequencies.

1.6 x 10¹⁹ Hz - 1.599 x 10¹⁹ Hz = 10¹⁶ Hz

E = hf = (6.6 x 10^-34 J∙s)(10 ¹⁶ Hz) = 6.6 x 10^-18 J

This quantity is NOT equal to the kinetic energy of the electron, as some energy is required simply to liberate the electron from lead’s poisonous grasp. The work function represents this amount of energy. We must first convert lead’s work function to joules, then subtract the work function from our calculated energy to get the kinetic energy. To make this calculation easier, use 4 instead of 4.14:

W_Pb = (4 eV)(1.6 x 10^-19 J) = 6.4 x 10^-19 J

KE = E_total - W_Pb = 6.6 x 10^-18 J - 6.4 x 10^-19 J = 6 x 10^-18 J

Question 23

A student places a vial of gas at the bottom of a graduated cylinder. A liquid sample is suspended in a chamber at the top of the cylinder. The vial is broken and the gas is allowed to diffuse throughout the cylinder. When the two compounds mix, a black precipitate is formed. Which gas should the student choose to minimize the time required for the precipitate to form?

A. Argon

B. Neon

C. Krypton

D. Radon

Answer 23

B. Neon

Rate of effusion is inversely related to the square root of molecular weight: rate∝ 1/√ MW). Thus, the smaller the molecular weight, the higher the rate of effusion. Examining the periodic table, we can see that neon has the lowest molecular weight.

Question 24

Which of the following has the smallest atomic/ionic radius?

A. K⁺

B. Ca²⁺

C. Ar

D. Cl^-

Answer 24

B. Ca²⁺

All four answer choices are isoelectronic: they have the exact same configuration and number of electrons. The main difference between them is the number of protons in their nuclei. Protons, which are positively charged, attract electrons. Calcium has the highest atomic number (i.e. the most protons), so its electrons will be pulled closest to the nucleus

Question 25

What are the units for k in the following rate law: rate = k[A]²[B]? Note that the concentration unit is mol/L.

A. L²mol^-2s^-1

B. s^-1

C. s

D. L mol^-1s^-1

Answer 25

A. L²mol^-2s^-1

We can see that the reaction is third order overall, and reaction rates are measured in M/s, where M = mol/L. Thus, we can set up our equation:

M/s = k (mol/L)³ = k (mol³/L³)

k = (M/s) (L³/mol³) = (mol/L•s) (L³/mol³)

k = L²/s•mol² = L²mol^-2s^-1