Medical Tests Medical Tests Certifications MCAT-TEST Questions & Answers

• Question 291:

  A continuous spectrum of light, sometimes called blackbody radiation, is emitted from a region of the Sun called the photosphere. Although the continuous spectrum contains light of all wavelengths, the intensity of the emitted light is much greater at some wavelengths than at others. The relationship between the most intense wavelength of blackbody radiation and the temperature of the emitting body is given by Wien's law, λ = 2.9 x 106/T, where λ is the wavelength in nanometers and T is the temperature in kelvins.

  As the blackbody radiation from the Sun passes through the cooler gases in the Sun's atmosphere, some of the photons are absorbed by the atoms in these gases. A photon will be absorbed if it has just enough energy to excite an electron from a lower energy state to a higher one. The absorbed photon will have an energy equal to the energy difference between these two states. The energy of a photon is given by E = hf = hc/λ where h = 6.63 x 10-34 J•s, Planck's constant, and c = 3 x 108 m/s, the speed of light in a vacuum.

  The Sun is composed primarily of hydrogen. Electron transitions in the hydrogen atom from energy state n = 2 to higher energy states are listed below along with the energy of the absorbed photon:

  Final Energy State Energy (x 10-19 J) n = 3

  3.02

  n = 4

  4.08

  n = 5

  4.57 n = 6

  4.84 n = ∞

  5.44

  Based on the data in the table, what is the approximate wavelength of a photon emitted in the electron transition from energy state n = 4 to energy state n = 3?

  A. 5 nm

  B. 30 nm

  C. 100 nm

  D. 2,000 nm

  Correct Answer: D

• Question 292:

  A continuous spectrum of light, sometimes called blackbody radiation, is emitted from a region of the Sun called the photosphere. Although the continuous spectrum contains light of all wavelengths, the intensity of the emitted light is much greater at some wavelengths than at others. The relationship between the most intense wavelength of blackbody radiation and the temperature of the emitting body is given by Wien's law, λ = 2.9 x 106 /T, where λ is the wavelength in nanometers and T is the temperature in kelvins.

  As the blackbody radiation from the Sun passes through the cooler gases in the Sun's atmosphere, some of the photons are absorbed by the atoms in these gases. A photon will be absorbed if it has just enough energy to excite an electron from a lower energy state to a higher one. The absorbed photon will have an energy equal to the energy difference between these two states. The energy of a photon is given by E = hf = hc/λ where h = 6.63 x 10-34 J•s, Planck's constant, and c = 3 x 108 m/s, the speed of light in a vacuum.

  The Sun is composed primarily of hydrogen. Electron transitions in the hydrogen atom from energy state n = 2 to higher energy states are listed below along with the energy of the absorbed photon:

  Final Energy State Energy (x 10-19 J)

  n = 3

  3.02

  n = 4

  4.08

  n = 5

  4.57 n = 6

  4.84 n = ∞

  5.44

  The energy absorbed by a hydrogen atom as its electron undergoes a transition from the n = 1 energy state to the n = ∞ state is: (Note: The n = 1 energy state is the ground state of hydrogen.)

  A. infinite.

  B. equal to the binding energy of the electron.

  C. equal to the energy of a zero-frequency photon.

  D. smaller than the energy absorbed in the n = 2 ton = transition.

  Correct Answer: B

• Question 293:

  If the noise level is increased by 30 decibels, what is the ratio of the new intensity to the original intensity?

  A. 10

  B. 30

  C. 100

  D. 1000

  Correct Answer: D

• Question 294:

  A given volume of a diprotic acid is completely neutralized by twice that volume of a 0.3 N NaOH solution. What is the molarity of the acid?

  A. 0.15 M

  B. 0.30 M

  C. 0.60 M

  D. 1.20 M

  Correct Answer: B

  You could find this answer by setting up a calculation equating the product of the number of acid equivalents and the acid volume to the number of base equivalents and the base volume, but that would get you into some messy calculations with variables for the diprotic acid molarity and the unknown volumes. However, it is much easier to think your way through to the answer. You know a certain volume of 0.3 N NaOH completely neutralizes half that volume of a diprotic acid. Since NaOH has one base equivalent per mole and a diprotic acid has two base equivalents per mole, you will need two moles of NaOH to neutralize one mole of the acid. Since the base volume is twice that of the acid volume and the neutralization is complete, you must be adding twice the number of moles of base as there are moles of acid in the acid solution. That means that the molarities of the two solutions are the same. Since NaOH releases one equivalent per mole, its normality and molarity are equal. That means that since the base is 0.3 N, it is also 0.3 M. Hence, the acid solution must also be 0.3 M, choice B. The important thing to do here was to determine how the number of equivalents per mole for each species correlated to the volumes needed for complete neutralization.

• Question 295:

  A fireman of mass m slides down a vertical pole with an average acceleration a. If the acceleration due to gravity is g, what is the average frictional force exerted by the fireman?

  A. mg

  B. m(g + a)

  C. m(g – a)

  D. ma

  Correct Answer: C

• Question 296:

  Which of the following is NOT an intermolecular force?

  A. Dispersion forces

  B. Resonance

  C. Hydrogen bonding

  D. Dipole interactions

  Correct Answer: B

  All intermolecular attractions involve some kind of electrostatic interaction. Hydrogen bonding is a strong dipole interaction between a positive hydrogen dipole in one molecule and another negative dipole. This definitely is an intermolecular attraction, so choice C isn't the answer. Dispersion forces, choice A, are slight momentary dipole interactions that occur between molecules as the electron clouds around them go through random variations, so this too is incorrect. Dipole interactions characterize the attractions between molecules that have permanent variations in electron density around the molecule, so choice D is also wrong. However, resonance, choice B, is something else entirely; it is the distribution of electrons among atoms within a molecule in more than one possible pattern. Resonance is not a form of intermolecular attraction because it only involves the effects the electron density has on the single molecule. It is an intramolecular phenomenon, so choice B is the correct answer.

• Question 297:

  The simple harmonic motion of a mass suspended from vertical springs is investigated in two experiments. The springs used in both experiments have a spring constant k and a natural length L0. The material used to make the springs has a

  Young's modulus of 2 x 1011 Pa.

  In the first experiment a mass m is suspended from a spring. The mass stretches the spring to a new length L, called the equilibrium length.

  In the second experiment the mass m is suspended from two identical springs as shown in Figure 2 below. When the mass m is in equilibrium, each spring is stretched from its natural length by the same amount xe.

  In both experiments the masses of the springs are negligible, and the elastic limits of the springs are never exceeded.

  

  In the first experiment, when a 5-kg mass is oscillating, the frequency of oscillation is 2 Hz. What is the value of the spring constant?

  A. 5/2 N/m

  B. 20 N/m

  C. 402 N/m

  D. 802 N/m

  Correct Answer: D

• Question 298:

  The simple harmonic motion of a mass suspended from vertical springs is investigated in two experiments. The springs used in both experiments have a spring constant k and a natural length L0. The material used to make the springs has a

  Young's modulus of 2 x 1011 Pa.

  In the first experiment a mass m is suspended from a spring. The mass stretches the spring to a new length L, called the equilibrium length.

  In the second experiment the mass m is suspended from two identical springs as shown in Figure 2 below. When the mass m is in equilibrium, each spring is stretched from its natural length by the same amount xe.

  In both experiments the masses of the springs are negligible, and the elastic limits of the springs are never exceeded.

  

  The two springs in Experiment 2 are replaced by a single spring having a spring constant k' such that the equilibrium length xe does not change. What is the ratio of k' to k?

  

  A. Option A

  B. Option B

  C. Option C

  D. Option D

  Correct Answer: D

• Question 299:

  The simple harmonic motion of a mass suspended from vertical springs is investigated in two experiments. The springs used in both experiments have a spring constant k and a natural length L0 . The material used to make the springs has a

  Young's modulus of 2 x 1011 Pa.

  In the first experiment a mass m is suspended from a spring. The mass stretches the spring to a new length L, called the equilibrium length.

  In the second experiment the mass m is suspended from two identical springs as shown in Figure 2 below. When the mass m is in equilibrium, each spring is stretched from its natural length by the same amount xe.

  In both experiments the masses of the springs are negligible, and the elastic limits of the springs are never exceeded.

  

  If the spring in Experiment 1 was suspended from the ceiling of an elevator accelerating with acceleration a, how would the equilibrium length of the spring compare to the equilibrium length of the spring when the elevator is stationary?

  A. The equilibrium length of the spring would be greater when the elevator is accelerating upward.

  B. The equilibrium length of the spring would be greater when the elevator is stationary.

  C. The equilibrium length of the spring would be greater when the elevator is accelerating downward with acceleration smaller than the acceleration due to gravity.

  D. The equilibrium length of the spring doesn't depend on the acceleration of the elevator.

  Correct Answer: A

• Question 300:

  The simple harmonic motion of a mass suspended from vertical springs is investigated in two experiments. The springs used in both experiments have a spring constant k and a natural length L0. The material used to make the springs has a Young's modulus of 2 x 1011 Pa.

  In the first experiment a mass m is suspended from a spring. The mass stretches the spring to a new length L, called the equilibrium length.

  In the second experiment the mass m is suspended from two identical springs as shown in Figure 2 below. When the mass m is in equilibrium, each spring is stretched from its natural length by the same amount xe.

  In both experiments the masses of the springs are negligible, and the elastic limits of the springs are never exceeded.

  

  In the first experiment, what is the mass of the object hanging from the spring?

  A. kL/g

  B. kL0/g

  C. k(L-L0)/g

  D. k/g

  Correct Answer: C