Medical Tests Medical Tests Certifications MCAT-TEST Questions & Answers

• Question 281:

  The lead-acid battery, also called a lead storage battery, is the battery of choice for starting automobiles. It contains 6 cells connected in series, each composed of a lead oxide cathode "sandwiched" between 2 lead anodes. Insulating separators are placed between the electrodes to prevent internal short-circuits. Aqueous sulfuric acid is the electrolyte.

  When the battery is being discharged, the following reaction takes place:

  

  Reaction 1

  The electrode reactions, both written as reductions, are shown in Table 1.

  Table 1

  Half-reaction

  E?V)

  PbO2(s) + SO42-(aq) + 4H+(aq) + 2ePbSO4(s) + 2H2O

  PbSO4(s) + 2e-Pb(s) + SO42-(aq)

  1.69

  ?.36

  As a car operates, the battery is recharged by electricity produced by the car's alternator, an AC generator whose ultimate power source is the car's internal combustion engine. In spite of this, batteries eventually lose their power. The battery is said to be "dead" when Reaction 1 has proceeded completely to the right.

  Often in cold weather the battery goes "dead". Thermodynamic data confirms that the voltage of most electrochemical cells decreases with decreasing temperature. If the battery is warmed to room temperature, it often recovers its ability to deliver normal power. The battery appeared "dead" because:

  I) the resistance of the electrolyte had decreased. II) the viscosity of the electrolyte had increased. III) the viscosity of the electrolyte had decreased.

  A. I only

  B. II only

  C. I and II only

  D. I and III only

  Correct Answer: B

  The reason that the battery "goes dead" with decreasing temperature is that the viscosity of the electrolyte increases. At higher viscosities the ions are moving much slower, which leads to an increase in resistance and a decrease in the power output. Roman numeral I states that the resistance of the electrolyte has decreased. This is not true, the resistance of the electrolyte increases with decreasing temperature. Since choice A, choice C, and choice D all contain Roman numeral I, they can be eliminated, leaving choice B as the correct answer.

• Question 282:

  The resistance of a resistor is defined as the ratio of the voltage drop across it to the current passing through it. The resistance of a resistor can be measured using the circuit illustrated in Figure 1.

  

  Figure 1

  In the above circuit, a variable voltage source with negligible internal resistance is connected to a resistor. The voltage across the resistor is measured by a voltmeter and the current through the resistor is measured by an ammeter.

  

  Additional resistors may be added to the circuit. The total resistance can be calculated as follows: If and are two resistances of two resistors, then the total resistance is given by = + when the resistors are connected in

  

  series, and by 1/ = 1/ + 1/ when the resistors are connected in parallel.

  

  Circuits similar to the one above are used in the common household appliance known as the toaster. The rate by which energy in the form of heat is dissipated by the resistor equals , where I is the current that passes through the resistor and R is the resistance of the resistor. Energy is dissipated in a resistor because moving electrons collide with atoms in the resistor, causing the atoms to vibrate.

  The variable voltage source in the circuit in Figure 1 is replaced by a battery connected in series with the resistor and ammeter. The battery has a small internal resistance. How will the circuit be affected?

  A. The current measured by the ammeter at a specific voltage will be greater in the circuit with the battery than in the old circuit.

  B. The current measured by the ammeter at a specific voltage will be smaller in the circuit with the battery than in the old circuit.

  C. The resistance of the resistor at a specific voltage will be greater in the circuit with the battery than in the old circuit.

  D. The resistance of the resistor at a specific voltage will be smaller in the circuit with the battery than in the old circuit.

  Correct Answer: B

  The question stem states that the voltage source, which according to the passage has negligible internal resistance, is replaced by a battery with a small, but significant internal resistance. A battery with significant internal resistance can be modeled as a voltage source connected in series with a resistor. Hence, the new circuit is the same as the one in Figure 1, except for the addition of a second resistor (call its resistance Rint) in series with the original resistor (call its resistance R0). To determine how this additional resistance affects the circuit, use Ohm's law V = IR, where V is the voltage, I is the current, and R is the resistance. Solving for I gives I = V/R. Thus, in the original circuit, the voltage source supplies a

  

  

  voltage V and the ammeter measures a current . In the new circuit, the total resistance is given by the addition rule for resistors in series, . Therefore, the current measured by the

  

  ammeter in the new circuit will be , which is clearly smaller than Ioriginal since its denominator is larger. So choice B is correct, and choice A is wrong. Choices C and D are wrong because the resistance of a resistor is a property that does not change with the addition of other resistors to the circuit. The resistance of a resistor remains constant when the voltage across it or the current through it changes.

• Question 283:

  The lead-acid battery, also called a lead storage battery, is the battery of choice for starting automobiles. It contains 6 cells connected in series, each composed of a lead oxide cathode "sandwiched" between 2 lead anodes. Insulating separators are placed between the electrodes to prevent internal short-circuits. Aqueous sulfuric acid is the electrolyte.

  When the battery is being discharged, the following reaction takes place:

  

  Reaction 1

  The electrode reactions, both written as reductions, are shown in Table 1.

  Table 1

  Half-reaction

  E?V)

  PbO2(s) + SO42-(aq) + 4H+(aq) + 2ePbSO4(s) + 2H2O

  PbSO4(s) + 2e-Pb(s) + SO42-(aq)

  1.69

  ?.36

  As a car operates, the battery is recharged by electricity produced by the car's alternator, an AC generator whose ultimate power source is the car's internal combustion engine. In spite of this, batteries eventually lose their power. The battery

  is said to be "dead" when Reaction 1 has proceeded completely to the right.

  The graph below shows the change in potential versus time of a 12-V lead storage battery during discharge.

  

  Which of the following is true?

  A. The electrolyte density at point A is greater than it is at point B.

  B. The electrolyte density at point A is less than it is at point B.

  C. The electrolyte density at point A is the same as it is at point B.

  D. The electrolyte density at points A and B cannot be compared without more information.

  Correct Answer: B

  As can be seen in Reaction 1, when the lead-acid battery is being discharged, sulfuric acid, a reactant, is being consumed. Since it is stated in the passage that sulfuric acid is the electrolyte, the density will decrease as the discharge progresses, so the density at point A is less than that at point B. Choice B is, therefore, the correct answer. Choice A is wrong because at point A there is less electrolyte present, not more. Choice C is wrong because the density decreases as the discharge progresses. Choice D is wrong because there is a direct relationship between the density of the electrolyte and the state of discharge.

• Question 284:

  The lead-acid battery, also called a lead storage battery, is the battery of choice for starting automobiles. It contains 6 cells connected in series, each composed of a lead oxide cathode "sandwiched" between 2 lead anodes. Insulating separators are placed between the electrodes to prevent internal short-circuits. Aqueous sulfuric acid is the electrolyte.

  When the battery is being discharged, the following reaction takes place:

  

  Reaction 1

  The electrode reactions, both written as reductions, are shown in Table 1.

  Table 1

  Half-reaction

  E?V)

  PbO2(s) + SO42-(aq) + 4H+(aq) + 2ePbSO4(s) + 2H2O

  PbSO4(s) + 2e-Pb(s) + SO42-(aq)

  ?.36

  As a car operates, the battery is recharged by electricity produced by the car's alternator, an AC generator whose ultimate power source is the car's internal combustion engine. In spite of this, batteries eventually lose their power. The battery

  is said to be "dead" when Reaction 1 has proceeded completely to the right.

  Where does oxidation occur in the lead storage battery?

  A. At the lead oxide cathodes

  B. At the lead oxide anodes

  C. At the lead cathodes

  D. At the lead anodes

  Correct Answer: D

  Oxidation occurs when a species' oxidation number increases; reduction occurs when a species' oxidation number decreases. Also discussed earlier, oxidation occurs at the anode and reduction occurs at the cathode. From the answer choices, it can be seen that lead oxide and lead are the only species that have to be investigated. In Reaction 1, Pb4+, in lead oxide, is going to Pb2+, in lead sulfate. Since lead's oxidation number has decreased, it has been reduced. Choice A, choice B, and choice C can, therefore, all be eliminated, leaving choice D as the correct answer. Choice D is correct because lead is being reduced at the anode -- where oxidation occurs -- from Pb to Pb2+.

• Question 285:

  The lead-acid battery, also called a lead storage battery, is the battery of choice for starting automobiles. It contains 6 cells connected in series, each composed of a lead oxide cathode "sandwiched" between 2 lead anodes. Insulating separators are placed between the electrodes to prevent internal short-circuits. Aqueous sulfuric acid is the electrolyte.

  When the battery is being discharged, the following reaction takes place:

  

  Reaction 1

  The electrode reactions, both written as reductions, are shown in Table 1.

  Table 1

  Half-reaction

  E?V)

  PbO2(s) + SO42-(aq) + 4H+(aq) + 2ePbSO4(s) + 2H2O

  PbSO4(s) + 2e-Pb(s) + SO42-(aq)

  ?.36

  As a car operates, the battery is recharged by electricity produced by the car's alternator, an AC generator whose ultimate power source is the car's internal combustion engine. In spite of this, batteries eventually lose their power. The battery

  is said to be "dead" when Reaction 1 has proceeded completely to the right.

  Which of the following occurs as the battery is being recharged?

  

  A. Option A

  B. Option B

  C. Option C

  D. Option D

  Correct Answer: A

  The battery is being recharged, so Reaction 1 is proceeding to the left. Since aqueous sulfuric acid, H2SO4, is one of the products of recharging, the concentration of H+ will increase, making choice A the correct answer. Choice B is wrong because PbSO4 (s) is serving as a reactant during recharging, and is therefore consumed during the reaction. Choice C is wrong because water is also a reactant during recharging. Choice D is wrong because the amount of lead oxide increases, not decreases as the battery is being recharged.

• Question 286:

  The lead-acid battery, also called a lead storage battery, is the battery of choice for starting automobiles. It contains 6 cells connected in series, each composed of a lead oxide cathode “sandwiched” between 2 lead anodes. Insulating

  separators are placed between the electrodes to prevent internal short-circuits. Aqueous sulfuric acid is the electrolyte.

  When the battery is being discharged, the following reaction takes place:

  Reaction 1

  The electrode reactions, both written as reductions, are shown in Table 1.

  Table 1

  Half-reaction

  E°(V)

  PbO2(s) + SO42-(aq) + 4H+(aq) + 2e-→

  PbSO4(s) + 2H2O

  PbSO4(s) + 2e-→ Pb(s) + SO42-(aq)

  –0.36

  As a car operates, the battery is recharged by electricity produced by the car's alternator, an AC generator whose ultimate power source is the car's internal combustion engine. In spite of this, batteries eventually lose their power. The battery

  is said to be “dead” when Reaction 1 has proceeded completely to the right.

  How many cells would be required to produce a 20-volt lead-acid battery of the type described in the passage?

  A. 5

  B. 10

  C. 15

  D. 20

  Correct Answer: B

  When the half-reactions are written as reductions--as they are here--the cell voltage is determined by the equation

  

  (Remember that Ecell must be greater than zero for the reaction to be spontaneous.) The cell voltage is therefore 2.05 V:

  1.69 V ?(?.36 V) = 2.05 V

  Since the potentials of cells connected in series are additive, a 20 V battery needs ten 2.05 V cells. Choice B is the correct answer. Choice A is a 10 V battery, choice C is a 30 V battery, and choice D is a 40 V battery.

• Question 287:

  The lead-acid battery, also called a lead storage battery, is the battery of choice for starting automobiles. It contains 6 cells connected in series, each composed of a lead oxide cathode "sandwiched" between 2 lead anodes. Insulating separators are placed between the electrodes to prevent internal short-circuits. Aqueous sulfuric acid is the electrolyte.

  When the battery is being discharged, the following reaction takes place:

  

  Reaction 1

  The electrode reactions, both written as reductions, are shown in Table 1.

  Table 1

  Half-reaction

  E?V)

  PbO2(s) + SO42-(aq) + 4H+(aq) + 2ePbSO4(s) + 2H2O

  PbSO4(s) + 2e-Pb(s) + SO42-(aq)

  1.69

  ?.36

  As a car operates, the battery is recharged by electricity produced by the car's alternator, an AC generator whose ultimate power source is the car's internal combustion engine. In spite of this, batteries eventually lose their power. The battery is said to be "dead" when Reaction 1 has proceeded completely to the right.

  Which reaction takes place at the anode as the battery is discharging?

  A. The first half-reaction, proceeding to the left

  B. The first half-reaction, proceeding to the right

  C. The second half-reaction, proceeding to the left

  D. The second half-reaction, proceeding to the right

  Correct Answer: C

  Oxidation--the loss of electrons--occurs at the anode. When both half-reactions are written as reductions, the following equation is used to determine the cell potential:

  

  (Again, remember that Ecell must be greater than zero for the reaction to be spontaneous.)

  The cell potential is positive when the reduction potential of the second reaction in Table 1 is subtracted from the first reaction, making the first reaction the cathode and the second reaction the anode. Since oxidation occurs at the anode, the

  reaction must proceed to the left, making choice C the correct answer. Choice A and choice B are wrong because the first reaction is the cathode reaction, not the anode reaction. Choice D is wrong because the second reaction proceeds to

  the left, not to the right.

• Question 288:

  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 × 10-34 J•s, Planck's constant,

  and c = 3 × 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

  At the center of the visible spectrum is light with a wavelength of 550 nm. What is the frequency of this light?

  A. 9.0 x 108 Hz

  B. 1.8 x 1012 Hz

  C. 5.4 x 1014 Hz

  D. 1.8 x 1016 Hz

  Correct Answer: C

  This question asks for the frequency f of light given its wavelength . If you didn't remember the relationship between the two, you could have figured it out from the formula hf = hc/ given in the passage. Dividing both sides by h, we obtain f =

  

  c/, where c is the speed of light in a vacuum. Plugging in = 550 nm, we obtain which most closely corresponds to choice C.

• Question 289:

  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, = 2.9 x 106 /T, where 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/ where h = 6.63 x 10-34 Js, 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

  If a star suddenly doubles in size but remains at the same temperature, how does its continuous spectrum change?

  A. The peak intensity occurs at the same wave-length.

  B. The peak intensity occurs at a longer wave-length.

  C. The peak intensity occurs at a shorter wave-length.

  D. The intensity peak narrows.

  Correct Answer: A

• Question 290:

  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

  If the temperature of the Sun's photosphere is 5800 K, what wavelength of radiation does the Sun emit with the greatest intensity?

  A. 2 nm

  B. 50 nm

  C. 500 nm

  D. 4,500 nm

  Correct Answer: C