Medical Tests Medical Tests Certifications MCAT-TEST Questions & Answers

• Question 261:

  Band theory explains the conductivity of certain solids by stating that the atomic orbitals of the individual atoms in the solid merge to produce a series of atomic orbitals comprising the entire solid. The closely-spaced energy levels of the orbitals form bands. The band corresponding to the outermost occupied subshell of the original atoms is called the valence band. If partially full, as in metals, it serves as a conduction band through which electrons can move freely. If the valence band is full, then electrons must be raised to a higher band for conduction to occur. The greater the band gap between the separate valence and conduction bands, the poorer the material's conductivity. Figure 1 shows the valence and conduction bands of a semiconductor, which is intermediate in conductivity between conductors and insulators.

  

  Figure 1

  When silicon, a semiconductor with tetrahedral covalent bonds, is heated, a few electrons escape into the conduction band. Doping the silicon with a few phosphorus atoms provides unbonded electrons that escape more easily, increasing conductivity. Doping with boron produces holes in the bonding structure, which may be filled by movement of nearby electrons within the lattice. When a semiconductor in an electric circuit has excess electrons on one side and holes on the other, electron flow occurs more easily from the side with excess electrons to the side with holes than in the reverse direction.

  

  Figure 2

  The energy of the band gap for pure silicon is about 1.1 electron volts. If a 1.5-volt electrical potential is connected across a sample of silicon:

  A. the electrons would jump to the conduction band and the silicon would conduct.

  B. the holes in the silicon lattice would move.

  C. the energy of the band gap would be lowered.

  D. the silicon would not conduct.

  Correct Answer: A

  Since the band gap is 1.1 eV, an applied voltage in excess of 1.1 V would give the electrons enough energy to cross the band gap and populate the conduction band. (You should know that an electron volt is equal to the energy required to move an electron through a potential difference of one volt.) Choice A is therefore the correct response. Choice B is wrong because pure, undoped silicon doesn't have any holes. The passage states that holes are produced when the silicon is doped with boron choice C is wrong because there is no information in the passage that suggests that the band gap is affected by an external potential. Choice D is wrong because the electrons do have enough energy to overcome the band gap and would move into the conduction band.

• Question 262:

  Band theory explains the conductivity of certain solids by stating that the atomic orbitals of the individual atoms in the solid merge to produce a series of atomic orbitals comprising the entire solid. The closely-spaced energy levels of the orbitals form bands. The band corresponding to the outermost occupied subshell of the original atoms is called the valence band. If partially full, as in metals, it serves as a conduction band through which electrons can move freely. If the valence band is full, then electrons must be raised to a higher band for conduction to occur. The greater the band gap between the separate valence and conduction bands, the poorer the material's conductivity. Figure 1 shows the valence and conduction bands of a semiconductor, which is intermediate in conductivity between conductors and insulators.

  

  Figure 1

  When silicon, a semiconductor with tetrahedral covalent bonds, is heated, a few electrons escape into the conduction band. Doping the silicon with a few phosphorus atoms provides unbonded electrons that escape more easily, increasing conductivity. Doping with boron produces holes in the bonding structure, which may be filled by movement of nearby electrons within the lattice. When a semiconductor in an electric circuit has excess electrons on one side and holes on the other, electron flow occurs more easily from the side with excess electrons to the side with holes than in the reverse direction.

  

  Figure 2

  Why do phosphorus and boron atoms enhance the conductivity of silicon?

  A. Their electronegativities differ from that of silicon.

  B. They have different numbers of valence electrons.

  C. Their semimetallic nature makes them good semiconductors.

  D. They are better conductors than silicon even as pure substances.

  Correct Answer: B

  The passage states that phosphorus atoms increase the conductivity of silicon because phosphorus provides the crystal with unbonded electrons, while boron atoms produce holes in the bonding structure. From the periodic table, it can be seen that boron has three valence electrons, silicon has four, and phosphorus has 5. Since the bonds in the silicon crystal are tetrahedral, a phosphorus atom surrounded by silicon atoms will be able to form covalent bonds with only four of its valence electrons, leaving the fifth one unbonded. Boron has only three valence electrons, so a boron atom surrounded by silicon atoms will only be able to form three covalent bonds, leaving a hole in the bonding structure of silicon. It is the presence of these unbonded electrons and holes that enhances the conductivity of phosphorus. Choice B is therefore the correct answer. Choice A is wrong because the difference in electronegativity between boron, phosphorus, and silicon is quite small. Choice C is wrong for two reasons: phosphorus isn't a semimetal and even if these two elements were both semimetals and good semiconductors, this in itself wouldn't explain the enhancement of silicon's conductivity produced by doping silicon with some of these elements. Choice D is wrong because, like choice C, it wouldn't explain the phenomenon the question asks about even if it were true.

• Question 263:

  Band theory explains the conductivity of certain solids by stating that the atomic orbitals of the individual atoms in the solid merge to produce a series of atomic orbitals comprising the entire solid. The closely-spaced energy levels of the orbitals form bands. The band corresponding to the outermost occupied subshell of the original atoms is called the valence band. If partially full, as in metals, it serves as a conduction band through which electrons can move freely. If the valence band is full, then electrons must be raised to a higher band for conduction to occur. The greater the band gap between the separate valence and conduction bands, the poorer the material's conductivity. Figure 1 shows the valence and conduction bands of a semiconductor, which is intermediate in conductivity between conductors and insulators.

  

  Figure 1

  When silicon, a semiconductor with tetrahedral covalent bonds, is heated, a few electrons escape into the conduction band. Doping the silicon with a few phosphorus atoms provides unbonded electrons that escape more easily, increasing conductivity. Doping with boron produces holes in the bonding structure, which may be filled by movement of nearby electrons within the lattice. When a semiconductor in an electric circuit has excess electrons on one side and holes on the other, electron flow occurs more easily from the side with excess electrons to the side with holes than in the reverse direction.

  

  Figure 2

  Why is iron a good conductor of electricity?

  A. Its 3d electrons only partially fill the valence band.

  B. The band gap is small.

  C. The 4s and 3d orbitals form a filled valence band.

  D. The energy levels of the atomic orbitals are closely separated.

  Correct Answer: A

  The passage mentions that metals have partially filled valence bands. This means that there are low energy unoccupied atomic orbitals in metals through which electrons may move freely. Therefore, the valence band for metals is the conduction band, and consequently, there is no band gap. Metals such as iron are good conductors of electricity because of these unoccupied low energy orbitals. All of this information is contained in the passage and requires little or no background knowledge. Choice A is therefore the correct response. Choice B is incorrect because metals, unlike semiconductors, do not have a band gap. Choice C is wrong because iron's 3d orbital is not filled; it has only six electrons, not ten. Choice D is true of many solids including metals, semiconductors, and insulators, but it does not answer the question and so is incorrect.

• Question 264:

  Band theory explains the conductivity of certain solids by stating that the atomic orbitals of the individual atoms in the solid merge to produce a series of atomic orbitals comprising the entire solid. The closely-spaced energy levels of the orbitals form bands. The band corresponding to the outermost occupied subshell of the original atoms is called the valence band. If partially full, as in metals, it serves as a conduction band through which electrons can move freely. If the valence band is full, then electrons must be raised to a higher band for conduction to occur. The greater the band gap between the separate valence and conduction bands, the poorer the material's conductivity. Figure 1 shows the valence and conduction bands of a semiconductor, which is intermediate in conductivity between conductors and insulators.

  

  Figure 1

  When silicon, a semiconductor with tetrahedral covalent bonds, is heated, a few electrons escape into the conduction band. Doping the silicon with a few phosphorus atoms provides unbonded electrons that escape more easily, increasing conductivity. Doping with boron produces holes in the bonding structure, which may be filled by movement of nearby electrons within the lattice. When a semiconductor in an electric circuit has excess electrons on one side and holes on the other, electron flow occurs more easily from the side with excess electrons to the side with holes than in the reverse direction.

  

  Figure 2

  How does heat increase the conductivity of a semiconductor?

  I) By reducing collisions between moving electrons

  II) By breaking covalent bonds

  III) By raising electrons to a higher energy level

  A. I only

  B. III only

  C. I and III only

  D. II and III only

  Correct Answer: D

  Statement I says that heat could be expected to reduce the frequency of collisions between moving electrons. This is not true: heat increases the kinetic energy of electrons, thereby increasing, not decreasing, the frequency of collisions. Choice A and choice C can be eliminated. Statement II says that heat breaks covalent bonds. This is true: sufficient energy -- in this case supplied by heat -- will break covalent bonds. The passage doesn't tell you explicitly that heat breaks covalent bonds in semiconductors, but it is implied in the description of the semiconductive properties of silicon. In the first sentence of the second paragraph, you are told that silicon forms tetrahedral covalent bonds, and that when heated, it will conduct. Since silicon is sp3 hybridized -- forming a filled valence band -- bonds must be broken in order to promote electrons to the conduction band. Choice B can therefore be eliminated, making choice D the correct response. The truth of Statement III should then be obvious: in order for the freed electrons to "jump" to the higher energy band gap, they must gain energy. Heat supplies the energy and permits this to happen.

• Question 265:

  In the figure below, aqueous solutions A and B are separated by a semipermeable membrane. Which of the following will occur?

  

  A. Glucose molecules will move from side B to side A.

  B. Glucose molecules will move from side A to side B.

  C. Both water and glucose molecules will move from side A to side B.

  D. Water molecules will move from side B to side A.

  Correct Answer: D

  A semipermeable membrane allows the movement of solvent molecules, but not solute molecules. This means that water molecules can move through the semipermeable membrane and that glucose molecules cannot. Choice A, choice B, and choice C can all be eliminated, leaving choice D as the correct response. You should be aware that osmosis is the net movement of solvent molecules through a semipermeable membrane from a pure solvent or a dilute solution to a more concentrated solution. In this case, solution A, which is 5% glucose, is more concentrated than solution B, which is only 1% glucose. So, there will be a movement from the more dilute solution, solution B, to one of higher concentration, which is solution A.

• Question 266:

  What is the normality of a solution containing 49 g of H3PO4(MW=98 g/mol) in 2,000 mL of solution?

  A. 0.25

  B. 0.50

  C. 0.75

  D. 1.50

  Correct Answer: C

  Normality is defined as the number of equivalents of solute per liter of solution. An equivalent of an acid is the amount of acid required to give one mole of H+. Phosphoric acid, which contains three moles of protons per mole of acid, therefore contains three equivalents per mole. A one molar solution of phosphoric acid is said to be 3 normal. The solution in question contains 49 grams of phosphoric acid, and the molar mass of phosphoric acid is 98. Since 49 is half of 98, the amount of phosphoric acid in this sample is half of the 3 equivalents in each mole of the acid. So there are 1.5 equivalents of phosphoric acid in this solution. Now, the total volume of the solution is 2000 milliliters, or 2 liters. So this solution contains 1.5 equivalents of acid per 2 liters. This comes to 0.75 equivalents per liter, which is a 0.75 N solution.

• Question 267:

  The mouthpiece of a telephone handset has a mass of 100 g, and the earpiece has a mass of 150 g. To balance the handset on one finger, that finger must be: (Note: Assume the bridge connecting the mouthpiece and the earpiece has a negligible mass.)

  A. one and one half times farther from the earpiece than from the mouthpiece.

  B. two times farther from the earpiece than from the mouthpiece

  C. one and one half times farther from the mouthpiece than from the earpiece.

  D. two times farther from the mouthpiece than from the earpiece

  Correct Answer: C

  Since rotation about a point is at issue, an understanding of torques is necessary to solve the problem. The torque , or rotational force on a rigid body resting on a pivot point, is given by = rF, when a force F acts at a distance r from the pivot in a direction perpendicular to the axis of rotation. Balance is achieved when the net torque is zero, i.e., the clockwise torque equals the counterclockwise torque. Draw a little sketch of the telephone handset, balanced horizontally on a pivot point. The weight of the mouthpiece is (100 grams)g, where g is the acceleration due to gravity. Call the distance between the point where this force acts and the pivot rm. Similarly, the weight of the earpiece is (150 grams)g, and acts a distance re from the pivot. To achieve balance, the opposing torques must be equal to each other, so (100 grams)g rm = (150 grams)g re. Solving for rm and canceling out g gives rm = (3/2) re. So the mouthpiece must be (3/2) times farther from the pivot than the earpiece, and choice C is correct.

• Question 268:

  The reaction below is NOT spontaneous at any temperature.

  2ICl(g) I2(g) + Cl2(g)

  Which of the following is true?

  

  A. Option A

  B. Option B

  C. Option C

  D. Option D

  Correct Answer: B

  The relationship between G , H, S, and temperature is G = H ?TS For a reaction to be spontaneous, G must be less than zero. If a reaction is not spontaneous at any temperature, H must be positive and S must be negative. No matter what the temperature, G will always be positive and the reaction will be nonspontaneous. Choice B is therefore the correct response. Choice A is for a reaction that would be spontaneous if the temperature was sufficiently high. Choice C is for a reaction that is spontaneous at ALL temperatures. Choice D is for a reaction that would be spontaneous if the temperature was sufficiently low.

• Question 269:

  Which of the following is the reason that water boils at a much higher temperature than does hydrogen sulfide?

  A. The intramolecular O–H bonds are stronger than the intramolecular S–H bonds.

  B. The enthalpy of vaporization of water is less than that of hydrogen sulfide.

  C. The relative molecular mass of water is less than that of hydrogen sulfide.

  D. The intermolecular O–H bonds are stronger than the intermolecular S–H bonds.

  Correct Answer: D

  To achieve boiling, intermolecular forces holding the liquid together have to be overcome so that the molecules of the substance can enter the vapor phase. Boiling points of substances, therefore, often reflect the strength of the intermolecular forces operating among the molecules. Choice D is correct because the intermolecular hydrogen-oxygen bonds in water are stronger than the intermolecular hydrogen-sulfur bonds in hydrogen sulfide. When hydrogen is bonded to a strongly electronegative element such as nitrogen, oxygen, or fluorine, very strong intermolecular forces exist. These forces are called hydrogen bonds. Choice A is wrong because, even though it is true that intramolecular oxygen-hydrogen bonds are stronger than intramolecular sulfurhydrogen bonds, it is not the reason that water boils at a higher temperature than hydrogen sulfide. Choice B is wrong because if the enthalpy of vaporization of water is less than that of hydrogen sulfide, less energy is required to evaporate a given quantity of water at a constant temperature than for hydrogen sulfide. This is not what happens: the heat of vaporization of water is greater than that of hydrogen sulfide, so water boils at a higher temperature. Choice C is wrong because it is the ability of water to hydrogen bond that makes it boil at a higher temperature.

• Question 270:

  It is critical for the human body blood to maintain its pH at approximately 7.4. Decreased or increased blood pH are called acidosis and alkalosis respectively; both are serious metabolic problems that can cause death. The table below lists the major buffers found in the blood and/or kidneys. Table 1 Buffer pKa of a typical conjugate acid:*

  

  + Histidine side chains

  

  Organic phosphates N-terminal amino groups

  

  7.0

  8.0

  9.2

  

  *For buffers in many of these categories, there is a range of actual values.

  

  The relationship between blood pH and the of any buffer can be described by the Henderson-Hasselbalch equation:

  

  pH = + log([conjugate base]/[conjugate acid]) Equation 1

  

  Bicarbonate, the most important buffer in the plasma, enters the blood in the form of carbon dioxide, a byproduct of metabolism, and leaves in two forms: exhaled and excreted bicarbonate. Blood pH can be adjusted rapidly by changes

  

  in the rate of exhalation. The reaction given below, which is catalyzed by carbonic anhydrase in the erythrocytes, describes how bicarbonate and interact in the blood.

  

  + + Reaction 1

  What would be the nature of the compensatory change that would take place in the respiratory system response to acidosis caused by organic acids?

  

  A. Option A

  B. Option B

  C. Option C

  D. Option D

  Correct Answer: B

  As the passage states, the immediate buffering effect of bicarbonate is controlled by changes in the breathing rate. When acidosis occurs, the concentration of H+ is too high. As discussed earlier, Le Chatelier's principle applies: In order to decrease the concentration of H+, the concentration of carbon dioxide must decrease. If the breathing rate increases, more carbon dioxide is exhaled and its concentration in the blood decreases. Since the effect of this rapid breathing is to remove carbon dioxide from the body, the ultimate effect is to decrease the total CO2/ HCO3?concentration in the blood. Therefore, choice B is correct. Choice A is wrong because the ratio of CO2/ HCO3?would decrease, not increase. Choice C and choice D are wrong because the breathing rate would increase, not decrease.