Medical Tests Medical Tests Certifications MCAT-TEST Questions & Answers

• Question 341:

  When light in the ultraviolet region of the spectrum is shone on a type of material known as a phosphor, it fluoresces and emits light in the visible region of the spectrum. Lamps that utilize this property, known as fluorescent lamps, are very efficient light sources. The arrangement of a typical fluorescent lamp is shown below. The lamp is a glass tube whose inside walls are covered with a phosphor. The tube has an appreciable length-to-diameter ratio so as to reduce the power losses at each end, and it is filled with argon gas mixed with mercury vapor. Inside each end of the tube are tungsten electrodes covered with an emission material.

  Electrons are liberated at the cathode and accelerated by an applied electric field. These free electrons encounter the gas mixture, ionizing some mercury atoms and exciting others. Since it requires more energy to ionize the atoms than to excite the electrons, more excitation than ionization occurs. When the excited electrons revert to their ground state, they radiate ultraviolet photons with a wavelength of 253.7 nm. These photons impinge on the phosphor coating of the tube and excite electrons in the phosphor to higher energy states. The excited electrons in the phosphor return to their ground state in two or more steps, producing radiation in the visible region of the spectrum. Not every fluorescent lamp emits the same color of radiation; the color is dependent on the relative percentages of different heavy metal compounds in the phosphor.

  The fluorescent lamp shown operates at 100 volts and draws 400 milliamps of current during normal operation. Of the total power that the lamp consumes, only 25% is converted to light, while the remaining 75% is dissipated as heat. This energy keeps the lamp at its optimum working temperature of 40°C. In the lamp shown, the phosphor coating is calcium metasilicate, which emits orange to yellow light.

  

  In the phosphor coating, an electron falls from an excited state to a lower energy state, emitting a photon with an energy of 2.07 eV. What is the wavelength of the light emitted by the fluorescent tube? (Note: Planck's constant h = 4.14?0-15 eV穝, and c = 3?08m/s.)

  A. 300 nm

  B. 600 nm

  C. 900 nm

  D. 1242 nm

  Correct Answer: B

  This question is a straightforward application of the equation energy equals hc/λ as given in the answer to question 92. Here, the energy is given in electron-volts instead of joules, but that should not bother you since Planck's constant, h, is given as 4.14×10-15 electron-volt seconds. The wavelength therefore equals (4.14×10-15)( 3×108)/2.07. Rounding everything to the nearest integer gives λ equals (12×10-7)/2, or 6×10-7 meters, choice B.

• Question 342:

  When light in the ultraviolet region of the spectrum is shone on a type of material known as a phosphor, it fluoresces and emits light in the visible region of the spectrum. Lamps that utilize this property, known as fluorescent lamps, are very efficient light sources. The arrangement of a typical fluorescent lamp is shown below. The lamp is a glass tube whose inside walls are covered with a phosphor. The tube has an appreciable length-to-diameter ratio so as to reduce the power losses at each end, and it is filled with argon gas mixed with mercury vapor. Inside each end of the tube are tungsten electrodes covered with an emission material.

  Electrons are liberated at the cathode and accelerated by an applied electric field. These free electrons encounter the gas mixture, ionizing some mercury atoms and exciting others. Since it requires more energy to ionize the atoms than to excite the electrons, more excitation than ionization occurs. When the excited electrons revert to their ground state, they radiate ultraviolet photons with a wavelength of 253.7 nm. These photons impinge on the phosphor coating of the tube and excite electrons in the phosphor to higher energy states. The excited electrons in the phosphor return to their ground state in two or more steps, producing radiation in the visible region of the spectrum. Not every fluorescent lamp emits the same color of radiation; the color is dependent on the relative percentages of different heavy metal compounds in the phosphor.

  The fluorescent lamp shown operates at 100 volts and draws 400 milliamps of current during normal operation. Of the total power that the lamp consumes, only 25% is converted to light, while the remaining 75% is dissipated as heat. This energy keeps the lamp at its optimum working temperature of 40°C. In the lamp shown, the phosphor coating is calcium metasilicate, which emits orange to yellow light.

  

  Some fluorescent light bulbs are observed to glow for a short period after their power supply has been turned off. This glow is generated mainly by:

  A. the incandescence of the hot ionic gas within the bulb surface.

  B. emission of light stored as vibrational kinetic energy in the phosphor coating.

  C. the dissipation of electric charge built up on the bulb's surface.

  D. electrons returning to the ground state from excited states after the power was shut off.

  Correct Answer: D

  We can eliminate the other answer choices. Choice A talks about the incandescence of the hot ionic gases in the bulb's surface. This is one of those answer choices which sounds impressive and very feasible, but is incorrect. Incandescence is light which is emitted due to heat and since we are told in the passage that the lamp works at an optimum temperature of 40°C, this choice cannot bet correct. 40°C is not nearly hot enough for incandescence to occur. Choice B states that the glow is due to energy stored in the coating molecules' vibrational kinetic energy. We can rule out this answer choice since although moving charges can radiate light, the energy of vibration is much too small to emit visible light. Choice C is also incorrect since the dissipation of electric charge cannot cause a steady glow.

• Question 343:

  When light in the ultraviolet region of the spectrum is shone on a type of material known as a phosphor, it fluoresces and emits light in the visible region of the spectrum. Lamps that utilize this property, known as fluorescent lamps, are very efficient light sources. The arrangement of a typical fluorescent lamp is shown below. The lamp is a glass tube whose inside walls are covered with a phosphor. The tube has an appreciable length-to-diameter ratio so as to reduce the power losses at each end, and it is filled with argon gas mixed with mercury vapor. Inside each end of the tube are tungsten electrodes covered with an emission material.

  Electrons are liberated at the cathode and accelerated by an applied electric field. These free electrons encounter the gas mixture, ionizing some mercury atoms and exciting others. Since it requires more energy to ionize the atoms than to excite the electrons, more excitation than ionization occurs. When the excited electrons revert to their ground state, they radiate ultraviolet photons with a wavelength of 253.7 nm. These photons impinge on the phosphor coating of the tube and excite electrons in the phosphor to higher energy states. The excited electrons in the phosphor return to their ground state in two or more steps, producing radiation in the visible region of the spectrum. Not every fluorescent lamp emits the same color of radiation; the color is dependent on the relative percentages of different heavy metal compounds in the phosphor.

  The fluorescent lamp shown operates at 100 volts and draws 400 milliamps of current during normal operation. Of the total power that the lamp consumes, only 25% is converted to light, while the remaining 75% is dissipated as heat. This energy keeps the lamp at its optimum working temperature of 40°C. In the lamp shown, the phosphor coating is calcium metasilicate, which emits orange to yellow light.

  

  As the excited electrons in the coating drop back to their ground states in more than one step, they will emit light of:

  A. higher frequency than the light absorbed.

  B. longer wavelength than the light absorbed.

  C. the same wavelength as the light absorbed.

  D. greater energy than the light absorbed.

  Correct Answer: B

  We have already discussed in the explanation for question 90 that when an electron falls from an excited energy state to a lower energy state light is emitted. The electrons in the phosphor atoms in the coating are excited to higher energy levels in one step by absorbing ultraviolet light. When these electrons return to the lower energy state from which they were excited, the total energy that is emitted must be equal to the energy of the excited level minus the energy of the lower level. If they return to the lower state in more than one transition, then the energy for each transition will be less than this total energy. In fact, the sum of the energies of the transitions must equal the total energy of the excitation because energy is conserved. Using the fact that the energy of a photon equals hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the light, we see that since the energy emitted in each step of the multistep transitions is smaller than the energy absorbed, the wavelengths of the photons emitted in multi-step transitions will be longer than the wavelength of the light absorbed. Once again, choice B is correct.

• Question 344:

  When light in the ultraviolet region of the spectrum is shone on a type of material known as a phosphor, it fluoresces and emits light in the visible region of the spectrum. Lamps that utilize this property, known as fluorescent lamps, are very efficient light sources. The arrangement of a typical fluorescent lamp is shown below. The lamp is a glass tube whose inside walls are covered with a phosphor. The tube has an appreciable length-to-diameter ratio so as to reduce the power losses at each end, and it is filled with argon gas mixed with mercury vapor. Inside each end of the tube are tungsten electrodes covered with an emission material.

  Electrons are liberated at the cathode and accelerated by an applied electric field. These free electrons encounter the gas mixture, ionizing some mercury atoms and exciting others. Since it requires more energy to ionize the atoms than to excite the electrons, more excitation than ionization occurs. When the excited electrons revert to their ground state, they radiate ultraviolet photons with a wavelength of 253.7 nm. These photons impinge on the phosphor coating of the tube and excite electrons in the phosphor to higher energy states. The excited electrons in the phosphor return to their ground state in two or more steps, producing radiation in the visible region of the spectrum. Not every fluorescent lamp emits the same color of radiation; the color is dependent on the relative percentages of different heavy metal compounds in the phosphor.

  The fluorescent lamp shown operates at 100 volts and draws 400 milliamps of current during normal operation. Of the total power that the lamp consumes, only 25% is converted to light, while the remaining 75% is dissipated as heat. This energy keeps the lamp at its optimum working temperature of 40°C. In the lamp shown, the phosphor coating is calcium metasilicate, which emits orange to yellow light.

  

  If the fluorescent light is left on for 4 hours, how much useful energy is emitted as light?

  A. 144 kJ

  B. 432 kJ

  C. 576 kJ

  D. 900 kJ

  Correct Answer: A

  Before answering this question, we must first remember a few facts given in the passage. In the passage we are told that only 25% of the operating power of the lamp is converted to light. We are also told that the lamp has an operating voltage of 100 volts, and that it draws 400 milliamps or 0.4 amps of current. From these facts, we can calculate the operating power of the lamp since P = VI, where P = power, V = voltage, and I = current. The operating power of this lamp, therefore, is (100 V)(0.4 A) or 40 watts. Since only 25% of this power gets converted to light, we get only (10 watts)(4 hours)(3600 seconds/hour), which equals 144 kilo-joules, choice A.

• Question 345:

  Many nutrients required by plants exist in soil as basic cations:

  

  A soil's cation-exchange capacity is a measure of its ability to adsorb these basic cations as well as exchangeable hydrogen and aluminum ions. The cation-exchange capacity of soil is derived from two sources: small clay particles called micelles consisting of alternating layers of alumina and silica crystals, and organic colloids.

  

  Replacement of + and + by other cations of lower valence creates a net negative charge within the inner layers of the micelles. This is called the soil's permanent charge. For example, replacement of an atom of aluminum by calcium within a section where the net charge was previously zero, as shown below, produces a net charge of ?, to which other cations can become adsorbed.

  

  Figure 1

  A pH-dependent charge develops when hydrogen dissociates from hydroxyl moieties on the outer surfaces of the clay micelles. This leaves negatively-charged oxygen atoms to which basic cations may adsorb. Likewise, a large pH-

  dependent charge develops when hydrogen dissociates from carboxylic acids and phenols in organic matter.

  In most clays, permanent charges brought about by substitution account for anywhere from half to nearly all of the total cation-exchange capacity. Soils very high in organic matter contain primarily pH-dependent charges. In a research study,

  three samples of soil were leached with a 1 N solution of neutral KCl, and the displaced A13+ and basic cations measured. The sample was then leached again with a buffered solution of BaCl2 and triethanolamine at pH 8.2, and the

  displaced H+ measured. Table 1 gives results for three soils tested by this method.

  Table 1

  

  Due to the buffering effect of the soil's cation exchange capacity, just measuring the soil solution's pH will not indicate how much base is needed to change the soil pH. In another experiment, measured amounts of acid and base were added to 10-gram samples of well-mixed soil that had been collected from various locations in a field. The volumes of the samples were equalized by adding water. The results were recorded in Figure 2.

  

  Figure 2.

  A body is dropped from a height of 30 m on Earth and hits the ground with a velocity. The body is then taken to the Moon, which has a gravitational acceleration 1/6 that of Earth. It is again dropped from a height of 30 m, hitting the Moon with a velocity of vm. What is the ratio of vm/ve?

  A. 1/6

  B. sqrt (1/6)

  C. 6

  D. 36

  Correct Answer: B

  The main concept needed to answer this question is that the acceleration of gravity on the moon is one sixth that on the earth. Since the object is dropped, its initial velocity is zero acceleration on both the earth and the moon. Therefore, we use the kinematic formula: v2 = 2ad, where a is the acceleration due to gravity and d is the distance. Dividing the equation for the velocity squared of the body on the moon by the equation for the velocity squared of the body on the earth gives

  

  . The distance d cancels out since it equals 30 meters in each case am/9 is 1/6 as given in the question stem. Taking the square root, we get which is choice B. If you forget to take the square root, you get choice A.

• Question 346:

  When light in the ultraviolet region of the spectrum is shone on a type of material known as a phosphor, it fluoresces and emits light in the visible region of the spectrum. Lamps that utilize this property, known as fluorescent lamps, are very efficient light sources. The arrangement of a typical fluorescent lamp is shown below. The lamp is a glass tube whose inside walls are covered with a phosphor. The tube has an appreciable length-to-diameter ratio so as to reduce the power losses at each end, and it is filled with argon gas mixed with mercury vapor. Inside each end of the tube are tungsten electrodes covered with an emission material.

  Electrons are liberated at the cathode and accelerated by an applied electric field. These free electrons encounter the gas mixture, ionizing some mercury atoms and exciting others. Since it requires more energy to ionize the atoms than to excite the electrons, more excitation than ionization occurs. When the excited electrons revert to their ground state, they radiate ultraviolet photons with a wavelength of 253.7 nm. These photons impinge on the phosphor coating of the tube and excite electrons in the phosphor to higher energy states. The excited electrons in the phosphor return to their ground state in two or more steps, producing radiation in the visible region of the spectrum. Not every fluorescent lamp emits the same color of radiation; the color is dependent on the relative percentages of different heavy metal compounds in the phosphor.

  The fluorescent lamp shown operates at 100 volts and draws 400 milliamps of current during normal operation. Of the total power that the lamp consumes, only 25% is converted to light, while the remaining 75% is dissipated as heat. This energy keeps the lamp at its optimum working temperature of 40°C. In the lamp shown, the phosphor coating is calcium metasilicate, which emits orange to yellow light.

  

  The photons emitted by the mercury vapor have energies:

  A. equal to the energies of the electric current.

  B. equal to the voltage across the tube.

  C. equal to the energy differences between electron orbitals in the mercury atom.

  D. less than or equal to the energy differences between the electron orbitals of the mercury atom.

  Correct Answer: C

  Photons are emitted from an atom when its electrons fall from an excited state to a lower energy state. Conservation of energy then tells us that the energy of the emitted photon must be equal to the energy of the excited state minus the energy of the lower state. Choice A is incorrect because the energies of the electric current have nothing to do with this process. Choice B is likewise unrelated to the physical process of an atom emitting a photon. Choice D violates conservation of energy.

• Question 347:

  Many nutrients required by plants exist in soil as basic cations:

  

  A soil's cation-exchange capacity is a measure of its ability to adsorb these basic cations as well as exchangeable hydrogen and aluminum ions. The cation-exchange capacity of soil is derived from two sources: small clay particles called micelles consisting of alternating layers of alumina and silica crystals, and organic colloids.

  

  Replacement of + and + by other cations of lower valence creates a net negative charge within the inner layers of the micelles. This is called the soil's permanent charge. For example, replacement of an atom of aluminum by calcium within a section where the net charge was previously zero, as shown below, produces a net charge of ?, to which other cations can become adsorbed.

  

  Figure 1

  A pH-dependent charge develops when hydrogen dissociates from hydroxyl moieties on the outer surfaces of the clay micelles. This leaves negatively-charged oxygen atoms to which basic cations may adsorb. Likewise, a large pH-

  dependent charge develops when hydrogen dissociates from carboxylic acids and phenols in organic matter.

  In most clays, permanent charges brought about by substitution account for anywhere from half to nearly all of the total cation-exchange capacity. Soils very high in organic matter contain primarily pH-dependent charges. In a research study,

  three samples of soil were leached with a 1 N solution of neutral KCl, and the displaced A13+ and basic cations measured. The sample was then leached again with a buffered solution of BaCl2 and triethanolamine at pH 8.2, and the

  displaced H+ measured. Table 1 gives results for three soils tested by this method.

  Table 1

  

  Due to the buffering effect of the soil's cationexchange capacity, just measuring the soil solution's pH will not indicate how much base is needed to change the soil pH. In another experiment, measured amounts of acid and base were added to 10-gram samples of well-mixed soil that had been collected from various locations in a field. The volumes of the samples were equalized by adding water. The results were recorded in Figure 2.

  

  Figure 2.

  An electron travels in the plane of the page from left to right, perpendicular to a magnetic field that points into the page. The direction of the resulting magnetic force on the electron will be in the plane of the page and:

  A. upwards.

  B. downwards.

  C. to the left.

  D. to the right.

  Correct Answer: B

  To answer this question, we simply apply the right-hand rule. Since the particle is an electron, the direction of qv is opposite to that of v or from right to left. When you point your thumb in this direction and your fingers into the page along the direction of the magnetic field, the palm of your right hand points in the direction of the force which is downwards. Therefore, B is the correct choice.

• Question 348:

  Many nutrients required by plants exist in soil as basic cations:

  

  A soil's cation-exchange capacity is a measure of its ability to adsorb these basic cations as well as exchangeable hydrogen and aluminum ions. The cation-exchange capacity of soil is derived from two sources: small clay particles called micelles consisting of alternating layers of alumina and silica crystals, and organic colloids.

  

  Replacement of + and + by other cations of lower valence creates a net negative charge within the inner layers of the micelles. This is called the soil's permanent charge. For example, replacement of an atom of aluminum by calcium within a section where the net charge was previously zero, as shown below, produces a net charge of ?, to which other cations can become adsorbed.

  

  Figure 1

  A pH-dependent charge develops when hydrogen dissociates from hydroxyl moieties on the outer surfaces of the clay micelles. This leaves negatively-charged oxygen atoms to which basic cations may adsorb. Likewise, a large pH-

  dependent charge develops when hydrogen dissociates from carboxylic acids and phenols in organic matter.

  In most clays, permanent charges brought about by substitution account for anywhere from half to nearly all of the total cation-exchange capacity. Soils very high in organic matter contain primarily pH-dependent charges. In a research study,

  three samples of soil were leached with a 1 N solution of neutral KCl, and the displaced A13+ and basic cations measured. The sample was then leached again with a buffered solution of BaCl2 and triethanolamine at pH 8.2, and the

  displaced H+ measured. Table 1 gives results for three soils tested by this method.

  Table 1

  

  Due to the buffering effect of the soil's cation exchange capacity, just measuring the soil solution's pH will not indicate how much base is needed to change the soil pH. In another experiment, measured amounts of acid and base were added to 10-gram samples of well-mixed soil that had been collected from various locations in a field. The volumes of the samples were equalized by adding water. The results were recorded in Figure 2.

  

  Figure 2.

  How much solid NaOH is required to neutralize 700 mL of 2 N HNO3?

  A. 40 g

  B. 48 g

  C. 56 g

  D. 64 g

  Correct Answer: C

  To solve this problem, you need to remember the definition of normality, and know how acids and bases interact to neutralize each other. Normality is defined as the number of equivalents per liter of solution. Here nitric acid has one equivalent, that is, it has one proton to donate pr mole. Thus, for this compound, normality equals molarity, and we have 2 moles in one liter, or 1.4 moles in 700 milliliters. Since acids and bases react in a 1 to 1 ration of equivalents, in order to neutralize 1.4 moles of nitric acid, you must have an equal number of equivalents of the neutralizing substance present. Since sodium hydroxide also has 1 equivalent per mole, we need the same number of moles of each. Therefore, we need 1.4 moles of sodium hydroxide. Sodium hydroxide has a molar weight of 40 grams, 1.4 moles will have a mass of 56 grams. This is answer C, which is the correct choice.

• Question 349:

  Many nutrients required by plants exist in soil as basic cations:

  

  A soil's cation-exchange capacity is a measure of its ability to adsorb these basic cations as well as exchangeable hydrogen and aluminum ions. The cation-exchange capacity of soil is derived from two sources: small clay particles called micelles consisting of alternating layers of alumina and silica crystals, and organic colloids.

  

  Replacement of + and + by other cations of lower valence creates a net negative charge within the inner layers of the micelles. This is called the soil's permanent charge. For example, replacement of an atom of aluminum by calcium within a section where the net charge was previously zero, as shown below, produces a net charge of ?, to which other cations can become adsorbed.

  

  Figure 1

  A pH-dependent charge develops when hydrogen dissociates from hydroxyl moieties on the outer surfaces of the clay micelles. This leaves negatively-charged oxygen atoms to which basic cations may adsorb. Likewise, a large pH-

  dependent charge develops when hydrogen dissociates from carboxylic acids and phenols in organic matter.

  In most clays, permanent charges brought about by substitution account for anywhere from half to nearly all of the total cation-exchange capacity. Soils very high in organic matter contain primarily pH-dependent charges. In a research study,

  three samples of soil were leached with a 1 N solution of neutral KCl, and the displaced A13+ and basic cations measured. The sample was then leached again with a buffered solution of BaCl2 and triethanolamine at pH 8.2, and the

  displaced H+ measured. Table 1 gives results for three soils tested by this method.

  Table 1

  

  Due to the buffering effect of the soil's cation exchange capacity, just measuring the soil solution's pH will not indicate how much base is needed to change the soil pH. In another experiment, measured amounts of acid and base were added to 10-gram samples of well-mixed soil that had been collected from various locations in a field. The volumes of the samples were equalized by adding water. The results were recorded in Figure 2.

  Figure 2.

  

  If 29 g of maleic acid () is dissolved in 500 g of ammonia (N), what is the molality of the resulting solution?

  A. 0.05 m

  B. 0.10 m

  C. 0.25 m

  D. 0.50 m

  Correct Answer: D

  This is a relatively simple concentration problem. Molality is defined as the number of moles of solute per kilogram of solvent. Here, you must first calculate the molecular weight of maleic acid, or 116 grams per mole, and determine the number of moles. Dividing actual grams by the grams per mole you get 0.25 moles. Since this is dissolved in 500 grams or 0.5 kilograms of ammonia you have a 0.5 molal solution, so choice D is the correct response.

• Question 350:

  Many nutrients required by plants exist in soil as basic cations:

  

  A soil's cation-exchange capacity is a measure of its ability to adsorb these basic cations as well as exchangeable hydrogen and aluminum ions. The cation-exchange capacity of soil is derived from two sources: small clay particles called micelles consisting of alternating layers of alumina and silica crystals, and organic colloids.

  

  Replacement of + and + by other cations of lower valence creates a net negative charge within the inner layers of the micelles. This is called the soil's permanent charge. For example, replacement of an atom of aluminum by calcium within a section where the net charge was previously zero, as shown below, produces a net charge of ?, to which other cations can become adsorbed.

  

  Figure 1

  A pH-dependent charge develops when hydrogen dissociates from hydroxyl moieties on the outer surfaces of the clay micelles. This leaves negatively-charged oxygen atoms to which basic cations may adsorb. Likewise, a large pH-

  dependent charge develops when hydrogen dissociates from carboxylic acids and phenols in organic matter.

  In most clays, permanent charges brought about by substitution account for anywhere from half to nearly all of the total cation-exchange capacity. Soils very high in organic matter contain primarily pH-dependent charges. In a research study,

  three samples of soil were leached with a 1 N solution of neutral KCl, and the displaced A13+ and basic cations measured. The sample was then leached again with a buffered solution of BaCl2 and triethanolamine at pH 8.2, and the

  displaced H+ measured. Table 1 gives results for three soils tested by this method.

  Table 1

  

  Due to the buffering effect of the soil's cation exchange capacity, just measuring the soil solution's pH will not indicate how much base is needed to change the soil pH. In another experiment, measured amounts of acid and base were added to 10-gram samples of well-mixed soil that had been collected from various locations in a field. The volumes of the samples were equalized by adding water. The results were recorded in Figure 2.

  Figure 2.

  

  A converging lens has a focal length of 8 cm. If the object is 10 cm to the left of the lens, what are the position of the image formed and the magnification of the lens?

  A. 0.025 cm to the right of the lens and 0.0025X

  B. 4.4 cm to the right of the lens and 0.4X

  C. 40 cm to the right of the lens and 4X

  D. 40 cm to the left of the lens and 4X

  Correct Answer: C

  

  In the question we are told that the object is a distance of 10 centimeters to the left of the lens. To determine the position of the image, simply plug the numbers into the equation: where f is the focal length of the lens, p is the

  

  object distance, and q is the image distance. This equation rearranges to give . Plugging in values for the focal length of 8 centimeters, and an object distance of 10 cm, we get an answer of for the reciprocal of the image distance, which is equivalent to an image distance of 40 centimeters. Since we have a positive image distance, the image is on the real side of the lens, which is the right side of the lens. To calculate the magnification, we use the equation:

  

  where M is the magnification. Putting the known values into the equation, magnification is equal to ?0 centimeters divided by 10 centimeters, or 4x. The minus sign indicates that the image is inverted. So the image is 40 centimeters to the right of the lens, with a magnification of 4x, answer choice C.