Chemistry (Grades 9–12)
Subtest 1 Sample Items
1. The following experiment has been designed to determine the effect of temperature on the rate of corrosion of an iron nail.
Control  Experimental Conditions 
Nail at 22°C cut into 2 pieces 
Intact nail at 5°C 
Intact nail at 35°C 
Intact nail at 45°C 
Three 2.5 g iron nails were used for the control condition and for each experimental condition. Individual nails were placed into open test tubes containing 10 mL of H2O and they were maintained at the specified temperatures for 48 hours. After the 48hour incubation period, the presence of corrosion was determined through the use of a chemical indicator. Which of the following best describes the flaw in this experimental design?
 Temperature has little effect on reaction rate.
 The experiment lacks a valid control for the experimental conditions.
 Iron is a poor choice of metal for this experiment.
 The experimental conditions are too difficult to maintain for 48 hours.
 Answer
 Correct Response: B. (Objective 0001) This question requires the examinee to demonstrate knowledge of the principles and procedures for designing and carrying out scientific investigations involving inorganic substances. An effective control is an important part of good experimental design. The control should be identical to the experimental condition in every way except for the variable being studied. This type of control helps a researcher make valid conclusions about the effect of the experimental condition being tested. In the described experiment, the control condition differs from the experimental conditions in two ways—temperature and surface area. The introduction of a second variable in the control condition complicates the analysis and interpretation of the results. An appropriate control for this experiment would have been an intact nail at 22°C.
Correct Response: B. (Objective 0001) This question requires the examinee to demonstrate knowledge of the principles and procedures for designing and carrying out scientific investigations involving inorganic substances. An effective control is an important part of good experimental design. The control should be identical to the experimental condition in every way except for the variable being studied. This type of control helps a researcher make valid conclusions about the effect of the experimental condition being tested. In the described experiment, the control condition differs from the experimental conditions in two ways—temperature and surface area. The introduction of a second variable in the control condition complicates the analysis and interpretation of the results. An appropriate control for this experiment would have been an intact nail at twenty two degrees celsius.
2. The table below shows data for the solubility of a gas in water over a range of temperatures.
Temperature (°C)  Gas Solubility (mM) 
0  2.0 
10  1.7 
20  1.5 
30  1.3 
40  1.1 
50  1.0 
A chemist wants to use these data to determine the solubility of the gas at 25°C. Which of the following methods of presenting data would be most helpful in making this determination?
 bar graph
 stemandleaf plot
 line graph
 frequency diagram
 Answer
 Correct Response: C. (Objective 0001) This question requires the examinee to recognize methods and criteria for analyzing and graphically presenting scientific data. A line graph is the best choice for presenting the solubility data. This method will assist in identifying any trends in gas solubility that exist as a function of water temperature. Plotting the experimental data as a line graph will allow the chemist to make a reasonable determination of the solubility of the gas at 25°C.
Correct Response: C. (Objective 0001) This question requires the examinee to recognize methods and criteria for analyzing and graphically presenting scientific data. A line graph is the best choice for presenting the solubility data. This method will assist in identifying any trends in gas solubility that exist as a function of water temperature. Plotting the experimental data as a line graph will allow the chemist to make a reasonable determination of the solubility of the gas at twenty five degress celsius.
3. The instructions for a titration experiment are shown below.
A chart with the title, Titrating a Weak Acid with a Strong Base. Materials: buret, clamp, ring stand, two hundred and fifty m upper L beaker, p upper H probe, p upper H meter, glass stirring rod, 1 M Upper N a Upper O upper H, Upper C upper H sub three Upper C Upper O Upper O upper H solution open parenthesis concentration unknown close parenthesis.
Method. bullet prepare a zero point one M Upper N a Upper O upper H solution from the stock solution bullet set up glassware including p upper H probe. bullet place fifty m upper L of upper C upper H sub three upper C upper O upper O upper H into beaker. bullet record the p upper H of the upper C upper H sub three upper C upper O upper O upper H solution. bullet fill the buret with zero point one m upper N a upper O upper H and record the volume. bullet slowly add zero point one m upper N a upper O upper H to beaker bullet stir gently and record the p upper H. bullet continue beyond the equivalence point safety notes. bullet protective eyewear and laboratory aprons must be worn at all times. bullet acid and bases must be disposed of properly
Which of the following represents the greatest potential safety hazard associated with this experiment?
 the process of preparing the 0.1 M NaOH solution
 the effect of prolonged exposure to CH_{3}COOH
 the product formed by combining the acid and the base
 the reactivity of the acid and the base with glassware
 Answer
 Correct Response: A. (Objective 0001) This question requires the examinee to demonstrate knowledge of hazards associated with chemical investigations. Preparing the 0.1 M NaOH solution requires dispensing a volume of the highly corrosive concentrated stock solution. Proper protocols need to be followed to avoid personal injury and spills.
Correct Response: A. (Objective 0001) This question requires the examinee to demonstrate knowledge of hazards associated with chemical investigations. Preparing the zero point one mole of upper n a upper o upper h solution requires dispensing a volume of the highly corrosive concentrated stock solution. Proper protocols need to be followed to avoid personal injury and spills.
4. Which of the components in the voltaic cell shown maintains the electrical neutrality of the two solutions?
Two beakers the first beaker has an anode solution and the second beaker has a cathode solution. There is an anode in the first beaker connected to a voltmeter that is connected to a cathode in the second beaker. There is also a salt bridge with each end in the beakers connecting the two solutions
 anode
 voltmeter
 cathode
 salt bridge
 Answer
 Correct Response: D. (Objective 0001) This question requires the examinee to demonstrate familiarity with laboratory tools and instruments routinely used in chemical investigations. As the processes of oxidation and reduction proceed, there is a change in the electrical charge of the solutions in the anode and cathode compartments. In order for the electrical circuit to continue to function, the two solutions must remain electrically neutral. This is accomplished by the salt bridge, which contains an inert electrolyte solution. As the reaction proceeds, ions in this electrolyte solution move into the two compartments and neutralize the accumulating charge.
Correct Response: D. (Objective 0001) This question requires the examinee to demonstrate familiarity with laboratory tools and instruments routinely used in chemical investigations. As the processes of oxidation and reduction proceed, there is a change in the electrical charge of the solutions in the anode and cathode compartments. In order for the electrical circuit to continue to function, the two solutions must remain electrically neutral. This is accomplished by the salt bridge, which contains an inert electrolyte solution. As the reaction proceeds, ions in this electrolyte solution move into the two compartments and neutralize the accumulating charge.
5. Which of the following mathematical equations would be most helpful to a chemist trying to determine the age of a bone found at an archeological site?
A. The natural log of the quantity A divided by A subscript naught equals negative k t
B. y equals m x plus b
C. delta G equals delta H minus T delta S
D. x equals a fraction. The numerator is negative b plus or minus the square root of the quantity b squared minus 4 a c. The denominator is 2 a.
 ln(A/A_{0}) = –kt
 y = mx + b
 ΔG = ΔH –TΔS
 x =
6. Which of the following types of chemistry problems is most likely to involve the use of the quadratic formula?
 calculating the pH of a weak acid with a known molarity and K_{a}
 determining the enthalpy of a reaction from bond enthalpy values
 calculating the number of moles in a given volume of a gas
 determining the spontaneity of a reaction using Gibbs free energy
 Answer
 Correct Response: A. (Objective 0002) This question requires the examinee to apply basic mathematical concepts to solve chemistry problems. Calculating the pH of a weak acid involves expressing the equilibrium concentrations of the starting acid and its ions in terms of the variable x (the change in concentration). These concentration terms can then be substituted into the equilibrium constant expression, which can then be expressed as a quadratic equation (ax^{2} + bx + c = 0). The quadratic formula, , is used to solve this type of equation. While the quadratic formula can be used to solve all problems of this type, there are cases in which it is not necessary to use it. For example, if a weak acid has a very small acid ionization value, its concentration at equilibrium will not be significantly different from the starting concentration. If this is the case, the equilibrium constant expression can be simplified into a form that does not require using the quadratic formula.
Correct Response: A. (Objective 0002) This question requires the examinee to apply basic mathematical concepts to solve chemistry problems. Calculating the pH of a weak acid involves expressing the equilibrium concentrations of the starting acid and its ions in terms of the variable x (the change in concentration). These concentration terms can then be substituted into the equilibrium constant expression, which can then be expressed as a quadratic equation open parenthesis a x squared plus b x plus c equals zero close parenthesis. The quadratic formula, x equals start fraction numerator negative b plus or minus start square root b squared minus four a c end root denominator two a end fraction, is used to solve this type of equation. While the quadratic formula can be used to solve all problems of this type, there are cases in which it is not necessary to use it. For example, if a weak acid has a very small acid ionization value, its concentration at equilibrium will not be significantly different from the starting concentration. If this is the case, the equilibrium constant expression can be simplified into a form that does not require using the quadratic formula.
7. In a laboratory exercise, students were asked to determine the boiling point of an unknown liquid. Each lab group performed the exercise multiple times and then calculated the mean boiling point temperature and the standard deviation of the results. Their results are shown in the table below.
 Group 1  Group 2  Group 3  Group 4 
Mean Boiling Point (°C) 
79.8 
81.7 
82.3 
79.4 
Standard Deviation 
0.8 
2.5 
1.1 
0.3 
According to these results, which lab group's boiling point data showed the least amount of variation across multiple trials?
 Group 1
 Group 2
 Group 3
 Group 4
 Answer
 Correct Response: D. (Objective 0002) This question requires the examinee to use statistics to analyze experimental data. Standard deviation is a measure of the variation within a set of data. A low standard deviation value indicates less variation within the data set. Since the standard deviation of Group 4's data is less than that of the other groups, this lab group's boiling point data were the most consistent across multiple trials.
Correct Response: D. (Objective 0002) This question requires the examinee to use statistics to analyze experimental data. Standard deviation is a measure of the variation within a set of data. A low standard deviation value indicates less variation within the data set. Since the standard deviation of Group 4's data is less than that of the other groups, this lab group's boiling point data were the most consistent across multiple trials.
8. A chemistry class has completed a lab exercise on the effect of temperature on reaction rate. Which of the following next steps would aid them most in understanding their results?
 reviewing the concepts of heat and temperature
 comparing technical difficulties experienced during the experiment
 viewing a computer simulation of collision theory
 searching the Internet for relevant mathematical models
 Answer
 Correct Response: C. (Objective 0002) This question requires the examinee to demonstrate the ability to use computers to analyze experimental data. Computer simulations of a variety of chemistry topics are widely available. The visual and often interactive nature of these simulations makes them a valuable supplement to the teaching of conceptually challenging topics. In the case of the lab exercise, viewing a computer simulation of collision theory will strengthen the class's understanding of why temperature affects reaction rate. This review will aid them in analyzing and interpreting their results.
Correct Response: C. (Objective 0002) This question requires the examinee to demonstrate the ability to use computers to analyze experimental data. Computer simulations of a variety of chemistry topics are widely available. The visual and often interactive nature of these simulations makes them a valuable supplement to the teaching of conceptually challenging topics. In the case of the lab exercise, viewing a computer simulation of collision theory will strengthen the class's understanding of why temperature affects reaction rate. This review will aid them in analyzing and interpreting their results.
9. A chemistry teacher plans the student activities listed below as part of a new unit of study.
• comparing new terminology with related terminology from previous units
• developing nonverbal representations (e.g., charts, illustrations) of new terminology
• classifying new terminology according to specific criteria
• generating analogies with new terminology
These activities are likely to promote students' reading comprehension related to this unit primarily in which of the following ways?
 by providing the students with strategies for determining the meaning of unfamiliar vocabulary as they read
 by promoting the students' ability to decode and spell new vocabulary words accurately
 by teaching the students how to use structural analysis as a strategy for building domainspecific vocabulary
 by broadening the students' understanding of new vocabulary words and their associated concepts
10. Some students in a chemistry class are English language learners and/or struggling readers who lack the basic reading skills necessary to fully comprehend assigned readings. Which of the following strategies would be most appropriate for the chemistry teacher to use to scaffold the reading comprehension of these students?
 providing the students with oral previews of text content and noting key concepts on the board
 having the students conduct alternative content research using online texts rather than printed ones
 teaching the students the basic reading skills they lack and reviewing key skills before assignments
 substituting lowerlevel chemistry texts for the students' reading assignments
 Answer
 Correct Response: A. (Objective 0003) This question requires the examinee to demonstrate the ability to plan instruction and select strategies that support all students' contentarea reading. Teachers need to provide scaffolding to English language learners and struggling readers who lack the basic reading skills necessary to comprehend assigned contentarea readings. An oral preview of the content covered in a class reading assignment provides scaffolding by activating students' prior knowledge related to a text and building needed vocabulary and background knowledge required to make inferences and repair comprehension during reading. Noting key concepts on the board during the oral preview both reinforces the new concepts and vocabulary and helps students become familiar with how these elements will appear in print.
Correct Response: A. (Objective 0003) This question requires the examinee to demonstrate the ability to plan instruction and select strategies that support all students' contentarea reading. Teachers need to provide scaffolding to English language learners and struggling readers who lack the basic reading skills necessary to comprehend assigned contentarea readings. An oral preview of the content covered in a class reading assignment provides scaffolding by activating students' prior knowledge related to a text and building needed vocabulary and background knowledge required to make inferences and repair comprehension during reading. Noting key concepts on the board during the oral preview both reinforces the new concepts and vocabulary and helps students become familiar with how these elements will appear in print.
11. Which of the following compounds is paired with its correct Lewis dot structure?
A. The Lewis dot structure is as follows; One central capital C atom is bonded to 3 individual capital O atoms. Two of the C O bonds are single bonds and one C O bond is a double bond. The whole structure is framed by a large set of brackets. On the outside of the brackets, on the upper right side, there is a negative 2.
B. The Lewis dot structure is as follows; The three atoms in this compound are arranged in a linear fashion in the order capital H, capital S, capital H. Each S H bond is a double bond.
C. The Lewis dot structure is as follows; One central capital A lowercase S atom is bonded to six individual capital F atoms. Each AsF bond is a single bond. Each F atom is surrounded by six dots. The whole structure is framed by a large set of brackets. On the outside of the brackets, on the upper right side, there is a negative one.
D. The Lewis dot structure is as follows; One central capital N atom is bonded to four individual capital H atoms. Each N H bond is a single bond. In addition to the four bond lines, the capital N atom also has one dot. The whole structure is framed by a large set of brackets. On the outside of the brackets, on the upper right side, there is a positive one.
12. An electron in a multielectron atom has the quantum numbers shown below.
Quantum Numbers 
n 
ℓ 
m_{ℓ} 
m_{s} 
3 
1 
0 

According to these numbers, what is known about this electron?
 It has acquired a positive charge.
 It is most likely found right next to the nucleus.
 It occupies a pshaped orbital.
 It moves among three orbitals in the same subshell.
 Answer
 Correct Response: C. (Objective 0004) This question requires the examinee to demonstrate knowledge of the principles of quantum mechanics. Quantum numbers describe the most probable location of an electron in an atom. The principle quantum number (n) provides information on the electron's distance from the nucleus and its energy level. The angular momentum quantum number or azimuthal quantum number (ℓ) describes the type of orbital. The magnetic quantum number (m_{ℓ}) describes the orbital's orientation in space and the electron spin quantum number (m_{s}) describes the direction of the electron's spin. The quantum numbers for the given electron indicate that this electron occupies a 3p orbital.
Correct Response: C. (Objective 0004) This question requires the examinee to demonstrate knowledge of the principles of quantum mechanics. Quantum numbers describe the most probable location of an electron in an atom. The principle quantum number open parenthesis n close parenthesis provides information on the electron's distance from the nucleus and its energy level. The angular momentum quantum number or azimuthal quantum number open parenthesis ℓ close parenthesis describes the type of orbital. The magnetic quantum number open parenthesis m sub ℓ close parenthesis describes the orbital's orientation in space and the electron spin quantum number open parenthesis m sub s close parenthesis describes the direction of the electron's spin. The quantum numbers for the given electron indicate that this electron occupies a 3 p orbital.
13. The electron configurations of four atoms are shown in the table below.
Atom 
Electron Configuration 
1 
1s^{2} 
2 
1s^{2}2s^{1} 
3 
1s^{2}2s^{2} 
4 
1s^{2}2s^{2}2p^{6} 
According to these electron configurations, which atom is most likely to be attracted by a magnetic field?
 1
 2
 3
 4
 Answer
 Correct Response: B. (Objective 0004) This question requires the examinee to use the principles of quantum mechanics to analyze chemical systems. Substances with one or more unpaired electrons will be attracted by a magnetic field. This property is known as paramagnetism. The electron configurations shown indicate that atom 2 is the only atom with an unpaired electron (in the 2s orbital). This suggests that atom 2 is the atom most likely to be attracted by a magnetic field.
Correct Response: B. (Objective 0004) This question requires the examinee to use the principles of quantum mechanics to analyze chemical systems. Substances with one or more unpaired electrons will be attracted by a magnetic field. This property is known as paramagnetism. The electron configurations shown indicate that atom 2 is the only atom with an unpaired electron (in the 2s orbital). This suggests that atom 2 is the atom most likely to be attracted by a magnetic field.
14. The structure of nitric acid (HNO_{3}) is best described as a resonance hybrid of two possible Lewis structures. Which of the following best represents one of these Lewis structures?
This structure shows from left to right a hydrogen singly bonded to an oxygen with two lone pairs, the oxygen is singly bonded to an nitrogen, which is doubly bonded to another oxygen with two lone pairs. In addition, the nitrogen is also singly binded to a third oxygen which has three lone pairs.
This structure shows from left to right a hydrogen singly bonded to an oxygen with two lone pairs, the oxygen is singly bonded to an nitrogen, which is singly bonded to another oxygen with three lone pairs. In addition, the nitrogen is also singly binded to a third oxygen which has three lone pairs.
This structure shows a central nitrogen atom. It is doubly bound on three of the sides to three oxygen atoms, and each oxygen atom has two lone pairs. In addition, the nitrogen is singly bound to a hydrogen atom.
This structure shows a central nitrogen atom. It is singly bound on three of the sides to three oxygen atoms, and each oxygen atom has three lone pairs. In addition, the nitrogen is singly bound to a hydrogen atom.
 Answer
 Correct Response: A. (Objective 0004) This question requires the examinee to analyze Lewis dot symbols. Nitric acid is a compound whose actual bond lengths and bond angles are not accurately described by a single Lewis structure. Instead, its structure more closely resembles a hybrid of two possible Lewis structures. One of those structures, represented by the correct response, is one in which the central N atom is surrounded by three O atoms (two with single bonds and one with a double bond). The other possible Lewis structure for nitric acid is one in which the double bond occurs between N and the other O atom not bonded to H.
Correct Response: A. (Objective 0004) This question requires the examinee to analyze Lewis dot symbols. Nitric acid is a compound whose actual bond lengths and bond angles are not accurately described by a single Lewis structure. Instead, its structure more closely resembles a hybrid of two possible Lewis structures. One of those structures, represented by the correct response, is one in which the central N atom is surrounded by three O atoms (two with single bonds and one with a double bond). The other possible Lewis structure for nitric acid is one in which the double bond occurs between upper N and the other upper O atom not bonded to upper H.
15. Which of the following is the orbital diagram for sulfur?
Orbital Diagram, reading left to right open bracket s close bracket, box labeled three s with one up arrow and one down arrow, set of three boxes each with one up arrow in them and they are labeled three p
Orbital Diagram, reading left to right box labeled one s with one up arrow and one down arrow, box labeled two s with one up arrow and one down arrow, set of three boxes each with one up arrow and one down arrow labeled two p and then one box labeled three s with one up arrow and one down arrow
Orbital Diagram, reading left to right open bracket upper n e close bracket box labeled two s with one up arrow and one down arrow set of three boxes labeled two p where the first box has an up arrow and a down arrow and the other two boxes have one up arrow in them
Orbital Diagram, reading left to right box labeled one s with one up arrow and one down arrow box labeled two s with one up arrow and one down arrow set of three boxes labeled two p where each box has one up arrow and one down arrow box labeled three s with one up arrow and one down arrow set of three boxes labeled three p where the first box has an up arrow and a down arrow and the other two boxes have one up arrow in them
 Answer
 Correct Response: D. (Objective 0004) This question requires the examinee to analyze orbital diagrams. Orbital diagrams are representations of an atom's electron configuration. The boxes represent the different orbitals and the single arrows represent individual electrons. For atoms with electrons in p, d, or f subshells, Hund's rule is used to determine the order of filling. The electron configuration of sulfur is 1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}. This electron configuration is best represented by the orbital diagram that includes four electrons in the 3p orbital.
Correct Response: D. (Objective 0004) This question requires the examinee to analyze orbital diagrams. Orbital diagrams are representations of an atom's electron configuration. The boxes represent the different orbitals and the single arrows represent individual electrons. For atoms with electrons in p, d, or f subshells, Hund's rule is used to determine the order of filling. The electron configuration of sulfur is 1 s squared 2 s squared 2 p sup six baseline 3 s squared 3 p sup four. This electron configuration is best represented by the orbital diagram that includes four electrons in the 3 p orbital.
16. An ionic compound is most likely to form when a Group 1 element is reacted with an element from Group:
 2.
 6.
 11.
 17.
17. Which of the following Period 2 elements has the greatest atomic radius?
 Li
 B
 C
 F
 Answer
 Correct Response: A. (Objective 0005) This question requires the examinee to predict periodic trends within periods of the periodic table. While the number of protons increases from left to right for elements within the same period, the number of inner shell electrons stays the same. As the number of protons increases across a period, the effective nuclear charge increases. As a result, the valence electrons are attracted more strongly to the positive charge of the nucleus and the atomic radius is less. This explains why the atomic radius of main group elements in the same period increases from right to left across the periodic table. Of the Period 2 elements listed, lithium will have the greatest atomic radius.
Correct Response: A. (Objective 0005) This question requires the examinee to predict periodic trends within periods of the periodic table. While the number of protons increases from left to right for elements within the same period, the number of inner shell electrons stays the same. As the number of protons increases across a period, the effective nuclear charge increases. As a result, the valence electrons are attracted more strongly to the positive charge of the nucleus and the atomic radius is less. This explains why the atomic radius of main group elements in the same period increases from right to left across the periodic table. Of the Period 2 elements listed, lithium will have the greatest atomic radius.
18. H, Li, Na, K
The Group 1 atoms shown above follow the general periodic trend in electron affinity. Which of these Group 1 atoms has the weakest affinity for electrons?
 H
 Li
 Na
 K
 Answer
 Correct Response: D. (Objective 0005) This question requires the examinee to predict periodic trends within periods and groups of the periodic table. Electron affinity is a measure of how easily an atom gains an electron to form an anion. In general, electron affinity increases from the bottom to the top of the periodic table and from left to right across the periodic table. These trends are related to the electron configurations of the elements. All of the Group 1 atoms listed have one valence electron in an s orbital (H = 1s, Li = 3s, Na = 2s, and K = 4s). As the principal quantum number increases, the relative distance between an added electron and the nucleus increases. The increased distance plus the presence of more core electrons results in a reduced attraction between the positive nucleus and an added electron. This reduced attraction between the nucleus and an added electron corresponds to a weaker electron affinity. Of the Group 1 atoms listed, K, which has the greatest principal quantum number and the most core electrons, has the weakest affinity for electrons.
Correct Response: D. (Objective 0005) This question requires the examinee to predict periodic trends within periods and groups of the periodic table. Electron affinity is a measure of how easily an atom gains an electron to form an anion. In general, electron affinity increases from the bottom to the top of the periodic table and from left to right across the periodic table. These trends are related to the electron configurations of the elements. All of the Group 1 atoms listed have one valence electron in an s orbital open parenthesis upper h equals 1 s, upper l i equals 3 s, upper n a equals 2 s, and upper K equals 4 s close parenthesis. As the principal quantum number increases, the relative distance between an added electron and the nucleus increases. The increased distance plus the presence of more core electrons results in a reduced attraction between the positive nucleus and an added electron. This reduced attraction between the nucleus and an added electron corresponds to a weaker electron affinity. Of the Group 1 atoms listed, upper K, which has the greatest principal quantum number and the most core electrons, has the weakest affinity for electrons.
19.
In the reaction above showing the nuclear decay of strontium90 to yttrium90, X represents:
 a positron.
 an alpha particle.
 a neutron.
 a beta particle.
 Answer
 Correct Response: D. (Objective 0006) This question requires the examinee to analyze the changes in matter and energy that occur in radioactive decay. The type of particle represented by X can be determined by first balancing the given nuclear equation. In a balanced nuclear equation, the sum of the mass numbers and the sum of the atomic numbers are the same on both sides of the equation. The mass number in this reaction is the same on both sides of the equation, but the atomic number changes from 38 to 39. Therefore, the mass number of X must be 0 and the atomic number of X must be –1, giving X the characteristics of a beta particle. Beta particles are highspeed electrons and are represented in nuclear equations by or .
Correct Response: D. (Objective 0006) This question requires the examinee to analyze the changes in matter and energy that occur in radioactive decay. The type of particle represented by X can be determined by first balancing the given nuclear equation. In a balanced nuclear equation, the sum of the mass numbers and the sum of the atomic numbers are the same on both sides of the equation. The mass number in this reaction is the same on both sides of the equation, but the atomic number changes from 38 to 39. Therefore, the mass number of X must be 0 and the atomic number of X must be negative 1, giving X the characteristics of a beta particle. Beta particles are highspeed electrons and are represented in nuclear equations by sup zero sub negative one baseline e or sup zero sub negative one baseline beta.
20. The graph below shows the number of protons and the number of neutrons present in various stable isotopes.
Graph showing the Neutrons vs. Protons for Stable Isotopes where the number of neutrons are on the y axis and the number of protons are on the x axis. the graph has a large number of data points starting close to the origin in the first quadrant and increases pretty steadily where when the number of protons increase so does the number of neutrons.
If an isotope with an atomic number greater than 30 has a neutrontoproton ratio that falls below this belt of stable isotopes, which of the following modes of radioactive decay is this isotope most likely to undergo?
 positron emission
 alpha emission
 gamma ray emission
 beta emission
 Answer
 Correct Response: A. (Objective 0006) This question requires the examinee to relate neutrontoproton ratios to nuclear stability. The graph of neutrons and protons in various stable atomic nuclei can be used to help predict the stability of an isotope and to predict the most probable mode of radioactive decay. In this case, the neutrontoproton ratio for the isotope falls below the belt of stability. This means its neutrontoproton ratio is lower than those of stable isotopes with the same number of protons. For the isotope to raise its neutrontoproton ratio, the number of protons must decrease while the atomic mass remains unchanged. There are two modes of radioactive decay that can achieve this result—positron emission and electron capture.
Correct Response: A. (Objective 0006) This question requires the examinee to relate neutrontoproton ratios to nuclear stability. The graph of neutrons and protons in various stable atomic nuclei can be used to help predict the stability of an isotope and to predict the most probable mode of radioactive decay. In this case, the neutrontoproton ratio for the isotope falls below the belt of stability. This means its neutrontoproton ratio is lower than those of stable isotopes with the same number of protons. For the isotope to raise its neutrontoproton ratio, the number of protons must decrease while the atomic mass remains unchanged. There are two modes of radioactive decay that can achieve this result—positron emission and electron capture.
21. The radioisotope tritium has a halflife of 12.3 years. How long will it take for a 100 μg sample of tritium to decay to 12.5% of its original activity?
 12.3 years
 24.6 years
 36.9 years
 49.2 years
 Answer
 Correct Response: C. (Objective 0006) This question requires the examinee to solve problems involving the halflife of radioactive particles. The activity of the tritium sample is directly related to the mass of sample remaining. This 100 μg sample will have 12.5% of its activity remaining when there are 12.5 μg of tritium left. Since each halflife reduces the mass of the sample by onehalf, the tritium sample will have 12.5% of its original activity after three halflives (i.e., 100 μg to 50 μg to 25 μg to 12.5 μg). Three halflives for tritium equal 36.9 years (i.e., 3 × 12.3 years).
Correct Response: C. (Objective 0006) This question requires the examinee to solve problems involving the halflife of radioactive particles. The activity of the tritium sample is directly related to the mass of sample remaining. This 100 microgram sample will have twelve point five percent of its activity remaining when there are 12.5 microgram of tritium left. Since each halflife reduces the mass of the sample by onehalf, the tritium sample will have twelve point five percent of its original activity after three halflives (i.e., 100 μ g to 50 μ g to 25 μ g to 12.5 μ g). Three halflives for tritium equal 36.9 years (i.e., 3 times 12.3 years).
22. A scientist adds 2.50 mL of toluene (C_{7}H_{8}) to a reaction mixture at 25°C. Given that the density of toluene is 0.8623 g/mL at 25°C, how many molecules of toluene were added to the mixture?
 1.26 × 10^{22}
 1.41 × 10^{22}
 1.63 × 10^{22}
 1.89 × 10^{22}
 Answer
 Correct Response: B. (Objective 0007) This question requires the examinee to demonstrate knowledge of the mole concept. The number of molecules of toluene added to the reaction mixture can be determined using the given volume of toluene, its density, its molar mass, and Avogadro's number. The mass of toluene added to the mixture is calculated by multiplying the volume times the density (2.50 mL × 0.8623 g/mL = 2.16 g). The molar mass of toluene is 92.15 g/mol and is calculated by adding the atomic masses of each atom type in 1 mol of toluene (7 C = 84.07 g, 8 H = 8.08 g). The number of moles of toluene can then be determined by dividing the mass of toluene by its molar mass (2.16 g ÷ 92.15 g/mol = 0.0234 mol of toluene). To convert from moles to molecules of toluene, 0.0234 mol is multiplied by Avogadro's number (6.02 × 10^{23} molecules/mol). These calculations determine that 1.41 × 10^{22} molecules of toluene were added to the reaction mixture.
Correct Response: B. (Objective 0007) This question requires the examinee to demonstrate knowledge of the mole concept. The number of molecules of toluene added to the reaction mixture can be determined using the given volume of toluene, its density, its molar mass, and Avogadro's number. The mass of toluene added to the mixture is calculated by multiplying the volume times the density open parenthesis 2.50 m L times 0.8623 g slash m L equals 2.16 g). The molar mass of toluene is 92.15 g slash mol and is calculated by adding the atomic masses of each atom type in 1 mol of toluene open parenthesis 7 upper C equals 84.07 g, 8 upper H equals 8.08 g close parenthesis. The number of moles of toluene can then be determined by dividing the mass of toluene by its molar mass open parenthesis 2.16 g divided by 92.15 g slash mol equals 0.0234 mol of toluene close parenthesis. To convert from moles to molecules of toluene, 0.0234 mol is multiplied by Avogadro's number open parenthesis 6.02 times 10 sup twenty three molecules slash mol). These calculations determine that 1.41 times 10 sup twenty two molecules of toluene were added to the reaction mixture.
23. A student wants to determine the molality of a 5.00% by mass aqueous solution of NaCl prepared using 10.0 g of NaCl and 190.0 mL of H_{2}O at 20°C. Given this information, what is the molality of the prepared NaCl solution?
 0.0162 m
 0.0526 m
 0.450 m
 0.900 m
 Answer
 Correct Response: D. (Objective 0007) This question requires the examinee to calculate the concentrations of solutions in terms of molality. The concentration of a solution can be expressed as molality, molarity, percent by mass, or mole fraction. Molality (m) is equal to moles of solute divided by the mass of solvent in kg. In this example, there is 0.171 mol of NaCl and 0.190 kg of H_{2}O . Given these values, the molality of the NaCl solution is 0.900 m .
Correct Response: D. (Objective 0007) This question requires the examinee to calculate the concentrations of solutions in terms of molality. The concentration of a solution can be expressed as molality, molarity, percent by mass, or mole fraction. Molality (m) is equal to moles of solute divided by the mass of solvent in k g. In this example, there is 0.171 mol of Upper N a upper C l open parenthesis ten point zero g Upper n a upper c l time open parenthesis start fraction numerator one mol upper n a upper c l denominator fifty eight point five g upper n a upper c l end fraction close parenthesis close parenthesis and 0.190 k g of Upper H sub two upper O open parenthesis one hundred ninety point zero m l upper h two upper o times start fraction numerator one point zero zero g upper h two upper o denominator m l upper h two upper o end fraction times start fraction numerator one k g denominator one thousand g end fraction close parenthesis. Given these values, the molality of the upper N a upper C l solution is 0.900 m open parenthesis start fraction numerator zero point one seven one mol upper n a upper c l denominator zero point one nine zero k g upper h two upper o end fraction close parenthesis.
24. How many grams of H are present in 52.5 g of carbonic acid (H_{2}CO_{3})?
 1.63 g
 1.71 g
 2.02 g
 3.26 g
 Answer
 Correct Response: B. (Objective 0007) This question requires the examinee to apply the mole concept to solve problems involving percent composition. The first step in determining the mass of H present in 52.5 g of H_{2}CO_{3} is determining the percent composition by mass of H. The chemical formula indicates that in 1 mol of H_{2}CO_{3} there are 2 mol or 2.02 g of H. The following calculations can be used to determine the percent by mass of H in 1 mol of H_{2}CO_{3}: . This means that there are 3.26 g of H in every 100 g of H_{2}CO_{3}. This information can be used to determine the mass of H in 52.5 g of H_{2}CO_{3} .
Correct Response: B. (Objective 0007) This question requires the examinee to apply the mole concept to solve problems involving percent composition. The first step in determining the mass of upper H present in 52.5 g of upper H sub two upper C upper O sub three is determining the percent composition by mass of upper H. The chemical formula indicates that in 1 mol of upper H sub two upper C upper O sub three there are 2 mol or 2.02 g of upper H. The following calculations can be used to determine the percent by mass of upper H in 1 mol of upper H sub two upper C upper O three: start fraction numerator two point zero two g upper h denominator sixty two point zero three g upper h two upper c upper o three end fraction times one hundred percent equals three point two six percent upper h. This means that there are 3.26 g of upper H in every 100 g of upper H sub two upper C upper O sub three. This information can be used to determine the mass of upper H in 52.5 g of upper H sub two upper C upper O sub three open parenthesis fifty two point five g upper h two upper c upper o three times start fraction numerator three point two six g upper h denominator one hundred point zero g upper h two upper c upper o three equals one point seven one g upper h.
25. A compound is 5.94% H by mass and 94.01% O by mass and has a molecular weight between 30.00 g and 35.00 g. What is the molecular formula of this compound?
 OH−
 H_{2}O
 HO_{2}
 H_{2}O_{2}
 Answer
 Correct Response: D. (Objective 0007) This question requires the examinee to apply the mole concept to solve problems involving molecular formulas. The molecular formula for a compound includes the numbers and types of atoms present in the compound. In this example, the percent composition by mass and information about the molecular weight of the compound are given. The percent composition of each atom can be used to determine the molar ratio of the two atoms. In a 100 g sample of the compound, there are 5.94 g of H and 94.01 g of O. When these mass values are converted to moles, the molar ratio of H:O is determined to be 5.88 mol H:5.88 mol O. This corresponds to a 1:1 ratio of H:O. However, HO is not the molecular formula for this compound because the molecular weight of HO is not between 30.00 g and 35.00 g. The molecular formula must exhibit a 1:1 ratio between the atom types and have a molecular weight within the given range. The molecular formula for hydrogen peroxide (H_{2}O_{2}) is the only listed compound that satisfies both of these criteria.
Correct Response: D. (Objective 0007) This question requires the examinee to apply the mole concept to solve problems involving molecular formulas. The molecular formula for a compound includes the numbers and types of atoms present in the compound. In this example, the percent composition by mass and information about the molecular weight of the compound are given. The percent composition of each atom can be used to determine the molar ratio of the two atoms. In a 100 g sample of the compound, there are 5.94 g of upper H and 94.01 g of upper O. When these mass values are converted to moles, the molar ratio of upper H : upper O is determined to be 5.88 mol upper H : 5.88 mol upper O. This corresponds to a 1:1 ratio of upper H : upper O. However, upper H upper O is not the molecular formula for this compound because the molecular weight of upper H upper O is not between 30.00 g and 35.00 g. The molecular formula must exhibit a 1:1 ratio between the atom types and have a molecular weight within the given range. The molecular formula for hydrogen peroxide open parenthesis upper H sub two upper O sub two close parenthesis is the only listed compound that satisfies both of these criteria.
26. CH_{4}(g) + 2O_{2}(g) → CO_{2}(g) + 2H_{2}O(ℓ) ΔH = –890.3 kJ
A chemical equation is shown and reads as follows;
Capital C capital H subscript four, gas, reacts with two capital O subscript two, gas, to produce capital C capital O subscript two, gas, and two capital H subscript two capital O, liquid.
The enthalpy change for the reaction is equal to negative eight hundred ninety point three kilojoules.
Given the balanced equation for the combustion of methane shown above, how many moles of methane would need to be reacted in order to produce 3561 kJthree thousand five hundred sixtyone kilojoules of energy?
 2
 4
 8
 16
27. C_{8}H_{18}(ℓ) + O2(g) → CO_{2}(g) + H_{2}O(g)
The chemical equation for the combustion of octane shown above is unbalanced. When this equation is balanced with the lowest set of whole number coefficients, which of the following is the stoichiometric coefficient for CO_{2}?
 8
 12
 16
 20
 Answer
 Correct Response: C. (Objective 0008) This question requires the examinee to demonstrate the ability to balance chemical equations. In a balanced chemical equation there are the same numbers and types of atoms on the reactant side as there are on the product side. Using the lowest set of whole number coefficients, the balanced chemical equation for the combustion of octane is as follows: 2C_{8}H_{18}(ℓ) + 25O_{2}(g) → 16CO_{2}(g) + 18H_{2}O(g). These coefficients correspond to 16 C atoms, 36 H atoms, and 50 O atoms on both sides of the reaction arrow.
Correct Response: C. (Objective 0008) This question requires the examinee to demonstrate the ability to balance chemical equations. In a balanced chemical equation there are the same numbers and types of atoms on the reactant side as there are on the product side. Using the lowest set of whole number coefficients, the balanced chemical equation for the combustion of octane is as follows: two upper C sub eight upper H sub eighteen open parenthesis ℓ close parenthesis plus twenty five upper o sub two open parenthesis g close parenthesis produces sixteen upper c upper o sub two open parenthesis g close parenthesis plus eighteen upper h sub two upper o open parenthesis g close parenthesis. These coefficients correspond to 16 upper C atoms, 36 upper H atoms, and 50 upper O atoms on both sides of the reaction arrow.
28. 2Ca(s) + O_{2}(g) 2CaO(s)
According to the oxidation reaction shown above, what mass of CaO will be produced when 120.3 g of Ca react with excess O_{2}?
 240.6 g
 168.3 g
 120.3 g
 112.2 g
 Answer
 Correct Response: B. (Objective 0008) This question requires the examinee to solve stoichiometric problems involving mass. The balanced chemical equation indicates a 1:1 stoichiometric ratio between Ca and CaO. Therefore, the oxidation of 120.3 g (or 3 mol) of Ca will produce 3 mol of CaO, which is equivalent to 168.3 g of CaO .
Correct Response: B. (Objective 0008) This question requires the examinee to solve stoichiometric problems involving mass. The balanced chemical equation indicates a 1:1 stoichiometric ratio between upper C a and upper C a upper O. Therefore, the oxidation of 120.3 g (or 3 mol) of upper C a will produce 3 mol of upper C a upper O, which is equivalent to 168.3 g of upper C a upper O open parenthesis three moles upper c a upper o times start fraction numerator fifty six point one zero g upper c a upper o denominator one mole upper c a upper o end fraction equals one hundred sixty eight point three g of upper c a upper o close parenthesis.
29. Pb(NO_{3})_{2}(aq) + 2KI(aq) → PbI_{2}(s) + 2KNO_{3}(aq)
According to the chemical equation shown above, how many moles of PbI_{2} will be formed when 2.0 mol of Pb(NO_{3})_{2} are reacted with 3.0 mol of KI?
 1.0 mol
 1.5 mol
 2.0 mol
 2.5 mol
 Answer
 Correct Response: B. (Objective 0008) This question requires the examinee to solve stoichiometric problems involving limiting reactants. Identifying the limiting reactant is the first step in determining the number of moles of PbI_{2} formed. The stoichiometric ratio between the reactants and the amount of each reactant present can be used to determine which reactant is limiting. In this example, KI is the limiting reactant. This is because 2.0 mol of Pb(NO_{3})_{2} require 4.0 mol of KI to react completely and only 3.0 mol of KI are available. Once the limiting reactant has been identified, the amount of product can be calculated using the stoichiometric ratio between the limiting reactant and the desired product. In this example, 1.5 mol of PbI_{2} will be formed when 3.0 mol of KI are reacted with 2 mol of Pb(NO_{3})_{2} .
Correct Response: B. (Objective 0008) This question requires the examinee to solve stoichiometric problems involving limiting reactants. Identifying the limiting reactant is the first step in determining the number of moles of upper P b upper I sub two formed. The stoichiometric ratio between the reactants and the amount of each reactant present can be used to determine which reactant is limiting. In this example, upper K upper I is the limiting reactant. This is because 2.0 mol of upper P b open parenthesis upper n upper o sub three close parenthesis sub two require 4.0 mol of upper K upper I to react completely and only 3.0 mol of upper K upper I are available. Once the limiting reactant has been identified, the amount of product can be calculated using the stoichiometric ratio between the limiting reactant and the desired product. In this example, 1.5 mol of upper P b upper I sub two will be formed when 3.0 mol of upper K upper I are reacted with 2 mol of upper P b open parenthesis upper n upper o sub three close parenthesis sub two open parenthesis three point zero mole upper k upper i time start fraction numerator one mole upper p b upper i sub two denominator two mole upper k upper i end fraction equals one point five mole upper p b upper i sub two close parenthesis.
30. 2Fe_{2}O_{3}(s) + 3C(s) → 4Fe(s) + 3CO_{2}(g)
When 1 mol of Fe_{2}O_{3} is reacted with excess C according to the chemical equation shown above, 91.5 g of Fe are produced. What is the percent yield of this reaction?
 54.8%
 41.0%
 81.7%
 91.9%
 Answer
 Correct Response: C. (Objective 0008) This question requires the examinee to calculate the percent yield for given chemical reactions. The percent yield for a reaction is defined as the . The actual yield for the reaction is given and the theoretical yield can be calculated using the 1:2 stoichiometric ratio between Fe_{2}O_{3} and Fe. When 1 mol of Fe_{2}O_{3} is reacted with excess C, 2 mol (or 112 g) of Fe will be produced. This mass of Fe represents the theoretical yield for the reaction. Using the formula for percent yield and the values for actual yield (91.5 g) and theoretical yield (112 g), the percent yield is calculated to be 81.7%.
Correct Response: C. (Objective 0008) This question requires the examinee to calculate the percent yield for given chemical reactions. The percent yield for a reaction is defined as the actual yield over theoretical yield multiplied by 100%. The actual yield for the reaction is given and the theoretical yield can be calculated using the 1:2 stoichiometric ratio between upper F e sub two upper O sub three and upper F e. When 1 mol of upper F e sub two upper O sub three is reacted with excess upper C, 2 mol (or 112 g) of upper F e will be produced. This mass of upper F e represents the theoretical yield for the reaction. Using the formula for percent yield and the values for actual yield (91.5 g) and theoretical yield (112 g), the percent yield is calculated to be eight one point seven percent.
Answer Key