04 Atomic Theory Problem Set
ATOMIC THEORY | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: An unknown element Z is found to have an average atomic mass of 107.9 amu, and is known to have only two naturally occurring isotopes. If 107 Z has an atomic mass of 106.905 amu and 109 Z has an atomic mass of 108.791 amu, the percent abundance of 107 Z is most close to: A. 47.32% B. 51.84% C. 67.28% D. 83.21%
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SOLUTION 1: The TOPIC of AVERAGE ATOMIC MASS is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.
We are given two ISOTOPES, 107 Z and 109 of an UNKNOWN ELEMENT βZβ, and the AVERAGE ATOMIC MASS of the ELEMENT, and are looking to SOLVE for the PERCENT ABUNDANCE of each ISOTOPE. As we are told they are ISOTOPES, we can ASSUME they have the same NUMBER of PROTONS, but have differing MASS NUMBERS representing the SUM of the PROTONS and NEUTRONS in each particular ISOTOPE. Given the AVERAGE ATOMIC MASS of element βZβ Carbon is 107.9 πππ’, we can relate this to the RELATIVE ATOMIC MASS of each ISOTOPE, and SOLVE for the PERCENT ABUNDANCE of each ISOTOPE. The FORMULA to CALCULATE AVERAGE ATOMIC MASS is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.
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To calculate the AVERAGE ATOMIC MASS from the known ISOTOPES of a given element, we use the equation for WEIGHTED AVERAGE based on the ABUNDANCE of each ISOTOPE: π΄π£πππππ π΄π‘ππππ πππ π = β(ππΌ Γ π΄πΌ )π Where: β’ ππΌ represents the MASS of the ISOTOPE β’ π΄πΌ represents the ABUNDANCE of the ISOTOPE, and WEIGHT of the AVERAGE β’ π is the NUMBER of ISOTOPES Using the FORMULA for WEIGHTED AVERAGE, we can re-write the FORMULA in TERMS of the ATOMIC MASS for the unknown element βZβ as: π΄π‘ππππ πππ π ππ π = (106.905 Γ π΄107 ) + (108.791 Γ π΄109 ) Each ISOTOPE has its own ABUNDANCE in NATURE, where the PERCENT ABUNDANCES of all the ISOTOPES of an ELEMENT have to equal 100%. Therefore, we can write a FORMULA for the PERCENT ABUNDANCE of each ISOTOPE as: π΄107 + π΄109 = 100% = 1 We can then write an EXPRESSION to SUBSTITUTE in for the ABUNDANCE of one of the ISOTOPES. Made with
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Letβs go AHEAD and write an EXPRESSION for the ABUNDANCE of the 107 ISOTOPE as: π΄107 = 1 β π΄109 Plugging in this EXPRESSION for the ABUNDANCE of the 107 ISOTOPE, and the given ATOMIC MASS VALUES, we can re-write the expression for the AVERAGE ATOMIC MASS of the unknown ELEMENT as: 107.9 πππ’ = (106.905 πππ’)(1 β π΄109 ) + (108.971 πππ’)(π΄109 ) SOLVING for the ABUNDANCE of the 109 ISOTOPE, we find: π΄109 = 0.4816 PLUGGING this VALUE into the EXPRESSION for the ABUNDANCE of the 107 ISOTOPE, we find our answer is calculated as: π΄107 = 1 β π΄109 = 1 β (0.4816) = 0.5184 = 51.84%
Therefore, the correct answer choice is B. ππ. ππ%
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PROBLEM 2: The element carbon has two naturally occurring isotopes, 12C and 13C, each of which has a relative atomic mass 12.00000 πππ’ and 13.00335 πππ’, respectively. Given the relative mass of each isotope, the percent abundance of each isotope is most close to: A. 96.9% 12C;1.0% 13C B. 97.9% 12C;1.1% 13C C. 98.9% 12C;1.1% 13C D. 99.9% 12C;1.4% 13C
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SOLUTION 2: The PERIODIC TABLE OF ELEMENTS can be referenced under the SUBJECT of CHEMISTRY on page 55 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem we are looking to solve for the PERCENT ABUNDANCE of each ISOTOPE of CARBON. As we are given the RELATIVE ATOMIC MASS for each ISOTOPE, our FIRST STEP is to determine the AVERAGE ATOMIC MASS of CARBON using the PERIODIC TABLE. Looking at the PERIODIC TABLE OF ELEMENTS in the reference handbook, we see that we are given the ATOMIC NUMBER, SYMBOL, and ATOMIC WEIGHT for every ELEMENT on the TABLE. Therefore, the first step in this PROBLEM is to simply IDENTIFY the SYMBOL for CARBON in the PERIODIC TABLE. In the Reference Handbook, we are not EXPLICITLY TOLD the corresponding SYMBOLS for each ELEMENT, so we must be able to IDENTIFY the SYMBOL of each ELEMENT from MEMORIZATION or using CONTEXT CLUES in the PERIODIC TABLE. When in doubt try to MATCH up the SYMBOL for the ELEMENT by looking at the SPELLING of the ELEMENT. This will NOT WORK for all ELEMENTS, but can help you to ELIMINATE answer choices and REVERSE ENGINEER the PERIODIC TABLE.
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Looking at the PERIODIC TABLE, we see that in COLUMN IV, there is the SYMBOL βCβ, which represents CARBON.
Referencing the information for CARBON in the periodic table, we find the associated ATOMIC WEIGHT for CARBON is given as 12.011 amu. The TOPIC of AVERAGE ATOMIC MASS is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.
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We are given two ISOTOPES, 12 C and 13 C of the element CARBON, and the RELATIVE ATOMIC MASS of each ISOTOPE. As we are told they are ISOTOPES, we can ASSUME they have the same NUMBER of PROTONS, but have differing MASS NUMBERS representing the SUM of the PROTONS and NEUTRONS in each particular ISOTOPE. Given the AVERAGE ATOMIC MASS of CARBON is 12.011 πππ’ from the PERIODIC TABLE, we can relate this to the RELATIVE ATOMIC MASS of each ISOTOPE, and SOLVE for the PERCENT ABUNDANCE of each ISOTOPE. The FORMULA to CALCULATE AVERAGE ATOMIC MASS is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. To calculate the AVERAGE ATOMIC MASS from the known ISOTOPES of a given element, we use the equation for WEIGHTED AVERAGE based on the ABUNDANCE of each ISOTOPE: π΄π£πππππ π΄π‘ππππ πππ π = β(ππΌ Γ π΄πΌ )π Where: β’ ππΌ represents the MASS of the ISOTOPE β’ π΄πΌ represents the ABUNDANCE of the ISOTOPE, and WEIGHT of the AVERAGE β’ π is the NUMBER of ISOTOPES
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Using the FORMULA for WEIGHTED AVERAGE, we can re-write the FORMULA in TERMS of the ATOMIC MASS for CARBON as: π΄π‘ππππ πππ π ππ πΆ = (12.00000 Γ π΄12 ) + (13.00335 Γ π΄13 ) Each ISOTOPE has its own ABUNDANCE in NATURE, where the PERCENT ABUNDANCES of all the ISOTOPES of an ELEMENT have to equal 100%. Therefore, we can write a FORMULA for the PERCENT ABUNDANCE of each ISOTOPE as: π΄12 + π΄13 = 100% = 1 We can then write an EXPRESSION to SUBSTITUTE in for the ABUNDANCE of one of the ISOTOPES. Letβs go AHEAD and assume we want to write an EXPRESSION for the ABUNDANCE of the CARBON-12 ISOTOPE as: π΄12 = 1 β π΄13 Plugging in this EXPRESSION for the ABUNDANCE of the CARBON-12 ISOTOPE, and the given ATOMIC MASS VALUES, we can re-write the expression for the AVERAGE ATOMIC MASS of the unknown ELEMENT as: 12.011 πππ’ = (12.00000 πππ’)(1 β π΄13 ) + (13.00335 πππ’)(π΄13 )
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SOLVING for the ABUNDANCE of the CARBON-13 ISOTOPE, we find: π΄13 = 0.01096 = 1.096% PLUGGING this VALUE into the EXPRESSION for the ABUNDANCE of the CARBON13 ISOTOPE, we the PERCENT ABUNDANCE for the CARBON-12 ISOTOPE is: π΄12 = 1 β π΄13 = 1 β (0.01096) = 0.98904 = 98.9%
Therefore, the correct answer choice is C. 98.9% 12C;1.1% 13C
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PROBLEM 3: Given a sample of solid sodium chloride, which of the following bonds is most prevalent? A. πππ‘πππππ π΅πππ B. πΌππππ π΅πππ C. πΆππ£πππππ‘ π΅πππ D. πππππ π΅πππ
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SOLUTION 3: The TOPIC OF PRIMARY BONDS can be referenced under SUBJECT of MATERIALS SCIENCE on page 60 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A sample of SODIUM CHLORIDE, both ELEMENTS are IONICALLY CHARGED to act as SODIUM and CHLORINE IONS. IONIC BONDING is based on the ELECTROSTATIC or COULOMBIC ATTRACTION between opposite ELECTRICALLY CHARGED ions, and the TRANSFER of ONE or more ELECTRONS from one ATOM to another. The ELECTROSTATIC ATTRACTION between the POSITIVE SODIUM ION and the NEGATIVE CHLORIDE IONS forms the basis of the IONIC BONDS that EXISTS in a SAMPLE of SODIUM CHLORIDE.
Therefore, the correct answer choice is B. π°ππππ π©πππ
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PROBLEM 4: Which of the following statements best describes valence electrons? A. Electrons located on the outermost energy level B. Electrons with a positive charge C. Electrons located within the nucleus D. Electrons that are most stable
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SOLUTION 4: The TOPIC of VALENCE ELECTRONS is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. ELECTRONS that are on the OUTERMOST ENERGY LEVEL are commonly referred to as VALENCE ELECTRONS. One of the properties that ELEMENTS within a GROUP share is the tendency to gain or lose a specific number of VALENCE ELECTRONS, which is referred to as the VALENCE or VALENCE STATE.
Therefore, the correct answer choice is A. Electrons located on the outermost energy level
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PROBLEM 5: Which of the following bond provides the strongest force? A. Ionic Bond B. Metallic Bond C. Ionic and Metallic Bonds D. Covalent Bond
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SOLUTION 5: The TOPIC OF PRIMARY BONDS can be referenced under the SUBJECT of MATERIALS SCIENCE on page 60 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. COVALENT BONDS are the STRONGEST ATTRACTIVE FORCES typically FORMED between ATOMS with half-filled ORBITALS, as well as between NON-METALS, or between NON-METALS and METALLOIDS.
Therefore, the correct answer choice is D. Covalent Bond
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PROBLEM 6: What is the difference between two isotopes of the same element? A. Number of Protons B. Number of Electrons C. Electrical Charge D. Number of Neutrons
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SOLUTION 6: The TOPIC of ISOTOPES is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. ISOTOPES are ATOMS of the same ELEMENT with the SAME NUMBER of PROTONS, but a different ATOMIC MASS NUMBER resulting from a DIFFERENCE in the varying number of NEUTRONS.
Therefore, the correct answer choice is D. Number of Neutrons
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PROBLEM 7: Which of the following statements best describes an isotropic substance? A. Properties vary by direction B. Properties are homogeneous in all directions C. Property vary by location D. All of the above
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SOLUTION 7: The TOPIC of ISOTROPIC SUBSTANCES is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.
An ISOTROPIC SUBSTANCE is a SUBSTANCE that has the SAME PHYSICAL PROPERTIES in ALL DIRECTIONS.
Therefore, the correct answer choice is B. Properties are homogeneous in all directions
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PROBLEM 8: The element tin (Sn) has 10 stable isotopes. The atomic nuclei of all ten isotopes have the same: A. Number of Neutrons B. Mass C. Number of Protons D. Number of Electrons
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SOLUTION 8: The TOPIC of ISOTOPES is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. ISOTOPES are ATOMS of the same ELEMENT with the SAME NUMBER of PROTONS, but a different ATOMIC MASS NUMBER resulting from a DIFFERENCE in the varying number of NEUTRONS. Answer choice A is INCORRECT as ISOTOPES have EXTRA NEUTRONS. Answer choice B is INCORRECT as NEUTRONS have the SAME MASS and PROTONS. Answer choice D is INCORRECT as ELECTRONS are not PART of the NUCLEUS
Therefore, the correct answer choice is C. Number of Protons
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PROBLEM 1: An unknown element Z is found to have an average atomic mass of 107.9 amu, and is known to have only two naturally occurring isotopes. If 107 Z has an atomic mass of 106.905 amu and 109 Z has an atomic mass of 108.791 amu, the percent abundance of 107 Z is most close to: A. 47.32% B. 51.84% C. 67.28% D. 83.21%
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SOLUTION 1: The TOPIC of AVERAGE ATOMIC MASS is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.
We are given two ISOTOPES, 107 Z and 109 of an UNKNOWN ELEMENT βZβ, and the AVERAGE ATOMIC MASS of the ELEMENT, and are looking to SOLVE for the PERCENT ABUNDANCE of each ISOTOPE. As we are told they are ISOTOPES, we can ASSUME they have the same NUMBER of PROTONS, but have differing MASS NUMBERS representing the SUM of the PROTONS and NEUTRONS in each particular ISOTOPE. Given the AVERAGE ATOMIC MASS of element βZβ Carbon is 107.9 πππ’, we can relate this to the RELATIVE ATOMIC MASS of each ISOTOPE, and SOLVE for the PERCENT ABUNDANCE of each ISOTOPE. The FORMULA to CALCULATE AVERAGE ATOMIC MASS is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.
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To calculate the AVERAGE ATOMIC MASS from the known ISOTOPES of a given element, we use the equation for WEIGHTED AVERAGE based on the ABUNDANCE of each ISOTOPE: π΄π£πππππ π΄π‘ππππ πππ π = β(ππΌ Γ π΄πΌ )π Where: β’ ππΌ represents the MASS of the ISOTOPE β’ π΄πΌ represents the ABUNDANCE of the ISOTOPE, and WEIGHT of the AVERAGE β’ π is the NUMBER of ISOTOPES Using the FORMULA for WEIGHTED AVERAGE, we can re-write the FORMULA in TERMS of the ATOMIC MASS for the unknown element βZβ as: π΄π‘ππππ πππ π ππ π = (106.905 Γ π΄107 ) + (108.791 Γ π΄109 ) Each ISOTOPE has its own ABUNDANCE in NATURE, where the PERCENT ABUNDANCES of all the ISOTOPES of an ELEMENT have to equal 100%. Therefore, we can write a FORMULA for the PERCENT ABUNDANCE of each ISOTOPE as: π΄107 + π΄109 = 100% = 1 We can then write an EXPRESSION to SUBSTITUTE in for the ABUNDANCE of one of the ISOTOPES. Made with
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Letβs go AHEAD and write an EXPRESSION for the ABUNDANCE of the 107 ISOTOPE as: π΄107 = 1 β π΄109 Plugging in this EXPRESSION for the ABUNDANCE of the 107 ISOTOPE, and the given ATOMIC MASS VALUES, we can re-write the expression for the AVERAGE ATOMIC MASS of the unknown ELEMENT as: 107.9 πππ’ = (106.905 πππ’)(1 β π΄109 ) + (108.971 πππ’)(π΄109 ) SOLVING for the ABUNDANCE of the 109 ISOTOPE, we find: π΄109 = 0.4816 PLUGGING this VALUE into the EXPRESSION for the ABUNDANCE of the 107 ISOTOPE, we find our answer is calculated as: π΄107 = 1 β π΄109 = 1 β (0.4816) = 0.5184 = 51.84%
Therefore, the correct answer choice is B. ππ. ππ%
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PROBLEM 2: The element carbon has two naturally occurring isotopes, 12C and 13C, each of which has a relative atomic mass 12.00000 πππ’ and 13.00335 πππ’, respectively. Given the relative mass of each isotope, the percent abundance of each isotope is most close to: A. 96.9% 12C;1.0% 13C B. 97.9% 12C;1.1% 13C C. 98.9% 12C;1.1% 13C D. 99.9% 12C;1.4% 13C
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SOLUTION 2: The PERIODIC TABLE OF ELEMENTS can be referenced under the SUBJECT of CHEMISTRY on page 55 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem we are looking to solve for the PERCENT ABUNDANCE of each ISOTOPE of CARBON. As we are given the RELATIVE ATOMIC MASS for each ISOTOPE, our FIRST STEP is to determine the AVERAGE ATOMIC MASS of CARBON using the PERIODIC TABLE. Looking at the PERIODIC TABLE OF ELEMENTS in the reference handbook, we see that we are given the ATOMIC NUMBER, SYMBOL, and ATOMIC WEIGHT for every ELEMENT on the TABLE. Therefore, the first step in this PROBLEM is to simply IDENTIFY the SYMBOL for CARBON in the PERIODIC TABLE. In the Reference Handbook, we are not EXPLICITLY TOLD the corresponding SYMBOLS for each ELEMENT, so we must be able to IDENTIFY the SYMBOL of each ELEMENT from MEMORIZATION or using CONTEXT CLUES in the PERIODIC TABLE. When in doubt try to MATCH up the SYMBOL for the ELEMENT by looking at the SPELLING of the ELEMENT. This will NOT WORK for all ELEMENTS, but can help you to ELIMINATE answer choices and REVERSE ENGINEER the PERIODIC TABLE.
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Looking at the PERIODIC TABLE, we see that in COLUMN IV, there is the SYMBOL βCβ, which represents CARBON.
Referencing the information for CARBON in the periodic table, we find the associated ATOMIC WEIGHT for CARBON is given as 12.011 amu. The TOPIC of AVERAGE ATOMIC MASS is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.
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We are given two ISOTOPES, 12 C and 13 C of the element CARBON, and the RELATIVE ATOMIC MASS of each ISOTOPE. As we are told they are ISOTOPES, we can ASSUME they have the same NUMBER of PROTONS, but have differing MASS NUMBERS representing the SUM of the PROTONS and NEUTRONS in each particular ISOTOPE. Given the AVERAGE ATOMIC MASS of CARBON is 12.011 πππ’ from the PERIODIC TABLE, we can relate this to the RELATIVE ATOMIC MASS of each ISOTOPE, and SOLVE for the PERCENT ABUNDANCE of each ISOTOPE. The FORMULA to CALCULATE AVERAGE ATOMIC MASS is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. To calculate the AVERAGE ATOMIC MASS from the known ISOTOPES of a given element, we use the equation for WEIGHTED AVERAGE based on the ABUNDANCE of each ISOTOPE: π΄π£πππππ π΄π‘ππππ πππ π = β(ππΌ Γ π΄πΌ )π Where: β’ ππΌ represents the MASS of the ISOTOPE β’ π΄πΌ represents the ABUNDANCE of the ISOTOPE, and WEIGHT of the AVERAGE β’ π is the NUMBER of ISOTOPES
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Using the FORMULA for WEIGHTED AVERAGE, we can re-write the FORMULA in TERMS of the ATOMIC MASS for CARBON as: π΄π‘ππππ πππ π ππ πΆ = (12.00000 Γ π΄12 ) + (13.00335 Γ π΄13 ) Each ISOTOPE has its own ABUNDANCE in NATURE, where the PERCENT ABUNDANCES of all the ISOTOPES of an ELEMENT have to equal 100%. Therefore, we can write a FORMULA for the PERCENT ABUNDANCE of each ISOTOPE as: π΄12 + π΄13 = 100% = 1 We can then write an EXPRESSION to SUBSTITUTE in for the ABUNDANCE of one of the ISOTOPES. Letβs go AHEAD and assume we want to write an EXPRESSION for the ABUNDANCE of the CARBON-12 ISOTOPE as: π΄12 = 1 β π΄13 Plugging in this EXPRESSION for the ABUNDANCE of the CARBON-12 ISOTOPE, and the given ATOMIC MASS VALUES, we can re-write the expression for the AVERAGE ATOMIC MASS of the unknown ELEMENT as: 12.011 πππ’ = (12.00000 πππ’)(1 β π΄13 ) + (13.00335 πππ’)(π΄13 )
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SOLVING for the ABUNDANCE of the CARBON-13 ISOTOPE, we find: π΄13 = 0.01096 = 1.096% PLUGGING this VALUE into the EXPRESSION for the ABUNDANCE of the CARBON13 ISOTOPE, we the PERCENT ABUNDANCE for the CARBON-12 ISOTOPE is: π΄12 = 1 β π΄13 = 1 β (0.01096) = 0.98904 = 98.9%
Therefore, the correct answer choice is C. 98.9% 12C;1.1% 13C
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PROBLEM 3: Given a sample of solid sodium chloride, which of the following bonds is most prevalent? A. πππ‘πππππ π΅πππ B. πΌππππ π΅πππ C. πΆππ£πππππ‘ π΅πππ D. πππππ π΅πππ
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SOLUTION 3: The TOPIC OF PRIMARY BONDS can be referenced under SUBJECT of MATERIALS SCIENCE on page 60 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A sample of SODIUM CHLORIDE, both ELEMENTS are IONICALLY CHARGED to act as SODIUM and CHLORINE IONS. IONIC BONDING is based on the ELECTROSTATIC or COULOMBIC ATTRACTION between opposite ELECTRICALLY CHARGED ions, and the TRANSFER of ONE or more ELECTRONS from one ATOM to another. The ELECTROSTATIC ATTRACTION between the POSITIVE SODIUM ION and the NEGATIVE CHLORIDE IONS forms the basis of the IONIC BONDS that EXISTS in a SAMPLE of SODIUM CHLORIDE.
Therefore, the correct answer choice is B. π°ππππ π©πππ
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PROBLEM 4: Which of the following statements best describes valence electrons? A. Electrons located on the outermost energy level B. Electrons with a positive charge C. Electrons located within the nucleus D. Electrons that are most stable
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SOLUTION 4: The TOPIC of VALENCE ELECTRONS is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. ELECTRONS that are on the OUTERMOST ENERGY LEVEL are commonly referred to as VALENCE ELECTRONS. One of the properties that ELEMENTS within a GROUP share is the tendency to gain or lose a specific number of VALENCE ELECTRONS, which is referred to as the VALENCE or VALENCE STATE.
Therefore, the correct answer choice is A. Electrons located on the outermost energy level
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PROBLEM 5: Which of the following bond provides the strongest force? A. Ionic Bond B. Metallic Bond C. Ionic and Metallic Bonds D. Covalent Bond
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SOLUTION 5: The TOPIC OF PRIMARY BONDS can be referenced under the SUBJECT of MATERIALS SCIENCE on page 60 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. COVALENT BONDS are the STRONGEST ATTRACTIVE FORCES typically FORMED between ATOMS with half-filled ORBITALS, as well as between NON-METALS, or between NON-METALS and METALLOIDS.
Therefore, the correct answer choice is D. Covalent Bond
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PROBLEM 6: What is the difference between two isotopes of the same element? A. Number of Protons B. Number of Electrons C. Electrical Charge D. Number of Neutrons
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SOLUTION 6: The TOPIC of ISOTOPES is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. ISOTOPES are ATOMS of the same ELEMENT with the SAME NUMBER of PROTONS, but a different ATOMIC MASS NUMBER resulting from a DIFFERENCE in the varying number of NEUTRONS.
Therefore, the correct answer choice is D. Number of Neutrons
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PROBLEM 7: Which of the following statements best describes an isotropic substance? A. Properties vary by direction B. Properties are homogeneous in all directions C. Property vary by location D. All of the above
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SOLUTION 7: The TOPIC of ISOTROPIC SUBSTANCES is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.
An ISOTROPIC SUBSTANCE is a SUBSTANCE that has the SAME PHYSICAL PROPERTIES in ALL DIRECTIONS.
Therefore, the correct answer choice is B. Properties are homogeneous in all directions
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PROBLEM 8: The element tin (Sn) has 10 stable isotopes. The atomic nuclei of all ten isotopes have the same: A. Number of Neutrons B. Mass C. Number of Protons D. Number of Electrons
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SOLUTION 8: The TOPIC of ISOTOPES is not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook. ISOTOPES are ATOMS of the same ELEMENT with the SAME NUMBER of PROTONS, but a different ATOMIC MASS NUMBER resulting from a DIFFERENCE in the varying number of NEUTRONS. Answer choice A is INCORRECT as ISOTOPES have EXTRA NEUTRONS. Answer choice B is INCORRECT as NEUTRONS have the SAME MASS and PROTONS. Answer choice D is INCORRECT as ELECTRONS are not PART of the NUCLEUS
Therefore, the correct answer choice is C. Number of Protons
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