CHEM 121 Final Exam
General Chemistry I
Test Guide Questions and Answers | Grade A | 100%
Aligned with 2026-2027 Portage Learning Curriculum Standards
Comprehensive Final Exam Preparation
Total Questions 100 Multiple Choice
Cognitive Levels 30% Recall | 45% Application | 25% Analysis
Question Format 4 Options (A-D), One Correct Answer
Content Style 65% Scenario-Based | 20% Direct Recall | 15% Calculation
Sections 10 Comprehensive Domains
COMPLETE EXAM - 100 Questions Generated
,CHEM 121 Final Exam | General Chemistry I | Portage Learning 100 Questions | 2026-2027
Section 1: Foundations of Chemistry (Q1-Q12)
Q1: A student measures the mass of a sample as 0.00450 g. How many significant figures are in this
measurement?
A. Two
B. Three [CORRECT]
C. Four
D. Five
Correct Answer: B
Rationale: Leading zeros are not significant; they only indicate the position of the decimal point. The significant figures
are 4, 5, and 0, giving three significant figures. This is a foundational rule in CHEM 121: leading zeros never count,
trailing zeros after a decimal point always count, and all nonzero digits are significant. Understanding this rule is
essential for proper rounding and reporting of calculated results throughout the course.
Q2: Which of the following is an example of an intensive property?
A. Mass
B. Volume
C. Density [CORRECT]
D. Weight
Correct Answer: C
Rationale: Intensive properties are independent of the amount of matter present. Density is the ratio of mass to volume
and remains the same regardless of sample size. Mass, volume, and weight are all extensive properties because they
depend on the quantity of the substance. Portage Learning CHEM 121 emphasizes the distinction between intensive and
extensive properties as a core concept in understanding matter classification and physical vs. chemical properties.
Q3: A student converts 25.0 degrees Celsius to Kelvin. What is the correct result with proper
significant figures?
A. 298 K [CORRECT]
B. 298.0 K
C. 298.15 K
D. 297 K
Correct Answer: A
Rationale: The Kelvin temperature is calculated by adding 273.15 to the Celsius value: 25.0 + 273.15 = 298.15.
However, the original measurement has three significant figures (25.0), and the conversion constant 273.15 is exact, so
the result should be reported to the tenths place: 298 K. In CHEM 121, students learn that when adding or subtracting,
the result is limited by the least precise measurement. Since 25.0 has its last digit in the tenths place, the sum 298.15
rounds to 298.
Q4: A compound is analyzed and found to contain 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by
mass. What is the empirical formula of this compound?
A. CHO
B. C2H4O2
C. CH2O [CORRECT]
D. C2H3O
Correct Answer: C
Rationale: To find the empirical formula, assume a 100 g sample: 40.0 g C, 6.7 g H, 53.3 g O. Convert to moles: C =
40.0/12.01 = 3.33 mol, H = 6.7/1.008 = 6.65 mol, O = 53.3/16.00 = 3.33 mol. Divide by the smallest (3.33): C = 1, H
= 2, O = 1, giving CH2O. This is a classic CHEM 121 stoichiometry problem testing the empirical formula calculation
Page 2
, CHEM 121 Final Exam | General Chemistry I | Portage Learning 100 Questions | 2026-2027
procedure, which requires converting mass percentages to moles and finding the simplest whole-number ratio.
Q5: Which of the following represents a chemical change rather than a physical change?
A. Dissolving sugar in water
B. Melting ice to form liquid water
C. Burning propane gas to produce carbon dioxide and water vapor [CORRECT]
D. Boiling water to produce steam
Correct Answer: C
Rationale: A chemical change involves the transformation of one or more substances into new substances with
different chemical compositions. Burning propane (C3H8 + 5O2 yielding 3CO2 + 4H2O) produces entirely new
compounds. Dissolving, melting, and boiling are physical changes because the chemical identity of the substance
remains the same; only the state or form changes. CHEM 121 emphasizes this distinction as students learn to classify
changes and understand the difference between chemical and physical properties of matter.
Q6: A piece of copper metal has a density of 8.96 g/cm3. If a sample has a mass of 45.2 g, what is its
volume in cm3?
A. 5.04 cm3 [CORRECT]
B. 0.198 cm3
C. 405 cm3
D. 5.05 cm3
Correct Answer: A
Rationale: Volume is calculated by dividing mass by density: V = m/d = 45.2 g / 8.96 g/cm3 = 5.0446... cm3. With
three significant figures in both the mass (45.2) and density (8.96), the answer is rounded to three significant figures:
5.04 cm3. This dimensional analysis problem tests CHEM 121 students on density calculations and proper rounding,
which are fundamental skills applied repeatedly in subsequent units on solutions, stoichiometry, and gas laws.
Q7: Which of the following is a homogeneous mixture?
A. Sand and water
B. Trail mix
C. Air [CORRECT]
D. Oil and water
Correct Answer: C
Rationale: A homogeneous mixture has uniform composition throughout; its components are not visibly distinguishable.
Air is a mixture of nitrogen, oxygen, argon, and other gases in a single uniform phase. Sand and water, trail mix, and
oil and water are all heterogeneous mixtures because their components are visibly distinct or form separate phases.
CHEM 121 covers classification of matter in depth, requiring students to distinguish between elements, compounds, and
mixtures, and further between homogeneous and heterogeneous mixtures.
Q8: Express the number 0.000378 in scientific notation with the correct number of significant figures.
A. 3.78 x 10-4 [CORRECT]
B. 3.78 x 104
C. 3.8 x 10-4
D. 37.8 x 10-5
Correct Answer: A
Rationale: In scientific notation, the coefficient must be between 1 and 10. Moving the decimal point four places to the
right gives 3.78, and the exponent is negative 4. All three digits (3, 7, 8) are significant because they are nonzero. The
correct form is 3.78 x 10-4. Option B has the wrong sign on the exponent, option C incorrectly rounds, and option D has
a coefficient outside the range of 1 to 10. Scientific notation is a critical skill in CHEM 121 for handling very large
and very small numbers encountered in atomic measurements and molar calculations.
Page 3
General Chemistry I
Test Guide Questions and Answers | Grade A | 100%
Aligned with 2026-2027 Portage Learning Curriculum Standards
Comprehensive Final Exam Preparation
Total Questions 100 Multiple Choice
Cognitive Levels 30% Recall | 45% Application | 25% Analysis
Question Format 4 Options (A-D), One Correct Answer
Content Style 65% Scenario-Based | 20% Direct Recall | 15% Calculation
Sections 10 Comprehensive Domains
COMPLETE EXAM - 100 Questions Generated
,CHEM 121 Final Exam | General Chemistry I | Portage Learning 100 Questions | 2026-2027
Section 1: Foundations of Chemistry (Q1-Q12)
Q1: A student measures the mass of a sample as 0.00450 g. How many significant figures are in this
measurement?
A. Two
B. Three [CORRECT]
C. Four
D. Five
Correct Answer: B
Rationale: Leading zeros are not significant; they only indicate the position of the decimal point. The significant figures
are 4, 5, and 0, giving three significant figures. This is a foundational rule in CHEM 121: leading zeros never count,
trailing zeros after a decimal point always count, and all nonzero digits are significant. Understanding this rule is
essential for proper rounding and reporting of calculated results throughout the course.
Q2: Which of the following is an example of an intensive property?
A. Mass
B. Volume
C. Density [CORRECT]
D. Weight
Correct Answer: C
Rationale: Intensive properties are independent of the amount of matter present. Density is the ratio of mass to volume
and remains the same regardless of sample size. Mass, volume, and weight are all extensive properties because they
depend on the quantity of the substance. Portage Learning CHEM 121 emphasizes the distinction between intensive and
extensive properties as a core concept in understanding matter classification and physical vs. chemical properties.
Q3: A student converts 25.0 degrees Celsius to Kelvin. What is the correct result with proper
significant figures?
A. 298 K [CORRECT]
B. 298.0 K
C. 298.15 K
D. 297 K
Correct Answer: A
Rationale: The Kelvin temperature is calculated by adding 273.15 to the Celsius value: 25.0 + 273.15 = 298.15.
However, the original measurement has three significant figures (25.0), and the conversion constant 273.15 is exact, so
the result should be reported to the tenths place: 298 K. In CHEM 121, students learn that when adding or subtracting,
the result is limited by the least precise measurement. Since 25.0 has its last digit in the tenths place, the sum 298.15
rounds to 298.
Q4: A compound is analyzed and found to contain 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by
mass. What is the empirical formula of this compound?
A. CHO
B. C2H4O2
C. CH2O [CORRECT]
D. C2H3O
Correct Answer: C
Rationale: To find the empirical formula, assume a 100 g sample: 40.0 g C, 6.7 g H, 53.3 g O. Convert to moles: C =
40.0/12.01 = 3.33 mol, H = 6.7/1.008 = 6.65 mol, O = 53.3/16.00 = 3.33 mol. Divide by the smallest (3.33): C = 1, H
= 2, O = 1, giving CH2O. This is a classic CHEM 121 stoichiometry problem testing the empirical formula calculation
Page 2
, CHEM 121 Final Exam | General Chemistry I | Portage Learning 100 Questions | 2026-2027
procedure, which requires converting mass percentages to moles and finding the simplest whole-number ratio.
Q5: Which of the following represents a chemical change rather than a physical change?
A. Dissolving sugar in water
B. Melting ice to form liquid water
C. Burning propane gas to produce carbon dioxide and water vapor [CORRECT]
D. Boiling water to produce steam
Correct Answer: C
Rationale: A chemical change involves the transformation of one or more substances into new substances with
different chemical compositions. Burning propane (C3H8 + 5O2 yielding 3CO2 + 4H2O) produces entirely new
compounds. Dissolving, melting, and boiling are physical changes because the chemical identity of the substance
remains the same; only the state or form changes. CHEM 121 emphasizes this distinction as students learn to classify
changes and understand the difference between chemical and physical properties of matter.
Q6: A piece of copper metal has a density of 8.96 g/cm3. If a sample has a mass of 45.2 g, what is its
volume in cm3?
A. 5.04 cm3 [CORRECT]
B. 0.198 cm3
C. 405 cm3
D. 5.05 cm3
Correct Answer: A
Rationale: Volume is calculated by dividing mass by density: V = m/d = 45.2 g / 8.96 g/cm3 = 5.0446... cm3. With
three significant figures in both the mass (45.2) and density (8.96), the answer is rounded to three significant figures:
5.04 cm3. This dimensional analysis problem tests CHEM 121 students on density calculations and proper rounding,
which are fundamental skills applied repeatedly in subsequent units on solutions, stoichiometry, and gas laws.
Q7: Which of the following is a homogeneous mixture?
A. Sand and water
B. Trail mix
C. Air [CORRECT]
D. Oil and water
Correct Answer: C
Rationale: A homogeneous mixture has uniform composition throughout; its components are not visibly distinguishable.
Air is a mixture of nitrogen, oxygen, argon, and other gases in a single uniform phase. Sand and water, trail mix, and
oil and water are all heterogeneous mixtures because their components are visibly distinct or form separate phases.
CHEM 121 covers classification of matter in depth, requiring students to distinguish between elements, compounds, and
mixtures, and further between homogeneous and heterogeneous mixtures.
Q8: Express the number 0.000378 in scientific notation with the correct number of significant figures.
A. 3.78 x 10-4 [CORRECT]
B. 3.78 x 104
C. 3.8 x 10-4
D. 37.8 x 10-5
Correct Answer: A
Rationale: In scientific notation, the coefficient must be between 1 and 10. Moving the decimal point four places to the
right gives 3.78, and the exponent is negative 4. All three digits (3, 7, 8) are significant because they are nonzero. The
correct form is 3.78 x 10-4. Option B has the wrong sign on the exponent, option C incorrectly rounds, and option D has
a coefficient outside the range of 1 to 10. Scientific notation is a critical skill in CHEM 121 for handling very large
and very small numbers encountered in atomic measurements and molar calculations.
Page 3
