CHEM 103 CUMULATIVE FINAL EXAM MODULES 1-6
2026/2027 | General Chemistry Foundations | Portage
Learning | 100% Verified Answers | Pass Guaranteed - A+
Graded
Module 1: Measurement, Units & Dimensional Analysis
(Questions 1–18)
Q1. A student measures the length of a metal rod using a metric ruler and records
15.20 cm. How many significant figures are in this measurement?
A. 2
B. 3
C. 4
D. 5
Correct Answer: C
Rationale: The measurement 15.20 cm contains four significant figures: the digits 1,
5, and 2 are always significant, and the trailing zero after the decimal point is
significant because it indicates precision to the hundredths place. Students often
mistakenly drop the trailing zero, thinking it is not significant. [CORRECT]
Q2. Convert 2.45 × 10⁻³ meters to millimeters.
A. 2.45 × 10⁻⁶ mm
B. 2.45 × 10⁻¹ mm
C. 2.45 mm
D. 24.5 mm
Correct Answer: C
Rationale: Since 1 m = 10³ mm, multiply 2.45 × 10⁻³ m by 10³ mm/m: (2.45 × 10⁻³) ×
,2
10³ = 2.45 mm. The exponents cancel to give 10⁰. A common error is dividing by 10³
instead of multiplying, yielding 2.45 × 10⁻⁶. [CORRECT]
Q3. The density of a liquid is measured as 0.875 g/mL. What is the mass of 250.0 mL
of this liquid?
A. 218.8 g
B. 285.7 g
C. 0.00350 g
D. 875 g
Correct Answer: A
Rationale: Using d = m/V, rearrange to m = d × V = 0.875 g/mL × 250.0 mL =
218.75 g, which rounds to 218.8 g with four significant figures (matching 250.0).
Choice B results from dividing instead of multiplying; choice C inverts the
relationship entirely. [CORRECT]
Q4. A temperature reading on a Fahrenheit thermometer is 98.6 °F. What is this
temperature in Kelvin?
A. 310.2 K
B. 37.0 K
C. 273.15 K
D. 371.1 K
Correct Answer: A
Rationale: First convert °F to °C: °C = (98.6 − 32)/1.8 = 37.0 °C. Then convert to
Kelvin: K = 37.0 + 273.15 = 310.15 K ≈ 310.2 K. Choice B incorrectly treats 37.0 as
Kelvin; choice D adds 273.15 to 98.6 directly. [CORRECT]
Q5. Express the number 0.004070 in proper scientific notation.
,3
A. 4.070 × 10⁻³
B. 4.07 × 10⁻³
C. 4.070 × 10³
D. 4070 × 10⁻⁶
Correct Answer: A
Rationale: The coefficient must be between 1 and 10, and all significant figures must
be preserved. 0.004070 has four significant figures (the trailing zero after the decimal
is significant), so 4.070 × 10⁻³ is correct. Choice B drops the significant trailing zero;
choice D is not proper scientific notation. [CORRECT]
Q6. Perform the following calculation and report the answer with the correct number
of significant figures: (3.42 × 5.1) / 2.876
A. 6.07
B. 6.1
C. 6
D. 6.066
Correct Answer: B
Rationale: For multiplication and division, the result carries the same number of
significant figures as the factor with the fewest significant figures. Here, 5.1 has two
significant figures (the least), so the result 6.066… rounds to 6.1. Choice A incorrectly
retains three sig figs; choice C rounds too aggressively. [CORRECT]
Q7. A graduated cylinder contains 45.6 mL of water. After a metal sample is added,
the volume reads 62.3 mL. If the metal has a mass of 150.2 g, what is its density?
A. 8.97 g/mL
B. 2.66 g/mL
C. 0.112 g/mL
D. 3.30 g/mL
, 4
Correct Answer: A
Rationale: Volume of metal = 62.3 mL − 45.6 mL = 16.7 mL. Density = mass/volume
= 150.2 g / 16.7 mL = 8.994 g/mL ≈ 8.97 g/mL (three significant figures, limited by
the subtraction result 16.7). Choice B uses the final volume instead of the displaced
volume. [CORRECT]
Q8. How many nanoseconds are in 3.5 milliseconds?
A. 3.5 × 10³ ns
B. 3.5 × 10⁶ ns
C. 3.5 × 10⁻⁶ ns
D. 3.5 × 10⁻³ ns
Correct Answer: B
Rationale: 1 ms = 10⁻³ s and 1 ns = 10⁻⁹ s, so 1 ms = 10⁶ ns. Therefore, 3.5 ms = 3.5
× 10⁶ ns. Alternatively: 3.5 ms × (10⁻³ s / 1 ms) × (1 ns / 10⁻⁹ s) = 3.5 × 10⁶ ns. Choice
A confuses micro- and nano- prefixes. [CORRECT]
Q9. The accepted value for the density of copper is 8.96 g/mL. A student
experimentally measures 8.72 g/mL. What is the percent error?
A. 2.68%
B. 2.75%
C. −2.68%
D. 24%
Correct Answer: A
Rationale: Percent error = |experimental − theoretical| / theoretical × 100% = |8.72
− 8.96| / 8.96 × 100% = 0..96 × 100% = 2.678…% ≈ 2.68%. The absolute value
makes the result positive. Choice B uses the experimental value in the denominator.
[CORRECT]