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CHM 101 EXAM 1 – DONOVAN 2026/2027 COMPLETE (100)
CURRENT TESTING QUESTIONS AND CORRECT
ANSWERS WITH DETAILED RATIONALES.
CHM 101
Prepare for CHM 101 Exam 1 – Donovan with a focused study resource covering
foundational chemistry concepts and essential principles introduced early in the
course. It is designed to support review of key topics, reinforce understanding, and
help you recognize areas that may require additional study. Use the practice material
to structure your preparation and build confidence before the exam. This resource is
best suited for CHM 101 students preparing for Exam 1 and reviewing introductory
chemistry concepts.
MULTIPLE CHOICE.
1. Convert 3.59 kg into grams.
• A) 0.00359 g
• B) 35.9 g
• C) 359 g
• D) 3590 g
Answer: D
Rationale: 1 kg = 1000 g, so 3.59 kg × 1000 = 3590 g. Option A is incorrect
because it divides instead of multiplies. Option B is off by a factor of 100.
Option C is off by a factor of 10.
2. Convert 2000.4 mm to kilometers.
• A) 0.0020004 km
• B) 2.0004 km
• C) 2000.4 km
• D) 2.0004 × 10⁶ km
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Answer: A
Rationale: 1 km = 1,000,000 mm, so 2000.4 mm ÷ 1,000,000 = 0.0020004
km. Option B uses meters (not kilometers). Option C ignores the conversion
entirely. Option D multiplies instead of dividing.
3. Convert 62 picometers to millimeters.
• A) 6.2 × 10⁻⁸ mm
• B) 6.2 × 10⁻⁶ mm
• C) 6.2 × 10⁻¹⁰ mm
• D) 6.2 × 10⁻⁵ mm
Answer: A
Rationale: 1 pm = 10⁻¹² m and 1 mm = 10⁻³ m, so 62 pm = 62 × 10⁻¹² m = 6.2
× 10⁻¹¹ m = 6.2 × 10⁻⁸ mm. Option B is off by 100. Option C is off by a factor of
100. Option D is off by 1000.
4. Convert 5.45 micrograms to femtograms.
• A) 5.45 × 10⁶ fg
• B) 5.45 × 10⁹ fg
• C) 5.45 × 10³ fg
• D) 5.45 × 10¹² fg
Answer: B
Rationale: 1 μg = 10⁻⁶ g and 1 fg = 10⁻¹⁵ g, so 1 μg = 10⁹ fg. 5.45 μg × 10⁹ =
5.45 × 10⁹ fg. Option A is off by a factor of 1000. Option C is off by a factor of
10⁶. Option D is off by a factor of 1000.
5. Convert 35 centigrams to grams.
• A) 0.035 g
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• B) 0.35 g
• C) 3.5 g
• D) 350 g
Answer: B
Rationale: 1 cg = 0.01 g, so 35 cg × 0.01 = 0.35 g. Option A is off by a factor of
10. Option C is off by a factor of 10. Option D multiplies instead of dividing.
6. Convert 0.067 microliters to liters.
• A) 6.7 × 10⁻⁸ L
• B) 6.7 × 10⁻⁷ L
• C) 6.7 × 10⁻⁶ L
• D) 6.7 × 10⁻⁵ L
Answer: A
Rationale: 1 μL = 10⁻⁶ L, so 0.067 μL = 0.067 × 10⁻⁶ L = 6.7 × 10⁻⁸ L. Option B
is off by a factor of 10. Option C is off by a factor of 100. Option D is off by a
factor of 1000.
7. How many significant figures are in the number 100?
• A) 1
• B) 2
• C) 3
• D) Ambiguous without a decimal point
Answer: A
Rationale: Trailing zeros in a whole number without a decimal point are
not significant. Only the "1" is significant, so 100 has 1 significant
figure. Option B is incorrect because trailing zeros are not counted. Option C
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would be correct only if written as "100." with a decimal point. Option D is
partially correct but the best answer is 1.
8. How many significant figures are in the number 0.00090?
• A) 1
• B) 2
• C) 4
• D) 5
Answer: B
Rationale: Leading zeros are not significant. The digits "9" and the trailing
zero after the decimal are significant, so 0.00090 has 2 significant
figures. Option A misses the trailing zero. Option C counts leading zeros
incorrectly. Option D counts all zeros.
9. How many significant figures are in the number 1.000803?
• A) 4
• B) 5
• C) 6
• D) 7
Answer: D
Rationale: All digits in 1.000803 are significant because zeros between
non-zero digits and trailing zeros after a decimal are significant. Count: 1,
0, 0, 0, 8, 0, 3 = 7 significant figures. Option C misses one zero. Options A
and B undercount.
10. How many significant figures are in the number 0.0000000001?
• A) 1
CHM 101 EXAM 1 – DONOVAN 2026/2027 COMPLETE (100)
CURRENT TESTING QUESTIONS AND CORRECT
ANSWERS WITH DETAILED RATIONALES.
CHM 101
Prepare for CHM 101 Exam 1 – Donovan with a focused study resource covering
foundational chemistry concepts and essential principles introduced early in the
course. It is designed to support review of key topics, reinforce understanding, and
help you recognize areas that may require additional study. Use the practice material
to structure your preparation and build confidence before the exam. This resource is
best suited for CHM 101 students preparing for Exam 1 and reviewing introductory
chemistry concepts.
MULTIPLE CHOICE.
1. Convert 3.59 kg into grams.
• A) 0.00359 g
• B) 35.9 g
• C) 359 g
• D) 3590 g
Answer: D
Rationale: 1 kg = 1000 g, so 3.59 kg × 1000 = 3590 g. Option A is incorrect
because it divides instead of multiplies. Option B is off by a factor of 100.
Option C is off by a factor of 10.
2. Convert 2000.4 mm to kilometers.
• A) 0.0020004 km
• B) 2.0004 km
• C) 2000.4 km
• D) 2.0004 × 10⁶ km
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Answer: A
Rationale: 1 km = 1,000,000 mm, so 2000.4 mm ÷ 1,000,000 = 0.0020004
km. Option B uses meters (not kilometers). Option C ignores the conversion
entirely. Option D multiplies instead of dividing.
3. Convert 62 picometers to millimeters.
• A) 6.2 × 10⁻⁸ mm
• B) 6.2 × 10⁻⁶ mm
• C) 6.2 × 10⁻¹⁰ mm
• D) 6.2 × 10⁻⁵ mm
Answer: A
Rationale: 1 pm = 10⁻¹² m and 1 mm = 10⁻³ m, so 62 pm = 62 × 10⁻¹² m = 6.2
× 10⁻¹¹ m = 6.2 × 10⁻⁸ mm. Option B is off by 100. Option C is off by a factor of
100. Option D is off by 1000.
4. Convert 5.45 micrograms to femtograms.
• A) 5.45 × 10⁶ fg
• B) 5.45 × 10⁹ fg
• C) 5.45 × 10³ fg
• D) 5.45 × 10¹² fg
Answer: B
Rationale: 1 μg = 10⁻⁶ g and 1 fg = 10⁻¹⁵ g, so 1 μg = 10⁹ fg. 5.45 μg × 10⁹ =
5.45 × 10⁹ fg. Option A is off by a factor of 1000. Option C is off by a factor of
10⁶. Option D is off by a factor of 1000.
5. Convert 35 centigrams to grams.
• A) 0.035 g
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• B) 0.35 g
• C) 3.5 g
• D) 350 g
Answer: B
Rationale: 1 cg = 0.01 g, so 35 cg × 0.01 = 0.35 g. Option A is off by a factor of
10. Option C is off by a factor of 10. Option D multiplies instead of dividing.
6. Convert 0.067 microliters to liters.
• A) 6.7 × 10⁻⁸ L
• B) 6.7 × 10⁻⁷ L
• C) 6.7 × 10⁻⁶ L
• D) 6.7 × 10⁻⁵ L
Answer: A
Rationale: 1 μL = 10⁻⁶ L, so 0.067 μL = 0.067 × 10⁻⁶ L = 6.7 × 10⁻⁸ L. Option B
is off by a factor of 10. Option C is off by a factor of 100. Option D is off by a
factor of 1000.
7. How many significant figures are in the number 100?
• A) 1
• B) 2
• C) 3
• D) Ambiguous without a decimal point
Answer: A
Rationale: Trailing zeros in a whole number without a decimal point are
not significant. Only the "1" is significant, so 100 has 1 significant
figure. Option B is incorrect because trailing zeros are not counted. Option C
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would be correct only if written as "100." with a decimal point. Option D is
partially correct but the best answer is 1.
8. How many significant figures are in the number 0.00090?
• A) 1
• B) 2
• C) 4
• D) 5
Answer: B
Rationale: Leading zeros are not significant. The digits "9" and the trailing
zero after the decimal are significant, so 0.00090 has 2 significant
figures. Option A misses the trailing zero. Option C counts leading zeros
incorrectly. Option D counts all zeros.
9. How many significant figures are in the number 1.000803?
• A) 4
• B) 5
• C) 6
• D) 7
Answer: D
Rationale: All digits in 1.000803 are significant because zeros between
non-zero digits and trailing zeros after a decimal are significant. Count: 1,
0, 0, 0, 8, 0, 3 = 7 significant figures. Option C misses one zero. Options A
and B undercount.
10. How many significant figures are in the number 0.0000000001?
• A) 1
