CHM 101 Exam 1 — Donovan
,Table of Contents
1. Measurements & Metric Conversions
2. Scientific Notation & Significant Figures
3. Density Calculations
4. Heat, Energy & Temperature
5. Matter & Chemical Changes
6. Elements, Compounds & Mixtures
7. Atomic Structure
8. Isotopes & Atomic Mass
9. Periodic Table & Periodic Trends
10. Ions & Ionic Charges
11. Chemical Formulas & Nomenclature
12. Mixed Review Questions
13. Answers & Rationales
CHM 101 Exam 1 — Donovan
1. A laboratory bottle contains 4.72 kg of sodium chloride. What is this mass in
grams?
A. 472 g
B. 4720 g
C. 47,200 g
D. 0.00472 g
Answer: B
Rationale: One kilogram equals 1000 grams.
1000 g
4.72 kg × = 4720 g
1 kg
The conversion factor is exact, so the original three significant figures are
retained.
,2. During a microscale experiment, 0.00635 L of solution is used. How many
microliters is this?
A. 6.35 µL
B. 63.5 µL
C. 635 µL
D. 6350 µL
Answer: D
Rationale: 1 𝐿 = 106 𝜇𝐿.
0.00635 × 106 = 6350 𝜇𝐿
The measured quantity contains three significant figures.
3. Convert 8.40 × 105 picometers to millimeters.
A. 8.40 × 10−4 mm
B. 8.40 × 10−7 mm
C. 8.40 × 10−9 mm
D. 8.40 × 102 mm
Answer: A
Rationale: A picometer is 10−12 m, while a millimeter is 10−3 m. Therefore,
1 pm = 10−9 mm
and
8.40 × 105 × 10−9 = 8.40 × 10−4 mm
4. A chemical sample has a mass of 0.285 mg. Express this quantity in nanograms.
A. 285 ng
B. 28,500 ng
C. 285,000 ng
D. 285,000,000 ng
Answer: C
Rationale: One milligram contains 106 nanograms.
0.285 mg × 106 = 285,000 ng
,5. How many significant figures are present in 0.004060?
A. 3
B. 4
C. 5
D. 6
Answer: B
Rationale: The zeros preceding 4 are placeholders and are not significant. The
zero between 4 and 6 is significant, and the final zero to the right of the decimal
is also significant.
Significant digits are:
4, 0, 6, 0
Therefore, there are 4 significant figures.
6. Evaluate the following measurement calculation using the proper decimal-
place rule:
12.36 + 0.8 + 4.127
A. 17.3
B. 17.29
C. 17.287
D. 17
Answer: A
Rationale: The unrounded sum is:
17.287
For addition and subtraction, the answer is rounded according to the
measurement with the fewest decimal places. Because 0.8 has one decimal
place, the result becomes:
17.3
7. A student calculates 5.62 × 3.4. What answer should be reported using the
correct number of significant figures?
A. 19.108
B. 19.11
,C. 19.1
D. 19
Answer: D
Rationale: The calculator result is:
5.62 × 3.4 = 19.108
The value 3.4 contains only 2 significant figures, so the final answer must contain
two significant figures:
19
8. Calculate
22.40 − 1.6
4.25
using appropriate significant-figure rules.
A. 4.8941
B. 4.9
C. 4.89
D. 5.00
Answer: C
Rationale: First perform subtraction:
22.40 − 1.6 = 20.8
The subtraction result is limited to one decimal place. Then:
20.8 ÷ 4.25 = 4.8941. ..
Both usable values contain three significant figures, so:
4.89
9. Which number represents 7.5 × 10−6 in standard decimal notation?
A. 0.000075
B. 0.0000075
C. 0.00000075
D. 7500000
Answer: B
,Rationale: A negative exponent means moving the decimal point to the left six
places:
7.5 × 10−6 = 0.0000075
10. Which scientific notation correctly represents 0.0008040?
A. 8.04 × 10−3
B. 8.4 × 10−4
C. 8.040 × 10−5
D. 8.040 × 10−4
Answer: D
Rationale: Moving the decimal four places to the right gives 8.040, so the
exponent is −4:
0.0008040 = 8.040 × 10−4
The final zero is retained because the original measurement contains four
significant figures.
11. Using 1 in = 2.54 cm, convert 8.00 inches to centimeters.
A. 20.3 cm
B. 20.32 cm
C. 203 cm
D. 3.15 cm
Answer: A
Rationale:
2.54 cm
8.00 in × = 20.32 cm
1 in
The conversion factor is exact, while 8.00 has three significant figures. Therefore:
20.3 cm
12. A road distance is 72.0 km. Using 1 km = 0.6214 mi, what is the equivalent
distance?
A. 115.9 mi
B. 44.74 mi
,C. 44.7 mi
D. 4.47 mi
Answer: C
Rationale:
72.0 × 0.6214 = 44.7408
The original measurement has three significant figures, so the answer is:
44.7 mi
13. A mineral sample has a mass of 18.6 g and occupies 6.2 cm³. Determine its
density.
A. 2.8 g/cm³
B. 3.0 g/cm³
C. 3.00 g/cm³
D. 115 g/cm³
Answer: B
Rationale: Density is:
𝑚
𝑑=
𝑉
Thus:
18.6
𝑑= = 3.0 g/cm3
6.2
Because 6.2 contains two significant figures, the answer must contain two
significant figures.
14. Copper has a density of 8.96 g/cm³. What is the mass of a copper sample
with a volume of 12.5 cm³?
A. 0.717 g
B. 71.7 g
C. 896 g
D. 112 g
Answer: D
Rationale: Rearrange the density relationship:
, 𝑚 = 𝑑𝑉
𝑚 = (8.96)(12.5) = 112.0 g
Both measurements contain three significant figures, giving:
112 g
15. A 14.58-g metal object raises the water level in a graduated cylinder from
31.4 mL to 36.8 mL. What is its density?
A. 2.7 g/mL
B. 0.37 g/mL
C. 5.4 g/mL
D. 78.7 g/mL
Answer: A
Rationale: First determine the object's volume:
36.8 − 31.4 = 5.4 mL
Then:
14.58
𝑑= = 2.7 g/mL
5.4
The volume has two significant figures, so density is reported with two significant
figures.
16. A laboratory thermometer reads 24°C. Approximately what is the
temperature in kelvins?
A. 249 K
B. 273 K
C. 297 K
D. 321 K
Answer: C
Rationale: Use:
𝐾 =∘ 𝐶 + 273.15
24 + 273.15 = 297.15 K
For this introductory calculation, the appropriate reported value is approximately:
297 K
,17. A patient's environment is maintained at 86°F. What is this temperature in
degrees Celsius?
A. 19°C
B. 30°C
C. 47°C
D. 118°C
Answer: B
Rationale: Use:
∘ ∘ 5
𝐶 = ( 𝐹 − 32)
9
5
(86 − 32) = 30∘ 𝐶
9
18. How much heat is required to raise the temperature of 125 g of water by
18.0°C? Use 𝑐 = 4.184 J/g°C.
A. 2.91 × 102 J
B. 2.35 × 103 J
C. 5.23 × 103 J
D. 9.41 × 103 J
Answer: D
Rationale: Apply:
𝑞 = 𝑚𝑐Δ𝑇
𝑞 = (125)(4.184)(18.0)
𝑞 = 9414 J
With three significant figures:
9.41 × 103 J
19. A quantity of water absorbs 5.23 × 103 J as its temperature rises by 12.5°C.
What mass of water was heated?
Use 𝑐 = 4.184 J/g°C.
A. 100 g
B. 43.8 g
, C. 156 g
D. 523 g
Answer: A
Rationale: Rearrange:
𝑞
𝑚=
𝑐Δ𝑇
5.23 × 103
𝑚=
(4.184)(12.5)
𝑚 = 100 g
20. During an experiment, electrical energy is converted into thermal energy.
Which principle explains why the total amount of energy remains constant?
A. Law of definite proportions
B. Law of conservation of mass
C. Law of conservation of energy
D. Periodic law
Answer: C
Rationale: The law of conservation of energy states that energy cannot be
created or destroyed; it can only be transferred or transformed from one form to
another.
21.
While calibrating a balance, a student records the mass of a sample as 0.07050 g.
How many significant figures are contained in this measurement?
A. 2
B. 3
C. 4
D. 5
Answer: C
Rationale: Leading zeros are not significant because they only locate the decimal
point. The zero between 7 and 5 and the final zero after 5 are significant.
0.07050 → 7, 0, 5, 0
Therefore, the measurement contains 4 significant figures.
,Table of Contents
1. Measurements & Metric Conversions
2. Scientific Notation & Significant Figures
3. Density Calculations
4. Heat, Energy & Temperature
5. Matter & Chemical Changes
6. Elements, Compounds & Mixtures
7. Atomic Structure
8. Isotopes & Atomic Mass
9. Periodic Table & Periodic Trends
10. Ions & Ionic Charges
11. Chemical Formulas & Nomenclature
12. Mixed Review Questions
13. Answers & Rationales
CHM 101 Exam 1 — Donovan
1. A laboratory bottle contains 4.72 kg of sodium chloride. What is this mass in
grams?
A. 472 g
B. 4720 g
C. 47,200 g
D. 0.00472 g
Answer: B
Rationale: One kilogram equals 1000 grams.
1000 g
4.72 kg × = 4720 g
1 kg
The conversion factor is exact, so the original three significant figures are
retained.
,2. During a microscale experiment, 0.00635 L of solution is used. How many
microliters is this?
A. 6.35 µL
B. 63.5 µL
C. 635 µL
D. 6350 µL
Answer: D
Rationale: 1 𝐿 = 106 𝜇𝐿.
0.00635 × 106 = 6350 𝜇𝐿
The measured quantity contains three significant figures.
3. Convert 8.40 × 105 picometers to millimeters.
A. 8.40 × 10−4 mm
B. 8.40 × 10−7 mm
C. 8.40 × 10−9 mm
D. 8.40 × 102 mm
Answer: A
Rationale: A picometer is 10−12 m, while a millimeter is 10−3 m. Therefore,
1 pm = 10−9 mm
and
8.40 × 105 × 10−9 = 8.40 × 10−4 mm
4. A chemical sample has a mass of 0.285 mg. Express this quantity in nanograms.
A. 285 ng
B. 28,500 ng
C. 285,000 ng
D. 285,000,000 ng
Answer: C
Rationale: One milligram contains 106 nanograms.
0.285 mg × 106 = 285,000 ng
,5. How many significant figures are present in 0.004060?
A. 3
B. 4
C. 5
D. 6
Answer: B
Rationale: The zeros preceding 4 are placeholders and are not significant. The
zero between 4 and 6 is significant, and the final zero to the right of the decimal
is also significant.
Significant digits are:
4, 0, 6, 0
Therefore, there are 4 significant figures.
6. Evaluate the following measurement calculation using the proper decimal-
place rule:
12.36 + 0.8 + 4.127
A. 17.3
B. 17.29
C. 17.287
D. 17
Answer: A
Rationale: The unrounded sum is:
17.287
For addition and subtraction, the answer is rounded according to the
measurement with the fewest decimal places. Because 0.8 has one decimal
place, the result becomes:
17.3
7. A student calculates 5.62 × 3.4. What answer should be reported using the
correct number of significant figures?
A. 19.108
B. 19.11
,C. 19.1
D. 19
Answer: D
Rationale: The calculator result is:
5.62 × 3.4 = 19.108
The value 3.4 contains only 2 significant figures, so the final answer must contain
two significant figures:
19
8. Calculate
22.40 − 1.6
4.25
using appropriate significant-figure rules.
A. 4.8941
B. 4.9
C. 4.89
D. 5.00
Answer: C
Rationale: First perform subtraction:
22.40 − 1.6 = 20.8
The subtraction result is limited to one decimal place. Then:
20.8 ÷ 4.25 = 4.8941. ..
Both usable values contain three significant figures, so:
4.89
9. Which number represents 7.5 × 10−6 in standard decimal notation?
A. 0.000075
B. 0.0000075
C. 0.00000075
D. 7500000
Answer: B
,Rationale: A negative exponent means moving the decimal point to the left six
places:
7.5 × 10−6 = 0.0000075
10. Which scientific notation correctly represents 0.0008040?
A. 8.04 × 10−3
B. 8.4 × 10−4
C. 8.040 × 10−5
D. 8.040 × 10−4
Answer: D
Rationale: Moving the decimal four places to the right gives 8.040, so the
exponent is −4:
0.0008040 = 8.040 × 10−4
The final zero is retained because the original measurement contains four
significant figures.
11. Using 1 in = 2.54 cm, convert 8.00 inches to centimeters.
A. 20.3 cm
B. 20.32 cm
C. 203 cm
D. 3.15 cm
Answer: A
Rationale:
2.54 cm
8.00 in × = 20.32 cm
1 in
The conversion factor is exact, while 8.00 has three significant figures. Therefore:
20.3 cm
12. A road distance is 72.0 km. Using 1 km = 0.6214 mi, what is the equivalent
distance?
A. 115.9 mi
B. 44.74 mi
,C. 44.7 mi
D. 4.47 mi
Answer: C
Rationale:
72.0 × 0.6214 = 44.7408
The original measurement has three significant figures, so the answer is:
44.7 mi
13. A mineral sample has a mass of 18.6 g and occupies 6.2 cm³. Determine its
density.
A. 2.8 g/cm³
B. 3.0 g/cm³
C. 3.00 g/cm³
D. 115 g/cm³
Answer: B
Rationale: Density is:
𝑚
𝑑=
𝑉
Thus:
18.6
𝑑= = 3.0 g/cm3
6.2
Because 6.2 contains two significant figures, the answer must contain two
significant figures.
14. Copper has a density of 8.96 g/cm³. What is the mass of a copper sample
with a volume of 12.5 cm³?
A. 0.717 g
B. 71.7 g
C. 896 g
D. 112 g
Answer: D
Rationale: Rearrange the density relationship:
, 𝑚 = 𝑑𝑉
𝑚 = (8.96)(12.5) = 112.0 g
Both measurements contain three significant figures, giving:
112 g
15. A 14.58-g metal object raises the water level in a graduated cylinder from
31.4 mL to 36.8 mL. What is its density?
A. 2.7 g/mL
B. 0.37 g/mL
C. 5.4 g/mL
D. 78.7 g/mL
Answer: A
Rationale: First determine the object's volume:
36.8 − 31.4 = 5.4 mL
Then:
14.58
𝑑= = 2.7 g/mL
5.4
The volume has two significant figures, so density is reported with two significant
figures.
16. A laboratory thermometer reads 24°C. Approximately what is the
temperature in kelvins?
A. 249 K
B. 273 K
C. 297 K
D. 321 K
Answer: C
Rationale: Use:
𝐾 =∘ 𝐶 + 273.15
24 + 273.15 = 297.15 K
For this introductory calculation, the appropriate reported value is approximately:
297 K
,17. A patient's environment is maintained at 86°F. What is this temperature in
degrees Celsius?
A. 19°C
B. 30°C
C. 47°C
D. 118°C
Answer: B
Rationale: Use:
∘ ∘ 5
𝐶 = ( 𝐹 − 32)
9
5
(86 − 32) = 30∘ 𝐶
9
18. How much heat is required to raise the temperature of 125 g of water by
18.0°C? Use 𝑐 = 4.184 J/g°C.
A. 2.91 × 102 J
B. 2.35 × 103 J
C. 5.23 × 103 J
D. 9.41 × 103 J
Answer: D
Rationale: Apply:
𝑞 = 𝑚𝑐Δ𝑇
𝑞 = (125)(4.184)(18.0)
𝑞 = 9414 J
With three significant figures:
9.41 × 103 J
19. A quantity of water absorbs 5.23 × 103 J as its temperature rises by 12.5°C.
What mass of water was heated?
Use 𝑐 = 4.184 J/g°C.
A. 100 g
B. 43.8 g
, C. 156 g
D. 523 g
Answer: A
Rationale: Rearrange:
𝑞
𝑚=
𝑐Δ𝑇
5.23 × 103
𝑚=
(4.184)(12.5)
𝑚 = 100 g
20. During an experiment, electrical energy is converted into thermal energy.
Which principle explains why the total amount of energy remains constant?
A. Law of definite proportions
B. Law of conservation of mass
C. Law of conservation of energy
D. Periodic law
Answer: C
Rationale: The law of conservation of energy states that energy cannot be
created or destroyed; it can only be transferred or transformed from one form to
another.
21.
While calibrating a balance, a student records the mass of a sample as 0.07050 g.
How many significant figures are contained in this measurement?
A. 2
B. 3
C. 4
D. 5
Answer: C
Rationale: Leading zeros are not significant because they only locate the decimal
point. The zero between 7 and 5 and the final zero after 5 are significant.
0.07050 → 7, 0, 5, 0
Therefore, the measurement contains 4 significant figures.
