Chem 103 Module 1 to 6 Exam & Final Exam Portage learning (Latest 2026
/ 2027) Most Comprehensive Qs & Ans - to Pass the Exam, 100% Verified |
INSTANT PDF DOWNLOAD
⭐⭐⭐⭐⭐ "Complete and thorough – all 6 modules"
– Sarah K., Pre-Med
"This is the ultimate CHEM 103 exam prep resource. It covers Modules 1 through 6
completely, including measurement conversions, atomic structure, thermochemistry, gas
laws, bonding, and colligative properties. Each question has a detailed solution showing
every step. I referenced this for every exam and finished the course with an A-."
Document Information
Course: CHEM 103 – General Chemistry
Institution: Portage Learning
Modules Covered: 1, 2, 3, 4, 5, 6
Academic Year: 2026/2027
Content: Exam questions with verified answers and detailed explanations
Format:Instant PDF Download
Module Rating Number of
Reviews
Module 1: Measurement & Conversions ⭐⭐⭐⭐⭐ 65+
(4.9)
Module 2: Stoichiometry & Reactions ⭐⭐⭐⭐⭐ 72+
(5.0)
Module 3: Thermochemistry & Gas Laws ⭐⭐⭐⭐⭐ 88+
(4.9)
Module 4: Atomic Structure & Trends ⭐⭐⭐⭐⭐ 54+
(4.8)
, Module 5: Bonding & Molecular ⭐⭐⭐⭐⭐ 61+
Structure (4.9)
Module 6: Solutions & Colligative ⭐⭐⭐⭐⭐ 79+
Properties (5.0)
MODULE 1 EXAM: Measurement, Conversions & Significant
Figures
Module 1 Overview
Module 1 covers fundamental concepts including scientific notation, unit conversions,
temperature conversions, density calculations, significant figures, classification of matter,
chemical nomenclature, and basic atomic structure .
Question 1: Scientific Notation
Convert 845.3 to exponential form and explain your answer.
Answer: 845.3 = 8.453 × 10²
Explanation: Since 845.3 is larger than 1, the exponent is positive. Moving the decimal 2
places to the left gives 8.453, requiring multiplication by 10² .
Question 2: Ordinary Form
Convert 3.21 × 10⁻⁵ to ordinary form and explain your answer.
Answer: 3.21 × 10⁻⁵ = 0.0000321
,Explanation: The negative exponent indicates the number is smaller than 1. Moving the
decimal 5 places to the left yields 0.0000321 .
Question 3: Unit Conversions
Using the following information, perform the conversions below, showing all
work:
Conversion Factors
1 ft = 12 inches
1 pound = 16 oz
1 gallon = 4 quarts
1 mile = 5280 feet
1 ton = 2000 pounds
1 quart = 2 pints
kilo = 1000
centi = 1/100
milli = 1/1000
deci = 1/10
3a. 24.6 grams = ? kg
Answer: 24.6 g × (1 kg / 1000 g) = 0.0246 kg
3b. 6.3 ft = ? inches
, Answer: 6.3 ft × (12 in / 1 ft) = 75.6 inches
Question 4: Temperature Conversions
Perform the following temperature conversions, showing all work:
4a. 28°C = ? K
Answer: 28°C + 273 = 301 K
4b. 158°F = ? °C
Answer: (158°F - 32) ÷ 1.8 = 70°C
4c. 343 K = ? °F
Answer: (343 K - 273) = 70°C; (70°C × 1.8) + 32 = 158°F
Question 5: Density Calculations
Be sure to show the correct number of significant figures in each calculation.
5a. Calculate the mass of an 18.6 mL sample of freon with a density of 1.49 g/mL.
Answer: M = D × V = 1.49 g/mL × 18.6 mL = 27.7 g
5b. Calculate the density of crude oil if 26.3 g occupies 30.5 mL.
Answer: D = M / V = 26.3 g / 30.5 mL = 0.862 g/mL
Question 6: Significant Figures
6a. How many significant figures does 3.0600 contain?
/ 2027) Most Comprehensive Qs & Ans - to Pass the Exam, 100% Verified |
INSTANT PDF DOWNLOAD
⭐⭐⭐⭐⭐ "Complete and thorough – all 6 modules"
– Sarah K., Pre-Med
"This is the ultimate CHEM 103 exam prep resource. It covers Modules 1 through 6
completely, including measurement conversions, atomic structure, thermochemistry, gas
laws, bonding, and colligative properties. Each question has a detailed solution showing
every step. I referenced this for every exam and finished the course with an A-."
Document Information
Course: CHEM 103 – General Chemistry
Institution: Portage Learning
Modules Covered: 1, 2, 3, 4, 5, 6
Academic Year: 2026/2027
Content: Exam questions with verified answers and detailed explanations
Format:Instant PDF Download
Module Rating Number of
Reviews
Module 1: Measurement & Conversions ⭐⭐⭐⭐⭐ 65+
(4.9)
Module 2: Stoichiometry & Reactions ⭐⭐⭐⭐⭐ 72+
(5.0)
Module 3: Thermochemistry & Gas Laws ⭐⭐⭐⭐⭐ 88+
(4.9)
Module 4: Atomic Structure & Trends ⭐⭐⭐⭐⭐ 54+
(4.8)
, Module 5: Bonding & Molecular ⭐⭐⭐⭐⭐ 61+
Structure (4.9)
Module 6: Solutions & Colligative ⭐⭐⭐⭐⭐ 79+
Properties (5.0)
MODULE 1 EXAM: Measurement, Conversions & Significant
Figures
Module 1 Overview
Module 1 covers fundamental concepts including scientific notation, unit conversions,
temperature conversions, density calculations, significant figures, classification of matter,
chemical nomenclature, and basic atomic structure .
Question 1: Scientific Notation
Convert 845.3 to exponential form and explain your answer.
Answer: 845.3 = 8.453 × 10²
Explanation: Since 845.3 is larger than 1, the exponent is positive. Moving the decimal 2
places to the left gives 8.453, requiring multiplication by 10² .
Question 2: Ordinary Form
Convert 3.21 × 10⁻⁵ to ordinary form and explain your answer.
Answer: 3.21 × 10⁻⁵ = 0.0000321
,Explanation: The negative exponent indicates the number is smaller than 1. Moving the
decimal 5 places to the left yields 0.0000321 .
Question 3: Unit Conversions
Using the following information, perform the conversions below, showing all
work:
Conversion Factors
1 ft = 12 inches
1 pound = 16 oz
1 gallon = 4 quarts
1 mile = 5280 feet
1 ton = 2000 pounds
1 quart = 2 pints
kilo = 1000
centi = 1/100
milli = 1/1000
deci = 1/10
3a. 24.6 grams = ? kg
Answer: 24.6 g × (1 kg / 1000 g) = 0.0246 kg
3b. 6.3 ft = ? inches
, Answer: 6.3 ft × (12 in / 1 ft) = 75.6 inches
Question 4: Temperature Conversions
Perform the following temperature conversions, showing all work:
4a. 28°C = ? K
Answer: 28°C + 273 = 301 K
4b. 158°F = ? °C
Answer: (158°F - 32) ÷ 1.8 = 70°C
4c. 343 K = ? °F
Answer: (343 K - 273) = 70°C; (70°C × 1.8) + 32 = 158°F
Question 5: Density Calculations
Be sure to show the correct number of significant figures in each calculation.
5a. Calculate the mass of an 18.6 mL sample of freon with a density of 1.49 g/mL.
Answer: M = D × V = 1.49 g/mL × 18.6 mL = 27.7 g
5b. Calculate the density of crude oil if 26.3 g occupies 30.5 mL.
Answer: D = M / V = 26.3 g / 30.5 mL = 0.862 g/mL
Question 6: Significant Figures
6a. How many significant figures does 3.0600 contain?
