Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 38m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

8. Thermochemistry

Endothermic & Exothermic Reactions

General Chemistry

8. Thermochemistry

Endothermic & Exothermic Reactions

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Multiple Choice

Suppose that 1.08 g of rubbing alcohol (C3H8O) evaporates from a 74.0 g aluminum block. If the aluminum block is initially at 25 °C, what is the final temperature of the block after the evaporation of the alcohol? Assume that the heat required for the evaporation is absorbed entirely from the aluminum block, and the heat of vaporization of rubbing alcohol is 45.4 kJ/mol. The specific heat capacity of aluminum is 0.897 J/g°C.

15.0 °C

25.0 °C

30.0 °C

20.5 °C

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Verified step by step guidance

Calculate the number of moles of rubbing alcohol (C3H8O) that evaporates. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). The molar mass of C3H8O is approximately 60.1 g/mol.

Determine the total heat absorbed by the evaporation of the alcohol using the heat of vaporization. Use the formula: \( q = \text{moles} \times \text{heat of vaporization} \). The heat of vaporization is given as 45.4 kJ/mol, so convert this to J/mol by multiplying by 1000.

Calculate the change in temperature of the aluminum block using the formula: \( q = m \cdot c \cdot \Delta T \), where \( q \) is the heat absorbed, \( m \) is the mass of the aluminum block, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.

Rearrange the formula to solve for \( \Delta T \): \( \Delta T = \frac{q}{m \cdot c} \). Substitute the values for \( q \), \( m \), and \( c \) (specific heat capacity of aluminum is 0.897 J/g°C).

Subtract the change in temperature \( \Delta T \) from the initial temperature of the aluminum block (25 °C) to find the final temperature.