Ch.11 - Liquids, Solids & Intermolecular ForcesWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: Calculate the heat energy (kJ) released when 23.6 g of liquid mercury at 25.00°C is converted to solid mercury at its melting point.

Solution: Calculate the heat energy (kJ) released when 23.6 g of liquid mercury at 25.00°C is converted to solid mercury at its melting point.

Problem

Calculate the heat energy (kJ) released when 23.6 g of liquid mercury at 25.00°C is converted to solid mercury at its melting point.

Solution

We’re being asked to calculate the amount of energy released when 23.6 g of mercury at 25.00 ˚C is converted to a solid at its melting point (234.32 K or –38.68 ˚C).


Note that cooling and freezing is an exothermic process, which means q is negative (–)

There are two heats involved in this problem:

1. q1 which is the heat in cooling 23.6 g of Hg from 25 ˚C to –38.68 ˚C

2. q2 which is the heat in freezing 23.6 g of Hg at –38.68 ˚C


We need to solve for each heat individually then add them together to get the final answer.


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