Ch.6 - Thermochemistry WorksheetSee 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: The addition of 3.15 g of Ba(OH) 2∙8H2O to a solution of 1.52 g of NH 4SCN in 100 g of water in a calorimeter caused the temperature to fall by 3.1°C. Assuming the specific heat of the solution and products is 4.20 J/g°C, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation:Ba(OH)2∙8H2O(s) + 2NH4SCN(aq) ⟶ Ba(SCN)2(aq) + 2NH3(aq) + 10H2O(l)

Solution: The addition of 3.15 g of Ba(OH) 2∙8H2O to a solution of 1.52 g of NH 4SCN in 100 g of water in a calorimeter caused the temperature to fall by 3.1°C. Assuming the specific heat of the solution and pr

Problem

The addition of 3.15 g of Ba(OH) 2∙8H2O to a solution of 1.52 g of NH 4SCN in 100 g of water in a calorimeter caused the temperature to fall by 3.1°C. Assuming the specific heat of the solution and products is 4.20 J/g°C, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation:

Ba(OH)2∙8H2O(s) + 2NH4SCN(aq) ⟶ Ba(SCN)2(aq) + 2NH3(aq) + 10H2O(l)