Textbook Question
Explain each of the following observations:
c. The boiling point of HF is much higher than those of the other hydrogen halides.
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Ch.22 - Chemistry of the Nonmetals
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232
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Table of contents
- Ch.1 - Introduction: Matter, Energy, and Measurement140
- Ch.2 - Atoms, Molecules, and Ions192
- Ch.3 - Chemical Reactions and Reaction Stoichiometry172
- Ch.4 - Reactions in Aqueous Solution156
- Ch.5 - Thermochemistry151
- Ch.6 - Electronic Structure of Atoms137
- Ch.7 - Periodic Properties of the Elements127
- Ch.8 - Basic Concepts of Chemical Bonding143
- Ch.9 - Molecular Geometry and Bonding Theories167
- Ch.10 - Gases147
- Ch.11 - Liquids and Intermolecular Forces97
- Ch.12 - Solids and Modern Materials129
- Ch.13 - Properties of Solutions126
- Ch.14 - Chemical Kinetics175
- Ch.15 - Chemical Equilibrium107
- Ch.16 - Acid-Base Equilibria127
- Ch.17 - Additional Aspects of Aqueous Equilibria133
- Ch.18 - Chemistry of the Environment88
- Ch.19 - Chemical Thermodynamics140
- Ch.20 - Electrochemistry137
- Ch.21 - Nuclear Chemistry99
- Ch.22 - Chemistry of the Nonmetals66
- Ch.23 - Transition Metals and Coordination Chemistry69
- Ch.24 - The Chemistry of Life: Organic and Biological Chemistry11
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232Not the one you use?Change textbook
Chapter 22, Problem 38
Complete the exercises below. Complete and balance the following equation: c. Na₂O₂ (s) + H₂O (l) →
Verified step by step guidance1
Step 1: Identify the reactants and products. The reactants are sodium peroxide (Na₂O₂) and water (H₂O). The products will include sodium hydroxide (NaOH) and oxygen gas (O₂).
Step 2: Write the unbalanced chemical equation using the identified reactants and products: Na₂O₂ (s) + H₂O (l) → NaOH (aq) + O₂ (g).
Step 3: Balance the sodium (Na) atoms. There are 2 Na atoms in Na₂O₂, so you need 2 NaOH molecules on the product side: Na₂O₂ (s) + H₂O (l) → 2 NaOH (aq) + O₂ (g).
Step 4: Balance the oxygen (O) atoms. There are 2 O atoms in Na₂O₂ and 1 O atom in H₂O, making a total of 3 O atoms on the reactant side. On the product side, there are 2 O atoms in 2 NaOH and 2 O atoms in O₂, making a total of 4 O atoms. Adjust the water to balance the oxygen: Na₂O₂ (s) + 2 H₂O (l) → 2 NaOH (aq) + O₂ (g).
Step 5: Balance the hydrogen (H) atoms. There are 4 H atoms on the reactant side (from 2 H₂O) and 4 H atoms on the product side (from 2 NaOH), so the equation is now balanced.
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Balancing Chemical Equations
Balancing chemical equations is the process of ensuring that the number of atoms for each element is the same on both sides of the equation. This is based on the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. To balance an equation, coefficients are adjusted in front of the chemical formulas to achieve equal atom counts.
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Oxidation-Reduction Reactions
Oxidation-reduction (redox) reactions involve the transfer of electrons between substances, leading to changes in oxidation states. In the reaction involving sodium peroxide (Na₂O₂) and water (H₂O), sodium peroxide acts as an oxidizing agent, and understanding the oxidation states of sodium, oxygen, and hydrogen is crucial for predicting the products formed during the reaction.
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Products of Sodium Peroxide and Water Reaction
When sodium peroxide reacts with water, it typically produces sodium hydroxide (NaOH) and hydrogen peroxide (H₂O₂). Recognizing the expected products of this reaction is essential for completing and balancing the equation correctly. This knowledge helps in determining the stoichiometry of the reaction and ensuring that the final balanced equation accurately reflects the reactants and products involved.
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01:23Reaction with Water