Ch.2 - Atoms & ElementsWorksheetSee 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: You are working in the laboratory preparing sodium chloride. Consider the following results for three preparations of the compound:Case 1:     39.34 g Na + 60.66 g Cl 2 ⟶ 100.00 g NaCl Case 2:    39.34 g Na + 70.00 g Cl 2 ⟶ 100.00 g NaCl + 9.34 g Cl 2 Case 3:    50.00 g Na + 50.00 g Cl 2 ⟶ 82.43 g NaCl + 17.57 g NaExplain these results in terms of the laws of conservation of mass and definite composition.

Solution: You are working in the laboratory preparing sodium chloride. Consider the following results for three preparations of the compound:Case 1:     39.34 g Na + 60.66 g Cl 2 ⟶ 100.00 g NaClCase 2:    39.34

Problem

You are working in the laboratory preparing sodium chloride. Consider the following results for three preparations of the compound:

Case 1:     39.34 g Na + 60.66 g Cl 2 ⟶ 100.00 g NaCl
Case 2:    39.34 g Na + 70.00 g Cl 2 ⟶ 100.00 g NaCl + 9.34 g Cl 2
Case 3:    50.00 g Na + 50.00 g Cl 2 ⟶ 82.43 g NaCl + 17.57 g Na

Explain these results in terms of the laws of conservation of mass and definite composition.