Your SlideShare is downloading. ×
0
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Chapter 1  lecture 1- chm115
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

Chapter 1 lecture 1- chm115

1,019

Published on

0 Comments
1 Like
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total Views
1,019
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
39
Comments
0
Likes
1
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide













































  • Replace picture with new Figure 1.2
  • Replace picture with new Figure 1.2
  • Replace picture with new Figure 1.2
  • Replace picture with new Figure 1.2
  • Replace picture with new Figure 1.2
  • Replace picture with new Figure 1.2
  • Replace picture with new Figure 1.2
  • Replace picture with new Figure 1.2
  • Replace picture with new Figure 1.2
  • Replace picture with new Figure 1.2



























































  • Answer: e
  • Answer: e
  • Answer: e
  • Answer: e
  • Answer: e














  • replace with new figure 1.8


















  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: d
  • Answer: d
  • Answer: d
  • Answer: d
  • Answer: d
  • Answer: d
  • Answer: d
  • Answer: d
  • Answer: d
  • Answer: d

























  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a
  • Answer: a

















  • replace picture with figure on bottom left of page 14
  • replace picture with figure on bottom left of page 14
  • replace picture with figure on bottom left of page 14
  • replace picture with figure on bottom left of page 14
  • replace picture with figure on bottom left of page 14
  • replace picture with figure on bottom left of page 14
  • replace picture with figure on bottom left of page 14
  • replace picture with figure on bottom left of page 14
  • replace picture with figure on bottom left of page 14
  • replace picture with figure on bottom left of page 14








  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
























































  • Answer: b
  • Answer: b
  • Answer: b
  • Answer: b
  • Answer: b
  • Answer: b
  • Answer: b
  • Answer: b
  • Answer: b
  • Answer: b






































  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: c
  • Answer: d
  • Answer: d
  • Answer: d
  • Answer: d
  • Answer: d
  • Answer: d
  • Answer: d
  • Answer: d
  • Answer: d
  • Answer: d
  • Transcript

    • 1. Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Chapter 1 Matter, Measurement, and Problem Solving Grand Valley State University Chemistry Department- CHM115-60 Fall 2010 Copyright © 2011 Pearson Education, Inc.
    • 2. Key Concepts Lecture 1 Science and Scientific Method (Hypothesis, law, theory) Matter (solid, liquid, gas) Crystalline and Amorphous solids Mixture, pure substances, Elements and Compounds, Heterogeneous and homogenous Mixture  Properties of Matter (Physical and chemical changes )  Energy Changes in Matter (Kinetic and potential energy) Unit of measurement (English, Metric and International System) Base Units, Temperature Units, Derived units Tro: Chemistry: A Molecular Approach, 2/e 2 Copyright © 2011 Pearson Education, Inc.
    • 3. Problem Assignments, Tro Copyright © 2011 Pearson Education, Inc.
    • 4. Problem Assignments, Tro • Page 36-41 Copyright © 2011 Pearson Education, Inc.
    • 5. Problem Assignments, Tro • Page 36-41 • Q37-50 Copyright © 2011 Pearson Education, Inc.
    • 6. Problem Assignments, Tro • Page 36-41 • Q37-50 • Q51-64 Copyright © 2011 Pearson Education, Inc.
    • 7. Problem Assignments, Tro • Page 36-41 • Q37-50 • Q51-64 • Q67, 69, 71, 72 Copyright © 2011 Pearson Education, Inc.
    • 8. Problem Assignments, Tro • Page 36-41 • Q37-50 • Q51-64 • Q67, 69, 71, 72 • Q73, 75, 78, 81, 84, 85, Copyright © 2011 Pearson Education, Inc.
    • 9. Tro: Chemistry: A Molecular Approach 4 Copyright © 2011 Pearson Education, Inc.
    • 10. Composition of Matter Atoms and Molecules Scientific Method Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc.
    • 11. Structure Determines Properties Tro: Chemistry: A Molecular Approach, 2/e 6 Copyright © 2011 Pearson Education, Inc.
    • 12. Structure Determines Properties • The properties of matter are determined by the atoms and molecules that compose it Tro: Chemistry: A Molecular Approach, 2/e 6 Copyright © 2011 Pearson Education, Inc.
    • 13. Structure Determines Properties • The properties of matter are determined by the atoms and molecules that compose it carbon monoxide 1. composed of one carbon atom and one oxygen atom 2. colorless, odorless gas 3. burns with a blue flame 4. binds to hemoglobin Tro: Chemistry: A Molecular Approach, 2/e 6 Copyright © 2011 Pearson Education, Inc.
    • 14. Structure Determines Properties • The properties of matter are determined by the atoms and molecules that compose it carbon monoxide carbon dioxide 1. composed of one carbon 1. composed of one carbon atom and one oxygen atom atom and two oxygen atoms 2. colorless, odorless gas 2. colorless, odorless gas 3. burns with a blue flame 3. incombustible 4. binds to hemoglobin 4. does not bind to hemoglobin Tro: Chemistry: A Molecular Approach, 2/e 6 Copyright © 2011 Pearson Education, Inc.
    • 15. Atoms and Molecules Tro: Chemistry: A Molecular Approach 2/e Approach, 7 Copyright © 2011 Pearson Education, Inc.
    • 16. Atoms and Molecules • Atoms Tro: Chemistry: A Molecular Approach 2/e Approach, 7 Copyright © 2011 Pearson Education, Inc.
    • 17. Atoms and Molecules • Atoms  are submicroscopic particles Tro: Chemistry: A Molecular Approach 2/e Approach, 7 Copyright © 2011 Pearson Education, Inc.
    • 18. Atoms and Molecules • Atoms  are submicroscopic particles  are the fundamental building blocks of ordinary matter Tro: Chemistry: A Molecular Approach 2/e Approach, 7 Copyright © 2011 Pearson Education, Inc.
    • 19. Atoms and Molecules • Atoms  are submicroscopic particles  are the fundamental building blocks of ordinary matter • Molecules Tro: Chemistry: A Molecular Approach 2/e Approach, 7 Copyright © 2011 Pearson Education, Inc.
    • 20. Atoms and Molecules • Atoms  are submicroscopic particles  are the fundamental building blocks of ordinary matter • Molecules  are two or more atoms attached together in a specific geometrical arrangement Tro: Chemistry: A Molecular Approach 2/e Approach, 7 Copyright © 2011 Pearson Education, Inc.
    • 21. Atoms and Molecules • Atoms  are submicroscopic particles  are the fundamental building blocks of ordinary matter • Molecules  are two or more atoms attached together in a specific geometrical arrangement  attachments are called bonds Tro: Chemistry: A Molecular Approach 2/e Approach, 7 Copyright © 2011 Pearson Education, Inc.
    • 22. Atoms and Molecules • Atoms  are submicroscopic particles  are the fundamental building blocks of ordinary matter • Molecules  are two or more atoms attached together in a specific geometrical arrangement  attachments are called bonds  attachments come in different strengths Tro: Chemistry: A Molecular Approach 2/e Approach, 7 Copyright © 2011 Pearson Education, Inc.
    • 23. Atoms and Molecules • Atoms  are submicroscopic particles  are the fundamental building blocks of ordinary matter • Molecules  are two or more atoms attached together in a specific geometrical arrangement  attachments are called bonds  attachments come in different strengths  come in different shapes and patterns Tro: Chemistry: A Molecular Approach 2/e Approach, 7 Copyright © 2011 Pearson Education, Inc.
    • 24. Atoms and Molecules • Atoms  are submicroscopic particles  are the fundamental building blocks of ordinary matter • Molecules  are two or more atoms attached together in a specific geometrical arrangement  attachments are called bonds  attachments come in different strengths  come in different shapes and patterns • Chemistry is the science that seeks to understand the behavior of matter by studying the behavior of atoms and molecules Tro: Chemistry: A Molecular Approach 2/e Approach, 7 Copyright © 2011 Pearson Education, Inc.
    • 25. The Scientific Approach to Knowledge Tro: Chemistry: A Molecular Approach 2/e Approach, 8 Copyright © 2011 Pearson Education, Inc.
    • 26. The Scientific Approach to Knowledge • Philosophers try to understand the universe by reasoning and thinking about “ideal” behavior Tro: Chemistry: A Molecular Approach 2/e Approach, 8 Copyright © 2011 Pearson Education, Inc.
    • 27. The Scientific Approach to Knowledge • Philosophers try to understand the universe by reasoning and thinking about “ideal” behavior • Scientists try to understand the universe through empirical knowledge gained through observation and experiment Tro: Chemistry: A Molecular Approach 2/e Approach, 8 Copyright © 2011 Pearson Education, Inc.
    • 28. Gathering Empirical Knowledge ─ Observation Tro: Chemistry: A Molecular Approach 2/e Approach, 9 Copyright © 2011 Pearson Education, Inc.
    • 29. Gathering Empirical Knowledge ─ Observation • Some observations are descriptions of the characteristics or behavior of nature ─ qualitative  “The soda pop is a liquid with a brown color and a sweet taste. Bubbles are seen floating up through it.” Tro: Chemistry: A Molecular Approach 2/e Approach, 9 Copyright © 2011 Pearson Education, Inc.
    • 30. Gathering Empirical Knowledge ─ Observation • Some observations are descriptions of the characteristics or behavior of nature ─ qualitative  “The soda pop is a liquid with a brown color and a sweet taste. Bubbles are seen floating up through it.” • Some observations compare a characteristic to a standard numerical scale ─ quantitative  “A 240 mL serving of soda pop contains 27 g of sugar.” Tro: Chemistry: A Molecular Approach 2/e Approach, 9 Copyright © 2011 Pearson Education, Inc.
    • 31. From Observation to Understanding Tro: Chemistry: A Molecular Approach 2/e Approach, 10 Copyright © 2011 Pearson Education, Inc.
    • 32. From Observation to Understanding • Hypothesis – a tentative interpretation or explanation for an observation  “The sweet taste of soda pop is due to the presence of sugar.” Tro: Chemistry: A Molecular Approach 2/e Approach, 10 Copyright © 2011 Pearson Education, Inc.
    • 33. From Observation to Understanding • Hypothesis – a tentative interpretation or explanation for an observation  “The sweet taste of soda pop is due to the presence of sugar.” • A good hypothesis is one that can be tested to be proved wrong!  falsifiable  one test may invalidate your hypothesis Tro: Chemistry: A Molecular Approach 2/e Approach, 10 Copyright © 2011 Pearson Education, Inc.
    • 34. Testing Ideas Tro: Chemistry: A Molecular Approach 2/e Approach, 11 Copyright © 2011 Pearson Education, Inc.
    • 35. Testing Ideas • Ideas in science are tested with experiments Tro: Chemistry: A Molecular Approach 2/e Approach, 11 Copyright © 2011 Pearson Education, Inc.
    • 36. Testing Ideas • Ideas in science are tested with experiments • An experiment is a set of highly controlled procedures designed to test whether an idea about nature is valid Tro: Chemistry: A Molecular Approach 2/e Approach, 11 Copyright © 2011 Pearson Education, Inc.
    • 37. Testing Ideas • Ideas in science are tested with experiments • An experiment is a set of highly controlled procedures designed to test whether an idea about nature is valid • The experiment generates observations that will either validate or invalidate the idea Tro: Chemistry: A Molecular Approach 2/e Approach, 11 Copyright © 2011 Pearson Education, Inc.
    • 38. From Specific to General Observations Tro: Chemistry: A Molecular Approach 2/e Approach, 12 Copyright © 2011 Pearson Education, Inc.
    • 39. From Specific to General Observations • A scientific law is a statement that summarizes all past observations and predicts future observations  Law of Conservation of Mass – “In a chemical reaction matter is neither created nor destroyed.” Tro: Chemistry: A Molecular Approach 2/e Approach, 12 Copyright © 2011 Pearson Education, Inc.
    • 40. From Specific to General Observations • A scientific law is a statement that summarizes all past observations and predicts future observations  Law of Conservation of Mass – “In a chemical reaction matter is neither created nor destroyed.” • A scientific law allows you to predict future observations  so you can test the law with experiments Tro: Chemistry: A Molecular Approach 2/e Approach, 12 Copyright © 2011 Pearson Education, Inc.
    • 41. From Specific to General Observations • A scientific law is a statement that summarizes all past observations and predicts future observations  Law of Conservation of Mass – “In a chemical reaction matter is neither created nor destroyed.” • A scientific law allows you to predict future observations  so you can test the law with experiments • Unlike state laws, you cannot choose to violate a scientific law! Tro: Chemistry: A Molecular Approach 2/e Approach, 12 Copyright © 2011 Pearson Education, Inc.
    • 42. From Specific to General Understanding Tro: Chemistry: A Molecular Approach 2/e Approach, 13 Copyright © 2011 Pearson Education, Inc.
    • 43. From Specific to General Understanding • A hypothesis is a potential explanation for a single or small number of observations Tro: Chemistry: A Molecular Approach 2/e Approach, 13 Copyright © 2011 Pearson Education, Inc.
    • 44. From Specific to General Understanding • A hypothesis is a potential explanation for a single or small number of observations • A scientific theory is a general explanation for why things in nature are the way they are and behave the way they do Tro: Chemistry: A Molecular Approach 2/e Approach, 13 Copyright © 2011 Pearson Education, Inc.
    • 45. From Specific to General Understanding • A hypothesis is a potential explanation for a single or small number of observations • A scientific theory is a general explanation for why things in nature are the way they are and behave the way they do  models Tro: Chemistry: A Molecular Approach 2/e Approach, 13 Copyright © 2011 Pearson Education, Inc.
    • 46. From Specific to General Understanding • A hypothesis is a potential explanation for a single or small number of observations • A scientific theory is a general explanation for why things in nature are the way they are and behave the way they do  models  pinnacle of scientific knowledge Tro: Chemistry: A Molecular Approach 2/e Approach, 13 Copyright © 2011 Pearson Education, Inc.
    • 47. From Specific to General Understanding • A hypothesis is a potential explanation for a single or small number of observations • A scientific theory is a general explanation for why things in nature are the way they are and behave the way they do  models  pinnacle of scientific knowledge  validated or invalidated by experiment and observation Tro: Chemistry: A Molecular Approach 2/e Approach, 13 Copyright © 2011 Pearson Education, Inc.
    • 48. Scientific Method Tro: Chemistry: A Molecular Approach 2/e Approach, 14 Copyright © 2011 Pearson Education, Inc.
    • 49. Scientific Method Careful noting and recording of natural phenomena Tro: Chemistry: A Molecular Approach 2/e Approach, 14 Copyright © 2011 Pearson Education, Inc.
    • 50. Scientific Method Tentative explanation of a single or small number of observations Tro: Chemistry: A Molecular Approach 2/e Approach, 14 Copyright © 2011 Pearson Education, Inc.
    • 51. Procedure Scientific Method designed to test an idea Tro: Chemistry: A Molecular Approach 2/e Approach, 14 Copyright © 2011 Pearson Education, Inc.
    • 52. Scientific Method Generally observed occurence in nature Tro: Chemistry: A Molecular Approach 2/e Approach, 14 Copyright © 2011 Pearson Education, Inc.
    • 53. Scientific Method General explanation of natural phenomena Tro: Chemistry: A Molecular Approach 2/e Approach, 14 Copyright © 2011 Pearson Education, Inc.
    • 54. Scientific Method Tro: Chemistry: A Molecular Approach 2/e Approach, 14 Copyright © 2011 Pearson Education, Inc.
    • 55. Relationships Between Pieces of the Scientific Method Tro: Chemistry: A Molecular Approach 2/e Approach, 15 Copyright © 2011 Pearson Education, Inc.
    • 56. Classification of Matter States of Matter Physical and Chemical Properties Physical and Chemical Changes Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc.
    • 57. Classification of Matter Tro: Chemistry: A Molecular Approach 2/e Approach, 17 Copyright © 2011 Pearson Education, Inc.
    • 58. Classification of Matter • Matter is anything that occupies space and has mass Tro: Chemistry: A Molecular Approach 2/e Approach, 17 Copyright © 2011 Pearson Education, Inc.
    • 59. Classification of Matter • Matter is anything that occupies space and has mass • We can classify matter based on its state and its composition Tro: Chemistry: A Molecular Approach 2/e Approach, 17 Copyright © 2011 Pearson Education, Inc.
    • 60. Classification of Matter • Matter is anything that occupies space and has mass • We can classify matter based on its state and its composition  whether it’s solid, liquid, or gas Tro: Chemistry: A Molecular Approach 2/e Approach, 17 Copyright © 2011 Pearson Education, Inc.
    • 61. Classification of Matter • Matter is anything that occupies space and has mass • We can classify matter based on its state and its composition  whether it’s solid, liquid, or gas  its basic components Tro: Chemistry: A Molecular Approach 2/e Approach, 17 Copyright © 2011 Pearson Education, Inc.
    • 62. Classifying Matter by Physical State Tro: Chemistry: A Molecular Approach 2/e Approach, 18 Copyright © 2011 Pearson Education, Inc.
    • 63. Classifying Matter by Physical State • Matter can be classified as solid, liquid, or gas based on the characteristics it exhibits Tro: Chemistry: A Molecular Approach 2/e Approach, 18 Copyright © 2011 Pearson Education, Inc.
    • 64. Classifying Matter by Physical State • Matter can be classified as solid, liquid, or gas based on the characteristics it exhibits Tro: Chemistry: A Molecular Approach 2/e Approach, 18 Copyright © 2011 Pearson Education, Inc.
    • 65. Solids Tro: Chemistry: A Molecular Approach 2/e Approach, 19 Copyright © 2011 Pearson Education, Inc.
    • 66. Solids • The particles in a solid are packed close together and are fixed in position Tro: Chemistry: A Molecular Approach 2/e Approach, 19 Copyright © 2011 Pearson Education, Inc.
    • 67. Solids • The particles in a solid are packed close together and are fixed in position  though they may vibrate Tro: Chemistry: A Molecular Approach 2/e Approach, 19 Copyright © 2011 Pearson Education, Inc.
    • 68. Solids • The particles in a solid are packed close together and are fixed in position  though they may vibrate • The close packing of the particles results in solids being incompressible Tro: Chemistry: A Molecular Approach 2/e Approach, 19 Copyright © 2011 Pearson Education, Inc.
    • 69. Solids • The particles in a solid are packed close together and are fixed in position  though they may vibrate • The close packing of the particles results in solids being incompressible • The inability of the particles to move around results in solids retaining their shape and volume when placed in a new container, and prevents the solid from flowing Tro: Chemistry: A Molecular Approach 2/e Approach, 19 Copyright © 2011 Pearson Education, Inc.
    • 70. Crystalline Solids Tro: Chemistry: A Molecular Approach 2/e Approach, 20 Copyright © 2011 Pearson Education, Inc.
    • 71. Crystalline Solids • Some solids have their particles arranged in patterns with long-range repeating order – we call these crystalline solids Tro: Chemistry: A Molecular Approach 2/e Approach, 20 Copyright © 2011 Pearson Education, Inc.
    • 72. Crystalline Solids • Some solids have their particles arranged in patterns with long-range repeating order – we call these crystalline solids  salt Tro: Chemistry: A Molecular Approach 2/e Approach, 20 Copyright © 2011 Pearson Education, Inc.
    • 73. Crystalline Solids • Some solids have their particles arranged in patterns with long-range repeating order – we call these crystalline solids  salt  diamonds Tro: Chemistry: A Molecular Approach 2/e Approach, 20 Copyright © 2011 Pearson Education, Inc.
    • 74. Crystalline Solids • Some solids have their particles arranged in patterns with long-range repeating order – we call these crystalline solids  salt  diamonds  sugar Tro: Chemistry: A Molecular Approach 2/e Approach, 20 Copyright © 2011 Pearson Education, Inc.
    • 75. Amorphous Solids Tro: Chemistry: A Molecular Approach 2/e Approach, 21 Copyright © 2011 Pearson Education, Inc.
    • 76. Amorphous Solids • Some solids have their particles randomly distributed without any long-range pattern – we call these amorphous solids Tro: Chemistry: A Molecular Approach 2/e Approach, 21 Copyright © 2011 Pearson Education, Inc.
    • 77. Amorphous Solids • Some solids have their particles randomly distributed without any long-range pattern – we call these amorphous solids  plastic Tro: Chemistry: A Molecular Approach 2/e Approach, 21 Copyright © 2011 Pearson Education, Inc.
    • 78. Amorphous Solids • Some solids have their particles randomly distributed without any long-range pattern – we call these amorphous solids  plastic  glass Tro: Chemistry: A Molecular Approach 2/e Approach, 21 Copyright © 2011 Pearson Education, Inc.
    • 79. Amorphous Solids • Some solids have their particles randomly distributed without any long-range pattern – we call these amorphous solids  plastic  glass  charcoal Tro: Chemistry: A Molecular Approach 2/e Approach, 21 Copyright © 2011 Pearson Education, Inc.
    • 80. Liquids Tro: Chemistry: A Molecular Approach 2/e Approach, 22 Copyright © 2011 Pearson Education, Inc.
    • 81. Liquids • The particles in a liquid are closely packed, but they have some ability to move around Tro: Chemistry: A Molecular Approach 2/e Approach, 22 Copyright © 2011 Pearson Education, Inc.
    • 82. Liquids • The particles in a liquid are closely packed, but they have some ability to move around • The close packing results in liquids being incompressible Tro: Chemistry: A Molecular Approach 2/e Approach, 22 Copyright © 2011 Pearson Education, Inc.
    • 83. Liquids • The particles in a liquid are closely packed, but they have some ability to move around • The close packing results in liquids being incompressible • The ability of the particles to move allows liquids to take the shape of their container and to flow – however, they don’t have enough freedom to escape or expand to fill the container Tro: Chemistry: A Molecular Approach 2/e Approach, 22 Copyright © 2011 Pearson Education, Inc.
    • 84. Gases Tro: Chemistry: A Molecular Approach 2/e Approach, 23 Copyright © 2011 Pearson Education, Inc.
    • 85. Gases • In the gas state, the particles have freedom of motion and are not held together Tro: Chemistry: A Molecular Approach 2/e Approach, 23 Copyright © 2011 Pearson Education, Inc.
    • 86. Gases • In the gas state, the particles have freedom of motion and are not held together • The particles are constantly flying around, bumping into each other and the container Tro: Chemistry: A Molecular Approach 2/e Approach, 23 Copyright © 2011 Pearson Education, Inc.
    • 87. Gases • In the gas state, the particles have freedom of motion and are not held together • The particles are constantly flying around, bumping into each other and the container • In the gas state, there is a lot of empty space between the particles Tro: Chemistry: A Molecular Approach 2/e Approach, 23 Copyright © 2011 Pearson Education, Inc.
    • 88. Gases • In the gas state, the particles have freedom of motion and are not held together • The particles are constantly flying around, bumping into each other and the container • In the gas state, there is a lot of empty space between the particles  on average Tro: Chemistry: A Molecular Approach 2/e Approach, 23 Copyright © 2011 Pearson Education, Inc.
    • 89. Gases Tro: Chemistry: A Molecular Approach 2/e Approach, 24 Copyright © 2011 Pearson Education, Inc.
    • 90. Gases • Because there is a lot of empty space, the particles can be squeezed closer together – therefore gases are compressible Tro: Chemistry: A Molecular Approach 2/e Approach, 24 Copyright © 2011 Pearson Education, Inc.
    • 91. Gases • Because there is a lot of empty space, the particles can be squeezed closer together – therefore gases are compressible • Because the particles are not held in close contact and are moving freely, gases expand to fill and take the shape of their container, and will flow Tro: Chemistry: A Molecular Approach 2/e Approach, 24 Copyright © 2011 Pearson Education, Inc.
    • 92. Classifying Matter by Composition Tro: Chemistry: A Molecular Approach 2/e Approach, 25 Copyright © 2011 Pearson Education, Inc.
    • 93. Classifying Matter by Composition • Another way to classify matter is to examine its composition Tro: Chemistry: A Molecular Approach 2/e Approach, 25 Copyright © 2011 Pearson Education, Inc.
    • 94. Classifying Matter by Composition • Another way to classify matter is to examine its composition • Composition includes Tro: Chemistry: A Molecular Approach 2/e Approach, 25 Copyright © 2011 Pearson Education, Inc.
    • 95. Classifying Matter by Composition • Another way to classify matter is to examine its composition • Composition includes  types of particles Tro: Chemistry: A Molecular Approach 2/e Approach, 25 Copyright © 2011 Pearson Education, Inc.
    • 96. Classifying Matter by Composition • Another way to classify matter is to examine its composition • Composition includes  types of particles  arrangement of the particles Tro: Chemistry: A Molecular Approach 2/e Approach, 25 Copyright © 2011 Pearson Education, Inc.
    • 97. Classifying Matter by Composition • Another way to classify matter is to examine its composition • Composition includes  types of particles  arrangement of the particles  attractions and attachments between the particles Tro: Chemistry: A Molecular Approach 2/e Approach, 25 Copyright © 2011 Pearson Education, Inc.
    • 98. Tro: Chemistry: A Molecular Approach 2/e Approach, 26 Copyright © 2011 Pearson Education, Inc.
    • 99. Classification of Matter by Composition Tro: Chemistry: A Molecular Approach 2/e Approach, 27 Copyright © 2011 Pearson Education, Inc.
    • 100. Classification of Matter by Composition • Matter whose composition does not change from one sample to another is called a pure substance Tro: Chemistry: A Molecular Approach 2/e Approach, 27 Copyright © 2011 Pearson Education, Inc.
    • 101. Classification of Matter by Composition • Matter whose composition does not change from one sample to another is called a pure substance  made of a single type of atom or molecule Tro: Chemistry: A Molecular Approach 2/e Approach, 27 Copyright © 2011 Pearson Education, Inc.
    • 102. Classification of Matter by Composition • Matter whose composition does not change from one sample to another is called a pure substance  made of a single type of atom or molecule  because the composition of a pure substance is always the same, all samples have the same characteristics Tro: Chemistry: A Molecular Approach 2/e Approach, 27 Copyright © 2011 Pearson Education, Inc.
    • 103. Classification of Matter by Composition • Matter whose composition does not change from one sample to another is called a pure substance  made of a single type of atom or molecule  because the composition of a pure substance is always the same, all samples have the same characteristics • Matter whose composition may vary from one sample to another is called a mixture Tro: Chemistry: A Molecular Approach 2/e Approach, 27 Copyright © 2011 Pearson Education, Inc.
    • 104. Classification of Matter by Composition • Matter whose composition does not change from one sample to another is called a pure substance  made of a single type of atom or molecule  because the composition of a pure substance is always the same, all samples have the same characteristics • Matter whose composition may vary from one sample to another is called a mixture  two or more types of atoms or molecules combined in variable proportions Tro: Chemistry: A Molecular Approach 2/e Approach, 27 Copyright © 2011 Pearson Education, Inc.
    • 105. Classification of Matter by Composition • Matter whose composition does not change from one sample to another is called a pure substance  made of a single type of atom or molecule  because the composition of a pure substance is always the same, all samples have the same characteristics • Matter whose composition may vary from one sample to another is called a mixture  two or more types of atoms or molecules combined in variable proportions  because composition varies, different samples have different characteristics Tro: Chemistry: A Molecular Approach 2/e Approach, 27 Copyright © 2011 Pearson Education, Inc.
    • 106. Classification of Matter by Composition Tro: Chemistry: A Molecular Approach 2/e Approach, 28 Copyright © 2011 Pearson Education, Inc.
    • 107. Classification of Matter by Composition 1. made of one type of particle Tro: Chemistry: A Molecular Approach 2/e Approach, 28 Copyright © 2011 Pearson Education, Inc.
    • 108. Classification of Matter by Composition 1. made of one type of particle 2. all samples show the same intensive properties Tro: Chemistry: A Molecular Approach 2/e Approach, 28 Copyright © 2011 Pearson Education, Inc.
    • 109. Classification of Matter by Composition 1. made of one type of 1. made of multiple particle types of particles 2. all samples show the same intensive properties Tro: Chemistry: A Molecular Approach 2/e Approach, 28 Copyright © 2011 Pearson Education, Inc.
    • 110. Classification of Matter by Composition 1. made of one type of 1. made of multiple particle types of particles 2. all samples show 2. samples may show the same intensive different intensive properties properties Tro: Chemistry: A Molecular Approach 2/e Approach, 28 Copyright © 2011 Pearson Education, Inc.
    • 111. Classification of Pure Substances  Elements Tro: Chemistry: A Molecular Approach 2/e Approach, 29 Copyright © 2011 Pearson Education, Inc.
    • 112. Classification of Pure Substances  Elements • Pure substances that cannot be decomposed into simpler substances by chemical reactions are called elements Tro: Chemistry: A Molecular Approach 2/e Approach, 29 Copyright © 2011 Pearson Education, Inc.
    • 113. Classification of Pure Substances  Elements • Pure substances that cannot be decomposed into simpler substances by chemical reactions are called elements  decomposed = broken down Tro: Chemistry: A Molecular Approach 2/e Approach, 29 Copyright © 2011 Pearson Education, Inc.
    • 114. Classification of Pure Substances  Elements • Pure substances that cannot be decomposed into simpler substances by chemical reactions are called elements  decomposed = broken down  basic building blocks of matter Tro: Chemistry: A Molecular Approach 2/e Approach, 29 Copyright © 2011 Pearson Education, Inc.
    • 115. Classification of Pure Substances  Elements • Pure substances that cannot be decomposed into simpler substances by chemical reactions are called elements  decomposed = broken down  basic building blocks of matter  composed of single type of atom Tro: Chemistry: A Molecular Approach 2/e Approach, 29 Copyright © 2011 Pearson Education, Inc.
    • 116. Classification of Pure Substances  Elements • Pure substances that cannot be decomposed into simpler substances by chemical reactions are called elements  decomposed = broken down  basic building blocks of matter  composed of single type of atom though those atoms may or may not be combined into molecules Tro: Chemistry: A Molecular Approach 2/e Approach, 29 Copyright © 2011 Pearson Education, Inc.
    • 117. Classification of Pure Substances  Compounds Tro: Chemistry: A Molecular Approach 2/e Approach, 30 Copyright © 2011 Pearson Education, Inc.
    • 118. Classification of Pure Substances  Compounds • Pure substances that can be decomposed are called compounds  chemical combinations of elements  composed of molecules that contain two or more different kinds of atoms  all molecules of a compound are identical, so all samples of a compound behave the same way Tro: Chemistry: A Molecular Approach 2/e Approach, 30 Copyright © 2011 Pearson Education, Inc.
    • 119. Classification of Pure Substances  Compounds • Pure substances that can be decomposed are called compounds  chemical combinations of elements  composed of molecules that contain two or more different kinds of atoms  all molecules of a compound are identical, so all samples of a compound behave the same way • Most natural pure substances are compounds Tro: Chemistry: A Molecular Approach 2/e Approach, 30 Copyright © 2011 Pearson Education, Inc.
    • 120. Classification of Pure Substances Tro: Chemistry: A Molecular Approach 2/e Approach, 31 Copyright © 2011 Pearson Education, Inc.
    • 121. Classification of Pure Substances 1. made of one type of atom (some elements found as multi- atom molecules in nature) Tro: Chemistry: A Molecular Approach 2/e Approach, 31 Copyright © 2011 Pearson Education, Inc.
    • 122. Classification of Pure Substances 1. made of one type of atom (some elements found as multi- atom molecules in nature) 2. combine together to make compounds Tro: Chemistry: A Molecular Approach 2/e Approach, 31 Copyright © 2011 Pearson Education, Inc.
    • 123. Classification of Pure Substances 1. made of one 1. made of one type of atom type of (some molecule, or elements an array of found as multi- ions atom molecules in nature) 2. combine together to make compounds Tro: Chemistry: A Molecular Approach 2/e Approach, 31 Copyright © 2011 Pearson Education, Inc.
    • 124. Classification of Pure Substances 1. made of one 1. made of one type of atom type of (some molecule, or elements an array of found as multi- ions atom 2. units contain molecules in two or more nature) different kinds 2. combine of atoms together to make compounds Tro: Chemistry: A Molecular Approach 2/e Approach, 31 Copyright © 2011 Pearson Education, Inc.
    • 125. Which of the following is a pure substance? Copyright © 2011 Pearson Education, Inc.
    • 126. Which of the following is a pure substance? a.sweat Copyright © 2011 Pearson Education, Inc.
    • 127. Which of the following is a pure substance? a.sweat b.beef stew Copyright © 2011 Pearson Education, Inc.
    • 128. Which of the following is a pure substance? a.sweat b.beef stew c.coffee Copyright © 2011 Pearson Education, Inc.
    • 129. Which of the following is a pure substance? a.sweat b.beef stew c.coffee d.apple juice Copyright © 2011 Pearson Education, Inc.
    • 130. Which of the following is a pure substance? a.sweat b.beef stew c.coffee d.apple juice e.ice Copyright © 2011 Pearson Education, Inc.
    • 131. Classification of Mixtures Tro: Chemistry: A Molecular Approach 2/e Approach, 33 Copyright © 2011 Pearson Education, Inc.
    • 132. Classification of Mixtures • Homogeneous mixtures are mixtures that have uniform composition throughout Tro: Chemistry: A Molecular Approach 2/e Approach, 33 Copyright © 2011 Pearson Education, Inc.
    • 133. Classification of Mixtures • Homogeneous mixtures are mixtures that have uniform composition throughout  every piece of a sample has identical characteristics, though another sample with the same components may have different characteristics Tro: Chemistry: A Molecular Approach 2/e Approach, 33 Copyright © 2011 Pearson Education, Inc.
    • 134. Classification of Mixtures • Homogeneous mixtures are mixtures that have uniform composition throughout  every piece of a sample has identical characteristics, though another sample with the same components may have different characteristics  atoms or molecules mixed uniformly Tro: Chemistry: A Molecular Approach 2/e Approach, 33 Copyright © 2011 Pearson Education, Inc.
    • 135. Classification of Mixtures • Homogeneous mixtures are mixtures that have uniform composition throughout  every piece of a sample has identical characteristics, though another sample with the same components may have different characteristics  atoms or molecules mixed uniformly • Heterogeneous mixtures are mixtures that do not have uniform composition throughout Tro: Chemistry: A Molecular Approach 2/e Approach, 33 Copyright © 2011 Pearson Education, Inc.
    • 136. Classification of Mixtures • Homogeneous mixtures are mixtures that have uniform composition throughout  every piece of a sample has identical characteristics, though another sample with the same components may have different characteristics  atoms or molecules mixed uniformly • Heterogeneous mixtures are mixtures that do not have uniform composition throughout  regions within the sample can have different characteristics Tro: Chemistry: A Molecular Approach 2/e Approach, 33 Copyright © 2011 Pearson Education, Inc.
    • 137. Classification of Mixtures • Homogeneous mixtures are mixtures that have uniform composition throughout  every piece of a sample has identical characteristics, though another sample with the same components may have different characteristics  atoms or molecules mixed uniformly • Heterogeneous mixtures are mixtures that do not have uniform composition throughout  regions within the sample can have different characteristics  atoms or molecules not mixed uniformly Tro: Chemistry: A Molecular Approach 2/e Approach, 33 Copyright © 2011 Pearson Education, Inc.
    • 138. Classification of Mixtures Tro: Chemistry: A Molecular Approach 2/e Approach, 34 Copyright © 2011 Pearson Education, Inc.
    • 139. Classification of Mixtures 1. made of multiple substances, whose presence can be seen Tro: Chemistry: A Molecular Approach 2/e Approach, 34 Copyright © 2011 Pearson Education, Inc.
    • 140. Classification of Mixtures 1. made of multiple substances, whose presence can be seen 2. portions of a sample have different composition and properties Tro: Chemistry: A Molecular Approach 2/e Approach, 34 Copyright © 2011 Pearson Education, Inc.
    • 141. Classification of Mixtures 1. made of 1. made of multiple multiple substances, substances, whose but appears to presence can be one be seen substance 2. portions of a sample have different composition and properties Tro: Chemistry: A Molecular Approach 2/e Approach, 34 Copyright © 2011 Pearson Education, Inc.
    • 142. Classification of Mixtures 1. made of 1. made of multiple multiple substances, substances, whose but appears to presence can be one be seen substance 2. portions of a 2. all portions of sample have an individual different sample have composition the same and composition properties and properties Tro: Chemistry: A Molecular Approach 2/e Approach, 34 Copyright © 2011 Pearson Education, Inc.
    • 143. Changes in Matter Tro: Chemistry: A Molecular Approach 2/e Approach, 35 Copyright © 2011 Pearson Education, Inc.
    • 144. Changes in Matter • Changes that alter the state or appearance of the matter without altering the composition are called physical changes Tro: Chemistry: A Molecular Approach 2/e Approach, 35 Copyright © 2011 Pearson Education, Inc.
    • 145. Changes in Matter • Changes that alter the state or appearance of the matter without altering the composition are called physical changes • Changes that alter the composition of the matter are called chemical changes Tro: Chemistry: A Molecular Approach 2/e Approach, 35 Copyright © 2011 Pearson Education, Inc.
    • 146. Changes in Matter • Changes that alter the state or appearance of the matter without altering the composition are called physical changes • Changes that alter the composition of the matter are called chemical changes  during the chemical change, the atoms that are present rearrange into new molecules, but all of the original atoms are still present Tro: Chemistry: A Molecular Approach 2/e Approach, 35 Copyright © 2011 Pearson Education, Inc.
    • 147. Physical Changes in Matter The boiling of water is a physical change. The water molecules are separated from each other, but their structure and composition do not change. Tro: Chemistry: A Molecular Approach 2/e Approach, 36 Copyright © 2011 Pearson Education, Inc.
    • 148. Chemical Changes in Matter The rusting of iron is a chemical change. The iron atoms in the nail combine with oxygen atoms from O2 in the air to make a new substance, rust, with a different composition. Tro: Chemistry: A Molecular Approach 2/e Approach, 37 Copyright © 2011 Pearson Education, Inc.
    • 149. Properties of Matter Tro: Chemistry: A Molecular Approach 2/e Approach, 38 Copyright © 2011 Pearson Education, Inc.
    • 150. Properties of Matter • Physical properties are the characteristics of matter that can be changed without changing its composition Tro: Chemistry: A Molecular Approach 2/e Approach, 38 Copyright © 2011 Pearson Education, Inc.
    • 151. Properties of Matter • Physical properties are the characteristics of matter that can be changed without changing its composition  characteristics that are directly observable Tro: Chemistry: A Molecular Approach 2/e Approach, 38 Copyright © 2011 Pearson Education, Inc.
    • 152. Properties of Matter • Physical properties are the characteristics of matter that can be changed without changing its composition  characteristics that are directly observable • Chemical properties are the characteristics that determine how the composition of matter changes as a result of contact with other matter or the influence of energy Tro: Chemistry: A Molecular Approach 2/e Approach, 38 Copyright © 2011 Pearson Education, Inc.
    • 153. Properties of Matter • Physical properties are the characteristics of matter that can be changed without changing its composition  characteristics that are directly observable • Chemical properties are the characteristics that determine how the composition of matter changes as a result of contact with other matter or the influence of energy  characteristics that describe the behavior of matter Tro: Chemistry: A Molecular Approach 2/e Approach, 38 Copyright © 2011 Pearson Education, Inc.
    • 154. Common Physical Changes Tro: Chemistry: A Molecular Approach 2/e Approach, 39 Copyright © 2011 Pearson Education, Inc.
    • 155. Common Physical Changes • Processes that cause changes in the matter that do not change its composition Tro: Chemistry: A Molecular Approach 2/e Approach, 39 Copyright © 2011 Pearson Education, Inc.
    • 156. Common Physical Changes • Processes that cause changes in the matter that do not change its composition • State changes Tro: Chemistry: A Molecular Approach 2/e Approach, 39 Copyright © 2011 Pearson Education, Inc.
    • 157. Common Physical Changes • Processes that cause changes in the matter that do not change its composition • State changes  boiling / condensing Tro: Chemistry: A Molecular Approach 2/e Approach, 39 Copyright © 2011 Pearson Education, Inc.
    • 158. Common Physical Changes • Processes that cause changes in the matter that do not change its composition • State changes  boiling / condensing  melting / freezing Tro: Chemistry: A Molecular Approach 2/e Approach, 39 Copyright © 2011 Pearson Education, Inc.
    • 159. Common Physical Changes • Processes that cause Subliming of dry ice changes in the matter that do not change its CO2(g) composition • State changes Dry Ice  boiling / condensing  melting / freezing  subliming CO2(s) Tro: Chemistry: A Molecular Approach 2/e Approach, 39 Copyright © 2011 Pearson Education, Inc.
    • 160. Common Physical Changes • Processes that cause Dissolving of sugar changes in the matter C H O (s) 12 22 11 that do not change its composition • State changes  boiling / condensing  melting / freezing  subliming • Dissolving C12H22O11(aq) Tro: Chemistry: A Molecular Approach 2/e Approach, 39 Copyright © 2011 Pearson Education, Inc.
    • 161. Common Chemical Changes Tro: Chemistry: A Molecular Approach 2/e Approach, 40 Copyright © 2011 Pearson Education, Inc.
    • 162. Common Chemical Changes • Processes that cause changes in the matter that change its composition Tro: Chemistry: A Molecular Approach 2/e Approach, 40 Copyright © 2011 Pearson Education, Inc.
    • 163. Common Chemical Changes • Processes that cause changes in the matter that change its composition • Rusting Tro: Chemistry: A Molecular Approach 2/e Approach, 40 Copyright © 2011 Pearson Education, Inc.
    • 164. Common Chemical Changes • Processes that cause changes in the matter that change its composition • Rusting • Burning Tro: Chemistry: A Molecular Approach 2/e Approach, 40 Copyright © 2011 Pearson Education, Inc.
    • 165. Common Chemical Changes • Processes that cause changes in the matter that change its composition • Rusting • Burning • Dyes fading or changing color C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) Tro: Chemistry: A Molecular Approach 2/e Approach, 40 Copyright © 2011 Pearson Education, Inc.
    • 166. Which of the following represents a chemical change? Copyright © 2011 Pearson Education, Inc.
    • 167. Which of the following represents a chemical change? • Freezing water to make ice cubes Copyright © 2011 Pearson Education, Inc.
    • 168. Which of the following represents a chemical change? • Freezing water to make ice cubes • Dry ice evaporating at room temperature Copyright © 2011 Pearson Education, Inc.
    • 169. Which of the following represents a chemical change? • Freezing water to make ice cubes • Dry ice evaporating at room temperature • Toasting a piece of bread Copyright © 2011 Pearson Education, Inc.
    • 170. Which of the following represents a chemical change? • Freezing water to make ice cubes • Dry ice evaporating at room temperature • Toasting a piece of bread • Dissolving sugar in hot coffee Copyright © 2011 Pearson Education, Inc.
    • 171. Which of the following represents a chemical change? • Freezing water to make ice cubes • Dry ice evaporating at room temperature • Toasting a piece of bread • Dissolving sugar in hot coffee • Crushing an aluminum can Copyright © 2011 Pearson Education, Inc.
    • 172. Which of the following represents a chemical change? Copyright © 2011 Pearson Education, Inc.
    • 173. Which of the following represents a chemical change? • Freezing water to make ice cubes Copyright © 2011 Pearson Education, Inc.
    • 174. Which of the following represents a chemical change? • Freezing water to make ice cubes • Dry ice evaporating at room temperature Copyright © 2011 Pearson Education, Inc.
    • 175. Which of the following represents a chemical change? • Freezing water to make ice cubes • Dry ice evaporating at room temperature • Toasting a piece of bread Copyright © 2011 Pearson Education, Inc.
    • 176. Which of the following represents a chemical change? • Freezing water to make ice cubes • Dry ice evaporating at room temperature • Toasting a piece of bread • Dissolving sugar in hot coffee Copyright © 2011 Pearson Education, Inc.
    • 177. Which of the following represents a chemical change? • Freezing water to make ice cubes • Dry ice evaporating at room temperature • Toasting a piece of bread • Dissolving sugar in hot coffee • Crushing an aluminum can Copyright © 2011 Pearson Education, Inc.
    • 178. Which of the following is a chemical property? Copyright © 2011 Pearson Education, Inc.
    • 179. Which of the following is a chemical property? a.squeezing oranges to make orange juice Copyright © 2011 Pearson Education, Inc.
    • 180. Which of the following is a chemical property? a.squeezing oranges to make orange juice b.melting butter for popcorn Copyright © 2011 Pearson Education, Inc.
    • 181. Which of the following is a chemical property? a.squeezing oranges to make orange juice b.melting butter for popcorn c.separating sand from gravel Copyright © 2011 Pearson Education, Inc.
    • 182. Which of the following is a chemical property? a.squeezing oranges to make orange juice b.melting butter for popcorn c.separating sand from gravel d.hydrogen peroxide decomposes to water and oxygen Copyright © 2011 Pearson Education, Inc.
    • 183. Which of the following is a chemical property? a.squeezing oranges to make orange juice b.melting butter for popcorn c.separating sand from gravel d.hydrogen peroxide decomposes to water and oxygen e.ozone is a gas at room temperature Copyright © 2011 Pearson Education, Inc.
    • 184. Which of the following is a chemical property? Copyright © 2011 Pearson Education, Inc.
    • 185. Which of the following is a chemical property? a.squeezing oranges to make orange juice Copyright © 2011 Pearson Education, Inc.
    • 186. Which of the following is a chemical property? a.squeezing oranges to make orange juice b.melting butter for popcorn Copyright © 2011 Pearson Education, Inc.
    • 187. Which of the following is a chemical property? a.squeezing oranges to make orange juice b.melting butter for popcorn c.separating sand from gravel Copyright © 2011 Pearson Education, Inc.
    • 188. Which of the following is a chemical property? a.squeezing oranges to make orange juice b.melting butter for popcorn c.separating sand from gravel d.hydrogen peroxide decomposes to water and oxygen Copyright © 2011 Pearson Education, Inc.
    • 189. Which of the following is a chemical property? a.squeezing oranges to make orange juice b.melting butter for popcorn c.separating sand from gravel d.hydrogen peroxide decomposes to water and oxygen e.ozone is a gas at room temperature Copyright © 2011 Pearson Education, Inc.
    • 190. Energy Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc.
    • 191. Energy Changes in Matter Tro: Chemistry: A Molecular Approach 2/e Approach, 46 Copyright © 2011 Pearson Education, Inc.
    • 192. Energy Changes in Matter • Changes in matter, both physical and chemical, result in the matter either gaining or releasing energy Tro: Chemistry: A Molecular Approach 2/e Approach, 46 Copyright © 2011 Pearson Education, Inc.
    • 193. Energy Changes in Matter • Changes in matter, both physical and chemical, result in the matter either gaining or releasing energy • Energy is the capacity to do work Tro: Chemistry: A Molecular Approach 2/e Approach, 46 Copyright © 2011 Pearson Education, Inc.
    • 194. Energy Changes in Matter • Changes in matter, both physical and chemical, result in the matter either gaining or releasing energy • Energy is the capacity to do work • Work is the action of a force applied across a distance Tro: Chemistry: A Molecular Approach 2/e Approach, 46 Copyright © 2011 Pearson Education, Inc.
    • 195. Energy Changes in Matter • Changes in matter, both physical and chemical, result in the matter either gaining or releasing energy • Energy is the capacity to do work • Work is the action of a force applied across a distance  a force is a push or a pull on an object Tro: Chemistry: A Molecular Approach 2/e Approach, 46 Copyright © 2011 Pearson Education, Inc.
    • 196. Energy Changes in Matter • Changes in matter, both physical and chemical, result in the matter either gaining or releasing energy • Energy is the capacity to do work • Work is the action of a force applied across a distance  a force is a push or a pull on an object  electrostatic force is the push or pull on objects that have an electrical charge Tro: Chemistry: A Molecular Approach 2/e Approach, 46 Copyright © 2011 Pearson Education, Inc.
    • 197. Energy of Matter Tro: Chemistry: A Molecular Approach 2/e Approach, 47 Copyright © 2011 Pearson Education, Inc.
    • 198. Energy of Matter • All matter possesses energy Tro: Chemistry: A Molecular Approach 2/e Approach, 47 Copyright © 2011 Pearson Education, Inc.
    • 199. Energy of Matter • All matter possesses energy • Energy is classified as either kinetic or potential Tro: Chemistry: A Molecular Approach 2/e Approach, 47 Copyright © 2011 Pearson Education, Inc.
    • 200. Energy of Matter • All matter possesses energy • Energy is classified as either kinetic or potential • Energy can be converted from one form to another Tro: Chemistry: A Molecular Approach 2/e Approach, 47 Copyright © 2011 Pearson Education, Inc.
    • 201. Energy of Matter • All matter possesses energy • Energy is classified as either kinetic or potential • Energy can be converted from one form to another • When matter undergoes a chemical or physical change, the amount of energy in the matter changes as well Tro: Chemistry: A Molecular Approach 2/e Approach, 47 Copyright © 2011 Pearson Education, Inc.
    • 202. Energy of Matter − Kinetic Tro: Chemistry: A Molecular Approach 2/e Approach, 48 Copyright © 2011 Pearson Education, Inc.
    • 203. Energy of Matter − Kinetic • Kinetic energy is energy of motion Tro: Chemistry: A Molecular Approach 2/e Approach, 48 Copyright © 2011 Pearson Education, Inc.
    • 204. Energy of Matter − Kinetic • Kinetic energy is energy of motion  motion of the atoms, molecules, and subatomic particles Tro: Chemistry: A Molecular Approach 2/e Approach, 48 Copyright © 2011 Pearson Education, Inc.
    • 205. Energy of Matter − Kinetic • Kinetic energy is energy of motion  motion of the atoms, molecules, and subatomic particles  thermal (heat) energy is a form of kinetic energy because it is caused by molecular motion Tro: Chemistry: A Molecular Approach 2/e Approach, 48 Copyright © 2011 Pearson Education, Inc.
    • 206. Energy of Matter − Potential Tro: Chemistry: A Molecular Approach 2/e Approach, 49 Copyright © 2011 Pearson Education, Inc.
    • 207. Energy of Matter − Potential • Potential energy is energy that is stored in the matter Tro: Chemistry: A Molecular Approach 2/e Approach, 49 Copyright © 2011 Pearson Education, Inc.
    • 208. Energy of Matter − Potential • Potential energy is energy that is stored in the matter  due to the composition of the matter and its position relative to other things Tro: Chemistry: A Molecular Approach 2/e Approach, 49 Copyright © 2011 Pearson Education, Inc.
    • 209. Energy of Matter − Potential • Potential energy is energy that is stored in the matter  due to the composition of the matter and its position relative to other things  chemical potential energy arises from electrostatic attractive forces between atoms, molecules, and subatomic particles Tro: Chemistry: A Molecular Approach 2/e Approach, 49 Copyright © 2011 Pearson Education, Inc.
    • 210. Conversion of Energy Tro: Chemistry: A Molecular Approach 2/e Approach, 50 Copyright © 2011 Pearson Education, Inc.
    • 211. Conversion of Energy • You can interconvert kinetic energy and potential energy Tro: Chemistry: A Molecular Approach 2/e Approach, 50 Copyright © 2011 Pearson Education, Inc.
    • 212. Conversion of Energy • You can interconvert kinetic energy and potential energy • Whatever process you do that converts energy from one type or form to another, the total amount of energy remains the same Tro: Chemistry: A Molecular Approach 2/e Approach, 50 Copyright © 2011 Pearson Education, Inc.
    • 213. Conversion of Energy • You can interconvert kinetic energy and potential energy • Whatever process you do that converts energy from one type or form to another, the total amount of energy remains the same  Law of Conservation of Energy Tro: Chemistry: A Molecular Approach 2/e Approach, 50 Copyright © 2011 Pearson Education, Inc.
    • 214. Spontaneous Processes Tro: Chemistry: A Molecular Approach 2/e Approach, 51 Copyright © 2011 Pearson Education, Inc.
    • 215. Spontaneous Processes • Materials that possess high potential energy are less stable Tro: Chemistry: A Molecular Approach 2/e Approach, 51 Copyright © 2011 Pearson Education, Inc.
    • 216. Spontaneous Processes • Materials that possess high potential energy are less stable • Processes in nature tend to occur on their own when the result is material with lower total potential energy Tro: Chemistry: A Molecular Approach 2/e Approach, 51 Copyright © 2011 Pearson Education, Inc.
    • 217. Spontaneous Processes • Materials that possess high potential energy are less stable • Processes in nature tend to occur on their own when the result is material with lower total potential energy  processes that result in materials with higher total potential energy can occur, but generally will not happen without input of energy from an outside source Tro: Chemistry: A Molecular Approach 2/e Approach, 51 Copyright © 2011 Pearson Education, Inc.
    • 218. Changes in Energy Tro: Chemistry: A Molecular Approach 2/e Approach, 52 Copyright © 2011 Pearson Education, Inc.
    • 219. Changes in Energy • If a process results in the system having less potential energy at the end than it had at the beginning, the “lost” potential energy was converted into kinetic energy, which is released to the environment Tro: Chemistry: A Molecular Approach 2/e Approach, 52 Copyright © 2011 Pearson Education, Inc.
    • 220. Changes in Energy • If a process results in the system having less potential energy at the end than it had at the beginning, the “lost” potential energy was converted into kinetic energy, which is released to the environment • During the conversion of form, energy that is released can be harnessed to do work Tro: Chemistry: A Molecular Approach 2/e Approach, 52 Copyright © 2011 Pearson Education, Inc.
    • 221. Potential to Kinetic Energy Tro: Chemistry: A Molecular Approach 2/e Approach, 53 Copyright © 2011 Pearson Education, Inc.
    • 222. Which of the following is correct for the material pictured? Copyright © 2011 Pearson Education, Inc.
    • 223. Which of the following is correct for the material pictured? • a gaseous pure substance Copyright © 2011 Pearson Education, Inc.
    • 224. Which of the following is correct for the material pictured? • a gaseous pure substance • a liquid pure substance Copyright © 2011 Pearson Education, Inc.
    • 225. Which of the following is correct for the material pictured? • a gaseous pure substance • a liquid pure substance • a gaseous mixture Copyright © 2011 Pearson Education, Inc.
    • 226. Which of the following is correct for the material pictured? • a gaseous pure substance • a liquid pure substance • a gaseous mixture • a solid mixture Copyright © 2011 Pearson Education, Inc.
    • 227. Which of the following is correct for the material pictured? • a gaseous pure substance • a liquid pure substance • a gaseous mixture • a solid mixture • none of the above Copyright © 2011 Pearson Education, Inc.
    • 228. Which of the following is correct for the material pictured? Copyright © 2011 Pearson Education, Inc.
    • 229. Which of the following is correct for the material pictured? • a gaseous pure substance Copyright © 2011 Pearson Education, Inc.
    • 230. Which of the following is correct for the material pictured? • a gaseous pure substance • a liquid pure substance Copyright © 2011 Pearson Education, Inc.
    • 231. Which of the following is correct for the material pictured? • a gaseous pure substance • a liquid pure substance • a gaseous mixture Copyright © 2011 Pearson Education, Inc.
    • 232. Which of the following is correct for the material pictured? • a gaseous pure substance • a liquid pure substance • a gaseous mixture • a solid mixture Copyright © 2011 Pearson Education, Inc.
    • 233. Which of the following is correct for the material pictured? • a gaseous pure substance • a liquid pure substance • a gaseous mixture • a solid mixture • none of the above Copyright © 2011 Pearson Education, Inc.
    • 234. Which of the following is true? Copyright © 2011 Pearson Education, Inc.
    • 235. Which of the following is true? a.Energy is always conserved in a physical or chemical change. Copyright © 2011 Pearson Education, Inc.
    • 236. Which of the following is true? a.Energy is always conserved in a physical or chemical change. b.Systems with low potential energy tend to change in a direction of high potential energy spontaneously. Copyright © 2011 Pearson Education, Inc.
    • 237. Which of the following is true? a.Energy is always conserved in a physical or chemical change. b.Systems with low potential energy tend to change in a direction of high potential energy spontaneously. c.Thermal energy is a form of potential energy. Copyright © 2011 Pearson Education, Inc.
    • 238. Which of the following is true? a.Energy is always conserved in a physical or chemical change. b.Systems with low potential energy tend to change in a direction of high potential energy spontaneously. c.Thermal energy is a form of potential energy. d.Objects with high potential energy are stable. Copyright © 2011 Pearson Education, Inc.
    • 239. Which of the following is true? a.Energy is always conserved in a physical or chemical change. b.Systems with low potential energy tend to change in a direction of high potential energy spontaneously. c.Thermal energy is a form of potential energy. d.Objects with high potential energy are stable. e.Chemical potential energy is a form of kinetic energy. Copyright © 2011 Pearson Education, Inc.
    • 240. Which of the following is true? Copyright © 2011 Pearson Education, Inc.
    • 241. Which of the following is true? a.Energy is always conserved in a physical or chemical change. Copyright © 2011 Pearson Education, Inc.
    • 242. Which of the following is true? a.Energy is always conserved in a physical or chemical change. b.Systems with low potential energy tend to change in a direction of high potential energy spontaneously. Copyright © 2011 Pearson Education, Inc.
    • 243. Which of the following is true? a.Energy is always conserved in a physical or chemical change. b.Systems with low potential energy tend to change in a direction of high potential energy spontaneously. c.Thermal energy is a form of potential energy. Copyright © 2011 Pearson Education, Inc.
    • 244. Which of the following is true? a.Energy is always conserved in a physical or chemical change. b.Systems with low potential energy tend to change in a direction of high potential energy spontaneously. c.Thermal energy is a form of potential energy. d.Objects with high potential energy are stable. Copyright © 2011 Pearson Education, Inc.
    • 245. Which of the following is true? a.Energy is always conserved in a physical or chemical change. b.Systems with low potential energy tend to change in a direction of high potential energy spontaneously. c.Thermal energy is a form of potential energy. d.Objects with high potential energy are stable. e.Chemical potential energy is a form of kinetic energy. Copyright © 2011 Pearson Education, Inc.
    • 246. Standard Units of Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc.
    • 247. Measurements What kind of measurements did you take today? Example: Checking your weight (Mass) Reading your watch ( Time) Checking your body temperature ( Temperature) Tro: Chemistry: A Molecular Approach, 2/e 59 Copyright © 2011 Pearson Education, Inc.
    • 248. Measurements in Chemistry Measurements have two important components 1)A number 2)A Unit For example mass: Unit Number 70.5 kilograms Different units exist in measuring the same quantity Tro: Chemistry: A Molecular Approach, 2/e 60 Copyright © 2011 Pearson Education, Inc.
    • 249. Standards of Measurements Measurements require: 1) Measuring tool ( thermometer, weighing scale) 2) A standard for comparison ( A kilogram is a mass of metal cylinder kept in France. Tro: Chemistry: A Molecular Approach, 2/e 61 Copyright © 2011 Pearson Education, Inc.
    • 250. Units in Metric System Metric system is a decimal system based on 10 It is widely used across the world by scientists Units is metric system Length: Meter (m) Volume: Liter (L) Mass: gram (g) Temperature: Celsius (°C ) Tro: Chemistry: A Molecular Approach, 2/e 62 Copyright © 2011 Pearson Education, Inc.
    • 251. The Standard Units Tro: Chemistry: A Molecular Approach 2/e Approach, 63 Copyright © 2011 Pearson Education, Inc.
    • 252. The Standard Units • Scientists have agreed on a set of international standard units for comparing all our measurements called the SI units Tro: Chemistry: A Molecular Approach 2/e Approach, 63 Copyright © 2011 Pearson Education, Inc.
    • 253. The Standard Units • Scientists have agreed on a set of international standard units for comparing all our measurements called the SI units  Système International = International System Tro: Chemistry: A Molecular Approach 2/e Approach, 63 Copyright © 2011 Pearson Education, Inc.
    • 254. The Standard Units • Scientists have agreed on a set of international standard units for comparing all our measurements called the SI units  Système International = International System Tro: Chemistry: A Molecular Approach 2/e Approach, 63 Copyright © 2011 Pearson Education, Inc.
    • 255. Length Tro: Chemistry: A Molecular Approach 2/e Approach, 64 Copyright © 2011 Pearson Education, Inc.
    • 256. Length • Measure of the two-dimensional distance an object covers Tro: Chemistry: A Molecular Approach 2/e Approach, 64 Copyright © 2011 Pearson Education, Inc.
    • 257. Length • Measure of the two-dimensional distance an object covers  often need to measure lengths that are very long (distances between stars) or very short (distances between atoms) Tro: Chemistry: A Molecular Approach 2/e Approach, 64 Copyright © 2011 Pearson Education, Inc.
    • 258. Length • Measure of the two-dimensional distance an object covers  often need to measure lengths that are very long (distances between stars) or very short (distances between atoms) • SI unit = meter Tro: Chemistry: A Molecular Approach 2/e Approach, 64 Copyright © 2011 Pearson Education, Inc.
    • 259. Length • Measure of the two-dimensional distance an object covers  often need to measure lengths that are very long (distances between stars) or very short (distances between atoms) • SI unit = meter  about 3.37 inches longer than a yard Tro: Chemistry: A Molecular Approach 2/e Approach, 64 Copyright © 2011 Pearson Education, Inc.
    • 260. Length • Measure of the two-dimensional distance an object covers  often need to measure lengths that are very long (distances between stars) or very short (distances between atoms) • SI unit = meter  about 3.37 inches longer than a yard 1 meter = distance traveled by light in a specific period of time Tro: Chemistry: A Molecular Approach 2/e Approach, 64 Copyright © 2011 Pearson Education, Inc.
    • 261. Length • Measure of the two-dimensional distance an object covers  often need to measure lengths that are very long (distances between stars) or very short (distances between atoms) • SI unit = meter  about 3.37 inches longer than a yard 1 meter = distance traveled by light in a specific period of time • Commonly use centimeters (cm) Tro: Chemistry: A Molecular Approach 2/e Approach, 64 Copyright © 2011 Pearson Education, Inc.
    • 262. Length • Measure of the two-dimensional distance an object covers  often need to measure lengths that are very long (distances between stars) or very short (distances between atoms) • SI unit = meter  about 3.37 inches longer than a yard 1 meter = distance traveled by light in a specific period of time • Commonly use centimeters (cm)  1 m = 100 cm Tro: Chemistry: A Molecular Approach 2/e Approach, 64 Copyright © 2011 Pearson Education, Inc.
    • 263. Length • Measure of the two-dimensional distance an object covers  often need to measure lengths that are very long (distances between stars) or very short (distances between atoms) • SI unit = meter  about 3.37 inches longer than a yard 1 meter = distance traveled by light in a specific period of time • Commonly use centimeters (cm)  1 m = 100 cm  1 cm = 0.01 m = 10 mm Tro: Chemistry: A Molecular Approach 2/e Approach, 64 Copyright © 2011 Pearson Education, Inc.
    • 264. Length • Measure of the two-dimensional distance an object covers  often need to measure lengths that are very long (distances between stars) or very short (distances between atoms) • SI unit = meter  about 3.37 inches longer than a yard 1 meter = distance traveled by light in a specific period of time • Commonly use centimeters (cm)  1 m = 100 cm  1 cm = 0.01 m = 10 mm  1 inch = 2.54 cm (exactly) Tro: Chemistry: A Molecular Approach 2/e Approach, 64 Copyright © 2011 Pearson Education, Inc.
    • 265. Mass Tro: Chemistry: A Molecular Approach 2/e Approach, 65 Copyright © 2011 Pearson Education, Inc.
    • 266. Mass • Measure of the amount of matter present in an object Tro: Chemistry: A Molecular Approach 2/e Approach, 65 Copyright © 2011 Pearson Education, Inc.
    • 267. Mass • Measure of the amount of matter present in an object  weight measures the gravitational pull on an object, which depends on its mass Tro: Chemistry: A Molecular Approach 2/e Approach, 65 Copyright © 2011 Pearson Education, Inc.
    • 268. Mass • Measure of the amount of matter present in an object  weight measures the gravitational pull on an object, which depends on its mass • SI unit = kilogram (kg) Tro: Chemistry: A Molecular Approach 2/e Approach, 65 Copyright © 2011 Pearson Education, Inc.
    • 269. Mass • Measure of the amount of matter present in an object  weight measures the gravitational pull on an object, which depends on its mass • SI unit = kilogram (kg)  about 2 lbs. 3 oz. Tro: Chemistry: A Molecular Approach 2/e Approach, 65 Copyright © 2011 Pearson Education, Inc.
    • 270. Mass • Measure of the amount of matter present in an object  weight measures the gravitational pull on an object, which depends on its mass • SI unit = kilogram (kg)  about 2 lbs. 3 oz. • Commonly measure mass in grams (g) or milligrams (mg) Tro: Chemistry: A Molecular Approach 2/e Approach, 65 Copyright © 2011 Pearson Education, Inc.
    • 271. Mass • Measure of the amount of matter present in an object  weight measures the gravitational pull on an object, which depends on its mass • SI unit = kilogram (kg)  about 2 lbs. 3 oz. • Commonly measure mass in grams (g) or milligrams (mg)  1 kg = 2.2046 pounds, 1 lb. = 453.59 g Tro: Chemistry: A Molecular Approach 2/e Approach, 65 Copyright © 2011 Pearson Education, Inc.
    • 272. Mass • Measure of the amount of matter present in an object  weight measures the gravitational pull on an object, which depends on its mass • SI unit = kilogram (kg)  about 2 lbs. 3 oz. • Commonly measure mass in grams (g) or milligrams (mg)  1 kg = 2.2046 pounds, 1 lb. = 453.59 g  1 kg = 1000 g = 103 g Tro: Chemistry: A Molecular Approach 2/e Approach, 65 Copyright © 2011 Pearson Education, Inc.
    • 273. Mass • Measure of the amount of matter present in an object  weight measures the gravitational pull on an object, which depends on its mass • SI unit = kilogram (kg)  about 2 lbs. 3 oz. • Commonly measure mass in grams (g) or milligrams (mg)  1 kg = 2.2046 pounds, 1 lb. = 453.59 g  1 kg = 1000 g = 103 g  1 g = 1000 mg = 103 mg Tro: Chemistry: A Molecular Approach 2/e Approach, 65 Copyright © 2011 Pearson Education, Inc.
    • 274. Mass • Measure of the amount of matter present in an object  weight measures the gravitational pull on an object, which depends on its mass • SI unit = kilogram (kg)  about 2 lbs. 3 oz. • Commonly measure mass in grams (g) or milligrams (mg)  1 kg = 2.2046 pounds, 1 lb. = 453.59 g  1 kg = 1000 g = 103 g  1 g = 1000 mg = 103 mg  1 g = 0.001 kg = 10−3 kg Tro: Chemistry: A Molecular Approach 2/e Approach, 65 Copyright © 2011 Pearson Education, Inc.
    • 275. Mass • Measure of the amount of matter present in an object  weight measures the gravitational pull on an object, which depends on its mass • SI unit = kilogram (kg)  about 2 lbs. 3 oz. • Commonly measure mass in grams (g) or milligrams (mg)  1 kg = 2.2046 pounds, 1 lb. = 453.59 g  1 kg = 1000 g = 103 g  1 g = 1000 mg = 103 mg  1 g = 0.001 kg = 10−3 kg  1 mg = 0.001 g = 10−3 g Tro: Chemistry: A Molecular Approach 2/e Approach, 65 Copyright © 2011 Pearson Education, Inc.
    • 276. Time Tro: Chemistry: A Molecular Approach 2/e Approach, 66 Copyright © 2011 Pearson Education, Inc.
    • 277. Time • Measure of the duration of an event Tro: Chemistry: A Molecular Approach 2/e Approach, 66 Copyright © 2011 Pearson Education, Inc.
    • 278. Time • Measure of the duration of an event • SI units = second (s) Tro: Chemistry: A Molecular Approach 2/e Approach, 66 Copyright © 2011 Pearson Education, Inc.
    • 279. Time • Measure of the duration of an event • SI units = second (s) • 1 s is defined as the period of time it takes for a specific number of radiation events of a specific transition from cesium–133 Tro: Chemistry: A Molecular Approach 2/e Approach, 66 Copyright © 2011 Pearson Education, Inc.
    • 280. Temperature Tro: Chemistry: A Molecular Approach 2/e Approach, 67 Copyright © 2011 Pearson Education, Inc.
    • 281. Temperature • Measure of the average amount of kinetic energy caused by motion of the particles Tro: Chemistry: A Molecular Approach 2/e Approach, 67 Copyright © 2011 Pearson Education, Inc.
    • 282. Temperature • Measure of the average amount of kinetic energy caused by motion of the particles  higher temperature = larger average kinetic energy Tro: Chemistry: A Molecular Approach 2/e Approach, 67 Copyright © 2011 Pearson Education, Inc.
    • 283. Temperature • Measure of the average amount of kinetic energy caused by motion of the particles  higher temperature = larger average kinetic energy • Heat flows from the matter that has high thermal energy into matter that has low thermal energy until they reach the same temperature Tro: Chemistry: A Molecular Approach 2/e Approach, 67 Copyright © 2011 Pearson Education, Inc.
    • 284. Temperature • Measure of the average amount of kinetic energy caused by motion of the particles  higher temperature = larger average kinetic energy • Heat flows from the matter that has high thermal energy into matter that has low thermal energy until they reach the same temperature  heat flows from hot object to cold Tro: Chemistry: A Molecular Approach 2/e Approach, 67 Copyright © 2011 Pearson Education, Inc.
    • 285. Temperature • Measure of the average amount of kinetic energy caused by motion of the particles  higher temperature = larger average kinetic energy • Heat flows from the matter that has high thermal energy into matter that has low thermal energy until they reach the same temperature  heat flows from hot object to cold  heat is exchanged through molecular collisions between the two materials Tro: Chemistry: A Molecular Approach 2/e Approach, 67 Copyright © 2011 Pearson Education, Inc.
    • 286. Which of the following is not a base unit? Copyright © 2011 Pearson Education, Inc.
    • 287. Which of the following is not a base unit? a.meter Copyright © 2011 Pearson Education, Inc.
    • 288. Which of the following is not a base unit? a.meter b.kilogram Copyright © 2011 Pearson Education, Inc.
    • 289. Which of the following is not a base unit? a.meter b.kilogram c.liter Copyright © 2011 Pearson Education, Inc.
    • 290. Which of the following is not a base unit? a.meter b.kilogram c.liter d.kelvin Copyright © 2011 Pearson Education, Inc.
    • 291. Which of the following is not a base unit? a.meter b.kilogram c.liter d.kelvin e.ampere Copyright © 2011 Pearson Education, Inc.
    • 292. Which of the following is not a base unit? Copyright © 2011 Pearson Education, Inc.
    • 293. Which of the following is not a base unit? a.meter Copyright © 2011 Pearson Education, Inc.
    • 294. Which of the following is not a base unit? a.meter b.kilogram Copyright © 2011 Pearson Education, Inc.
    • 295. Which of the following is not a base unit? a.meter b.kilogram c.liter Copyright © 2011 Pearson Education, Inc.
    • 296. Which of the following is not a base unit? a.meter b.kilogram c.liter d.kelvin Copyright © 2011 Pearson Education, Inc.
    • 297. Which of the following is not a base unit? a.meter b.kilogram c.liter d.kelvin e.ampere Copyright © 2011 Pearson Education, Inc.
    • 298. Temperature Scales Tro: Chemistry: A Molecular Approach 2/e Approach, 70 Copyright © 2011 Pearson Education, Inc.
    • 299. Temperature Scales • Fahrenheit scale, °F Tro: Chemistry: A Molecular Approach 2/e Approach, 70 Copyright © 2011 Pearson Education, Inc.
    • 300. Temperature Scales • Fahrenheit scale, °F  used in the U.S. Tro: Chemistry: A Molecular Approach 2/e Approach, 70 Copyright © 2011 Pearson Education, Inc.
    • 301. Temperature Scales • Fahrenheit scale, °F  used in the U.S. • Celsius scale, °C Tro: Chemistry: A Molecular Approach 2/e Approach, 70 Copyright © 2011 Pearson Education, Inc.
    • 302. Temperature Scales • Fahrenheit scale, °F  used in the U.S. • Celsius scale, °C  used in all other countries Tro: Chemistry: A Molecular Approach 2/e Approach, 70 Copyright © 2011 Pearson Education, Inc.
    • 303. Temperature Scales • Fahrenheit scale, °F  used in the U.S. • Celsius scale, °C  used in all other countries • Kelvin scale, K Tro: Chemistry: A Molecular Approach 2/e Approach, 70 Copyright © 2011 Pearson Education, Inc.
    • 304. Temperature Scales • Fahrenheit scale, °F  used in the U.S. • Celsius scale, °C  used in all other countries • Kelvin scale, K  absolute scale Tro: Chemistry: A Molecular Approach 2/e Approach, 70 Copyright © 2011 Pearson Education, Inc.
    • 305. Temperature Scales • Fahrenheit scale, °F  used in the U.S. • Celsius scale, °C  used in all other countries • Kelvin scale, K  absolute scale  no negative numbers Tro: Chemistry: A Molecular Approach 2/e Approach, 70 Copyright © 2011 Pearson Education, Inc.
    • 306. Temperature Scales • Fahrenheit scale, °F  used in the U.S. • Celsius scale, °C  used in all other countries • Kelvin scale, K  absolute scale  no negative numbers  directly proportional to average amount of kinetic energy Tro: Chemistry: A Molecular Approach 2/e Approach, 70 Copyright © 2011 Pearson Education, Inc.
    • 307. Temperature Scales • Fahrenheit scale, °F  used in the U.S. • Celsius scale, °C  used in all other countries • Kelvin scale, K  absolute scale  no negative numbers  directly proportional to average amount of kinetic energy  0 K = absolute zero Tro: Chemistry: A Molecular Approach 2/e Approach, 70 Copyright © 2011 Pearson Education, Inc.
    • 308. Fahrenheit vs. Celsius Tro: Chemistry: A Molecular Approach 2/e Approach, 71 Copyright © 2011 Pearson Education, Inc.
    • 309. Fahrenheit vs. Celsius • A Celsius degree is 1.8 times larger than a Fahrenheit degree Tro: Chemistry: A Molecular Approach 2/e Approach, 71 Copyright © 2011 Pearson Education, Inc.
    • 310. Fahrenheit vs. Celsius • A Celsius degree is 1.8 times larger than a Fahrenheit degree • The standard used for 0° on the Fahrenheit scale is a lower temperature than the standard used for 0° on the Celsius scale Tro: Chemistry: A Molecular Approach 2/e Approach, 71 Copyright © 2011 Pearson Education, Inc.
    • 311. Kelvin vs. Celsius Tro: Chemistry: A Molecular Approach 2/e Approach, 72 Copyright © 2011 Pearson Education, Inc.
    • 312. Kelvin vs. Celsius • The size of a “degree” on the Kelvin scale is the same as on the Celsius scale Tro: Chemistry: A Molecular Approach 2/e Approach, 72 Copyright © 2011 Pearson Education, Inc.
    • 313. Kelvin vs. Celsius • The size of a “degree” on the Kelvin scale is the same as on the Celsius scale  though technically, we don’t call the divisions on the Kelvin scale degrees; we call them kelvins! Tro: Chemistry: A Molecular Approach 2/e Approach, 72 Copyright © 2011 Pearson Education, Inc.
    • 314. Kelvin vs. Celsius • The size of a “degree” on the Kelvin scale is the same as on the Celsius scale  though technically, we don’t call the divisions on the Kelvin scale degrees; we call them kelvins!  so 1 kelvin is 1.8 times larger than 1°F Tro: Chemistry: A Molecular Approach 2/e Approach, 72 Copyright © 2011 Pearson Education, Inc.
    • 315. Kelvin vs. Celsius • The size of a “degree” on the Kelvin scale is the same as on the Celsius scale  though technically, we don’t call the divisions on the Kelvin scale degrees; we call them kelvins!  so 1 kelvin is 1.8 times larger than 1°F • The 0 standard on the Kelvin scale is a much lower temperature than on the Celsius scale Tro: Chemistry: A Molecular Approach 2/e Approach, 72 Copyright © 2011 Pearson Education, Inc.
    • 316. Example 1.2: Convert 40.00 °C into K and °F • Find the equation that Given: relates the given quantity to Find: the quantity you want to find Equation: • Because the equation is solved for the quantity you want to find, substitute and compute • Find the equation that Given: relates the given quantity to Find: the quantity you want to find Equation: • Solve the equation for the quantity you want to find • Substitute and compute Tro: Chemistry: A Molecular Approach 2/e Approach, 73 Copyright © 2011 Pearson Education, Inc.
    • 317. Example 1.2: Convert 40.00 °C into K and °F • Find the equation that Given: 40.00 °C relates the given quantity to Find: the quantity you want to find Equation: • Because the equation is solved for the quantity you want to find, substitute and compute • Find the equation that Given: relates the given quantity to Find: the quantity you want to find Equation: • Solve the equation for the quantity you want to find • Substitute and compute Tro: Chemistry: A Molecular Approach 2/e Approach, 73 Copyright © 2011 Pearson Education, Inc.
    • 318. Example 1.2: Convert 40.00 °C into K and °F • Find the equation that Given: 40.00 °C relates the given quantity to Find: K the quantity you want to find Equation: • Because the equation is solved for the quantity you want to find, substitute and compute • Find the equation that Given: relates the given quantity to Find: the quantity you want to find Equation: • Solve the equation for the quantity you want to find • Substitute and compute Tro: Chemistry: A Molecular Approach 2/e Approach, 73 Copyright © 2011 Pearson Education, Inc.
    • 319. Example 1.2: Convert 40.00 °C into K and °F • Find the equation that Given: 40.00 °C relates the given quantity to Find: K the quantity you want to find Equation: K = °C + 273.15 • Because the equation is solved for the quantity you want to find, substitute and compute • Find the equation that Given: relates the given quantity to Find: the quantity you want to find Equation: • Solve the equation for the quantity you want to find • Substitute and compute Tro: Chemistry: A Molecular Approach 2/e Approach, 73 Copyright © 2011 Pearson Education, Inc.
    • 320. Example 1.2: Convert 40.00 °C into K and °F • Find the equation that Given: 40.00 °C relates the given quantity to Find: K the quantity you want to find Equation: K = °C + 273.15 • Because the equation is K = °C + 273.15 solved for the quantity you want to find, substitute and compute • Find the equation that Given: relates the given quantity to Find: the quantity you want to find Equation: • Solve the equation for the quantity you want to find • Substitute and compute Tro: Chemistry: A Molecular Approach 2/e Approach, 73 Copyright © 2011 Pearson Education, Inc.
    • 321. Example 1.2: Convert 40.00 °C into K and °F • Find the equation that Given: 40.00 °C relates the given quantity to Find: K the quantity you want to find Equation: K = °C + 273.15 • Because the equation is K = °C + 273.15 solved for the quantity you K = 40.00 + 273.15 want to find, substitute and compute • Find the equation that Given: relates the given quantity to Find: the quantity you want to find Equation: • Solve the equation for the quantity you want to find • Substitute and compute Tro: Chemistry: A Molecular Approach 2/e Approach, 73 Copyright © 2011 Pearson Education, Inc.
    • 322. Example 1.2: Convert 40.00 °C into K and °F • Find the equation that Given: 40.00 °C relates the given quantity to Find: K the quantity you want to find Equation: K = °C + 273.15 • Because the equation is K = °C + 273.15 solved for the quantity you K = 40.00 + 273.15 want to find, substitute and K = 313.15 K compute • Find the equation that Given: relates the given quantity to Find: the quantity you want to find Equation: • Solve the equation for the quantity you want to find • Substitute and compute Tro: Chemistry: A Molecular Approach 2/e Approach, 73 Copyright © 2011 Pearson Education, Inc.
    • 323. Example 1.2: Convert 40.00 °C into K and °F • Find the equation that Given: 40.00 °C relates the given quantity to Find: K the quantity you want to find Equation: K = °C + 273.15 • Because the equation is K = °C + 273.15 solved for the quantity you K = 40.00 + 273.15 want to find, substitute and K = 313.15 K compute • Find the equation that Given: 40.00 °C relates the given quantity to Find: the quantity you want to find Equation: • Solve the equation for the quantity you want to find • Substitute and compute Tro: Chemistry: A Molecular Approach 2/e Approach, 73 Copyright © 2011 Pearson Education, Inc.
    • 324. Example 1.2: Convert 40.00 °C into K and °F • Find the equation that Given: 40.00 °C relates the given quantity to Find: K the quantity you want to find Equation: K = °C + 273.15 • Because the equation is K = °C + 273.15 solved for the quantity you K = 40.00 + 273.15 want to find, substitute and K = 313.15 K compute • Find the equation that Given: 40.00 °C relates the given quantity to Find: °F the quantity you want to find Equation: • Solve the equation for the quantity you want to find • Substitute and compute Tro: Chemistry: A Molecular Approach 2/e Approach, 73 Copyright © 2011 Pearson Education, Inc.
    • 325. Example 1.2: Convert 40.00 °C into K and °F • Find the equation that Given: 40.00 °C relates the given quantity to Find: K the quantity you want to find Equation: K = °C + 273.15 • Because the equation is K = °C + 273.15 solved for the quantity you K = 40.00 + 273.15 want to find, substitute and K = 313.15 K compute • Find the equation that Given: 40.00 °C relates the given quantity to Find: °F the quantity you want to find Equation: • Solve the equation for the quantity you want to find • Substitute and compute Tro: Chemistry: A Molecular Approach 2/e Approach, 73 Copyright © 2011 Pearson Education, Inc.
    • 326. Example 1.2: Convert 40.00 °C into K and °F • Find the equation that Given: 40.00 °C relates the given quantity to Find: K the quantity you want to find Equation: K = °C + 273.15 • Because the equation is K = °C + 273.15 solved for the quantity you K = 40.00 + 273.15 want to find, substitute and K = 313.15 K compute • Find the equation that Given: 40.00 °C relates the given quantity to Find: °F the quantity you want to find Equation: • Solve the equation for the quantity you want to find • Substitute and compute Tro: Chemistry: A Molecular Approach 2/e Approach, 73 Copyright © 2011 Pearson Education, Inc.
    • 327. Example 1.2: Convert 40.00 °C into K and °F • Find the equation that Given: 40.00 °C relates the given quantity to Find: K the quantity you want to find Equation: K = °C + 273.15 • Because the equation is K = °C + 273.15 solved for the quantity you K = 40.00 + 273.15 want to find, substitute and K = 313.15 K compute • Find the equation that Given: 40.00 °C relates the given quantity to Find: °F the quantity you want to find Equation: • Solve the equation for the quantity you want to find • Substitute and compute Tro: Chemistry: A Molecular Approach 2/e Approach, 73 Copyright © 2011 Pearson Education, Inc.
    • 328. Practice – Convert 0.0°F into Kelvin Tro: Chemistry: A Molecular Approach 2/e Approach, 74 Copyright © 2011 Pearson Education, Inc.
    • 329. Practice – Convert 0.0°F into Kelvin • Sort Given: information Find: • Strategize Concept Plan: Equations: • Follow the Solution: concept plan to solve the problem • Sig. figs. and Round: round • Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 75 Copyright © 2011 Pearson Education, Inc.
    • 330. Practice – Convert 0.0°F into Kelvin • Sort Given: 0.0 °F information Find: • Strategize Concept Plan: Equations: • Follow the Solution: concept plan to solve the problem • Sig. figs. and Round: round • Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 75 Copyright © 2011 Pearson Education, Inc.
    • 331. Practice – Convert 0.0°F into Kelvin • Sort Given: 0.0 °F information Find: Kelvin • Strategize Concept Plan: Equations: • Follow the Solution: concept plan to solve the problem • Sig. figs. and Round: round • Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 75 Copyright © 2011 Pearson Education, Inc.
    • 332. Practice – Convert 0.0°F into Kelvin • Sort Given: 0.0 °F information Find: Kelvin • Strategize Concept °F °C K Plan: Equations: • Follow the Solution: concept plan to solve the problem • Sig. figs. and Round: round • Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 75 Copyright © 2011 Pearson Education, Inc.
    • 333. Practice – Convert 0.0°F into Kelvin • Sort Given: 0.0 °F information Find: Kelvin • Strategize Concept °F °C K Plan: Equations: • Follow the Solution: concept plan to solve the problem • Sig. figs. and Round: round • Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 75 Copyright © 2011 Pearson Education, Inc.
    • 334. Practice – Convert 0.0°F into Kelvin • Sort Given: 0.0 °F information Find: Kelvin • Strategize Concept °F °C K Plan: Equations: • Follow the Solution: concept plan to solve the problem • Sig. figs. and Round: round • Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 75 Copyright © 2011 Pearson Education, Inc.
    • 335. Practice – Convert 0.0°F into Kelvin • Sort Given: 0.0 °F information Find: Kelvin • Strategize Concept °F °C K Plan: Equations: • Follow the Solution: concept plan to solve the problem • Sig. figs. and Round: round • Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 75 Copyright © 2011 Pearson Education, Inc.
    • 336. Practice – Convert 0.0°F into Kelvin • Sort Given: 0.0 °F information Find: Kelvin • Strategize Concept °F °C K Plan: Equations: • Follow the Solution: concept plan to solve the problem • Sig. figs. and Round: round • Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 75 Copyright © 2011 Pearson Education, Inc.
    • 337. Practice – Convert 0.0°F into Kelvin • Sort Given: 0.0 °F information Find: Kelvin • Strategize Concept °F °C K Plan: Equations: • Follow the Solution: concept plan to solve the problem • Sig. figs. and Round: 255.37 K = 255 K round • Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 75 Copyright © 2011 Pearson Education, Inc.
    • 338. Practice – Convert 0.0°F into Kelvin • Sort Given: 0.0 °F information Find: Kelvin • Strategize Concept °F °C K Plan: Equations: • Follow the Solution: concept plan to solve the problem • Sig. figs. and Round: 255.37 K = 255 K round • Check Check: Because kelvin temperatures are always positive and generally between 250 and 300, the answer makes sense Tro: Chemistry: A Molecular Approach 2/e Approach, 75 Copyright © 2011 Pearson Education, Inc.
    • 339. Related Units in the SI System Tro: Chemistry: A Molecular Approach 2/e Approach, 76 Copyright © 2011 Pearson Education, Inc.
    • 340. Related Units in the SI System • All units in the SI system are related to the standard unit by a power of 10 Tro: Chemistry: A Molecular Approach 2/e Approach, 76 Copyright © 2011 Pearson Education, Inc.
    • 341. Related Units in the SI System • All units in the SI system are related to the standard unit by a power of 10 • The power of 10 is indicated by a prefix multiplier Tro: Chemistry: A Molecular Approach 2/e Approach, 76 Copyright © 2011 Pearson Education, Inc.
    • 342. Related Units in the SI System • All units in the SI system are related to the standard unit by a power of 10 • The power of 10 is indicated by a prefix multiplier • The prefix multipliers are always the same, regardless of the standard unit Tro: Chemistry: A Molecular Approach 2/e Approach, 76 Copyright © 2011 Pearson Education, Inc.
    • 343. Related Units in the SI System • All units in the SI system are related to the standard unit by a power of 10 • The power of 10 is indicated by a prefix multiplier • The prefix multipliers are always the same, regardless of the standard unit • Report measurements with a unit that is close to the size of the quantity being measured Tro: Chemistry: A Molecular Approach 2/e Approach, 76 Copyright © 2011 Pearson Education, Inc.
    • 344. Common Prefix Multipliers in the SI System Tro: Chemistry: A Molecular Approach 2/e Approach, 77 Copyright © 2011 Pearson Education, Inc.
    • 345. Volume Tro: Chemistry: A Molecular Approach 2/e Approach, 78 Copyright © 2011 Pearson Education, Inc.
    • 346. Volume • Derived unit Tro: Chemistry: A Molecular Approach 2/e Approach, 78 Copyright © 2011 Pearson Education, Inc.
    • 347. Volume • Derived unit  any length unit cubed Tro: Chemistry: A Molecular Approach 2/e Approach, 78 Copyright © 2011 Pearson Education, Inc.
    • 348. Volume • Derived unit  any length unit cubed • Measure of the amount of space occupied Tro: Chemistry: A Molecular Approach 2/e Approach, 78 Copyright © 2011 Pearson Education, Inc.
    • 349. Volume • Derived unit  any length unit cubed • Measure of the amount of space occupied • SI unit = cubic meter (m3) Tro: Chemistry: A Molecular Approach 2/e Approach, 78 Copyright © 2011 Pearson Education, Inc.
    • 350. Volume • Derived unit  any length unit cubed • Measure of the amount of space occupied • SI unit = cubic meter (m3) • Commonly measure solid volume in cubic centimeters (cm3) Tro: Chemistry: A Molecular Approach 2/e Approach, 78 Copyright © 2011 Pearson Education, Inc.
    • 351. Volume • Derived unit  any length unit cubed • Measure of the amount of space occupied • SI unit = cubic meter (m3) • Commonly measure solid volume in cubic centimeters (cm3)  1 m3 = 106 cm3 Tro: Chemistry: A Molecular Approach 2/e Approach, 78 Copyright © 2011 Pearson Education, Inc.
    • 352. Volume • Derived unit  any length unit cubed • Measure of the amount of space occupied • SI unit = cubic meter (m3) • Commonly measure solid volume in cubic centimeters (cm3)  1 m3 = 106 cm3  1 cm3 = 10−6 m3 = 0.000 001 m3 Tro: Chemistry: A Molecular Approach 2/e Approach, 78 Copyright © 2011 Pearson Education, Inc.
    • 353. Volume • Derived unit  any length unit cubed • Measure of the amount of space occupied • SI unit = cubic meter (m3) • Commonly measure solid volume in cubic centimeters (cm3)  1 m3 = 106 cm3  1 cm3 = 10−6 m3 = 0.000 001 m3 • Commonly measure liquid or gas volume in milliliters (mL) Tro: Chemistry: A Molecular Approach 2/e Approach, 78 Copyright © 2011 Pearson Education, Inc.
    • 354. Volume • Derived unit  any length unit cubed • Measure of the amount of space occupied • SI unit = cubic meter (m3) • Commonly measure solid volume in cubic centimeters (cm3)  1 m3 = 106 cm3  1 cm3 = 10−6 m3 = 0.000 001 m3 • Commonly measure liquid or gas volume in milliliters (mL)  1 L is slightly larger than 1 quart Tro: Chemistry: A Molecular Approach 2/e Approach, 78 Copyright © 2011 Pearson Education, Inc.
    • 355. Volume • Derived unit  any length unit cubed • Measure of the amount of space occupied • SI unit = cubic meter (m3) • Commonly measure solid volume in cubic centimeters (cm3)  1 m3 = 106 cm3  1 cm3 = 10−6 m3 = 0.000 001 m3 • Commonly measure liquid or gas volume in milliliters (mL)  1 L is slightly larger than 1 quart  1 L = 1 dm3 = 1000 mL = 103 mL Tro: Chemistry: A Molecular Approach 2/e Approach, 78 Copyright © 2011 Pearson Education, Inc.
    • 356. Volume • Derived unit  any length unit cubed • Measure of the amount of space occupied • SI unit = cubic meter (m3) • Commonly measure solid volume in cubic centimeters (cm3)  1 m3 = 106 cm3  1 cm3 = 10−6 m3 = 0.000 001 m3 • Commonly measure liquid or gas volume in milliliters (mL)  1 L is slightly larger than 1 quart  1 L = 1 dm3 = 1000 mL = 103 mL  1 mL = 0.001 L = 10−3 L Tro: Chemistry: A Molecular Approach 2/e Approach, 78 Copyright © 2011 Pearson Education, Inc.
    • 357. Common Units and Their Equivalents Length 1 kilometer (km) = 0.6214 mile (mi) 1 meter (m) = 39.37 inches (in.) 1 meter (m) = 1.094 yards (yd) 1 foot (ft) = 30.48 centimeters (cm) 1 inch (in.) = 2.54 centimeters (cm) exactly Tro: Chemistry: A Molecular Approach 2/e Approach, 79 Copyright © 2011 Pearson Education, Inc.
    • 358. Common Units and Their Equivalents Mass 1 kilogram (km) = 2.205 pounds (lb) 1 pound (lb) = 453.59 grams (g) 1 ounce (oz) = 28.35 grams (g) Volume 1 liter (L) = 1000 milliliters (mL) 1 liter (L) = 1000 cubic centimeters (cm3) 1 liter (L) = 1.057 quarts (qt) 1 U.S. gallon (gal) = 3.785 liters (L) Tro: Chemistry: A Molecular Approach 2/e Approach, 80 Copyright © 2011 Pearson Education, Inc.
    • 359. Practice — which of the following units would be best used for measuring the diameter of a quarter? a) kilometer b) meter c) centimeter d) micrometer e) megameters Tro: Chemistry: A Molecular Approach 2/e Approach, 81 Copyright © 2011 Pearson Education, Inc.
    • 360. Practice — which of the following units would be best used for measuring the diameter of a quarter? a) kilometer b) meter c) centimeter d) micrometer e) megameters Tro: Chemistry: A Molecular Approach 2/e Approach, 81 Copyright © 2011 Pearson Education, Inc.
    • 361. Chlorine vaporizes at –34.4 °C. What is this temperature in degrees Fahrenheit? Copyright © 2011 Pearson Education, Inc.
    • 362. Chlorine vaporizes at –34.4 °C. What is this temperature in degrees Fahrenheit? • –34.4 °F Copyright © 2011 Pearson Education, Inc.
    • 363. Chlorine vaporizes at –34.4 °C. What is this temperature in degrees Fahrenheit? • –34.4 °F • –29.9 °F Copyright © 2011 Pearson Education, Inc.
    • 364. Chlorine vaporizes at –34.4 °C. What is this temperature in degrees Fahrenheit? • –34.4 °F • –29.9 °F • 238.75 °F Copyright © 2011 Pearson Education, Inc.
    • 365. Chlorine vaporizes at –34.4 °C. What is this temperature in degrees Fahrenheit? • –34.4 °F • –29.9 °F • 238.75 °F • 307.55 °F Copyright © 2011 Pearson Education, Inc.
    • 366. Chlorine vaporizes at –34.4 °C. What is this temperature in degrees Fahrenheit? • –34.4 °F • –29.9 °F • 238.75 °F • 307.55 °F • 273.15 °F Copyright © 2011 Pearson Education, Inc.
    • 367. Chlorine vaporizes at –34.4 °C. What is this temperature in degrees Fahrenheit? Copyright © 2011 Pearson Education, Inc.
    • 368. Chlorine vaporizes at –34.4 °C. What is this temperature in degrees Fahrenheit? • –34.4 °F Copyright © 2011 Pearson Education, Inc.
    • 369. Chlorine vaporizes at –34.4 °C. What is this temperature in degrees Fahrenheit? • –34.4 °F • –29.9 °F Copyright © 2011 Pearson Education, Inc.
    • 370. Chlorine vaporizes at –34.4 °C. What is this temperature in degrees Fahrenheit? • –34.4 °F • –29.9 °F • 238.75 °F Copyright © 2011 Pearson Education, Inc.
    • 371. Chlorine vaporizes at –34.4 °C. What is this temperature in degrees Fahrenheit? • –34.4 °F • –29.9 °F • 238.75 °F • 307.55 °F Copyright © 2011 Pearson Education, Inc.
    • 372. Chlorine vaporizes at –34.4 °C. What is this temperature in degrees Fahrenheit? • –34.4 °F • –29.9 °F • 238.75 °F • 307.55 °F • 273.15 °F Copyright © 2011 Pearson Education, Inc.
    • 373. Density Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc.
    • 374. Intensive and Extensive Properties Tro: Chemistry: A Molecular Approach 2/e Approach, 85 Copyright © 2011 Pearson Education, Inc.
    • 375. Intensive and Extensive Properties • Extensive properties are properties whose value depends on the quantity of matter  extensive properties cannot be used to identify what type of matter something is if you are given a large glass containing 100 g of a clear, colorless liquid and a small glass containing 25 g of a clear, colorless liquid, are both liquids the same stuff? Tro: Chemistry: A Molecular Approach 2/e Approach, 85 Copyright © 2011 Pearson Education, Inc.
    • 376. Intensive and Extensive Properties • Extensive properties are properties whose value depends on the quantity of matter  extensive properties cannot be used to identify what type of matter something is if you are given a large glass containing 100 g of a clear, colorless liquid and a small glass containing 25 g of a clear, colorless liquid, are both liquids the same stuff? • Intensive properties are properties whose value is independent of the quantity of matter Tro: Chemistry: A Molecular Approach 2/e Approach, 85 Copyright © 2011 Pearson Education, Inc.
    • 377. Mass & Volume Volume vs. Mass of Brass 160 140 120 Mass, g 100 y = 8.38x 80 60 40 20 0 0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 Volume, cm3 Tro: Chemistry: A Molecular Approach 2/e Approach, 86 Copyright © 2011 Pearson Education, Inc.
    • 378. Mass & Volume • Two main physical properties of matter Volume vs. Mass of Brass 160 140 120 Mass, g 100 y = 8.38x 80 60 40 20 0 0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 Volume, cm3 Tro: Chemistry: A Molecular Approach 2/e Approach, 86 Copyright © 2011 Pearson Education, Inc.
    • 379. Mass & Volume • Two main physical properties of matter • Mass and volume are extensive properties Volume vs. Mass of Brass 160 140 120 Mass, g 100 y = 8.38x 80 60 40 20 0 0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 Volume, cm3 Tro: Chemistry: A Molecular Approach 2/e Approach, 86 Copyright © 2011 Pearson Education, Inc.
    • 380. Mass & Volume • Two main physical properties of matter • Mass and volume are extensive properties • Even though mass and volume are individual properties, for a given type of matter they are related to each other! Volume vs. Mass of Brass 160 140 120 Mass, g 100 y = 8.38x 80 60 40 20 0 0.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 Volume, cm3 Tro: Chemistry: A Molecular Approach 2/e Approach, 86 Copyright © 2011 Pearson Education, Inc.
    • 381. Density Tro: Chemistry: A Molecular Approach 2/e Approach, 87 Copyright © 2011 Pearson Education, Inc.
    • 382. Density • Density is the ratio of mass to volume Tro: Chemistry: A Molecular Approach 2/e Approach, 87 Copyright © 2011 Pearson Education, Inc.
    • 383. Density • Density is the ratio of mass to volume  is an intensive property Tro: Chemistry: A Molecular Approach 2/e Approach, 87 Copyright © 2011 Pearson Education, Inc.
    • 384. Density • Density is the ratio of mass to volume  is an intensive property • Solids = g/cm3 Tro: Chemistry: A Molecular Approach 2/e Approach, 87 Copyright © 2011 Pearson Education, Inc.
    • 385. Density • Density is the ratio of mass to volume  is an intensive property • Solids = g/cm3  1 cm3 = 1 mL Tro: Chemistry: A Molecular Approach 2/e Approach, 87 Copyright © 2011 Pearson Education, Inc.
    • 386. Density • Density is the ratio of mass to volume  is an intensive property • Solids = g/cm3  1 cm3 = 1 mL • Liquids = g/mL Tro: Chemistry: A Molecular Approach 2/e Approach, 87 Copyright © 2011 Pearson Education, Inc.
    • 387. Density • Density is the ratio of mass to volume  is an intensive property • Solids = g/cm3  1 cm3 = 1 mL • Liquids = g/mL • Gases = g/L Tro: Chemistry: A Molecular Approach 2/e Approach, 87 Copyright © 2011 Pearson Education, Inc.
    • 388. Density • Density is the ratio of mass to volume  is an intensive property • Solids = g/cm3  1 cm3 = 1 mL • Liquids = g/mL • Gases = g/L • Volume of a solid can be determined by water displacement – Archimedes principle Tro: Chemistry: A Molecular Approach 2/e Approach, 87 Copyright © 2011 Pearson Education, Inc.
    • 389. Density • Density is the ratio of mass to volume  is an intensive property • Solids = g/cm3  1 cm3 = 1 mL • Liquids = g/mL • Gases = g/L • Volume of a solid can be determined by water displacement – Archimedes principle • Density : solids > liquids >>> gases Tro: Chemistry: A Molecular Approach 2/e Approach, 87 Copyright © 2011 Pearson Education, Inc.
    • 390. Density • Density is the ratio of mass to volume  is an intensive property • Solids = g/cm3  1 cm3 = 1 mL • Liquids = g/mL • Gases = g/L • Volume of a solid can be determined by water displacement – Archimedes principle • Density : solids > liquids >>> gases  except ice is less dense than liquid water! Tro: Chemistry: A Molecular Approach 2/e Approach, 87 Copyright © 2011 Pearson Education, Inc.
    • 391. Density • Density is the ratio of mass to volume  is an intensive property • Solids = g/cm3  1 cm3 = 1 mL • Liquids = g/mL • Gases = g/L • Volume of a solid can be determined by water displacement – Archimedes principle • Density : solids > liquids >>> gases  except ice is less dense than liquid water! Tro: Chemistry: A Molecular Approach 2/e Approach, 87 Copyright © 2011 Pearson Education, Inc.
    • 392. Density Tro: Chemistry: A Molecular Approach 2/e Approach, 88 Copyright © 2011 Pearson Education, Inc.
    • 393. Density • For equal volumes, denser object has larger mass Tro: Chemistry: A Molecular Approach 2/e Approach, 88 Copyright © 2011 Pearson Education, Inc.
    • 394. Density • For equal volumes, denser object has larger mass • For equal masses, denser object has smaller volume Tro: Chemistry: A Molecular Approach 2/e Approach, 88 Copyright © 2011 Pearson Education, Inc.
    • 395. Density • For equal volumes, denser object has larger mass • For equal masses, denser object has smaller volume • Heating an object generally causes it to expand, therefore the density changes with temperature Tro: Chemistry: A Molecular Approach 2/e Approach, 88 Copyright © 2011 Pearson Education, Inc.
    • 396. Example 1.3: Decide if a ring with a mass of 3.15 g that displaces 0.233 cm3 of water is platinum • Write down the given Given: quantities and the quantity you want to find Find: • Find the equation that Equation: relates the given quantity to the quantity you want to find • Solve the equation for the quantity you want to find, check the units are correct, then substitute and compute • Compare to accepted value of the intensive property Tro: Chemistry: A Molecular Approach 2/e Approach, 89 Copyright © 2011 Pearson Education, Inc.
    • 397. Example 1.3: Decide if a ring with a mass of 3.15 g that displaces 0.233 cm3 of water is platinum • Write down the given Given: mass = 3.15 g quantities and the quantity you want to find Find: • Find the equation that Equation: relates the given quantity to the quantity you want to find • Solve the equation for the quantity you want to find, check the units are correct, then substitute and compute • Compare to accepted value of the intensive property Tro: Chemistry: A Molecular Approach 2/e Approach, 89 Copyright © 2011 Pearson Education, Inc.
    • 398. Example 1.3: Decide if a ring with a mass of 3.15 g that displaces 0.233 cm3 of water is platinum • Write down the given Given: mass = 3.15 g quantities and the quantity volume = 0.233 cm3 you want to find Find: • Find the equation that Equation: relates the given quantity to the quantity you want to find • Solve the equation for the quantity you want to find, check the units are correct, then substitute and compute • Compare to accepted value of the intensive property Tro: Chemistry: A Molecular Approach 2/e Approach, 89 Copyright © 2011 Pearson Education, Inc.
    • 399. Example 1.3: Decide if a ring with a mass of 3.15 g that displaces 0.233 cm3 of water is platinum • Write down the given Given: mass = 3.15 g quantities and the quantity volume = 0.233 cm3 you want to find Find: density, g/cm3 • Find the equation that Equation: relates the given quantity to the quantity you want to find • Solve the equation for the quantity you want to find, check the units are correct, then substitute and compute • Compare to accepted value of the intensive property Tro: Chemistry: A Molecular Approach 2/e Approach, 89 Copyright © 2011 Pearson Education, Inc.
    • 400. Example 1.3: Decide if a ring with a mass of 3.15 g that displaces 0.233 cm3 of water is platinum • Write down the given Given: mass = 3.15 g quantities and the quantity volume = 0.233 cm3 you want to find Find: density, g/cm3 • Find the equation that Equation: relates the given quantity to the quantity you want to find • Solve the equation for the quantity you want to find, check the units are correct, then substitute and compute • Compare to accepted value of the intensive property Tro: Chemistry: A Molecular Approach 2/e Approach, 89 Copyright © 2011 Pearson Education, Inc.
    • 401. Example 1.3: Decide if a ring with a mass of 3.15 g that displaces 0.233 cm3 of water is platinum • Write down the given Given: mass = 3.15 g quantities and the quantity volume = 0.233 cm3 you want to find Find: density, g/cm3 • Find the equation that Equation: relates the given quantity to the quantity you want to find • Solve the equation for the quantity you want to find, check the units are correct, then substitute and compute • Compare to accepted value of the intensive property Tro: Chemistry: A Molecular Approach 2/e Approach, 89 Copyright © 2011 Pearson Education, Inc.
    • 402. Example 1.3: Decide if a ring with a mass of 3.15 g that displaces 0.233 cm3 of water is platinum • Write down the given Given: mass = 3.15 g quantities and the quantity volume = 0.233 cm3 you want to find Find: density, g/cm3 • Find the equation that Equation: relates the given quantity to the quantity you want to find • Solve the equation for the quantity you want to find, check the units are correct, then substitute and compute • Compare to accepted value Density of platinum = of the intensive property 21.4 g/cm3 therefore not platinum Tro: Chemistry: A Molecular Approach 2/e Approach, 89 Copyright © 2011 Pearson Education, Inc.
    • 403. Calculating Density Tro: Chemistry: A Molecular Approach 2/e Approach, 90 Copyright © 2011 Pearson Education, Inc.
    • 404. Calculating Density • What is the density of a brass sample if 100.0 g added to a cylinder of water causes the water level to rise from 25.0 mL to 36.9 mL? Tro: Chemistry: A Molecular Approach 2/e Approach, 90 Copyright © 2011 Pearson Education, Inc.
    • 405. Practice — What is the density of the brass sample? Sort Given: information Find: Strategize Concept Plan: Equation: Solve the equation for the unknown variable Sig. figs. and Round: round Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 91 Copyright © 2011 Pearson Education, Inc.
    • 406. Practice — What is the density of the brass sample? Sort Given: mass = 100 g information Find: Strategize Concept Plan: Equation: Solve the equation for the unknown variable Sig. figs. and Round: round Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 91 Copyright © 2011 Pearson Education, Inc.
    • 407. Practice — What is the density of the brass sample? Sort Given: mass = 100 g information vol displ: 25.0 → 36.9 mL Find: Strategize Concept Plan: Equation: Solve the equation for the unknown variable Sig. figs. and Round: round Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 91 Copyright © 2011 Pearson Education, Inc.
    • 408. Practice — What is the density of the brass sample? Sort Given: mass = 100 g information vol displ: 25.0 → 36.9 mL Find: d, g/cm3 Strategize Concept Plan: Equation: Solve the equation for the unknown variable Sig. figs. and Round: round Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 91 Copyright © 2011 Pearson Education, Inc.
    • 409. Practice — What is the density of the brass sample? Sort Given: mass = 100 g information vol displ: 25.0 → 36.9 mL Find: d, g/cm3 Strategize Concept m, V d Plan: Equation: Solve the equation for the unknown variable Sig. figs. and Round: round Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 91 Copyright © 2011 Pearson Education, Inc.
    • 410. Practice — What is the density of the brass sample? Sort Given: mass = 100 g information vol displ: 25.0 → 36.9 mL Find: d, g/cm3 Strategize Concept m, V d Plan: Equation: Solve the equation for the unknown variable Sig. figs. and Round: round Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 91 Copyright © 2011 Pearson Education, Inc.
    • 411. Practice — What is the density of the brass sample? Sort Given: mass = 100 g information vol displ: 25.0 → 36.9 mL Find: d, g/cm3 Strategize Concept m, V d Plan: Equation: Solve the Solution: equation for V = 36.9−25.0 the unknown = 11.9 mL variable Sig. figs. and Round: round Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 91 Copyright © 2011 Pearson Education, Inc.
    • 412. Practice — What is the density of the brass sample? Sort Given: mass = 100 g information vol displ: 25.0 → 36.9 mL Find: d, g/cm3 Strategize Concept m, V d Plan: Equation: Solve the Solution: equation for V = 36.9−25.0 the unknown = 11.9 mL variable = 11.9 cm3 Sig. figs. and Round: round Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 91 Copyright © 2011 Pearson Education, Inc.
    • 413. Practice — What is the density of the brass sample? Sort Given: mass = 100 g information vol displ: 25.0 → 36.9 mL Find: d, g/cm3 Strategize Concept m, V d Plan: Equation: Solve the Solution: equation for V = 36.9−25.0 the unknown = 11.9 mL variable = 11.9 cm3 Sig. figs. and Round: round Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 91 Copyright © 2011 Pearson Education, Inc.
    • 414. Practice — What is the density of the brass sample? Sort Given: mass = 100 g information vol displ: 25.0 → 36.9 mL Find: d, g/cm3 Strategize Concept m, V d Plan: Equation: Solve the Solution: equation for V = 36.9−25.0 the unknown = 11.9 mL variable = 11.9 cm3 Sig. figs. and Round: round 8.4033 g/cm3 = 8.40 g/cm3 Check Check: Tro: Chemistry: A Molecular Approach 2/e Approach, 91 Copyright © 2011 Pearson Education, Inc.
    • 415. Practice — What is the density of the brass sample? Sort Given: mass = 100 g information vol displ: 25.0 → 36.9 mL Find: d, g/cm3 Strategize Concept m, V d Plan: Equation: Solve the Solution: equation for V = 36.9−25.0 the unknown = 11.9 mL variable = 11.9 cm3 Sig. figs. and Round: round 8.4033 g/cm3 = 8.40 g/cm3 Check Check: units and number make sense Tro: Chemistry: A Molecular Approach 2/e Approach, 91 Copyright © 2011 Pearson Education, Inc.
    • 416. Which of the following has the largest density? Copyright © 2011 Pearson Education, Inc.
    • 417. Which of the following has the largest density? • a material that has a mass of 10.0 g and a volume of 2.00 L Copyright © 2011 Pearson Education, Inc.
    • 418. Which of the following has the largest density? • a material that has a mass of 10.0 g and a volume of 2.00 L • a material that has a mass of 5.00 g and a volume of 10.0 cm3 Copyright © 2011 Pearson Education, Inc.
    • 419. Which of the following has the largest density? • a material that has a mass of 10.0 g and a volume of 2.00 L • a material that has a mass of 5.00 g and a volume of 10.0 cm3 • a material that sinks in ethanol but floats on water Copyright © 2011 Pearson Education, Inc.
    • 420. Which of the following has the largest density? Copyright © 2011 Pearson Education, Inc.
    • 421. Which of the following has the largest density? • a material that has a mass of 10.0 g and a volume of 2.00 L Copyright © 2011 Pearson Education, Inc.
    • 422. Which of the following has the largest density? • a material that has a mass of 10.0 g and a volume of 2.00 L • a material that has a mass of 5.00 g and a volume of 10.0 cm3 Copyright © 2011 Pearson Education, Inc.
    • 423. Which of the following has the largest density? • a material that has a mass of 10.0 g and a volume of 2.00 L • a material that has a mass of 5.00 g and a volume of 10.0 cm3 • a material that sinks in ethanol but floats on water Copyright © 2011 Pearson Education, Inc.
    • 424. Which of the following would NOT be considered an intensive property describing an unknown sample? Copyright © 2011 Pearson Education, Inc.
    • 425. Which of the following would NOT be considered an intensive property describing an unknown sample? • It is a solid at 25 °C. Copyright © 2011 Pearson Education, Inc.
    • 426. Which of the following would NOT be considered an intensive property describing an unknown sample? • It is a solid at 25 °C. • It has a density of 1.38 g/ cm3. Copyright © 2011 Pearson Education, Inc.
    • 427. Which of the following would NOT be considered an intensive property describing an unknown sample? • It is a solid at 25 °C. • It has a density of 1.38 g/ cm3. • It melts at 62.0 °C. Copyright © 2011 Pearson Education, Inc.
    • 428. Which of the following would NOT be considered an intensive property describing an unknown sample? • It is a solid at 25 °C. • It has a density of 1.38 g/ cm3. • It melts at 62.0 °C. • It has a volume of 0.52 cm3. Copyright © 2011 Pearson Education, Inc.
    • 429. Which of the following would NOT be considered an intensive property describing an unknown sample? • It is a solid at 25 °C. • It has a density of 1.38 g/ cm3. • It melts at 62.0 °C. • It has a volume of 0.52 cm3. • It is shiny. Copyright © 2011 Pearson Education, Inc.
    • 430. Which of the following would NOT be considered an intensive property describing an unknown sample? Copyright © 2011 Pearson Education, Inc.
    • 431. Which of the following would NOT be considered an intensive property describing an unknown sample? • It is a solid at 25 °C. Copyright © 2011 Pearson Education, Inc.
    • 432. Which of the following would NOT be considered an intensive property describing an unknown sample? • It is a solid at 25 °C. • It has a density of 1.38 g/ cm3. Copyright © 2011 Pearson Education, Inc.
    • 433. Which of the following would NOT be considered an intensive property describing an unknown sample? • It is a solid at 25 °C. • It has a density of 1.38 g/ cm3. • It melts at 62.0 °C. Copyright © 2011 Pearson Education, Inc.
    • 434. Which of the following would NOT be considered an intensive property describing an unknown sample? • It is a solid at 25 °C. • It has a density of 1.38 g/ cm3. • It melts at 62.0 °C. • It has a volume of 0.52 cm3. Copyright © 2011 Pearson Education, Inc.
    • 435. Which of the following would NOT be considered an intensive property describing an unknown sample? • It is a solid at 25 °C. • It has a density of 1.38 g/ cm3. • It melts at 62.0 °C. • It has a volume of 0.52 cm3. • It is shiny. Copyright © 2011 Pearson Education, Inc.

    ×