1. © 2015 Pearson Education, Inc. Clicker Questions Chapter 1 Barbara Mowery York College

2. © 2015 Pearson Education, Inc. Apple juice is an example of a(an) a.element. b.compound. c.homogeneous mixture. d.heterogeneous mixture.

3. © 2015 Pearson Education, Inc. Apple juice is an example of a(an) a.element. b.compound. c.homogeneous mixture. d.heterogeneous mixture.

4. © 2015 Pearson Education, Inc. Which of the following is not a pure substance? a.water b.carbon dioxide c.carbon d.air

5. © 2015 Pearson Education, Inc. Which of the following is not a pure substance? a.water b.carbon dioxide c.carbon d.air

6. © 2015 Pearson Education, Inc. Solutions may be a.solids. b.liquids. c.gases. d.All of the above

7. © 2015 Pearson Education, Inc. Solutions may be a.solids. b.liquids. c.gases. d.All of the above

8. © 2015 Pearson Education, Inc. Descriptions of a sample that are based on the amount of substance are called a.physical properties. b.chemical properties. c.intensive properties. d.extensive properties.

9. © 2015 Pearson Education, Inc. Descriptions of a sample that are based on the amount of substance are called a.physical properties. b.chemical properties. c.intensive properties. d.extensive properties.

10. © 2015 Pearson Education, Inc. Properties that describe the way a substance reacts to form other substances are called a.physical properties. b.chemical properties. c.homogeneous properties. d.heterogeneous properties.

11. © 2015 Pearson Education, Inc. Properties that describe the way a substance reacts to form other substances are called a.physical properties. b.chemical properties. c.homogeneous properties. d.heterogeneous properties.

12. © 2015 Pearson Education, Inc. Acetic acid that was liquid when stored was solid when a student looked for it on a cold morning. This illustrates a.a change in solubility. b.a chemical change. c.an equilibrium state. d.a physical change.

13. © 2015 Pearson Education, Inc. Acetic acid that was liquid when stored was solid when a student looked for it on a cold morning. This illustrates a.a change in solubility. b.a chemical change. c.an equilibrium state. d.a physical change.

14. © 2015 Pearson Education, Inc. A copper wire placed in a silver nitrate solution turns dark and “fuzzy.” This illustrates a.an accident. b.a chemical reaction. c.an extensive property. d.a physical change.

15. © 2015 Pearson Education, Inc. A copper wire placed in a silver nitrate solution turns dark and “fuzzy.” This illustrates a.an accident. b.a chemical reaction. c.an extensive property. d.a physical change.

16. © 2015 Pearson Education, Inc. Which statement explains the difference between a scientific law (X) and a scientific theory (Y)? a.X is proven; Y is not proven. b.X is not proven; Y is proven. c.X tells what happens; Y explains why things happen. d.X explains why things happen; Y tells what happens.

17. © 2015 Pearson Education, Inc. Which statement explains the difference between a scientific law (X) and a scientific theory (Y)? a.X is proven; Y is not proven. b.X is not proven; Y is proven. c.X tells what happens; Y explains why things happen. d.X explains why things happen; Y tells what happens.

18. © 2015 Pearson Education, Inc. When nitric acid is mixed with copper metal, a brown gas forms. This is an example of a.an accident. b.a chemical reaction. c.a physical property. d.an extensive property.

19. © 2015 Pearson Education, Inc. When nitric acid is mixed with copper metal, a brown gas forms. This is an example of a.an accident. b.a chemical reaction. c.a physical property. d.an extensive property.

20. © 2015 Pearson Education, Inc. A solution with a boiling point of 105 degrees Celsius contains either sugar or salt. How would you determine which is present? a.distill the solution b.filter the solution c.use chromatography d.taste the solution

21. © 2015 Pearson Education, Inc. A solution with a boiling point of 105 degrees Celsius contains either sugar or salt. How would you determine which is present? a.distill the solution b.filter the solution c.use chromatography d.taste the solution

22. © 2015 Pearson Education, Inc. Heat flows from an object a.at high temperature to an object at low temperature. b.at low temperature to an object at high temperature. c.to another object at the same temperature. d.at high elevation to an object at low elevation.

23. © 2015 Pearson Education, Inc. Heat flows from an object a.at high temperature to an object at low temperature. b.at low temperature to an object at high temperature. c.to another object at the same temperature. d.at high elevation to an object at low elevation.

24. © 2015 Pearson Education, Inc. In three trials, a student measures a sample’s mass to be to be 0.100 g, 0.600 g, and 0.300 g. The accepted value is 0.340 g. The student’s data have a.good accuracy and good precision. b.good accuracy, but poor precision. c.poor accuracy, but good precision. d.poor accuracy and poor precision.

25. © 2015 Pearson Education, Inc. a.good accuracy and good precision. b.good accuracy, but poor precision. c.poor accuracy, but good precision. d.poor accuracy and poor precision. In three trials, a student measures a sample’s mass to be to be 0.100 g, 0.600 g, and 0.300 g. The accepted value is 0.340 g. The student’s data have

26. © 2015 Pearson Education, Inc. The measured quantity 0.0860 g contains _____ significant figures. a.three b.four c.five d.six

27. © 2015 Pearson Education, Inc. a.three b.four c.five d.six The measured quantity 0.0860 g contains _____ significant figures.

28. © 2015 Pearson Education, Inc. 4.1 g + 7.08 g = _____ g a.11.180 b.11.18 c.11.2 d.11

29. © 2015 Pearson Education, Inc. 4.1 g + 7.08 g = _____ g a.11.180 b.11.18 c.11.2 d.11

30. © 2015 Pearson Education, Inc. 6.9 g – 5.07 g = _____ g a.1.830 b.1.83 c.1.8 d.1

31. © 2015 Pearson Education, Inc. 6.9 g – 5.07 g = _____ g a.1.830 b.1.83 c.1.8 d.1

32. © 2015 Pearson Education, Inc. 5.2 cm × 7.01 cm = ____ cm 2 a.36 b.36.4 c.36.45 d.36.452

33. © 2015 Pearson Education, Inc. 5.2 cm × 7.01 cm = ____ cm 2 a.36 b.36.4 c.36.45 d.36.452

34. © 2015 Pearson Education, Inc. 6.03 g ÷ 7.1 mL = _____ g/mL a.0.8 b.0.85 c.0.849 d.0.849257

35. © 2015 Pearson Education, Inc. 6.03 g ÷ 7.1 mL = _____ g/mL a.0.8 b.0.85 c.0.849 d.0.849257

36. © 2015 Pearson Education, Inc. 120 mL = _____ fl oz a.2 b.4 c.6 d.8

37. © 2015 Pearson Education, Inc. 120 mL = _____ fl oz a.2 b.4 c.6 d.8

38. © 2015 Pearson Education, Inc. A temperature of –40 degrees Celsius is the same as (X) degrees Fahrenheit and (Y) kelvins. a.X = –40, Y = 233 b.X = 233, Y = –40 c.X = 10, Y = 233 d.X = 233, Y = 10

39. © 2015 Pearson Education, Inc. A temperature of –40 degrees Celsius is the same as (X) degrees Fahrenheit and (Y) kelvins. a.X = –40, Y = 233 b.X = 233, Y = –40 c.X = 10, Y = 233 d.X = 233, Y = 10

40. © 2015 Pearson Education, Inc. Which measurement is expressed in non-SI units? a.12.9 Angstroms (Å) b.47.2 milliliters (mL) c.16.8 kilograms (kg) d.42.6 micrometers (µm)

41. © 2015 Pearson Education, Inc. Which measurement is expressed in non-SI units? a.12.9 Angstroms (Å) b.47.2 milliliters (mL) c.16.8 kilograms (kg) d.42.6 micrometers (µm)

42. © 2015 Pearson Education, Inc. Which unit is not an SI base unit? a.kilogram b.degree Celsius c.meter d.second

43. © 2015 Pearson Education, Inc. Which unit is not an SI base unit? a.kilogram b.degree Celsius c.meter d.second