Chapter 2—Matter and Energy - Simeon Career Academy



Chapter 2—Matter and Energy

MULTIPLE CHOICE

1. The energy that results from the breaking or formation of chemical bonds is

|a. |temperature. |c. |chemical energy. |

|b. |potential energy. |d. |kinetic energy. |

ANS: C PTS: 1 DIF: II OBJ: 2.1.1

2. “In any chemical or physical process, energy is neither created nor destroyed” is a statement of the

|a. |law of conservation of energy. |c. |law of physical and chemical change. |

|b. |law of conservation of mass. |d. |system. |

ANS: A PTS: 1 DIF: I OBJ: 2.1.2

3. The specific heat of a given substance

|a. |depends on the mass of the substance. |

|b. |varies with temperature. |

|c. |is the same as that for any other substance. |

|d. |is unique to that substance. |

ANS: D PTS: 1 DIF: I OBJ: 2.1.3

4. A temperature of 323 K is equivalent to

|a. |50°C. |c. |−50°C. |

|b. |1.2°C. |d. |596°C. |

ANS: A PTS: 1 DIF: I OBJ: 2.1.4

5. The temperature that is equivalent to 20°C is

|a. |253 K. |c. |−293 K. |

|b. |293 K. |d. |13.7 K. |

ANS: B PTS: 1 DIF: I OBJ: 2.1.3

6. For every investigation, the scientific method

|a. |is the same set of procedures. |

|b. |is a logical set of procedures. |

|c. |must be abandoned if there are unexpected results. |

|d. |helps to predict what results will be obtained. |

ANS: B PTS: 1 DIF: I OBJ: 2.2.1

7. The reason for organizing, analyzing, and classifying data is

|a. |so that computers can be used. |

|b. |to prove a law. |

|c. |to find relationships among the data. |

|d. |to separate qualitative and quantitative data. |

ANS: C PTS: 1 DIF: I OBJ: 2.2.1

[pic]

8. There are _____ variables represented in the weather graphs.

|a. |four |c. |six |

|b. |five |d. |seven |

ANS: A PTS: 1 DIF: I OBJ: 2.2.2

[pic]

9. Of the items depicted, only _____ is not a model.

|a. |figure a |c. |figure c |

|b. |figure b |d. |figure d |

ANS: D PTS: 1 DIF: II OBJ: 2.2.3

10. A plausible explanation of a body of observed natural phenomena is a scientific

|a. |principle. |c. |law. |

|b. |experiment. |d. |theory. |

ANS: D PTS: 1 DIF: I OBJ: 2.2.3

[pic]

11. The matter diagram represents a(n)

|a. |law. |c. |assumption. |

|b. |theory. |d. |hypothesis. |

ANS: B PTS: 1 DIF: I OBJ: 2.2.3

12. A theory is an accepted explanation of an observed phenomenon until

|a. |one study conflicts with the theory. |

|b. |repeated data and observation conflict with the theory. |

|c. |scientists disagree about the research method used to gather data. |

|d. |an eminent scientist feels that it is inadequate. |

ANS: B PTS: 1 DIF: II OBJ: 2.2.3

13. Poor precision in scientific measurement may arise from

|a. |the standard being too strict. |

|b. |human error. |

|c. |limitations of the measuring instrument. |

|d. |both human error and the limitations of the measuring instrument. |

ANS: D PTS: 1 DIF: II OBJ: 2.3.1

14. Precision pertains to all of the following except

|a. |reproducibility of measurements. |

|b. |agreement among numerical values. |

|c. |sameness of measurements. |

|d. |closeness of a measurement to an accepted value. |

ANS: D PTS: 1 DIF: I OBJ: 2.3.1

15. Five darts strike near the center of the target. Whoever threw the darts is

|a. |accurate. |c. |both accurate and precise. |

|b. |precise. |d. |neither accurate nor precise. |

ANS: C PTS: 1 DIF: II OBJ: 2.3.1

16. A chemist who frequently carries out a complex experiment is likely to have high

|a. |accuracy, but low precision. |c. |precision. |

|b. |accuracy. |d. |precision, but low accuracy. |

ANS: C PTS: 1 DIF: II OBJ: 2.3.1

17. When applied to scientific measurements, the words accuracy and precision

|a. |are used interchangeably. |c. |can cause uncertainty in experiments. |

|b. |have limitations. |d. |have distinctly different meanings. |

ANS: D PTS: 1 DIF: I OBJ: 2.3.1

18. When determining the number of significant digits in a measurement,

|a. |all zeros are significant. |

|b. |all nonzero digits are significant. |

|c. |all zeros between two nonzero digits are not significant. |

|d. |all nonzero digits are not significant. |

ANS: B PTS: 1 DIF: I OBJ: 2.3.2

19. To two significant figures, the measurement 0.0255 g should be reported as

|a. |0.02 g. |c. |0.026 g. |

|b. |0.025 g. |d. |2.5 × 102 g. |

ANS: C PTS: 1 DIF: II OBJ: 2.3.2

20. In division and multiplication, the answer must not have more significant figures than the

|a. |number in the calculation with the fewest significant figures. |

|b. |number in the calculation with the most significant figures. |

|c. |average number of significant figures in the calculation. |

|d. |total number of significant figures in the calculation. |

ANS: A PTS: 1 DIF: I OBJ: 2.3.2

21. The number of significant figures in the measurement 170.040 km is

|a. |3. |c. |5. |

|b. |4. |d. |6. |

ANS: D PTS: 1 DIF: II OBJ: 2.3.2

22. The measurement that has been expressed to four significant figures is

|a. |0.0020 mm. |c. |30.00 mm. |

|b. |0.004 02 mm. |d. |402.10 mm. |

ANS: C PTS: 1 DIF: II OBJ: 2.3.2

23. What is the density of 37.72 g of matter whose volume is 6.80 cm3?

|a. |0.18 g/cm3 |c. |30.92 g/cm3 |

|b. |5.55 g/cm3 |d. |256.4 g/cm3 |

ANS: B PTS: 1 DIF: III OBJ: 2.3.2

24. The dimensions of a rectangular solid are measured to be 1.27 cm, 1.3 cm, and 2.5 cm. The volume should be recorded as

|a. |4.128 cm3. |c. |4.13 cm3. |

|b. |4.12 cm3. |d. |4.1 cm3. |

ANS: D PTS: 1 DIF: III OBJ: 2.3.2

25. Three samples of 0.12 g, 1.8 g, and 0.562 g are mixed together. The combined mass of all three samples, expressed to the correct number of significant figures, should be recorded as

|a. |2.4 g. |c. |2.482 g. |

|b. |2.48 g. |d. |2.5 g. |

ANS: D PTS: 1 DIF: III OBJ: 2.1.2

26. Expressed in scientific notation, 0.0930 m is

|a. |93 × 10–3 m. |c. |9.30 × 10–2 m. |

|b. |9.3 × 10–3 m. |d. |9.30 × 10–4 m. |

ANS: C PTS: 1 DIF: III OBJ: 2.3.4

27. When 6.02 × 1023 is multiplied by 9.1 × 10–31, the product is

|a. |5.5 × 10–8. |c. |5.5 × 10–7. |

|b. |5.5 × 1054. |d. |5.5 × 10–53. |

ANS: C PTS: 1 DIF: III OBJ: 2.3.4

COMPLETION

1. The energy that matter has because of its motion is called ____________________.

ANS: kinetic energy

PTS: 1 DIF: I OBJ: 2.1.1

2. The capacity to do work or produce light is known as ____________________.

ANS: energy

PTS: 1 DIF: I OBJ: 2.1.1

3. A chemical reaction that releases energy is a(n) ____________________ reaction.

ANS: exothermic

PTS: 1 DIF: I OBJ: 2.1.1

4. The fact that mass and energy cannot be created or destroyed is the fundamental concept of the theme of ____________________.

ANS: conservation

PTS: 1 DIF: I OBJ: 2.1.2

5. A measure of the average kinetic energy of the particles in a sample of matter is its ____________________.

ANS: temperature

PTS: 1 DIF: I OBJ: 2.1.3

6. Specific heat is the amount of heat energy needed to raise the temperature of ____________________ of a substance 1 K.

ANS: one gram

PTS: 1 DIF: I OBJ: 2.1.3

7. In using the scientific method, after analyzing results, the next step is to ____________________.

ANS: draw conclusions

PTS: 1 DIF: II OBJ: 2.2.1

8. A proposed explanation of observations is a(n) ____________________.

ANS: hypothesis

PTS: 1 DIF: I OBJ: 2.2.1

9. A factor that could affect the results of an experiment is a(n) ____________________.

ANS: variable

PTS: 1 DIF: I OBJ: 2.2.2

10. A well-tested explanation of observations is a(n) ____________________.

ANS: theory

PTS: 1 DIF: I OBJ: 2.2.3

11. A statement or mathematical expression that reliably describes a behavior of the natural world is a(n) ____________________.

ANS: law

PTS: 1 DIF: I OBJ: 2.2.3

12. A testable statement used for making predictions and carrying out further experiments is a(n) ____________________.

ANS: hypothesis

PTS: 1 DIF: I OBJ: 2.2.3

13. When a candle burns, the total mass of the candle and the oxygen equals the total mass of the gases and ashes produced. This reaction illustrates the law of ____________________.

ANS: conservation of mass

PTS: 1 DIF: II OBJ: 2.2.3

14. A laboratory balance is not adjusted correctly, and each mass measured is 0.54 g too high. A student finds the mass of a sample several times, and the masses are close to each other. The results are ____________________ but not ____________________ .

ANS: precise, accurate

PTS: 1 DIF: II OBJ: 2.3.1

15. A measurement that closely agrees with accepted values is said to be ____________________.

ANS: accurate

PTS: 1 DIF: I OBJ: 2.3.1

16. If some measurements agree closely with each other but differ widely from the actual value, these measurements are ____________________ but not ____________________.

ANS: precise, accurate

PTS: 1 DIF: II OBJ: 2.3.1

17. The number of significant figures in the measured value 0.003 20 g is ____________________.

ANS: 3

PTS: 1 DIF: II OBJ: 2.3.2

18. The number of significant figures in the measurement 210 cm is ____________________.

ANS: 2

PTS: 1 DIF: II OBJ: 2.3.2

19. Using a metric ruler with 1-mm divisions, you find the sides of a rectangular piece of plywood are 3.54 cm and 4.85 cm. You calculate that the area is 17.1690 cm2. To the correct number of significant figures, the result should be expressed as ____________________.

ANS: 17.2 cm2

PTS: 1 DIF: II OBJ: 2.3.2

20. When 64.4 is divided by 2.00, the correct number of significant figures in the result is ____________________.

ANS: 3

PTS: 1 DIF: II OBJ: 2.3.2

21. The sum of 314.53 km and 32 km is correctly expressed as ____________________.

ANS: 347 km

PTS: 1 DIF: II OBJ: 2.3.2

22. Rounded to four significant figures, 1.245 633 501 × 108 is ____________________.

ANS: 1.246 × 108

PTS: 1 DIF: II OBJ: 2.3.2

23. Divide 5.7 m by 2 m. The quotient is correctly reported as ____________________.

ANS: 3

PTS: 1 DIF: II OBJ: 2.3.2

24. The quantity of heat transferred during a temperature change depends on the ____________________ and mass of the material and the ____________________ of the temperature change.

ANS: nature, size

PTS: 1 DIF: II OBJ: 2.3.3

25. The speed of light is 300 000 km/s. In scientific notation, this speed is ____________________.

ANS: 3 × 105 km / s

PTS: 1 DIF: II OBJ: 2.3.4

26. The average distance between Earth and the moon is 386 000 km. Expressed in scientific notation, this distance is ____________________.

ANS: 3.86 × 105

PTS: 1 DIF: II OBJ: 2.3.4

27. An analytical balance can measure mass to the nearest 1/10 000 of a gram, 0.0001 g. In scientific notation, the accuracy of the balance would be expressed as ____________________.

ANS: 1 × 10–4

PTS: 1 DIF: II OBJ: 2.3.4

28. The result of dividing 107 by 10–3 is ____________________.

ANS: 1010

PTS: 1 DIF: II OBJ: 2.3.4

SHORT ANSWER

1. Describe the energy change involved when an egg cooks.

ANS:

Energy is absorbed by the raw egg. The absorbed energy is stored as chemical energy in the cooked egg.

PTS: 1 DIF: II OBJ: 2.1.1

2. Explain the difference in an endothermic process and an exothermic process.

ANS:

An endothermic process absorbs energy, and an exothermic process releases energy.

PTS: 1 DIF: I OBJ: 2.1.1

3. Explain how energy is conserved when a piece of paper burns.

ANS:

The total energy contained in the bonds of the paper and oxygen equals the total energy contained in the bonds of the ashes and gases produced and the heat energy released.

PTS: 1 DIF: III OBJ: 2.1.2

4. Use pouring hot chocolate into a cold cup to explain the difference in heat and temperature.

ANS:

The hot chocolate has a higher average kinetic energy of its particles because it has a higher temperature. The energy that is transferred from the particles in the hot chocolate to the particles of the cup is heat.

PTS: 1 DIF: III OBJ: 2.1.3

5. Explain how to change from kelvins to degrees Celsius.

ANS:

Kelvins are the same size as degrees Celsius. The zero point on the Kelvin scale is 273 units below that on the Celsius scale, so to change from Kelvin’s to degrees Celsius, subtract 273 from the Kelvin temperature.

PTS: 1 DIF: II OBJ: 2.1.4

6. Why should a scientist record all observations, even those that appear insignificant?

ANS:

Many important discoveries are made through mistakes or by accident. If seemingly insignificant observations are not recorded, such discoveries cannot be reproduced.

PTS: 1 DIF: II OBJ: 2.1.1

7. What are the first steps scientists take to analyze the cause of a disease?

ANS:

Scientists must first observe the symptoms of the disease. Then they form a hypothesis and design an experiment with a control and variables to test their hypothesis about the probable cause of the disease.

PTS: 1 DIF: II OBJ: 2.2.1

8. Explain what is meant by a scientific method.

ANS:

A scientific method is a logical, systematic approach to problem solving. The exact steps followed might vary from problem to problem.

PTS: 1 DIF: II OBJ: 2.2.1

9. What is the purpose of a control in an experiment?

ANS:

Answers might include that controls eliminate error due to unforeseen variations in the system, serve as a comparison or baseline for other data, and increase confidence in collected data.

PTS: 1 DIF: I OBJ: 2.2.2

10. Explain why pictures of atoms and molecules are models.

ANS:

The pictures represent the atoms and molecules and are consistent with what we know about their behavior. They are not true representations because atoms and molecules are not hard, colored spheres.

PTS: 1 DIF: II OBJ: 2.2.3

11. Differentiate between a theory and a law.

ANS:

A theory is a broad generalization based on observations, data, and reasoning, that is used to explain a phenomenon. A law is a mathematical expression or statement that describes the behavior of some part of the natural world.

PTS: 1 DIF: III OBJ: 2.2.4

12. The mass of a 3.45-g piece of aluminum was measured several times. The measured masses were 2.67 g, 2.59 g, 2.60 g, and 2.64 g. Use these results to explain the difference between accuracy and precision.

ANS:

Because the masses were close to each other, they were precise. Because they were not close to the accepted value of the mass, they were not accurate.

PTS: 1 DIF: II OBJ: 2.3.1

13. What zeros are significant in 2.4070 L? In 0.00304 mL?

ANS:

Both zeros in 2.4070 L are significant. Only the zero between 3 and 4 is significant in 0.00304 mL.

PTS: 1 DIF: II OBJ: 2.3.2

14. Explain how to determine the number of significant figures in the product of 34.02 m and 0.029 m.

ANS:

The product has the number of significant figures in the measure with the smallest number of significant figures. Because 34.02 g has four significant figures and 0.029 has only two, the product can have only two significant figures.

PTS: 1 DIF: II OBJ: 2.3.2

15. The specific heat of iron is 0.449 J/(g•K), and that of aluminum is 0.897 J/(g•K). Explain how you know which metal will absorb and release more energy under the same temperature change. Assume samples of equal mass.

ANS:

Because the specific heat of aluminum is greater than that of iron, aluminum will absorb and release more energy.

PTS: 1 DIF: II OBJ: 2.3.3

16. Explain why the number 62.605 × 106 is not in correct scientific notation. What is the correct scientific notation for this number?

ANS:

The first part of the number should be a number between one and ten. The correct form is 6.2605 × 107.

PTS: 1 DIF: II OBJ: 2.3.4

PROBLEM

1. A 10.0 g sample of cadmium absorbed 43 J of energy as it was heated. The initial temperature was 285 K. What was the final temperature of the metal if the specific heat of cadmium is 0.232 J/(g•K)?

ANS:

[pic]

final temperature = initial temperature + • T = 285 K + 14 K = 299 K

PTS: 1 DIF: III OBJ: 2.3.3

2. The specific heat of tin is 0.228 J/(g•K). A sample of tin absorbs 92 J as it is heated from 271 K to 295 K. What is the mass of the sample?

ANS:

[pic]

PTS: 1 DIF: III OBJ: 2.3.3

3. The specific heat of carbon is 0.709 J/(g•K). How much energy is absorbed if a 7.34 g sample is heated until its temperature increases by 12 K?

ANS:

q = • T × m × cp= 12 K × 7.34 g × 0.709 J/(g •K) = 62 J

PTS: 1 DIF: III OBJ: 2.3.3

4. If a 12.6 g sample of a material is heated from 181 K to 199 K, 46 J of energy is absorbed. What is the specific heat of the material?

ANS:

[pic]

PTS: 1 DIF: III OBJ: 2.3.3

ESSAY

|Mass and Volume Data |

|Sample |Mass |Volume |

|1 |6.02 g |2.23 mL |

|2 |18.42 g |2.34 mL |

|3 |35.15 g |3.10 mL |

1. Explain how the samples in the table above could be identified.

ANS:

The density of each sample could be calculated to the correct number of significant figures. Then each value could be compared to known values to identify the substances.

PTS: 1 DIF: III OBJ: 2.2.1

2. Compare a model with a written theory.

ANS:

A written theory is a broad generalization used to explain observations. A model is a visual, verbal, or mathematical theory that illustrates or explains abstract concepts.

PTS: 1 DIF: II OBJ: 2.2.3

[pic]

3. Evaluate the models in the figure above. Describe any ways that the models differ from the real objects.

ANS:

The model of the sun is accurate in showing that the sun is round and has a fiery surface. The model of an atom shows that the atom is a particle. The sun model is inaccurate because it is only two-dimensional, and it is smaller than the real sun. In addition, it does not show the sun’s composition. The atomic model is inaccurate because it is larger than a real atom, because it is only two-dimensional, and because it does not depict the atom’s composition.

PTS: 1 DIF: III OBJ: 2.2.3

4. Differentiate between a law in science and a law in government.

ANS:

A scientific law explains a natural phenomenon and is a statement of scientific truth that cannot be broken. A governmental law is a guideline for societal behavior.

PTS: 1 DIF: III OBJ: 2.2.3

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