1. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.1 Classifying Matter Solution: Begin by examining the alphabetical listing of pure elements inside the back cover of this text. If the substance appears in that table, it is a pure substance and an element. If it is not in the table but is a pure substance, then it is a compound. If the substance is not a pure substance, then it is a mixture. Use your knowledge about the mixture to determine whether it is homogeneous or heterogeneous. (a) Lead is listed in the table of elements. It is a pure substance and an element. (b) Seawater is composed of several substances, including salt and water; it is a mixture. It has a uniform composition, so it is a homogeneous mixture. (c) Distilled water is not listed in the table of elements, but it is a pure substance (water); therefore, it is a compound. (d) Italian salad dressing contains a number of substances and is therefore a mixture. It usually separates into at least two distinct regions with different composition and is therefore a heterogeneous mixture. Classify each of the following as a pure substance or a mixture. If it is a pure substance, classify it as an element or a compound; if it is a mixture, classify it as homogeneous or heterogeneous. (a) a lead weight (b) seawater (c) distilled water (d) Italian salad dressing

2. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.1 Classifying Matter Classify each of the following as a pure substance or a mixture. If it is a pure substance, classify it as an element or a compound. If it is a mixture, classify it as homogeneous or heterogeneous. (a) mercury in a thermometer (b) exhaled air (c) minestrone soup (d) sugar SKILLBUILDER 3.1 Classifying Matter Note: The answers to all Skillbuilders appear at the end of the chapter. FOR MORE PRACTICE Example 3.12; Problems 29, 30, 31, 32, 33, 34. Continued

3. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.2 Physical and Chemical Properties Solution: (a) Copper turns green because it reacts with gases in air to form compounds; this is a chemical property. (b) Automobile paint dulls over time because it can fade (decompose) due to sunlight or it can react with oxygen in air. In either case, this is a chemical property. (c) Gasoline evaporates quickly because it has a low boiling point; this is a physical property. (d) Aluminum’s low mass (for a given volume) relative to other metals is due to its low density; this is a physical property. Determine whether each of the following is a physical or chemical property. (a) the tendency of copper to turn green when exposed to air (b) the tendency of automobile paint to dull over time (c) the tendency of gasoline to evaporate quickly when spilled (d) the low mass (for a given volume) of aluminum relative to other metals Determine whether each of the following is a physical or chemical property. (a) the explosiveness of hydrogen gas (b) the bronze color of copper (c) the shiny appearance of silver (d) the ability of dry ice to sublime (change from solid directly to vapor) SKILLBUILDER 3.2 Physical and Chemical Properties FOR MORE PRACTICE Example 3.13; Problems 35, 36, 37, 38.

4. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.3 Physical and Chemical Changes Solution: (a) Iron rusts because it reacts with oxygen in air to form iron oxide; therefore, this is a chemical change. (b) When fingernail-polish remover (acetone) evaporates, it changes from liquid to gas, but it remains acetone; therefore, this is a physical change. (c) Coal burns because it reacts with oxygen in air to form carbon dioxide; this is a chemical change. (d) A carpet fades on repeated exposure to sunlight because the molecules that give the carpet its color are decomposed by sunlight; this is a chemical change. Determine whether each of the following is a physical or chemical change. (a) the rusting of iron (b) the evaporation of fingernail-polish remover (acetone) from the skin (c) the burning of coal (d) the fading of a carpet upon repeated exposure to sunlight Determine whether each of the following is a physical or chemical change. (a) copper metal forming a blue solution when it is dropped into colorless nitric acid (b) a passing train flattening a penny placed on a railroad track (c) ice melting into liquid water (d) a match igniting a firework SKILLBUILDER 3.3 Physical and Chemical Changes FOR MORE PRACTICE Example 3.14; Problems 39, 40, 41, 42.

5. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.4 Conservation of Mass Solution: The sum of the masses of the potassium and iodine is: 3.9 g + 12.7 g = 16.6 g The sum of the masses of potassium and iodine equals the mass of the product, potassium iodide. The results are consistent with the law of conservation of mass. A chemist forms 16.6 g of potassium iodide by combining 3.9 g of potassium with 12.7 g of iodine. Show that these results are consistent with the law of conservation of mass. Suppose 12 g of natural gas combines with 48 g of oxygen in a flame. The chemical change produces 33 g of carbon dioxide and how many grams of water? SKILLBUILDER 3.4 Conservation of Mass FOR MORE PRACTICE Example 3.15; Problems 43, 44, 45, 46, 47, 48.

6. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Solve this problem using the procedure to solve numerical problems from Chapter 2 (Section 2.10). Here you are given energy in Calories and asked to convert it to energy in joules. EXAMPLE 3.5 Conversion of Energy Units Given: 225 Cal Find : J Conversion Factors : 1000 calories = 1 Cal 4.184 J = 1 cal A candy bar contains 225 Cal of nutritional energy. How many joules does it contain? Draw a solution map. Begin with Cal, convert to cal, and then convert to J. The complete combustion of a small wooden match produces approximately 512 cal of heat. How many kilojoules are produced? SKILLBUILDER 3.5 Conversion of Energy Units Follow the solution map to solve the problem and round to the correct number of significant figures. SKILLBUILDER PLUS Convert 2.75 ×10 4 kJ to calories. FOR MORE PRACTICE Example 3.16; Problems 49, 50, 51, 52, 53, 54, 55, 56. Solution Map : Solution :

7. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.6 Exothermic and Endothermic Processes Solution: (a) When wood burns, it emits heat into the surroundings. Therefore, the process is exothermic. (b) When ice melts, it absorbs heat from the surroundings. For example, when ice melts in a glass of water, it cools the water as the melting ice absorbs heat from the water. Therefore, the process is endothermic. Identify each of the following changes as exothermic or endothermic. (a) wood burning in a fire (b) ice melting Identify each of the following changes as exothermic or endothermic. (a) water freezing into ice (b) natural gas burning SKILLBUILDER 3.6 Exothermic and Endothermic Processes FOR MORE PRACTICE Problems 59, 60, 61, 62.

8. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Set up the problem in the normal way. You are given a temperature in °C and asked to find K. Use the equation that shows the relationship between the given quantity (°C) and the find quantity (K). EXAMPLE 3.7 Converting between Celsius and Kelvin Temperature Scales Given: -25 °C Find : K Equation : K = °C + 273 Convert –25°C to kelvins. Build the solution map. Convert 358 K to Celsius. SKILLBUILDER 3.7 Converting between Celsius and Kelvin Temperature Scales Follow the solution map to solve the problem by substituting the correct value for °C and computing the answer to the correct number of significant figures. Solution : K = °C + 273 K = –25 °C + 273 = 248 K FOR MORE PRACTICE Example 3.17; Problems 63c, 64d. Solution Map :

9. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Set up the problem in the normal way. You are given a temperature in °F and asked to find °C. Use the equation that shows the relationship between the given quantity (°C) and the find quantity (K). EXAMPLE 3.8 Converting between Fahrenheit and Celsius Temperature Scales Convert 55 °F to Celsius. Build the solution map. Convert 139 °C to Fahrenheit. SKILLBUILDER 3.8 Converting between Fahrenheit and Celsius Temperature Scales Substitute the given value into the equation and compute the answer to the correct number of significant figures. FOR MORE PRACTICE Example 3.18; Problems 63a, 64a,c. Given: 55 °F Find : °C Equation : Solution Map : Solution :

10. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Set up the problem in the normal way. You are given a temperature in K and asked to find °F. This problem requires two equations: one relating K and °C and the other relating °C and °F. EXAMPLE 3.9 Converting between Fahrenheit and Kelvin Temperature Scales Convert 310 K to Fahrenheit. Build the solution map, which requires two steps: one to convert K to °C and one to convert °C to °F. The first equation must be solved for °C. Substitute into this equation to convert K to °C. The second equation must be solved for °F. Substitute into this equation to convert °C to °F and compute the answer to the correct number of significant figures. Given: 310 K Find : °F Equation : Solution Map : Solution :

11. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.9 Converting between Fahrenheit and Kelvin Temperature Scales Continued Convert –321 °F to kelvins. SKILLBUILDER 3.9 Converting between Fahrenheit and Kelvin Temperature Scales FOR MORE PRACTICE Problems 63b, d, 64b.

12. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro You are given the mass of gallium and its initial and final temperatures and asked to find the amount of heat absorbed. The equation that relates the given and find quantities is the specific heat capacity equation. EXAMPLE 3.10 Relating Heat Energy to Temperature Changes Gallium is a solid metal at room temperature but melts at 29.9 °C. If you hold gallium in your hand, it melts from body heat. How much heat must 2.5 g of gallium absorb from your hand to raise its temperature from 25.0 °C to 29.9 °C? The specific heat capacity of gallium is 0.372 J/g °C. The solution map shows that the specific heat capacity equation relates the given and find quantities. Before solving the problem, you must gather the necessary quantities—C, m, and ∆T—in the correct units. Then substitute the correct variables into the equation, canceling units, and compute the answer to the right number of significant figures. Given: 2.5 g gallium T i = 25.0 °C T f = 29.9 °C C = 0.372 J/g °C Find :q Equation : Solution Map : Solution :

13. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.10 Relating Heat Energy to Temperature Changes Continued You find a copper penny (pre-1982) in the snow and pick it up. How much heat is absorbed by the penny as it warms from the temperature of the snow, –5.0 °C, to the temperature of your body, 37.0 °C? Assume the penny is pure copper and has a mass of 3.10 g. You can find the heat capacity of copper in Table 3.4 (p. 71). SKILLBUILDER 3.10 Relating Heat Energy to Temperature Changes * This is the amount of heat required to raise the temperature to the melting point. Actually melting the gallium requires additional heat. FOR MORE PRACTICE Example 3.19; Problems 71, 72, 73, 74. SKILLBUILDER PLUS The temperature of a lead fishing weight rises from 26 °C to 38 °C as it absorbs 11.3 J of heat. What is the mass of the fishing weight in grams?

14. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro You are given the mass of the “gold” rock, the amount of heat absorbed, and the initial and final temperature. You are asked to find the heat capacity. The equation that relates the given and find quantities is the heat capacity equation EXAMPLE 3.11 Relating Heat Capacity to Temperature Changes A chemistry student finds a shiny rock that she suspects is gold. She weighs the rock on a balance and obtains the mass, 14.3 g. She then finds that the temperature of the rock rises from 25 °C to 52 °C upon absorption of 174 J of heat. Find the heat capacity of the rock and determine whether the value is consistent with the heat capacity of gold. The solution map shows how the heat capacity equation relates the given and find quantities. Before solving the problem, you must gather the necessary quantities—m, q, and ΔT—in the correct units. Solution : m = 14.3 g q = 174 J ΔT = 52 °C 25 °C = 27 °C Solution Map : Given: 14.3 g 174 J of heat absorbed T i = 25 °C T f = 52 °C Find : C Equation :

15. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Solve the equation for C and then substitute the correct variables into the equation, canceling units, and compute the answer to the right number of significant figures. EXAMPLE 3.11 Relating Heat Capacity to Temperature Changes Continued A 328-g sample of water absorbs 5.78 × 10 3 J of heat. Find the change in temperature for the water. If the water is initially at 25.0 °C, what is its final temperature? SKILLBUILDER 3.11 Relating Heat Capacity to Temperature Changes FOR MORE PRACTICE Problems 81, 82, 83, 84. By comparing the computed value of the heat capacity (0.45 J/g °C) with the heat capacity of gold from Table 3.4 (0.128 J/g °C), we conclude that the rock could not be pure gold.

16. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.12 Classifying Matter Solution: (a) Pure element; silver appears in the element table. (b) Homogeneous mixture; pool water contains at least water and chlorine, and it is uniform throughout. (c) Compound; dry ice is a pure substance (carbon dioxide), but it is not listed in the table. (d) Heterogeneous mixture; a blueberry muffin is a mixture of several things and has nonuniform composition. Classify each of the following as a pure substance or a mixture. If it is a pure substance, classify it as an element or compound. If it is a mixture, classify it as homogeneous or heterogeneous. (a) pure silver (b) swimming-pool water (c) dry ice (solid carbon dioxide) (d) blueberry muffin

17. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.13 Physical and Chemical Properties Solution: (a) Physical; scratched platinum is still platinum. (b) Chemical; the acid chemically reacts with the skin to produce the burn. (c) Chemical; the hydrogen peroxide chemically reacts with hair to bleach it. (d) Physical; the heaviness can be felt without changing the lead into anything else. Determine whether each of the following is a physical or chemical property. (a) the tendency for platinum jewelry to scratch easily (b) the ability of sulfuric acid to burn the skin (c) the ability of hydrogen peroxide to bleach hair (d) the density of lead relative to other metals

18. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.14 Physical and Chemical Changes Solution: (a) Chemical; the gunpowder reacts with oxygen during the explosion. (b) Physical; the liquid gold is still gold. (c) Chemical; the bubbling is a result of a chemical reaction between the two substances to form new substances, one of which is carbon dioxide released as bubbles. (d) Physical; the bubbling is due to liquid water turning into gaseous water, but it is still water. Determine whether each of the following is a physical or chemical change. (a) the explosion of gunpowder in the barrel of a gun (b) the melting of gold in a furnace (c) the bubbling that occurs upon mixing baking soda and vinegar (d) the bubbling that occurs when water boils

19. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.15 Conservation of Mass Solution: The total mass after the chemical change is: 132 g + 34 g = 166 g The total mass before the change must also be 166 g. 47 g + oxygen = 166 g So, the mass of oxygen consumed is the total mass (166 g) minus the mass of gasoline (47 g). grams of oxygen = 166 g – 47 g = 119 g An automobile runs for 10 minutes and burns 47 g of gasoline. The gasoline combined with oxygen from air and formed 132 g of carbon dioxide and 34 g of water. How much oxygen was consumed in the process?

20. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.16 Conservation of Energy Units Convert 1.7 × 10 3 kWh (the amount of energy used by the average U.S. citizen in one week) into calories. Given: 1.7 × 10 3 kWh Find : cal Conversion Factors : 1 kWh = 3.60 × 10 6 J 1 cal = 4.18 J Solution Map : Solution : The unit of the answer, cal, is correct. The magnitude of the answer makes sense since cal is a smaller unit than kWh.

21. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.17 Converting between Celsius and Kelvin Temperature Scales Convert 257 K to Celsius. Given: 257 K Find : °C Equation : K = °C + 273 Solution Map : Solution : K = °C + 273 °C = K – 273 °C = 257 – 273 = –16 °C The answer has the correct unit, and its magnitude seems correct (see Figure 3.17).

22. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.18 Converting between Fahrenheit and Celsius Temperature Scales Convert 62.0 °C to Fahrenheit. Given: 62.0 °C Find : °F Equation : Solution Map : Solution : The answer has the correct unit, and its magnitude seems correct (see Figure 3.17).

23. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.19 Energy, Temperature Change, and Heat Capacity Calculations What is the temperature change in 355 mL of water upon absorption of 34 kJ of heat? Given: 355 mL water; 34 kJ of heat Find : Δ T Equation : q = m · C · ΔT Solution Map : The value for q must be converted from kJ to J: The value for m must be converted from milliliters to grams; use the density of water, 1.0 g/mL, to convert milliliters to grams.

24. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 3.19 Energy, Temperature Change, and Heat Capacity Calculations Continued Solution : The answer has the correct units, and the magnitude seems correct. If the magnitude of the answer were a huge number—3 × 10 6 for example—we would go back and look for a mistake. If water were to go above 100 °C, it would boil, so such a large answer would be unlikely.