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Example 3.1 Classifying Matter

Example 3.1 Classifying Matter. Classify each type of matter 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

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Example 3.1 Classifying Matter

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  1. Example 3.1Classifying Matter Classify each type of matter 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 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. Refer to your everyday experience with each mixture to determine if 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.

  2. Example 3.1Classifying Matter Continued SKILLBUILDER 3.1 Classifying Matter Classify each type of matter 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 Answers: (a) pure substance, element (b) mixture, homogeneous (c) mixture, heterogeneous (d) pure substance, compound For More Practice Example 3.12; Problems 31, 32, 33, 34, 35, 36. Note: The answers to all Skillbuilders appear at the end of the chapter.

  3. Example 3.2Physical and Chemical Properties Determine whether each property is physical or chemical. (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. 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. SKILLBUILDER 3.2 Physical and Chemical Properties Determine whether each property is physical or chemical. (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) For More Practice Example 3.13; Problems 37, 38, 39, 40. Answers: (a) chemical (b) physical (c) physical (d) physical

  4. Example 3.3Physical and Chemical Changes Determine whether each change is physical or chemical. (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 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

  5. Example 3.3Physical and Chemical Changes Continued SKILLBUILDER 3.3 Physical and Chemical Changes Determine whether each change is physical or chemical. (a) copper metal forming a blue solution when it is dropped into colorless nitric acid (b) a train flattening a penny placed on a railroad track (c) ice melting into liquid water (d) a match igniting a firework Answers: (a) chemical (b) physical (c) physical (d) chemical For More Practice Example 3.14; Problems 41, 42, 43, 44.

  6. Example 3.4Conservation 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. 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. SKILLBUILDER 3.4Conservation 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. How many grams of water form? Answer: 27 g For More PracticeExample 3.15; Problems 45, 46, 47, 48, 49, 50.

  7. Example 3.5Conversion of Energy Units A candy bar contains 225 Cal of nutritional energy. How many joules does it contain? GIVEN:225 Cal FIND:J SORT Begin by sorting the information in the problem. Here you are given energy in Calories and asked to find energy in joules. STRATEGIZE Draw a solution map. Begin with Cal, convert to cal, and then convert to J. SOLUTION MAP RELATIONSHIPS USED 1000 calories = 1 Cal (Table 3.2) 4.184 J = 1 cal (Table 3.2)

  8. Example 3.5 Conversion of Energy Units Continued SOLVE Follow the solution map to solve the problem. Begin with 225 Cal and multiply by the appropriate conversion factors to arrive at J. Round the answer to the correct number of significant figures (in this case, three because of the three significant figures in 225 Cal). CHECK Check your answer. Are the units correct? Does the answer make physical sense? SOLUTION The units of the answer (J) are the desired units. The magnitude of the answer makes sense because the J is a smaller unit than the Cal; therefore, the quantity of energy in J should be greater than the quantity in Cal. SKILLBUILDER PLUSConvert 2.75  104 kJ to calories Answer: 6.57  106 cal SKILLBUILDER 3.5 Conversion of Energy UnitsThe complete combustion of a small wooden match produces approximately 512 cal of heat. How many kilojoules are produced? Answer: 2.14 kJ For More Practice Example 3.16; Problems 51, 52, 53, 54, 55, 56, 57, 58.

  9. Example 3.6Exothermic and Endothermic Processes Identify each change as exothermic or endothermic. (a) wood burning in a fire (b) ice melting 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. SKILLBUILDER 3.6Exothermic and Endothermic Processes Identify each change as exothermic or endothermic. (a) water freezing into ice (b) natural gas burning Answers: (a) exothermic (b) exothermic For More Practice Problems 61, 62, 63, 64.

  10. Example 3.7Converting between Celsius and Kelvin Temperature Scales Convert 25 C to kelvins. SORT You are given a temperature in degrees Celsius and asked to find the value of the temperature in kelvins. GIVEN:25 C FIND:K STRATEGIZE Draw a solution map. Use the equation that relates the temperature in kelvins to the temperature in Celsius to convert from the given quantity to the quantity you are asked to find. SOLVE Follow the solution map to solve the problem bysubstituting the correct value for C and calculating theanswer to the correct number of significant figures. SOLUTION MAP RELATIONSHIPS USED K = C + 273.15 (presented in this section) SOLUTION K = C + 273.15 K = 25 C + 273.15 = 248 K

  11. Example 3.7 Converting between Celsius and Kelvin Temperature Scales Continued CHECK Check your answer. Are the units correct? Does the answer make physical sense? The units (K) are correct. The answer makes sense because the value in kelvins should be a more positive number than the value in degrees Celsius. SKILLBUILDER 3.7 Converting between Celsius and Kelvin Temperature ScalesConvert 358 K to Celsius. Answer: 85 C For More Practice Example 3.17; Problems 65c, 66d.

  12. Example 3.8Converting between Fahrenheit and Celsius Temperature Scales Convert 55 F to Celsius. SORT You are given a temperature in degrees Fahrenheit and asked to find the value of the temperature in degrees Celsius. STRATEGIZE Draw the solution map. Use the equation that shows the relationship between the given quantity (F) and the find quantity (C). SOLVE Substitute the given value into the equation and calculate the answer to the correct number of significant figures. GIVEN: 55 F FIND:C SOLUTION MAP RELATIONSHIPS USED (presented in this section) SOLUTION

  13. Example 3.8 Converting between Fahrenheit and Celsius Temperature Scales Continued Check Check your answer. Are the units correct? Does the answer make physical sense? The units (C) are correct. The value of the answer (13 C)is smaller than the value in degrees Fahrenheit. For positive temperatures, the value of a temperature in degrees Celsius will always be smaller than the value in degrees Fahrenheit because the Fahrenheit degree is smaller than the Celsius degree and the Fahrenheit scale is offset by 32 degrees (see Figure 3.17). FIGURE 3.17Comparison of the Fahrenheit, Celsius, and Kelvin temperature scales.

  14. Example 3.8 Converting between Fahrenheit and Celsius Temperature Scales Continued SKILLBUILDER 3.8 Converting between Fahrenheit and Celsius Temperature Scales Convert 139 C to Fahrenheit. Answer: 282 F For More Practice Example 3.18; Problems 65a, 66a, c.

  15. Example 3.9Converting between Fahrenheit and Kelvin Temperature Scales Convert 310 K to Fahrenheit. SORT You are given a temperature in kelvins and asked to find the value of the temperature in degrees Fahrenheit. STRATEGIZE Build the solution map, which requires two steps: one to convert kelvins to degrees Celsius and one to convert degrees Celsius to degrees Fahrenheit. GIVEN: 310 K FIND:F SOLUTION MAP RELATIONSHIPS USED

  16. Example 3.9Converting between Fahrenheit and Kelvin Temperature Scales Continued SOLVE Solve the first equation for C and substitute the given quantity in K to convert it to C. Solve the second equation for F. Substitute thevalue of the temperature in C (from the previousstep) to convert it to F and round the answer to thecorrect number of significant figures. SOLUTION CHECK Check your answer. Are the units correct? Does theanswer make physical sense? The units (F) are correct. The magnitude of the answer is a bit trickier to judge. In this temperature range, a temperature in Fahrenheit should indeed be smaller than the temperature in kelvins. However, because the Fahrenheit degree is smaller, temperatures in Fahrenheit become larger than temperatures in kelvins above 575 F.

  17. Example 3.9Converting between Fahrenheit and Kelvin Temperature Scales Continued SKILLBUILDER 3.9 Converting between Fahrenheit and Kelvin Temperature Scales Convert – 321 F to kelvins Answer: 77K For More Practice Problems 65b, d, 66b.

  18. Example 3.10Relating 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 your 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. SORT You are given the mass of gallium, its initial and final temperatures, and its specific heat capacity, and are asked to find the amount of heat absorbed. STRATEGIZEThe equation that relates the given and find quantities is the specific heat capacity equation. The solution map indicates that this equation takes you from the given quantities to the quantity you are asked to find. GIVEN: 2.5 g gallium (m) Ti = 25.0 C Tf = 29.9 C C = 0.372 J/g C FIND:q SOLUTION MAP RELATIONSHIPS USED

  19. Example 3.10Relating Heat Energy to Temperature Changes Continued SOLVE Before solving the problem, you must gather the necessary quantities—C, m, and T—in the correct units. Substitute C, m, and T into the equation, canceling units, and calculate the answer to the correct number of significant figures. CHECKCheck your answer. Are the units correct? Does the answer make physical sense? SOLUTION * This is the amount of heat required to raise the temperature to the melting point. Actually melting the gallium requires additional heat. The units (J) are correct. The magnitude of the answer makes sense because it takes almost 1 J to heat the 2.5 g sample of the metal by 1 C; therefore, it should take about 5 J to heat the sample by 5 C.

  20. Example 3.10Relating Heat Energy to Temperature Changes Continued SKILLBUILDER 3.10 Relating Heat Energy to Temperature Changes 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. 74). Answer: 50.1 J 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? Answer: 7.4 g For More Practice Example 3.19; Problems 75, 76, 77, 78.

  21. Example 3.11Relating 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. SORT 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. STRATEGIZE The solution map shows how the heat capacity equation relates the given and find quantities. GIVEN: 14.3 g 174 J of heat absorbed Ti = 25 C Tf = 52 C FIND: C SOLUTION MAP RELATIONSHIPS USED q= m · C · T (presented in this section)

  22. Example 3.11Relating Heat Capacity to Temperature Changes continued SOLUTION m = 14.3 g q = 174 J T= 52 C –25 C = 27 C By comparing the calculated value of the specific heat capacity (0.45 J/g C) with the specific heat capacity of gold from Table 3.4 (0.128 J/g C), we conclude that the rock could not be pure gold. SOLVE First, gather the necessary quantities—m, q, and T —in the correct units. Then solve the equation for C and substitute thecorrect variables into the equation. Finally,calculate the answer to the right number ofsignificant figures.

  23. Example 3.11Relating Heat Capacity to Temperature Changes Continued CHECK Check your answer. Are the units correct? Does theanswer make physical sense? The units of the answer are those of specific heat capacity, so they are correct. The magnitude of the answer falls in the range of specific heat capacities given in Table 3.4. A value of heat capacity that falls far outside this range would immediately be suspect. SKILLBUILDER 3.11 Relating Heat Capacity to Temperature Changes A 328-g sample of water absorbs 5.78  103 J of heat. Calculate the change in temperature for the water. If the water is initially at 25.0 C, what is its final temperature? Answer:T = 4.21 C; Tf = 29.2 C For More Practice Problems 85, 86, 87, 88.

  24. Example 3.12Classifying Matter Classify each type of matter 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 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.

  25. Example 3.13Physical and Chemical Properties Determine whether each property is physical or chemical. (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 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.

  26. Example 3.14Physical and Chemical Changes Determine whether each change is physical or chemical. (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 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.

  27. Example 3.15Conservation of Mass 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? 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

  28. Example 3.16Conversion of Energy Units Convert 1.7  103 kWh (the amount of energy used by the average U.S. citizen in one week) into calories. GIVEN:1.7  103 kWh FIND:cal SOLUTION MAP: RELATIONSHIPS USED 1 kWh = 3.60  106 J (Table 3.2) 1cal = 4.18J (Table 3.2) SOLUTION The unit of the answer, cal, is correct. The magnitude of the answer makes sense since cal is a smaller unit than kWh;therefore, the value in cal should be larger than the value in kWh.

  29. Example 3.17Converting between Celsius and Kelvin Temperature Scales Convert 257 K to Celsius. GIVEN:257 K FIND:ºC SOLUTION MAP: RELATIONSHIPS USED K = ºC + 273.15 (Section 3.10) SOLUTION K = ºC + 273.15 ºC = K –273.15 ºC = 257 – 273.15 = –16 ºC The answer has the correct unit, and its magnitude seems correct (see Figure 3.17). FIGURE 3.17 Comparison of the Fahrenheit, Celsius, and Kelvin temperature scales

  30. Example 3.18Converting between Fahrenheit And Celsius Temperature Scales Convert 62.0 ºC to Fahrenheit. GIVEN:62.0 ºC FIND:ºF SOLUTION MAP: RELATIONSHIPS USED SOLUTION FIGURE 3.17 Comparison of the Fahrenheit, Celsius, and Kelvin temperature scales The answer is in the correct units, and its magnitude seems correct (see Figure 3.17).

  31. Example 3.19Energy, 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:T SOLUTION MAP: RELATIONSHIPS USED q= m · C · T

  32. Example 3.19Energy, 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  106, for example—we would go back and look for a mistake. Above 100 ºC, water boils, so such a large answer would be unlikely.

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