Solution manual for chemistry the molecular science 5th edition by moore

Solution Manual for Chemistry The Molecular Science 5th Edition by Moore Chapter 1: The Nature of Chemistry Chapter 1: The Nature of Chemistry Teaching for Conceptual Understanding Chemistry – The Molecular Science encourages students to think about chemistry at three different levels: the macroscopic, the particulate or nanoscale, and the symbolic Science education research has shown that an understanding of just one level does not imply an understanding of the others Whenever possible use all three levels in your teaching and assessment of student learning Figure 1.17 illustrates the macroscopic and particulate levels for solid ice, liquid water, and water vapor Have students visualize or draw other simple molecules, such as bromine or mercury, at the particulate level in three different states Section 1-3 discusses how science is done, i.e., how observations lead to a hypothesis, how the hypothesis leads to more observations or experimentation, which in turn can lead to a law or theory Students sometimes get the idea that science is a step-by-step procedure that takes place in a laboratory instead of a process that people can use everyday to solve problems and understand the world around them It would be helpful to explain the process of science by using examples with which students can identify Relevant examples will change over time, so be diligent in coming up with new examples from year to year Learning chemistry is like learning a foreign language because of the extensive use of new terminology Students will make quicker and stronger associations between terms and concepts when the root or origin of the terms is explained For example, in this chapter the terms nanoscale and microscale are introduced This is a good time to reinforce the size of the metric prefixes “micro-” and “nano-” as explanations for the terminology Suggestions for Effective Learning Keep in mind that students are excited and ready to work the first day of class It is best not to waste this valuable moment by simply reviewing the syllabus and dismissing the students Have demonstrations, multimedia material, and activities ready to engage them in learning If you plan on using demonstrations during the course, show some of your favorites during the first class A few easy demonstrations that always grab students’ attention are: Igniting hydrogen filled balloons, freezing flowers in liquid nitrogen, and dropping pieces of sodium or potassium into water In addition to the cooperative learning activities suggested below, consider having the students write briefly on their impressions of chemistry, or what properties solids, liquids, and gases have, or how they use the scientific process in solving their everyday problems Finally, take a few minutes to explain something about yourself The students will respect and engage with you better when they feel they understand who you are, why you’re there, and what you care about Their college experience as well as their relationship with you is enriched by a small amount of personal information Cooperative Learning Activities Whether you intend to use cooperative learning activities during class time or not, it is very important for students to get acquainted with each other Set aside at least five minutes during your first class for students to meet others seated around them In addition to exchanging personal information, e.g., name, hometown, major, have them share their feelings about taking a college chemistry course Some students feel anxious about taking chemistry; however, knowing how others feel can help them see that they are not alone Questions, problems, and topics that can be used for Cooperative Learning Exercises and other group work are: • Questions for Review and Thought: 4, 5, 7-8, 11-12, 37-38, 47-48, 63-64, 98-99, 117, 118 • Conceptual Challenge Problems: CP1.A, CP1.B, CP1.C, CP1.D, CP1.E, CP1.F, CP1.G, and CP1.H • Choose one or two statements from Section 1-1 that are most relevant to your students Have the students spend five minutes writing down their ideas, then have them share those ideas with others around them in a short discussion Another group activity is to have students list three to five ways that they have used chemicals today Solution Manual for Chemistry The Molecular Science 5th Edition by Moore Chapter 1: The Nature of Chemistry End-of-Chapter Solutions for Chapter Summary Problem ×1015 Result: J; water is a compound, none are mixtures; See nanoscale diagram below, CO2 (g) + H2 (g) CH4 (g) + O2 (g) Analyze and Plan: Given the number of electric vehicles, the average miles each drives per year, the relationship between miles and kilometers, the energy efficiency of each vehicle, and some relationships between energy units, calculate the total energy needed for these vehicles Execute: ⎛ 10, 000 mi / vehicle ⎞ ⎛ 1.6 km ⎞ ⎛ kW h ⎞ ⎛ 1000 W h ⎞ ⎛ 3.6 PJ ⎞ e × e ×⎜ ⎟ = PJ 300, 000 vehicles × yr × ⎜ ⎟× ⎜ ⎟× ⎜ ⎟ ⎜ ⎟ yr ⎠ ⎝ mi ⎠ ⎝ km ⎠ ⎝ kWe h ⎠ ⎜⎝ 1012 We h ⎟⎠ ⎝ Because 1PJ = ×1015 J, this is × 1015 J of energy
Reasonable Result Check: The answer in petajoules is a convenient number, 3, which is consistent with PJ being commonly used to measure such energies Explanation: Given information about reactions involving hydrogen, water, and oxygen, determine if these are elements, compounds, or mixtures Because the substance called water is formed using hydrogen and hydrogen is recreated when water is decomposed, it is certain that water is a compound The information provided does not make it clear whether the substances labeled hydrogen or oxygen are compounds or elements, but they are not mixtures because each time one destroyed or created they can be isolated and characterized as a pure substance Explanation: Given chemical formulas and information about a reaction write nanoscale and symbolic representations of the reaction that are consistent with modern atomic theory The parts of modern atomic theory that applies to this example are: Compounds are formed by the chemical combination of two or more different kinds of atoms and a chemical reaction involves joining, separating, or rearranging atoms A molecule of methane is composed of one atom of carbon combined with four atoms of hydrogen A molecule of oxygen is composed of two atoms of oxygen combined together These two substances together represent the reactants in the first nanoscale box and on the left side of the symbolic representations shown below A molecule of carbon dioxide is composed of one atom of carbon combined with two atoms of oxygen A molecule of hydrogen is composed of two atoms of hydrogen combined together One carbon dioxide molecule and two hydrogen molecules represent the products in the second nanoscale box and on the right side of the symbolic representations shown below Because the reactants include four atoms of hydrogen, there are two molecules of H2 represented in the products to account for all the hydrogen atoms present initially methane and oxygen gas CH4(g) + O2(g) carbon dioxide gas and hydrogen gas CO2(g) + H2(g) Solution Manual for Chemistry The Molecular Science 5th Edition by Moore Chapter 1: The Nature of Chemistry End-of-Chapter Solutions for Chapter Questions for Review and Thought Review Questions Result/Explanation: The structure of a molecule refers to the way atoms are connected together in the molecule and to the three dimensional arrangement of the atoms relative to one another Structure is important because it is the key to the properties and reactivity of a molecule Result/Explanation: Quantitative observations of a piece of electronics must include numerical information; whereas qualitative observations not involve numerical details For example, a television: Quantitative observations might include the dimensions of the screen and case, the various control settings (volume level, channel selection, etc.), the screen resolution, the power supply requirements, its position in the room, on-off, etc The qualitative observations might include the color of the case, the types and locations of control buttons or knobs, the quality of the audio output, the quality of the video image, and the type of connection to a video signal (antenna, cable, satellite, internet, etc.) Result/Explanation: A scientific law (a) summarizes and explains a wide range of experimental results, (b) has not been contradicted by experiments, and (c) is capable of predicting unknown results Some laws are described in Atomic Theory Two of these laws are: the law of conservation of mass (i.e., there is no detectable change in mass during an ordinary chemical reaction.) and the law of constant composition (i.e., A chemical compound always contains the same elements in the same proportion by mass.) Result/Explanation: A theory is a unifying principle that explains a body of facts and the laws based on them – hence a theory is our reason for believing in the law; whereas a law gives just a summary conclusion of a wide range of experimental results Models are used to make theories more concrete, often in physical or mathematical form – hence a model is our way of looking at the theory in detail Result/Explanation: Chemists build the bridge between the nanoscale world into the microscale world The details of the nanoscale world often profoundly affect the activity of a chemical in the micro- and macroscopic worlds A specific example is the determination of the mechanism of paclitaxel activity in the search for improved anticancer drugs Result/Explanation: (Described in Section 1-7) Two examples of when purity of a substance is important: The purity of the elemental silicon is important to the production of electronic chips To properly characterize the properties of a substance, it is necessary to test the substance in its pure state Topical Questions Why Care About Chemistry? (Section 1-1) Result/Explanation: Important issues change from year to year, and sometimes even from month to month or even day to day A good website for seeking up-to-date information on a wide range of scientific topics is http://www.sciencedaily.com/ In summer of 2013, the following issues were big: • forensics • energy technology, renewable energy, alternative fuels • nanotechnology • environmental issues: climate, air quality, pollution, global warming, acid rain • cancer, stem cells, brain tumors, heart disease, medical technology • transportation sciences • nature of water • materials in electronic devices (batteries, display, functionality, etc.) Result/Explanation: Several questions related to chemistry and science phenomena are given in Section 1-1 Solution Manual for Chemistry The Molecular Science 5th Edition by Moore Chapter 1: The Nature of Chemistry How Science is Done (Section 1-3) Result: (a) Qualitative (b) Quantitative and qualitative (c) Quantitative and qualitative (d) Qualitative Explanation: (a) The details of the appearance of a substance (silvery-white) and information about the specific element it contains (sodium) are both qualitative (b) The temperature at which a solid melts (660 °C) is quantitative information Information about what element it is (aluminum) is qualitative (c) The mass percentage of an element in the human body (about 23%) is quantitative information Information about what element it is (carbon) is qualitative (d) The allotropic forms of an element (graphite, diamond, and fullerenes) and information about what element it is (carbon) are both qualitative 10 Result: (a) Quantitative and qualitative (b) Qualitative (c) Quantitative and qualitative (d) Qualitative Explanation: (a) The atomic mass of an element (12.011 atomic mass units) is quantitative information Information about what element it is (carbon) is qualitative (b) The purity, the details of the appearance of a substance (silvery-white), the lack of magnetic capabilities (nonmagnetic), the relative density (low), the inability to produce sparks when struck, and information about the specific element it contains (aluminum) are all qualitative (c) The density of a substance (0.968 g/mL) is quantitative information Information about what element it is (sodium) is qualitative (d) The identity of an extracellular element (sodium), the element’s ionic form (cation), and the biological importance of the element (nerve function) are all qualitative Identifying Matter: Physical Properties (Section 1-4) 11 Result/Explanation: Bromine is a reddish-brown liquid Sulfur is a chalky yellowish solid They appear to have no property in common The physical phase, shape, color, and appearance are different 12 Result/Explanation: Calcite is a clear, colorless, cubic solid Calcium appears as a metallic solid with a white coating Carbon is a black, granular solid The property they have in common is that they are solids The shape, color, and appearance of all three are different 13 Result: The solid will melt because your body temperature of 37°C is above the melting point of 29.76°C Analyze and Plan: Many Americans only remember the human body temperature in the Fahrenheit scale That is 98.6 °F If that is the case, we can quickly apply the °F to °C conversion equation, so we can compare it to the melting point Execute:
C = × ⎛⎜
F − 32 ⎞⎟ = × 98.6 − 32 = 37.0
C ⎝ ⎠ ( ) If the sample melts at a temperature of 29.76 °C and your hand is 37 °C, the liquid will boil when exposed to the heat energy emitted by your hand when you hold the sample 14 Result: Charlotte is the warmest city Montreal is the coldest city Analyze: We use these questions about temperature using different scales to help us become familiar with the Celsius scale So, it is useful to try estimating the answer before using equations using some key connections between the Celsius scale and the Fahrenheit scale Plan: Compare two thermometers side-by-side, one calibrated with a Celsius scale and one calibrated with a Fahrenheit scale using body temperature (37 °C, 98.9 °F) and the freezing point of water (0 °C, 32 °F) as reference points Solution Manual for Chemistry The Molecular Science 5th Edition by Moore Chapter 1: The Nature of Chemistry Execute: 98.7 °F 37 °C Body temperature 20 °C 40 °F °C 32 °F 28 °F Water freezes –10 °C This comparison shows that Charlotte (at 20 °C) is the warmest city and Montreal (at – 10 °C) is the coldest To be quantitative, convert 40 °F to °C to make sure it’s lower than 20 °C, and convert 28 °F to °C to make sure it’s higher than –10 °C
9 ⎛ ⎞ C = × ⎜
F − 32 ⎟ ⎠ ⎝ ⎛ ⎞ × ⎜ 40
F − 32⎟ =
C lower than 20 °C (confirms estimate) ⎝ ⎠ ⎛ ⎞ × ⎜28
F − 32⎟ = −
C higher than – 10 °C (confirms estimate) ⎝ ⎠ Measurements, Units, and Calculations (Section 1-5) 15 Result: 0.00283 kg, Ca and F Analyze and Plan: Given the mass of the crystal as 2.83 grams, find the mass in kilograms Using the conversion factor between grams and kilograms, determine the mass in kilograms Execute: 2.83 g × kg = 0.00283 kg 1000 g The symbols for the elements in this crystal are Ca (calcium) and F (fluorine)
Reasonable Result Check: Kilograms are larger units than grams, so the number of kilograms should be smaller than the number of grams 16 Result: 7.8 × 102 m3, 7.8 × 108 cm3, 7.8 × 105 L Analyze: Given the length, width, and height of a room, determine the volume in cubic meters, cubic centimeters, and liters Plan: Using V = (length) × (width) × (height) calculate the volume in cubic meters Use conversion factors between centimeters and meters to determine the volume in cubic centimeters Use the conversion factor between cubic centimeters and milliliters and between milliliters and liters to determine the volume in liters Execute: V = (length) × (width) × (height) = (18 m) × (15 m) × (2.9 m) = 7.8 × 102 m3 ⎛ 100 cm ⎞3 7.8 × 10 m3 × ⎜ ⎟ = 7.8 × 10 cm ⎝ 1m ⎠ 7.8 × 10 cm3 × mL cm3 × 1L = 7.8 × 105 L 1000 mL
Reasonable Result Check: A liter is smaller than a cubic meter but larger than a cubic centimeter, so the number of m3 should be the smallest number, then the number of liters, then the number of cm3 Solution Manual for Chemistry The Molecular Science 5th Edition by Moore Chapter 1: The Nature of Chemistry 17 Result: No, 23.4 mi/hr < 25 mi/hr Analyze: Given the length of the track and the time to run its length, determine the miles per hour and compare to 25 mi/hr Plan: Divide the meters by the seconds, then use conversion factors between meters and centimeters, centimeters and inches, inches and feet, and feet and miles to determine the distance in miles And use the conversion factor between second and hours to determine the time in hours Execute: 100 m ⎛ 100 cm ⎞ ⎛ in ⎞ ⎛ ft ⎞ ⎛ mi ⎞ ⎛ 3600 s ⎞ mi ×⎜ ⎟× ⎜ ⎟× ⎜ ⎟× ⎜ ⎟×⎜ ⎟ = 23.4 9.58 s ⎝ m ⎠ ⎝ 2.54 cm ⎠ ⎝ 12 in ⎠ ⎝ 5280 ft ⎠ ⎝ hr ⎠ hr No, this runner could not be arrested for exceeding the 25 mi/hr speed limit, since 23.4 mi/hr < 25 mi/hr
Reasonable Result Check: No one on foot could chase and catch up with a car going 25 mph 18 Result: (a) BMI = 21.4 kg/m2 (b) No, because 21.4 kg/m2 < 25 kg/m2 Analyze: (a) Given mass in pounds and height in inches, use conversion factors to calculate mass in kg and height in meters, then calculate the body mass index (BMI) as described in the question Plan: Using conversion factors relating pounds to grams and grams to kilograms, determine the mass in kg Use inches to centimeters and centimeters to meters to determine the height in meters BMI = (kg)/[(m)]2 Execute: ⎛ 453.59 g ⎞ ⎛ kg ⎞ 176 lb × ⎜ ⎟× ⎜ ⎟ = 79.8 kg ⎝ lb ⎠ ⎝ 1000 g ⎠ ⎛ 2.54 cm ⎞ ⎛ m ⎞ 76.0 in × ⎜ ⎟× ⎜ ⎟ = 1.93 m ⎝ in ⎠ ⎝ 100 cm ⎠ BMI = 79.8 kg (1.93 m) = 21.4 kg m2 (b) No, this person should not be considered overweight, since 21.4 kg/m2 < 25 kg/m2
Reasonable Result Check: This person is fairly tall (over six feet), so his 176 lbs is distributed properly 19 Result: (a) Four (b) Three (c) Four (d) Four (e) Three (f) Four Analyze: Given several measured quantities, determine the number of significant figures Plan: Use rules given in Section 1-5, summarized here: All non-zeros are significant Zeros that precede (sit to the left of) non-zeros are never significant (e.g., 0.003) Zeros trapped between non-zeros are always significant (e.g., 3.003) Zeros that follow (sit to the right of non-zeros are (a) significant, if a decimal point is explicitly given (e.g., 3300.) OR (b) not significant, if a decimal point is not specified (e.g., 3300) Execute: (a) 1374 kg has four significant figures The 1, 3, 7, and digits are each significant (b) 0.00348 s has three significant figures The 3, 4, and digits are each significant The zeros are all before the first non-zero-digit and therefore they are not significant (c) 5.619 mm has four significant figures The 5, 6, 1, and digits are each significant (d) 2.475 × 10–3 cm has four significant figures The 2, 4, 7, and digits are each significant (e) 33.1 mL has three significant figures The 3, 3, and digits are each significant (f) 2300 m has four significant figures The 2, 3, 0, and digits are each significant
Reasonable Result Check: Only answers have zeros in them Those in (b) are to the left of the first nonzero digit, so none of the zeros there were significant Those in (f) are in a number with a decimal point, so all of them were significant Solution Manual for Chemistry The Molecular Science 5th Edition by Moore Chapter 1: The Nature of Chemistry 20 Result: (a) Four (b) Two (c) Two (d) Three (e) Three (f) Two Analyze and Plan: Given several measured quantities, determine the number of significant figures Use the rules given in Section 1-5, summarized in the solution to Question 19 Execute: (a) 1.022 × 102 km has four significant figures (The 1, 0, 2, and digits are each significant The zero is trapped between the non-zeros and and therefore is significant.) (b) 34 m2 has two significant figures (The and are each significant.) (c) 0.042 L has two significant figures (The and digits are each significant The zeros are all before the first non-zero-digit and therefore they are not significant.) (d) 28.2 °C has three significant figures (The 2, 8, and digits are each significant.) (e) 323 mg has three significant figures (The 3, 2, and digits are each significant.) (f) 420 mg has two significant figures (The and digits are each significant The zero follows the non-zerodigits in a number with no decimal point The ones place is unknown.)
Reasonable Result Check: Three answers had zeros in them In (a), it was trapped and significant; in (c), they were to the left of the first non-zero digit, so not significant; in (f), the number has a zero that is not significant because there is no decimal point 21 Result: (a) 1.9 g/mL (b) 291.2 cm3 (c) 0.0217 (d) 5.21 × 10–5 Analyze: Given numbers combined in calculations, determine the result with proper significant figures Plan: Perform the mathematical steps according to order of operations, applying the proper significant figures (addition and subtraction retains the least number of decimal places in the result; multiplication and division retain the least number of significant figures in the result) Notice: if operations are combined that use different rules, we must stop and determine the intermediate result any time the rule switches Execute: (a) 4.850 g − 2.34 g 1.3 mL The numerator uses the subtraction rule The first number has three decimal places (the 8, 5, and are all decimal places digit that follow the decimal point to the right) and the second number has two decimal places (the and the are both decimal places), so the result of the subtraction has two decimal places 2.51 g 1.3 mL The ratio uses the division rule The numerator has three significant figures and the denominator has two significant figures, so the answer will have two significant figures Therefore, the answer is 1.9 g/mL (b) V =4/3 πr3 = 4/3 × (3.1415926) × (4.112 cm)3 This whole calculation uses the multiplication rule, with four significant figures, limited by the measurement of r The numerals and are exact, in this context The value of π must be carried to more than four significant figures, such as 3.1415926… The answer comes out 291.2 cm3 (c) (4.66 × 10–3) × 4.666 This calculation uses the multiplication rule The first number, 4.66 × 10–3, has three significant figures and the second number, 4.666, has four significant figures, so the answer has three significant figures 0.0217 (d) 0.003400 65.2 This calculation uses the division rule The numerator has four significant figures and the denominator has three significant figures, so the answer has three significant figures 0.0000521 or 5.21 × 10–5
Reasonable Result Check: The significant figures, size, and units of the answers are appropriate Solution Manual for Chemistry The Molecular Science 5th Edition by Moore Chapter 1: The Nature of Chemistry 22 Result: (a) 1.20× × 103 (b) 0.690 (c) 81.7 cm2 (d) 1.7×102 Analyze: Given some numbers in calculations, determine the result with proper significant figures Plan: Perform the mathematical steps according to the order of operations, applying the proper significant figures (Remember: addition and subtraction retains the least number of decimal places in the result; multiplication and division retain the least number of significant figures in the result) Notice: if operations are combined that use different rules, it is important to stop and determine the intermediate result any time the rule switches Execute: (a) 3256.5 3.20 The ratio uses the division rule The numerator has five significant figures and the denominator has three significant figures, so the result of division will have three significant figures Therefore, we get: 2221.05 − 2221.05 − 1.02 × 103 = 2.22105 × 103 − 1.02 × 103 The next calculation uses the subtraction rule To count decimal places, both numbers must have the same power of ten: 2.22105 × 103 − 1.02 × 103 In the ×103 power of ten, the first number has five decimal places (the 2, 2, 1, 0, and are all decimal places) and the second number has two decimal places (the and the are both decimal places) The result × 103 of the subtraction has two decimal places, in the × 103 power of ten: 1.20× Another way to look at this is write both numbers 2221.05 – 1020 without scientific notation The uncertainty in the second number is in the tens place, so the answer must be rounded to the tens place (b) (343.2) × (2.01×10–3) This calculation uses the multiplication rule The first number, 343.2, has four significant figures and the second number, 2.01×10–3, has three significant figures, so the answer has three significant figures 0.690 (c) S = 4πr2 = × (3.1415926) × (2.55 cm)2 This whole calculation uses the multiplication rule, with three significant figures, limited by the measurement of r The numeral is exact, in this context The value of π should be carried to more than three significant figures, such as 3.1415926… The answer comes out 81.7 cm2 (d) 2802 − 0.0024 × 10, 000 15 The ratio uses the division rule The numerator has four significant figures and the denominator has two significant figures, so the result has two significant figures The multiplication also uses the same rule The first number has two significant figures and the second number has significant figures So, the result of the multiplication will have two significant figures ( ) 1.9 × 102 – 25 The difference uses the subtraction rule We need these numbers in the same power of ten to be able to count decimal places 1.9 × 102 – 0.25 × 102 In the ×102 power of ten, the first number has one decimal places (the 9) and the second number has two decimal places (the and the are both decimal places) The result of the subtraction (1.65 × 102) has to be rounded to one decimal place in the ×102 power of ten: 1.7 × 102 Another way to look at this is that the uncertainty in the first number (190) is in the tens place, so the answer (165) must be rounded to the tens place, 170, with the zero being non-significant
Reasonable Result Check: The significant figures, size, and units of the answers are appropriate Solution Manual for Chemistry The Molecular Science 5th Edition by Moore Chapter 1: The Nature of Chemistry 23 Result: Copper Analyze and Plan: We have the mass of the metal and some volume information We need to determine the density Use the initial and final volumes to find the volume of the metal piece, then use the mass and the volume to get the density The metal piece displaces the water when it sinks, making the volume level in the graduated cylinder rise Execute: The metal piece volume is the difference between the starting volume and the final volume: Vmetal = Vfinal – Vinitial = (37.2 mL) – (25.4 mL) = 11.8 mL d= m 105.5 g g = = 8.94 V 11.8 mL mL According to Table 1.1, this is very close to the density of copper (d = 8.93 g/mL)
Reasonable Result Check: The metal piece sinks, so the density of the metal piece must be higher than water (Table 1.1 gives water density as 0.998 g/mL.) 24 Result: 30.9 mL Analyze and Plan: We have the mass of the lead and the initial volume information We need to determine the final volume after the lead is submerged The metal piece displaces the water when it sinks, making the volume level in the graduated cylinder rise The difference between the starting volume and the final volume, must be the volume of the lead Since we know the metal is made of lead, look up the density Use the density and the mass information to find the volume of the lead, then use the volume of the lead and the initial volume of the ethanol to get the final volume Execute: The density of lead is 11.34 g/mL Use conversion factors to convert the lead mass into mL of lead: mL lead 10.0 g lead × = 0.882 mL lead 11.34 g lead The difference between the starting volume and the final volume must be the volume of the metal piece, so adding the volume of the metal piece to the starting volume gives the final volume: Vfinal = Vmetal + Vinitial = (30.0 mL) + (0.882 mL) = 30.882 mL Rounding to one decimal place, the limit of the uncertainty, gives a final volume of 30.9 mL
Reasonable Result Check: The sample mass is close to the density (the number of grams in one mL), so the ethanol level should rise about one mL 25 Result: Aluminum Analyze: We have the three linear dimensions of a regularly shaped piece of metal 10.0 cm long 1.0 cm thick 2.0 cm wide We also have its mass We have a table of densities (Table 1.1) We need to determine the identity of the metal Plan: Use the three linear dimensions to find the volume of the metal piece Use the volume and the mass to find the density Use the table of densities to find the identity of the metal Execute: V = (thickness) × (width) × (length) = (1.0 cm) × (2.0 cm) × (10.0 cm) = 20 cm Using conversion factors, find the volume in mL: 20 cm3 × Find the density: mL = 20 mL cm3 g m 54.0 g = 2.7 d= = mL V 20 mL According to Table 1.1, the density that most closely matches this one is aluminum (d = 2.70 g/mL)
Reasonable Result Check: The mass is larger than the volume so d is larger than Solution Manual for Chemistry The Molecular Science 5th Edition by Moore 10 Chapter 1: The Nature of Chemistry 26 Result: 15.7 mL Analyze and Plan: We have the mass of the sample, and the identity We are asked to find the volume Use density of bromobenzene and mass to find the volume Execute: Density of bromobenzene is 1.49 g/mL 23.4 g × mL = 15.7 mL 1.49 g
Reasonable Result Check: A density larger than 1g/mL tells us that the volume of a sample of bromobenzene will be smaller than the mass 27 Result: 3.9 × 103 g Analyze: We have the three linear dimensions of a regularly shaped sodium chloride crystal: 12 cm long 10 cm thick 15 cm wide We have a table of densities (Table 1.1) We need to determine the mass of the crystal Plan and Execute: Use the three linear dimensions to find the volume of the crystal Use the volume and density (Table 1.1) to find the mass V = (thickness) × (width) × (length) = (10 cm) × (15 cm) × (12 cm) = 1800 cm3 = 1.8 × 103 cm3 Using conversion factors, find the volume in mL: mL 1.8 × 103 cm3 × = 1.8 × 103 mL cm Find the mass using the density: 1.8 × 10 mL × 2.16 g = 3.9 × 10 g mL Notice: We’re carrying two significant figures since the length data was only that precise
Reasonable Result Check: This crystal is pretty large So, while the mass calculated is a large number, the volume is still about half the mass, consistent with a density around two 28 Result: 0.551 mL Analyze: We have the mass of a liquid We need to determine the volume Plan: Use the mass and density (Table 1.1) to find the volume Execute: Table 1.1 gives the density of iron (d = 7.86 g/mL) Find the volume using the density as a ratio (mL/g) so that the mass units (g) will cancel when multiplied by the mass in grams: 4.33 g × mL = 0.551 mL 7.86 g
Reasonable Result Check: The density is less than 1, so the volume should be larger than the mass Here, the calculated volume of 4.92 mL is a larger number than the original mass (4.33 g) Chemical Change and Chemical Properties (Section 1-6) 29 Result: (a) Physical (b) Chemical (c) Chemical (d) Physical Explanation: (a) The normal color of bromine is a physical property Determining the color of a substance does not change Solution Manual for Chemistry The Molecular Science 5th Edition by Moore 22 Chapter 1: The Nature of Chemistry 81 Result/Explanation: If the density of solid calcium is almost twice that of solid potassium, but their masses are approximately the same size, then the volume must account for the difference This suggests that the atoms of calcium are smaller than the atoms of potassium: solid calcium smaller atoms closer packed smaller volume solid potassium larger atoms less closely packed larger volume 82 Result/Explanation: The volume of a gas is independent of the identity of the gas molecule, because there is so much empty space between one molecule and another in the gas state Therefore, the density of a gas is directly proportional to the mass of the substance of which the gas is composed In this case, because the mass of an argon atom is almost exactly ten times the mass of a helium atom, we see that the density of the argon gas is almost exactly ten times the density of helium gas 83 Result: 508 m Analyze: We have the mass of a spool of aluminum wire with known diameter Assuming the wire is a cylinder, find the length () of wire in meters Use the density to find the volume from the mass, then use the given volume equation and the known diameter to find the length 10.0 lb × 453.59 g mL cm3 × × = 1680 cm3 2.70 g mL lb The radius is half the diameter Determine the radius in centimeters: R= 2.54 cm × 0.0808 in × = 0.103 cm in ( ) Rearrange V = πr2 to solve for , plug in the known values and convert to meters:
= V πr = 1680 cm3 ( ) (3.14159) × 0.103 cm × 1m = 508 m 100 cm
Reasonable Result Check: It makes sense that a quantity of wire that weighs 10 pound would be several hundred feet long 84 Result: 1.7 × 10 –3 cm Analyze: We have the length, width, and mass of a sample of aluminum foil Find the thickness of the foil in centimeters Plan and Execute: Use the density to find the volume from the mass: 0.83 kg × 1000 g mL cm3 × × = 307.4 cm3 (2 sig figs) 2.70 g mL kg Convert the known linear dimensions to centimeters: length = 66 ft 12 in 2.54 cm yards × × × = 6096 cm (3 sig figs) yard ft in width = 12 in × 2.54 cm = 30.48 cm (2 sig figs) in Solution Manual for Chemistry The Molecular Science 5th Edition by Moore Chapter 1: The Nature of Chemistry 23 Then use V = (thickness) × (width) × (length), to find the thickness in centimeters: (thickness) = V 307.4 cm3 = 1.7 × 10 –3 cm = (width) × (length) 30.48 cm × 6096 cm ( ) ( )
Reasonable Result Check: Aluminum foil is quite thin, so this small number for thickness makes sense 85 Result/Explanation: The highest density materials will sink to the bottom, with increasingly less dense materials floating on top The solid material with the highest density is the Teflon plastic pieces (d = 2.3 g/cm3), so those pieces will be found at the bottom of the graduated cylinder sitting in the liquid perfluorohexane, which has the highest density of the liquids (d = 1.669 g/cm3) Floating on the surface of the perfluorohexane will be the liquid water (d = 1.00 g/cm3) Floating on the water will be the pieces of HDPE plastic (d = 0.97 g/cm3) and the liquid hexane (d = 0.766 g/cm3) Hexane HDPE pieces Water Perfluorohexane Teflon pieces 86 Result/Explanation: The density of the HDPE is 0.97 g/mL It will float in any liquid with a density greater than 0.97 g/mL Examining the given chart, HDPE will float in ethylene glycol (d = 1.113 g/mL), water (d = 0.9982 g/mL), acetic acid (1.0498 g/mL) and glycerol (d = 1.2611 g/mL) 87 Result: He, Ne, Ar, Kr, Xe, Rn, H2 , N2, O2, Cl2, F2; found at top and/or right side of the periodic table Explanation: There is not one place in the textbook where all the gaseous elements at room temperature are identified, but throughout the text there are a number of places where we can find the identity of these eleven gaseous elements Some are identified in Section 1-12 when discussing examples of gaseous non-metals: H2, N2, and Cl2 and the noble gases: He, Ne, Ar, Kr, Xe, and Rn Hydrogen, H 2, is identified as a gas in Section 1-6, Figure 1.6, and Section 1-8 Oxygen, O2, is shown as a gas in Figure 1.22 The halogen chlorine, Cl2, is depicted as a gas in Figure 1.24, and because halogen fluorine, F2, is an even smaller molecule than chlorine, it is logical to believe it will be a gas, also Lastly, it is possible to use the boiling points for various elements given in Table 9.2 in Chapter to confirm that several of these elements (Ne, Ar, Xe, F2, and Cl2) are above their boiling point and therefore gaseous at room temperature Except hydrogen, all these elements are located at the far top right side of the periodic table 88 Result: (a) K (b) Ar (c) Cu (d) Ge (e) H (f) Ca (g) Br (h) P Explanation: Use the periodic table and information given in Section 1-13 (a) K is an alkali metal (Group 1A) (b) Ar is a noble gas (Group 8A) (c) Cu is a transition metal (Group 1B) (d) Ge is a metalloid (Group 4A) Solution Manual for Chemistry The Molecular Science 5th Edition by Moore 24 Chapter 1: The Nature of Chemistry (e) H is a group nonmetal (f) Ca is an alkaline earth metal (Group 2A) (g) Br is a halogen (Group 7A) (h) P is a nonmetal that is a solid (Group 5A) 89 Result/Explanation: Look at the periodic table given (a) A colorless gas is a non-metal Those gases are found in the lavender area (b) A solid that is ductile and malleable are metals Metals are found in the gray and blue areas (c) Non-metals and metalloids are poor electrical conductors Solids with this characteristic are found in the orange and lavender areas 90 Result/Explanation: Relate the characteristics of elements to locations on the periodic table (a) A shiny solid that conducts electricity would be a metal, so it will be in the grey and blue areas (b) A gas whose molecules consist of single atoms will be a nonmetal, so it will be in the lavender area (c) An element that is a semiconductor will be a metalloid, so it will be in the orange area (d) A yellow solid with a low electrical conductivity will be a nonmetal, so it will be in the lavender area 91 Result/Explanation: Se and S have the greatest similarities in physical and chemical properties because they are both in the same periodic group (Group 6A) 92 Result/Explanation: Mg and Sr have the greatest similarities in physical and chemical properties because they are both in the same periodic group (Group 2A) 93 Result/Explanation: A substance that can be broken down is not an element A series of tests will result in a confirmation with one positive test To prove that something is an element requires a battery of tests that all have negative results A hypothesis that the substance is an element and cannot be broken down is more difficult to prove (Section 1-3) 94 Result: (a) Mixture; homogeneous (b) 58.1 kg (c) See explanation below (a) Explanation: Fat is a mixture, solution of lye is a mixture of water (a substance that is a compound) and lye (sodium hydroxide, a compound) The mixtures are probably both homogeneous (b) Analyze: Conservation of mass says that the soap will have a mass that is the sum of all three component substances Plan: Convert the volume to kilograms and add all the masses Execute: 50.0 L water × 1000 mL 0.998 g water kg × × = 49.9 kg 1L mL water 1000 g 3.24 kg + 49.9 kg + 5.0 kg = 58.1 kg (c) Explanation: Physical processes that occur include the mixing of substances to make solutions Chemical processes involved are the formation of carboxylate salts (soap molecules) and glycerol from fat and lye
Reasonable Result Check: The density of water is close to one, so the mass and the volume are close to the same value If everything used is converted to soap, it make sense that the mass is additive 95 Result: (a) Nickel, lead and magnesium (b) Titanium Explanation: (a) According to the table of densities, a metal will float if the density is lower That means that nickel, lead and magnesium will float on liquid mercury (b) The more different the densities, the smaller the fraction of the floating element will be below the surface That means that titanium will float highest on mercury Solution Manual for Chemistry The Molecular Science 5th Edition by Moore Chapter 1: The Nature of Chemistry 25 96 Result: 6.02 × 10 –29 m3 Analyze and Plan: We have the length of the edge of a cube Find the volume of the cube in m3 392 pm long 392 pm thick 392 pm wide Use the linear dimensions to find the volume, then convert the volume to m3 using metric conversions Execute: V = (thickness) × (width) × (length) = (392 pm) × (392 pm) × (392 pm) = 6.02 × 107 pm3 Using conversion factors, find the volume in m3: ⎛ −12 ⎞3 10 m⎟ 6.02 × 10 pm × ⎜ = 6.02 × 10−29 m3 ⎜ pm ⎟ ⎝ ⎠
Reasonable Result Check: The cube is from the nanoscale, so it makes sense that it would be a very small volume using a macroscale unit of measure 97 Result: 566 pm Analyze and Plan: Given the volume of the cube in cm3, find the length of a side in picometers:  pm long  pm thick  pm wide Use the volume to find the three linear side dimensions of the cube, then convert the volume to pm using metric conversions Execute: V = 3  = V = 1.81 × 10−22 cm3 = 5.66 × 10−8 cm Using conversion factors, find the length in picometers: 5.66 × 10−8 cm × 10−2 m pm × = 566 pm cm 10−12 m
Reasonable Result Check: The cube is from the nanoscale, so it makes sense that it would be a number with convenient size using a nanoscale unit of measure Applying Concepts 98 Result/Explanation: Separating the components of a mixture containing sand, sugar, and sulfur requires employing variable physical properties Adding the mixture to water, would dissolve all the sugar The sugar can be reclaimed by separating the solution from the un-dissolved solid using filtration, then evaporating the water The remaining solid could be added to liquid carbon disulfide to dissolve the sulfur The sulfur could be reclaimed by separating the solution from the un-dissolved sand via filtration, then evaporating off the liquid carbon disulfide Drying the sand leaves it separate from the sugar and sulfur 99 Result/Explanation: Separating the components of a mixture containing sand, salt, and naphthalene requires employing variable physical properties Adding the mixture to water, would dissolve all the salt The salt could be reclaimed by separating the solution from the un-dissolved solid using filtration, then evaporating the water The remaining solid could be added to liquid benzene to dissolve the naphthalene Naphthalene could be Solution Manual for Chemistry The Molecular Science 5th Edition by Moore 26 Chapter 1: The Nature of Chemistry reclaimed by separating the solution from the un-dissolved sand via filtration, then evaporating the liquid benzene Drying the sand leaves it separate from the salt and naphthalene 100 Result: (a) Bromobenzene sample (b) Gold sample (c) Lead sample Explanation: (a) (Table 1.1) density of butane = 0.579 g/mL; density of bromobenzene = 1.49 g/mL mL butane weighs less than mL of bromobenzene so, 20 mL butane weighs less than 20 mL of bromobenzene The bromobenzene sample has a larger mass (b) (Table 1.1) density of benzene = 0.880 g/mL; density of gold = 19.32 g/mL There are 0.880 grams of benzene in mL of benzene, so there are 8.80 grams of benzene in 10 mL of benzene Since 1.0 mL of gold has a mass of 19.32 grams that means the gold sample has a larger mass (c) (Table 1.1) density of copper = 8.93 g/mL; density of lead = 11.34 g/mL Any volume of lead has a larger mass than the same volume of copper That means the lead sample has a larger mass 101 Result: (a) Ethanol sample (b) Water sample (c) Magnesium sample Explanation: (a) (Table 1.1) density of ethanol = 0.789 g/mL; density of bromobenzene = 1.49 g/mL gram of ethanol has a volume more than mL and gram of bromobenzene has a volume less than mL The ethanol sample has a larger volume (b) (Table 1.1) density of aluminum = 2.70 g/mL; density of water = 0.998 g/mL There are 2.70 grams of aluminum in mL aluminum That means there are 27.0 grams of aluminum in 10 mL aluminum, and therefore 10 grams of aluminum will occupy significantly less than 10 mL On the other hand, there is about gram of water in mL water, so there are about 12 grams of water in 12 mL water That means the water sample has a larger volume (c) (Table 1.1) density of gold = 19.32 g/mL; density of magnesium = 1.74 g/mL Approximately 20 grams of gold occupies a little bit more than mL Any more than grams of magnesium occupies more volume than 20 grams of gold, so a 40-gram sample occupies significantly more than mL The magnesium sample has a larger volume 102 (a) Result: 2.7 × 102 mL ice (b) Deformed, overflowing, or broken Analyze: Use the volume of the bottle and the densities of water and ice to determine the volume of ice formed from a fixed amount of water Plan: Use the volume of the bottle and the density of water to determine the mass of water frozen, then calculate the volume of the ice Execute: At 25 °C, density of water is 0.997 g/mL At °C, density of ice = 0.917 g/mL 250 mL water × mL ice 0.997 g H2O(
) g H2O(s) × × = 2.7 × 102 mL ice mL water g H2O(
) 0.917 g H2O(s)
Reasonable Result Check: The density of water is larger than the density of ice It makes sense that the volume of the ice produced is larger than the volume of water (b) Result/Explanation: If the bottle is made of flexible plastic, it might be deformed and bulging if not cracked or broken If the bottle is made of glass and the top came off, there would be ice (approximately 20 mL of it) oozing out of the top Worst case scenario: if the bottle was glass and the top did not come off, it would be broken 103 Result/Explanation: When water freezes, it expands, because the density of liquid water is greater than that of solid water When the water in the engine block freezes, the increased volume of ice creates pressure on the brittle cast iron engine block Prolonged pressure could stress the cast iron to the point of cracking 104 Result/Explanation: (a) (Table 1.1) density of water = 0.998 g/mL, density of bromobenzene = 1.49 g/mL Since water does not dissolve in bromobenzene, the lower density water will be the top layer of the immiscible layers Solution Manual for Chemistry The Molecular Science 5th Edition by Moore Chapter 1: The Nature of Chemistry 27 (b) If poured slowly and carefully, the ethanol will float on top of the water and slowly dissolve in the water Both ethanol and water will float on the bromobenzene (c) Stirring will speed up the ethanol dissolving with the water to make one phase Assuming the new mixture has the average density of the original liquids, the water/ethanol layer (average density is 0.894 g/mL) will sit on top of the bromobenzene layer (density is 1.49 g/mL) 105 Result/Explanation: (a) (Table 1.1) density of water = 0.998 g/mL, density of bromobenzene = 1.49 g/mL Since water does not dissolve in bromobenzene, a layer of the lower density water will be sitting above the bromobenzene layer (b) (Table 1.1) density of benzene = 0.880 g/mL Benzene doesn’t dissolve in water and since it has a lower density, it will sit on top of the water layer (c) The benzene and bromobenzene will mix into one phase Assuming the new mixture has the average density of the original liquids, the water layer (with a density of 0.998 g/mL) will sit on top of the benzene/bromobenzene layer (with an average density of 1.185 g/mL) 106 Result: Drawing (b) Explanation: The 90 °C mercury atoms will be a little bit further apart and moving somewhat more than the 10 °C mercury, though they still would be the same size atoms The individual atoms in (c) are bigger – that doesn’t happen The individual atoms in (d) are smaller – that doesn’t happen, either 107 Result: (a) Figures (a), (b), and (f) (b) Figures (h) and (i) (c) Figures (c), (d), and (g) (d) Figures (a), (d) and (f) (e) Figures (c) and (f) (f) Figures (a), (c), (d), and (f) (g) Figures (b), (e), (g), (h), and (i) Explanation: (a) Figures (a), (b), and (f) look like they are in the solid state The atoms or molecules are relatively close together and in a regular arrangement (b) Figures (h) and (i) look like they are in the liquid state The atoms are still pretty close together but they are more randomly spaced and no longer in a regular arrangement (c) Figures (c), (d), and (g) look like they are in the gas state The atoms are fairly far apart and they are randomly spaced (d) Figures (a) and (d) look like they are samples of elements The particles are atoms separated from each other, and they are all the same color, indicating that they are the same type of atom Figure (f) also shows diatomic molecules that are all the same color, this could also represent a diatomic element (e) Figures (c) and (f) look like they are samples of compounds The atoms are grouped in pairs resembling diatomic molecules that are all alike (f) Figures (a), (c), (d), and (f) look like they are samples of pure substances The atoms or atom groups are all the same (g) Figures (b), (e), (g), (h), and (i) look like they represent mixtures They have different colors, different atom combinations, and/or different phases 108 Result/Explanation: Total of ten molecules, eight (80%) are N2 molecules (white dots) and two are O2 (20%) molecules (black dots) Solution Manual for Chemistry The Molecular Science 5th Edition by Moore 28 Chapter 1: The Nature of Chemistry 109 Result/Explanation: In metals, the specific orientation of the atoms in the solid is not limited by a crystalline structure Non-metal atoms are arranged in a delicate crystal structure, which is more fragile and susceptible to breaking 110 Result: × 105 yr