Chapter 1: The Nature of Chemistry - University of...

Chapter 1: The Nature of Chemistry

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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 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 1.16 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 1.2 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. Don’t 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.

• Discussion Questions: Section 1.12 and Questions for Review and Thought 8, 9, 66, and 67 all contain good small-group discussion questions. Choose one or two statements from Section 1.12 that are most relevant to your students for discussion. 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.


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• Concept map terms: chemical change, chemical compound, chemical property, chemical reaction, chemistry, element, gas, liquid, macroscale, microscale, nanoscale, physical changes, physical properties, solid.

Solutions for Chapter 1

Questions for Review and Thought

Review Questions

1. 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.

2. The large enzyme molecules have a three dimensional structure, including “pockets” where various molecules “fit.” The shape and orientation of these pockets dictate what drugs are able to activate or inhibit an enzyme's function.

3. Quantitative observations of a piece of electronics must include numerical information; whereas qualitative observations do not involve numerical details. For example, on a CD player: Quantitative observations might include the number of disks it can hold, the number of speakers it has, the distance each of those speakers are from the unit, the length of the wires to the speakers, the various control settings (volume, base, treble, speaker balance, etc.), the power supply requirements, etc. The qualitative observations might include the color of the unit, the types of control buttons or knobs, the quality of the sound, the location in the room, the purpose for which it is used, etc.

4. 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 (There is no detectable change in mass during an ordinary chemical reaction.) and the law of constant composition (A chemical compound always contains the same elements in the same proportion by mass.).

5. 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.

6. The first suggestion that aspirin works by blocking the release of prostaglandins was a hypothesis. Remember that a successful hypothesis is a theory, so as these scientists tested the hypothesis and found it to be successful, the statement “aspirin works by blocking the release of prostaglandins” became a theory.

7. 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 described in the search for a better painkiller by connecting the structure of nanoscale molecules to the macroscale management of pain in humans.

8. (Described in Section 1.5) Two examples of when purity of a substance is important: The purity of 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.

Why Care About Chemistry


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9. Students are asked to relate chemistry to their area of study. It really depends on the particular students taking the course what their answers will be to this question. It is important that they see SOMETHING of chemistry in their future; however, they might not see it without help. Sometimes it’s useful to consult other faculty to find out what is in their immediate coursework that relates, as well as the traditional “future” information.

10. Fourteen questions related to chemistry and science phenomena are given in Section 1.1 of the textbook.

11. Important issues change from year to year, and sometimes even from month to month or even day to day. The following is sample list composed late in the year 2003:

• Alternate fuel source to replace fossil fuels and/or natural gas: hydrogen, alcohol from plants, solar energy, biomass.

• Engineering electric vehicles.

• Elective and life saving surgeries: laser eye surgery and organ transplants.

• Advances in cloning.

• Atmospheric environmental conditions: air quality, ozone depletion, global warming.

• Cancer, AIDS, and other life threatening conditions.

• Food substitutes (Olestra, aspartame, etc.)

• Antioxidants and other vitamins: pills, fortified cereals, and other sources.

• The West Nile encephalitis outbreak in Eastern U.S.

• Bio-terrorism

How Science is Done

12. (a) The temperature at which an element melts (29.8 °C) is quantitative information. Information about what element it is (gallium) is qualitative .

(b) The color of a compound (blue) and information about the specific elements it contains (cobalt and chlorine) are both qualitative .

(c) The fact that a metal conducts electricity and information about what element it is (aluminum) are qualitative .

(d) The temperature at which a compound boils (79 °C) is quantitative information. Information about what compound it is (ethanol) is qualitative .

(e) The details of the appearance of the crystals of a compound (shiny, plate-like, and yellow) and information about the specific elements it contains (lead and sulfur) are both qualitative .

13. Some qualitative observations: The object is solid and roughly cubic in shape. It appears to have cubic sub-units. It appears to be a slightly gold or silvery shiny metal.

Some quantitative observations: The side of the cube is about 1.5 inches or a little more than 4 centimeters across.

14. (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


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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 .

15. (a) The atomic mass of an element (12.011 amu) 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 primary location of an element in the animal body (extracellular), the element’s ionic form (cation), the biological importance of an element (nerve function), and what element it is (sodium) are all qualitative .

Identifying Matter: Physical Properties

16. 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.

17. 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 to the 29.8 °C melting point.

oC =

59

× oF − 32

=

59

× 98.6 −32( )= 37.0oC

Gallium is a solid at room temperature (22 °C), but it melts at 29.8 °C, so it will melt into the liquid state when you hold it in your hand.

18. One reason for these questions about temperature using different scales is to help us get familiar with the Celsius scale. So, it is important to try estimating the answer before using equations. For a while it is useful to have a few connection points to a familiar scale. For those reasons, let’s make some key connections between the Celsius scale and the Fahrenheit scale

Let’s look at two thermometers, one calibrated with a Celsius scale and one calibrated with a Fahrenheit scale:


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100 ¡C

37 ¡C

0 ¡C

212 ¡F

98.6 ¡F

32 ¡F

(a) 20 °C (above freezing) is higher than 20 °F (below freezing).

(b) 100 °C (at boiling) is higher than 180 °F (below boiling)

(c) 100 °F is close to body temperature, which is around 40 °C. Therefore 60 °C is higher than 100 °F.

(d) – 12 °C and 20 °F are both below freezing, so let’s use the conversion equation to figure this one out:

oC =

59

× oF − 32

=

59

× 20 oF − 32

= − 6.7 oC

– 6.7 °C is warmer than –12 °C, so 20 °F is a higher temperature than – 12 °C.

19. Look at the two scales side by side:

100 ¡C

37 ¡C

0 ¡C

212 ¡F

98.6 ¡F

32 ¡F

It’s easy to tell that the sample of water at 65 °C is warmer than the sample of water at 65 °F. The 65 °C sample is above body temperature and the 65 °F sample is below body temperature. If you’re still skeptical, a quick conversion will confirm it.

o F =

95

oC

+ 32 =

95

65 oC

+ 32 = 149 oF

149 °F is warmer than 65 °F, so 65 °C is a higher temperature than 65 °F.

20. Use the side-by-side thermometers again, using body temperature and the freezing point of water as reference points:


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37 ¡C

20 ¡C

0 ¡C

G10 ¡C

98.7 ¡F

40 ¡F

32 ¡F 28 ¡F

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.

oC =

59

× oF − 32

59

× 40 oF − 32

= 4 oC lower than 20 °C (confirms estimate)

59

× 28 oF − 32

= − 2 oC higher than – 10 °C (confirms estimate)

21. Define the problem: We have the mass of the metal and some volume information. We need to determine the density.

Develop a plan: 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.

Execute the plan: 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 metal piece:

Vmetal = Vfinal – Vinitial = (37.2 mL) – (25.4 mL) = 11.8 mL

d =

mV

=105.5 g11.8 mL

= 8.94g

mL

According to Table 1.1, this is very close to the density of copper (d = 8.93 g/mL).

Check your answer: The metal piece sinks, so the density of the metal piece needed to be higher than water. (Table 1.1 gives water density as 0.998 g/mL.) So, the answer seems right.

22. Define the problem: We have the mass of the lead and the initial volume information. We need to determine the final volume after the lead is submerged.

Develop a plan: The metal piece dis places 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.


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Execute the plan: For lead, the density is 11.34 g/mL. Use dimensional analysis to convert the grams of lead into mL of lead:

10 .0 g lead ×

1 mL lead11.34 g lead

= 0.882 mL 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.

Check your answer: The sample mass is close to the density (the number of grams in one mL), so the ethanol level should rise about one mL. This answer seems right.

23. Define the problem: 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.

Develop a 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 the plan:

V = (thickness) × (width) × (length) = (1.0 cm) × (2.0 cm) × (10.0 cm) = 20. cm3

Using dimensional analysis, find the volume in mL:

20. cm3 ×

1 mL

1 cm3= 20. mL

Find the density: d =

mV

=54.0 g20. mL

= 2.7g

mL

According to Table 1.1, the density that most closely matches this one is aluminum (density = 2.70 g/mL).

Check your answer: An object whose mass is larger than its volume will have a density larger than one. The mass of this object is between two and three times larger than the volume. This answer looks right.

24. Define the problem: We have the mass of the sample, and the identity. We are asked to find the volume.

Develop a plan: Find density of bromobenzene in Table 1.1. Use the density and the mass to find the volume.

Execute the plan: Density of bromobenzene is 1.49 g/mL.


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23 .4 g ×

1 mL1.49 g

= 15.7 mL

Check your answer: A density larger than one tells us that the mass of a sample of bromobenzene will be larger than the volume. This answer looks right.

25. Define the problem: 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.

Develop a plan: Use the three linear dimensions to find the volume of the crystal. Use the volume and density (Table 1.1) to find the mass.

Execute the plan:

V = (thickness) × (width) × (length) = (10. cm) × (15 cm) × (12 cm) = 1800 cm3 =1.8×103 cm3

Using dimensional analysis, find the volume in mL:

1.8 ×10 3 cm3 ×

1 mL

1 cm 3= 1.8 ×10 3 mL

Find the mass using the density: 1.8 ×10 3 mL ×

2.16 g1 mL

= 3.9 ×10 3 g

Note: We’re carrying two significant figures since the length data was only that precise.

Check your answer: This crystal is pretty big. So, while the mass calculated is a large number, the volume is still about half the mass, in keeping with a density around two.

Chemical Changes and Chemical Properties

26. Many people may not think of all of these things.

• Gathering the wood caused me to sweat and feel warm. That was due to metabolic processes involving complex biochemical changes resulting from my body expending energy. Water may have been a product of those metabolic processes– hence, a chemical change.

• My body also loses heat when the sweat evaporates off my skin – a physical change involving the transfer of heat.

• Once I stopped collecting wood, I started feeling cold again, since my warm body heat was escaping to the cold night air– again a physical change involving the transfer of heat – so I lit a fire.

• The wood, when lighted, underwent a chemical change known as combustion. This energy-producing chemical change produced heat that warmed up the air nearby, a physical change. Some of the heat was absorbed by me, also a physical change, to replenish the heat I was losing to the cold night air.


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That made me feel warmer.

• Later, I stopped the combustion reaction in the fire pit by separating the reactants of the combustion reaction. Adding water to the coals – a physical change – prevented it from continuing to interact with the oxygen.

• In addition, the cold water absorbed the heat from the fire, warming the water up – a physical change. Some of the water absorbed so much heat that it was converted from liquid form to vapor form – a physical change. That was the steam I saw.

• When I got into my sleeping bag, the heat lost from my body was retained near me by the reflective fabric of the bag. That reduced the loss of heat from my warm body into the cold night air, so I was able to keep myself warm all night.

27. (a) The normal color of bromine is a physical property. Determining the color of a substance does not change its chemical form.

(b) The fact that iron can be transformed into rust is a chemical property. Iron, originally in the elemental metallic state, is incorporated into a compound, rust, when it is observed to undergo this transformation.

(c) The fact that dynamite can explode is a chemical property. The dynamite is chemically changed when it is observed to explode.

(d) Observing the shininess of aluminum does not change it, so this is a physical property. Melting aluminum does not change it to a different substance, though it does change its physical state. It is still aluminum, so melting at 387 °C is a physical property of aluminum.

28. (a) Subliming Dry Ice (converting it from a solid directly into a gas without going through the liquid state, first) does not change it to a different substance, though the physical state does change. It is still Dry Ice. Therefore, subliming at – 78 °C is a physical property of Dry Ice.

(b) The fact that methanol burns in air is a chemical property. The methanol compound is converted into combustion products when it is observed burning.

(c) Sugar’s solubility in water is a physical property. Dissolving sugar does not change it to a different substance, though it does change its physical state. It is still sugar. Evaporating the water would recreate the sugar crystals.

(d) Hydrogen peroxide’s ability to decompose into oxygen and water is a chemical property. The hydrogen peroxide compound is converted into oxygen and water when it is observed decomposing.

29. (a) Bleaching clothes from purple to pink is a chemical change. The purple substance in the clothing reacts with the bleach to make a pink substance. The purple color cannot be brought back nor can the bleach.

(b) The burning of fuel in the space shuttle (hydrogen and oxygen) to form water and energy is a chemical change. The two elements react to form a compound.

(c) The ice cube melting in the lemonade is a physical change. The H2O molecules do not change to a different form in the physical state change.

30. (a) Salt dissolving when you add water is a physical change. The salt crystals would be reformed if the water were evaporated off.

(b) Food is digested and metabolized in your body by way of a chemical change. The molecules composing the food are decomposed to provide energy for activities and functions in your body.


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(c) Sugar stays sugar when it is ground from a crystalline state into a powdered state, so this is a physical change.

(d) When potassium is put in water, there is a purplish-pink flame and the water becomes basic (alkaline). This is a chemical change of the elemental potassium and water into a compound in the water that makes it basic.

31. (a) The conversion of excess food into fat molecules is the body’s way of storing energy for doing work later. So, this represents the outside source of energy (from the food we eat) is forcing a chemical reaction to occur (the production of fat).

(b) Sodium reacts with water rather violently. It produces a lot of heat and causes work to be done.

(c) Sodium azide in an automobile’s airbag decomposes causing the bag to inflate. This uses a chemical reaction to release energy and cause work to be done (inflation of the air bag).

(d) The process of hard-boiling an egg on your stove uses energy from the stove to cause a chemical reaction to occur (the coagulation of the white and yolk of the egg).

Classifying Matter: Substances and Mixtures

32. It is clear by visual inspection that the mixture is non-uniform (heterogeneous) at the macroscopic level. The iron could be separated from the sand using a magnet, since iron is attracted to magnets and the sand is not.

33. A solution of sugar in water would be homogeneous once the sugar is completely dissolved. No amount of optical scrutiny would make it possible to see the separate components of this mixture. The sugar could be recrystallized by evaporating the water at a low enough temperature not to burn the sugar – perhaps by using a vacuum. The water could then be recondensed into another container, separate from the sugar.

34. The terminology “heterogeneous” and “homogeneous” is somewhat subjective. In Section 1.5, the terms are described. The heterogeneous mixture is described as one whose uneven texture can be seen without magnification or with a microscope. The homogeneous mixture is defined as one that is completely uniform, wherein no amount of optical magnification will reveal variations in the mixture’s properties. Note that there is a “gap” between these two terms. The question “how close do I look?” comes to mind.

The bottom line is this: These terms were designed to HELP us classify things, not to create trick questions. If you explain your answer with a valid defense, the answer ought to be right; however, don’t go out of your way to imagine unusual circumstances that make the question more difficult to classify. Think about what you would SEE. Identify whether what you see has variations, then make a case for the proper term.

(a) Vodka is classified as a homogenous mixture. It is a clear, colorless solution of alcohol, water, and probably some other minor ingredients.

(b) Blood appears smooth in texture and to our eyes most likely appears to be homogeneous. Upon closer examination it is found to have various particles within the liquid, and might, for that reason, be called heterogeneous.

(c) Cowhide is heterogeneous. Even folks who have never seen a cowhide might be able to imagine that brown cows, white cows, black cows and spotted cows probably have different coloration to their hide. There are probably pores where the hair grows out, making it rough in texture. The hide of a young cow is probably softer and more pliable than the hide of an old cow. Unless presented with a sample that visually changed our perception of what constitutes a cowhide, it is quite safe to call the cowhide heterogeneous.


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(d) Bread is heterogeneous. The crust is a different color. Some parts of the bread have bigger bubbles than other parts. Some breads have whole grains in them. Some breads are composed of different colors (like the rye/white swirl breads). In general, most samples of bread have identifiable regions that are different from other regions.

35. The terminology “heterogeneous” and “homogeneous” is somewhat subjective. In Section 1.5, the terms are described. The heterogeneous mixture is described as one whose uneven texture can be seen without magnification or with a microscope. The homogeneous mixture is defined as one that is completely uniform, wherein no amount of optical magnification will reveal variations in the mixture’s properties. Note that there is a “gap” between these two terms. The question “how close do I look?” comes to mind. The bottom line is this: These terms were designed to HELP us classify things, not to create trick questions. If you explain your answer with a valid defense, the answer ought to be right; however, don’t go out of your way to imagine unusual circumstances that make the question more difficult to classify. Think about what you would SEE. Identify whether what you see has variations, then make a case for the proper term.

(a) An asphalt blacktop is classified as a heterogeneous mixture. Driving down the road, the blacktop looks pretty uniform, but if you look at it closely, the mixture is made up of a black goopy material along with other substances that are more like black or dark dull rocks. These visual variations lead us to conclude that the mixture is heterogeneous.

(b) Ocean water, if it were clear, would be considered to be a homogeneous mixture. Dissolved in the water, we know that there is at least salt and other dissolved minerals. However, if the solution is clear, then we rule out the possibility of seaweed or sand suspended in it – those might make us call it heterogeneous.

(c) Ice tea with ice cubes is definitely heterogeneous. The portions representing ice are far different from the aqueous solution of tea particles, both in composition and in physical state. There is no question that this is an example of a heterogeneous mixture.

(d) This filtered apple juice is a homogeneous solution of sugar, other flavors, and water. If the solution is filtered sufficiently to remove all the pulp, which might qualify it as heterogeneous, it is probably safe to call this mixture homogeneous.

Classifying Matter: Elements and Compounds

36. (a) A blue powder turns white and loses mass. The loss of mass is most likely due to the creation of a gaseous product. That suggests that the original material was a compound that decomposed into the white substance (a compound or an element) and a gas (a compound or an element).

(b) If three different gases were formed, that suggests that the original material was a compound that decomposed into three compounds or gases.

37. (a) A reddish metal is placed in a flame. It turns black and the black material has a higher mass than the original reddish material. That suggests that the material produced is a compound, and either a compound or an element combined with something in air (oxygen or nitrogen) to make the product.

(b) A white solid is heated in oxygen and forms two gases. The mass of the gases produced is the same as the solid and the oxygen. Oxygen is an element. That suggests that the product gases are compounds and the elemental oxygen combined with another element or compound to make one or more compounds.

38. (a) A piece of newspaper is a heterogeneous mixture. Paper and ink are distributed in a non-uniform fashion at the macroscopic level.


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(b) Solid granulated sugar is a pure compound. Sugar is made of two or more elements.

(c) Fresh squeezed orange juice is a heterogeneous mixture. The presence of unfiltered pulp makes part of the mixture solid and part of the mixture liquid.

(d) Gold jewelry is a homogeneous mixture, most of the time. Most jewelry is less than 24 carat – pure gold. If it is 18 carat, 14 carat, 10 carat gold, that means that other metals are mixed in with the gold (usually to make it more durable and cheaper). The combining of metals in a fashion that prevents us from seeing variations in the texture or properties of the metal qualifies it as a homogeneous mixture.

39. (a) A cup of coffee is a homogeneous mixture of colors, flavors and water. Some people’s coffee might be heterogeneous, in cases where they don’t dissolve all their sugar or if they add bad milk and it curdles, but most of the time, the solution in a coffee cup is homogenous.

(b) A soft drink such as Coke or Pepsi is a heterogeneous mixture. Freshly poured soft drinks have carbonation bubbles, which makes the properties in one region of the mixture different from the properties in another. If the soft drink is still in the bottle or can, the carbonation is completely dissolved, and the solution would appear homogeneous. However, to enjoy the soft drink, one must pour it, and upon pouring it, the bubbles make it heterogeneous.

(c) A piece of dry ice is a pure compound composed of CO2 in the solid state.

40. (a) Chunky peanut butter is definitely a heterogeneous mixture. The uncrushed peanut chunks do not have the same properties as the smooth, sweetened part of the mixture.

(b) Distilled water is a pure compound. The distillation process removes other minerals and substances from water, leaving it just water.

(c) Platinum is an element with the symbol “Pt”.

(d) Air is usually considered to be a homogeneous mixture. Now, sometimes air has enough variable properties to qualify as heterogeneous, such as near the tailpipe of a diesel truck. However, most of the time, the gases in a sample of air are sufficiently well mixed such that there is no visible difference in the properties of various regions of that air sample.

41. (a) Table salt (sodium chloride) is a compound. Some brands of table salt are “iodized” which suggests that some of the crystals of salt in the mixture are NaI along with NaCl. In those instances, it is appropriate to call table salt a mixture. However, table salt is generally considered to be just the compound NaCl.

(b) Methane is a compound. The combustion products (carbon dioxide and water) suggest that methane must contain at least carbon and hydrogen, making it a compound.

(c) Chocolate chip cookies are heterogeneous mixtures . It is clear that the properties of the regions including a chocolate chip are quite different from the regions including the cooked cookie dough.

(d) Silicon is an element, with the symbol “Si”.

Nanoscale Theories and Models

42. The crystal of halite pictured is in the macroscopic world. Its shape is cubic. It could be expected that the arrangement of particles (atoms and/or ions) in the nanoscale world are also arranged in a cubic fashion.

43. The shape of the galena crystals appears to be cubic. One would expect a cubic arrangement for the atoms deep inside the crystal.

44. The bacterium is in the microscale world. We need some enhanced magnification to see.


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45. The three states of matter are solid, liquid, and gas. A solid has a fixed shape and fixed volume. The particles comprising a solid are held tightly in a regular arrangement. Solid particles vibrate about their average positions and seldom translate. A liquid has a fixed volume, but takes on the shape of its container. The particles comprising a liquid are arranged more randomly than those in a solid are. They exert moderate forces on each other. Liquid particles are in constant random motion. A gas has no fixed volume or shape. Gases expand to fill their container. Gas particles are in continuous random motion. They exert small forces on one another.

46. When we open a can of a soft drink, the carbon dioxide gas expands rapidly as it rushes out of the can. At the nanoscale, this can be explained as large number of carbon dioxide molecules crowded into the unopened can. When the can is opened, the molecules that were about to hit the surface where the hole was made continue forward through the hole. A large number of the carbon dioxide particles that were contained within the can very quickly escape through the same hole.

47. Wet clothes dry on the line in the sun. At the nanoscale, this is described by the relatively slow liquid water molecules being made to move more quickly by the heat of the sun, as well as wind taking molecules off the surface of the liquid. Once the molecules get free of the liquid state, they don’t remember being part of a condensed phase. Once sufficient warming or evaporation occurs, the clothes feel dry. The water undergoes only a change in physical state, thus the change is a physical change, not a chemical one.

48. The atoms in the solid sucrose molecules start off at a relatively low energy and compose a rather complex molecule. A significant amount of heat must be added to increase the motion of these atoms so that they are able to break free of the bonds that hold them together in the sugar molecule and to interact with each other to make the “caramelization” products.

49. (a) 32.75 km ×

1000 m1 km

= 3.275 × 104 meters

(b)

0.0342 mm ×

10−3 m1 mm

×1 nm

10−9 m= 3.42 × 104 nanometers

(c)

1.21 × 10−12 km ×

103 m1 km

×1 µm

10−6 m= 1.21 × 10−3 micrometers

50. (a) 0.00572 kg ×

1000 g1 kg

= 5.72 grams

(b) 8.347 ×107 nL ×

10−9 L1 nL

= 8.347 ×10−2 liter

(c)

423.7 g ×

1 kg

103 g= 0.4237 kilograms

Atomic Theory

51. Conservation of mass is easy to see from the point of view of atomic theory. A chemical change is described as the rearrangement of atoms. Since the atoms in the starting materials must all be accounted for in the substances produced, then there would be no change in the mass.

52. Atomic theory can explain the constant composition of chemical compounds. Compounds are composed of


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a specific number and type of atoms. Atoms of each element all weigh the same. The sum of all the atoms’ masses is mass of the compound. The percentage of mass represented by one element in the compound must be constant, since all the atoms of that element are the same mass, and the compound contains a fixed number of those element’s atoms.

53. Four Postulates of Modern Atomic Theory

(I) All matter is composed of atoms, which are extremely tiny.

(II) All atoms of a given element have the same chemical properties.

(III) Compounds are formed by the chemical combination of two or more different kinds of atoms.

(IV) A chemical reaction involves joining separating or rearranging atoms.

54. The law of multiple proportions can be explained using atomic theory. Consider two compounds that both contain the same two elements. The proportion of the two elements in these two compounds must be different. Compounds are composed of a specific number and type of atoms. If you pick samples of each of these compounds such that they contain the same number of atoms of the first element, then you’ll find that the ratio of atoms on the other element is a small integer. Consider a concrete example. Fe2O3 and Fe3O4.

Compare a sample containing three Fe2O3 to a sample containing two Fe3O4. These samples both have six Fe atoms. The first sample has nine oxygen atoms and the second sample has 8 oxygen atoms. Since all oxygen atoms weigh the same, the ratio of mass will also be the ratio of atoms; in this example, the ratio of

mass and the ratio of atoms is 98

.

55. The law of multiple proportions says that if two compounds contain the same elements and samples of those two compounds both contain the same mass of one element, then the ratio of the masses of the other elements will be a small whole number.

56. Define the problem: These three compounds all contain chromium and oxygen. The mass of chromium in each compound is given for 100-gram samples of the respective compounds. Determine if these data conform to the law of multiple proportions.54. The law of multiple proportions says that if two compounds contain the same elements and samples of those two compounds both contain the same mass of one element, then the ratio of the masses of the other elements will be a small whole number.

Develop a plan: Find the masses of Cr and O in 100.0 gram samples. Scale the sample sizes of each compound so that all three of them contain the same amount of oxygen. Determine the mass of chromium in each of those samples. Take ratios of the chromium masses to see if they can be represented by ratios of small whole numbers.

Execute the plan: The masses of Cr in 100.0 gram samples are given. The mass of O must be the difference between the total compound mass and the mass of Cr:

Compound 1: 100.0 g of compound 1 – 52.0 g Cr = 48.0 g O



Determine the scale factor by determining a multiplier that makes every sample have the same oxygen mass as the first 100.0 gram sample:

100.0 g of compound 1 has 48.0 g O

48.0 g48.0 g

= 1.00


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48.0 g31.6 g

= 1.52


48.0 g23.5 g

= 2.04

Scale the samples by these factors to determine the mass of Cr in each sample.

100.0 g of compound 1 has 48.0 g O and 52.0 g Cr

(1.52) × 100.0 g of compound 2 has (1.52) × 31.6 g O and (1.52) × 68.4 g Cr

So, 151.9 g of compound 2 has 48.0 g O and 104 g Cr

(2.04) × 100.0 g of compound 3 has (2.04) × 23.5 g O and (2.04) × 76.5 g Cr

So, 204 g of compound 3 has 48.0 g O and 156 g Cr

Now, take ratios of the chromium masses:

g Cr in compound 2g Cr in compound 1

=104 g52.0 g

= 2.00 =21

g Cr in compound 3

g Cr in compound 1=

156.3 g52.0 g

= 3.01 =31

g Cr in compound 3

g Cr in compound 2=

156.3 g104 g

=1.50 =32

These simple whole-number ratios show that the law of multiple proportions is upheld among these compounds.

Check your answer: A law is a summary conclusion from a large number of different experiments. It is satisfying that these results also uphold the law.

The Chemical Elements

57. Many responses are equally valid here. Below are a few common examples. These lists are not comprehensive; many other answers are also right. The periodic table on the inside cover of your text is color coded to indicate metals, non-metals and metalloids.

(a) Common metallic elements: Fe, iron; Au, gold; Pb, lead; Cu, copper; Al, aluminum.

(b) Common non-metallic elements: C, carbon; H, hydrogen; O, oxygen; N, nitrogen

(c) Metalloids: B, boron; Si, silicon; Ge, germanium; As, arsenic; Sb, antimony; Te, tellurium.

(d) Elements that are diatomic molecules: nitrogen, N2; oxygen, O2; hydrogen, H2; fluorine, F2; chlorine,

Cl2; bromine, Br2; iodine, I2

58. Many responses are equally valid here. Below are a few common examples. These lists are not comprehensive; many other answers are also right.

(a) Some elements that are gases at room temperature: N2, nitrogen; O2, oxygen; H2, hydrogen; He, helium; Ne, neon; Ar, argon; Kr, krypton.


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(b) Some elements that are solids at room temperature: Fe, iron; Au, gold; Pb, lead; Cu, copper; Al, aluminum; C, carbon; B, boron; Si, silicon.

(c) Some elements that do not consist of molecules: Fe, iron; Au, gold; Pb, lead; Cu, copper; Al, aluminum; C, carbon; B, boron; Si, silicon.

(d) Some elements that have different allotropic forms: O oxygen (O2 and O3); C carbon (graphite, diamond, and fullerenes).

Communicating Chemistry: Symbolism

59. Formula for each substance and nanoscale picture:

(a) Water H2O (b) Nitrogen N2

(c) Neon Ne (d) Chlorine Cl2

60. Formula for the substance and nanoscale picture:

(a) Iodine I2 (b) ozone O3

(c) Helium He (d) Carbon dioxide CO2


17

61. 2 H2 (g) + O2 (g) 2 H2O (g)

Hydrogen and oxygen gas water vapor

62. 2 CO (g) + O2 (g) 2 CO2 (g)

carbon monoxide and oxygen gas carbon dioxide gas

63. I2 (s) I2 (g)

solid iodine iodine gas


18

64. Br2 (l) Br2 (g)

liquid bromine bromine gas

General Questions

65. (a) The mass of the compound (1.456 grams) is quantitative and relates to a physical property. The color (white), the fact that it reacts with a dye, and the color change in the dye (red to colorless) are all qualitative. The colors are related to physical properties. The reaction with the dye is related to a chemical property.

(b) The mass of the metal (0.6 grams) is quantitative and related to a physical property. The identity of the metal (lithium) and the identities of the chemicals it reacts with and produces (water, lithium hydroxide, and hydrogen) are all qualitative information. The fact that a chemical reaction occurs when the metal is added to water is qualitative information and related to a chemical property.

66. (a) The volume of the water (one liter) is quantitative and relates to a physical property. The identities of the materials (water, charcoal) are qualitative information. The fact that the charcoal adsorbs the dye is qualitative and related to a physical property, since the adsorption process is reversed later and the dye’s color (the one property we have to identify it) is retained.

(b) All of the information in these statements is qualitative. The colors, the fact that the white powder dissolved, the container used to hold the solution, the relative temperature change, and the physical state of the substance formed are related to physical properties. The fact that there was a transformation of the dissolved substance into an insoluble substance is related to a chemical property.

67. 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 solid potassium smaller atoms larger atoms closer packed less closely packed smaller volume larger volume

68. How the atoms are organized and bonded together is different between different allotropes. That is how allotropes are different at the nanoscale.


19

69. These molecules need to have similar nanoscale sizes and even similar structural details so that they are able to affect similar biological chemistry, both primary and secondary (side) effects. Biological chemistry is heavily reliant on enzyme chemistry, so these molecules probably will not be very different in size.

Applying Concepts

70. (a) (Table 1.1) density of butane = 0.579 g/mL; density of bromobenzene = 1.49 g/mL

1 mL butane weighs less than 1 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 1 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.

71. (a) (Table 1.1) density of ethanol = 0.789 g/mL; density of bromobenzene = 1.49 g/mL

1 gram of ethanol has a volume more than 1 mL and 1 gram of bromobenzene has a volume less than 1 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 1 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 1 gram of water in 1 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 1 mL. Any more than approximately 2 grams of magnesium will occupy more volume than 20 grams of gold, so a 40 gram sample occupies significantly more than 1 mL. The magnesium sample has a larger volume.

72. (a) Define the problem: 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.

Develop a plan: Use the volume of the bottle and the density of water to figure out the mass of water frozen, then calculate the volume of the ice.

Execute the plan: At 25 °C, density of water is 0.997 g/mL.

At 0 °C, density of ice = 0.917 g/mL

250 mL water ×

0.997 g H2O(l)1 mL water

×1 g H2O(s)1 g H2O(l)

×1 mL ice

0.917 g H2O(s)=2.7 ×10 2 mL ice

Check your answer: 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) If the bottle is made of flexible plastic, it might be deformed and bulging if not cracked and leaking ice. If the bottle is made of glass and the top came off, there would be ice (approximately 20 mL of it) oozing


20

out of the top of the bottle. Worst case scenario would be if the bottle was glass and the top did not come off, so it broke.

73. All substances with a density less than that of mercury’s will float. The only substance in Table 1.1 with a density greater than that of mercury is gold. Prospectors in California used to use liquid mercury to separate and clean gold from rocks and dirt. The small particles of dirt and rocks floated in the mercury and could be skimmed off. The mercury then could be poured carefully off, leaving just the gold behind.

74. (See Figure 1.2) Water at the top has the lowest density. Mercury at the bottom has the highest density.

75. (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 two imiscible layers.

(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) (Table 1.1) density of ethanol = 0.789 g/mL. Ethanol doesn’t dissolve in benzene, and since it has a lower density, it will sit on top of the benzene layer.

(d) The ethanol and water will mix and make one phase. The benzene and bromobenzene will also mix into one phase. Assuming the new mixtures have the average density of the original liquids, the water/ethanol layer (average density is 0.894 g/mL) will sit on top of the benzene/bromobenzene layer. (average density is 1.185 g/mL)

76. The melting point is the temperature at which a solid changes into a liquid. In the nanoscale picture, the liquid molecules are moving faster and they are farther apart. That motion spreads out the molecules and makes the volume larger, even though the collection of molecules have the same mass. Hence, the density of the liquid is smaller than the density of the solid.

The boiling point is the temperature at which the liquid molecules change into gas-phase molecules. In the nanoscale picture, the gas molecules are moving faster and they are farther apart. That motion spreads out the molecules and makes the volume larger, even though the collection of molecules have the same mass. Hence, the density of the gas is smaller than the density of the liquid.

77. The best answer is (b). 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.

78. (a) The diagrams (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) The diagrams (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) The diagrams (c), (d), and (g) look like they are in the gas state. The atoms are fairly far apart and they are randomly spaced.

(d) The diagrams (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. Diagram (f) also shows diatomic molecules that are all the same color, this could also represent a diatomic element.

(e) The diagrams (c) and (f) look like they are samples of compounds. The atoms are grouped in pairs resembling diatomic molecules that are all alike.


21

(f) The diagrams (a), (c), (d), and (f) look like they are samples of pure substances. The atoms or atom groups are all the same.

(g) The diagrams (b), (e), (h), and (i) look like they represent mixtures. They have different colors, different atom combinations, and/or different phases.

More Challenging Questions

79. Define the problem: 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

Develop a plan: Use the three linear dimensions to find the volume of the crystal, then convert the volume

to m3 using metric conversions.

Execute the plan: V = (thickness) × (width) × (length) = (392 pm) × (392 pm) × (392 pm)

= 6.02 ×107 pm3

Using dimensional analysis, find the volume in m3

6.02 × 107 pm3 ×10−12 m

1 pm

3

= 6.02 × 10−29 m 3

Check your answer: The cube is from the nanoscale, so it makes sense that it would be a very small volume using a macroscale unit of measure.

80. Define the problem: Given the volume of the cube in cm3, find the length of a side in picometers: l pm long

l pm thick

l pm wide

Develop a plan: Use the volume to find the three linear side dimensions of the cube, then convert the

volume to m3 using metric conversions.

Execute the plan: V = l3 l = V3 = 1.81 × 10−22 cm33= 5.66 × 10−8 cm

Using dimensional analysis, find the length in picometers:

5.66 × 10−8 cm ×

10−2 m1 cm

×1 pm

10−12 m= 566 pm

Check your answer: The cube is from the nanoscale, so it makes sense that it would be a reasonable-size


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number using a nanoscale unit of measure.

81. (a) Metals are in the blue area.

(b) Nonmetals are in the lavender area.

(c) Metalloids are in the orange area.

82. (a) A shiny solid that conducts electricity would be a metal, so it will be in the blue area.

(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 blue area.

83. 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. (see Section 1.3)

84. (a) Fat is a mixture, solution of lye is a mixture of water (a substance that is a compound) and lye (sodium hydroxide, a substance). The mixtures are probably both homogeneous.

(b) Define the problem: Conservation of mass says that the soap will have a mass that is the sum of all three component substances.

Develop a plan: Convert the volume to kilograms and add all the masses.

Execute the plan: 50.0 L water ×

1000 mL1 L

×0.998 g water1 mL water

×1 kg

1000 g= 49.9 kg

32.4 kg + 49.9 kg + 5.0 kg = 58.1 kg

Check your answer: 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.

(c) Physical processes that occur include the mixing of substances to make solutions.

Chemical processes involved are the formation of carboxylate salts (soap molecules) and gycerol from fat and lye.

85. (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.

86. Define the problem: Given the mass of a flask filled with a fixed volume of water, the mass of a flask filled with a given number of shots and the rest water, and the mass of the container with the same number of shots, (a) determine what pure metal the shots are made of. and (b) determine the volume occupied by 500 shots.

Develop a plan: Using the density of water, determine the mass of the water in the flask. From the combined mass of the flask+water, determine the mass of the flask. Use the combined mass of the flask+shots+water the mass of the shots and the density of water to determine what volume of water was displaced by the shots. Subtract this volume from 100.0 mL to determine the volume of the water displaced by the shots. (a) Take a ratio of the mass of the shots to their volume to determine the density, then use Table 1.1 to determine what the metal is. (b) Use the volume of 20 shots to determine the volume of 500


23

shots.

Execute the plan: (a) 100.0 mL water ×

0.998 g water1 mL water

= 99.8 g water

122.3 g − 99.8 g = 22.5 g flask

159.9 g − 42.3 g − 22.5 g = 95.1 g water in the container with shots

95 .1 g water ×

1 mL water0.998 g water

= 95.3 mL water

Volume of water displaced = Volume of water without shots − Volume of water with shorts

Volume of water displaced = 100.0 mL − 95.3 mL = 4.7 mL

d =

mV

=42.3 g4.7 mL

= 9.0 g / mL

The closest pure metal element listed in Table 1.1 to this value is copper (d = 8.93 g/mL)

(b) 500 shots×

4.7 mL20 shots

= 120 mL

Check your answers: (a) The density of the metal is fairly close to copper. It would have been useful if the scientific observer would have reported the color of the metal to distinguish it from possibly being nickel (with d = 8.90 g/mL), instead. (b) When the number of shots increases, the volume increases. This makes sense.

Media Questions

87. Look through magazines, local and national newspapers, and advertisements.

88. Over 13 billion web sites are identified using a typical search engine and the keyword “Chemistry”.

89. Aspirin web sites give many different answers for this question.

90. Celecoxib and meloxicam web sites give many different answers for this question.

91. Over 235 thousand web sites are identified using a typical search engine and the keyword “nanoscale.” Over one billion web sites are identified using the keyword “nanotechnology.” This is a hot business area.

Conceptual Challenge Problems

CP-1.A Students should be challenged to articulate what makes an expression a law of nature. The answers will vary depending on the student’s background and sophistication with natural and life sciences. The answers may include references to the conditions under which living things grow or not grow. The students are referred to Section .13, when asked to consider this answer.

CP-1.B The answers will vary depending on the student’s background and sophistication with natural and life sciences. Depending on the population, students may answer these questions in many ways. The students are referred to Section 1.3, when asked to consider these answer.

CP-1.C The answers will vary depending on the ability of the students to understand the mechanics of motion imagine the motion of microscopic (invisible) things.


24

(a) Correlating the motion to macroscopic events, gives answers such as:

diffusion of gases being related to smelling smelly things remote from the source.

dilution of solutes related to the dye swirls with food coloring is dripped into water.

(b) Correlating the motion to macroscopic motion: the paths should be straight until the molecules hit one another or the walls if the container, in which case they would bounce off and fly away in a new direction on a straight path. More sophisticated students might realize and that these collisions might result in chemical reactions or other inelastic processes.

CP-1.D Students may do this task in a variety of ways. The following is an example.

A nanobe’s maximum cylindrical volume is estimated = 2πr2d = 2π(10 nm)2 × 50 nm long =

3.1 × 104nm3/nanobe.

A hydrogen’s radius is 37 pm or 0.037 nm (as described in Chapter 7) so the volume of a hydrogen

atom is =

43

πr3 =

43

π(0.037 nm)3 = 2.1 × 10-4 nm3/atom.

If hydrogen atoms could completely fill the space inside the cylinder, then there would be:

Number of atoms =

3.1× 104 nm3 / nanobe

2.1× 10 −4 nm3 /atom= 1.5× 108 atoms / nanobe

According to Figure 1.14, protein molecules can have a diameter as small as 4 nm. Assuming the

proteins are spheres with radius of 2 nm, the volume of a hydrogen atom is =

43

πr3 =

43

π(2 nm)3 =30

nm3/protein.

Number of proteins =

3.1× 104 nm3 / nanobe

34 nm3 / protein= 910 proteins/ nanobe

This number of proteins does not rule out the possibility that a nanobe could be living, since over 900 proteins can fit into this space; however, this number allows no room for the proteins to function as enzymes, and no room for water, nutrients, DNA and other biological molecules.

CP-1.E Students will answer this question in a variety of ways. Most will answer (a) affirmatively, attributing science (broadly defined) with the power to continue to improve medicine and eliminate hazard. The answer to (b) will be varied depending on what students deem influential in their own experience. Certainly many students will point to understanding of health and wellness, understanding medicines and vaccines for illness and disease, understanding anatomy and physiology for improved transplantation and repairs.

Chapter 1: The Nature of Chemistry - University of...

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