This Document Contains Experiments 19 to 26 Name____________________________________________________Section________________Date___________ Experiment 19: Measurement of pH Invitation to Inquiry There are many common things that can be used as acid and base indicators. For example, from foods you could try boiled purple cabbage juice, grape juice, blackberry juice, ordinary tea, and others. From the office supply store certain construction papers might show color changes. To test foods, try soaking strips of filter paper in juices or solutions, then allowing it to dry completely. Materials such as construction paper can be tested directly. Plan tests to find what color each indicator will appear when exposed to solutions of acids or bases, and test other materials, too. Show how your collections of indicators will identify the pH from a wide range of possibilities. Think of some way to measure the “strength” of an acid or a base without using an indicator. Taking proper precautions, do experiments to compare several methods of measuring strength. Background Acids and bases are classes of chemical compounds and each class has certain properties in common. Acids have characteristic properties of reacting with active metals to release hydrogen gas, neutralizing bases to form a salt, and changing the color of certain substances such as litmus. Bases, on the other hand, have characteristic properties of converting plant and animal tissues to soluble substances, neutralizing acids to form a salt, and reversing the color changes that were caused by acids. Some chemical compounds are acidic, such as vinegar, and other compounds are basic, such as lye; but the terms acidic and basic are very general and inexact. The pH scale is a concise and quantitative way of expressing the strength of an acid or a base. On this scale a neutral solution (neither acidic nor basic properties) has a pH of 7.0. Acid solutions have pH values below 7 and smaller numbers mean greater acidic properties. Basic solutions have pH values above 7 and larger numbers mean greater basic properties. The pH scale is logarithmic, so a pH of 2 is 10 times more acidic than a pH of 3. Likewise, a pH of 8 is 10 times 10, or 100 times more basic than a pH of 6. Many plant extracts and synthetic dyes change colors when mixed with acids or bases. Such substances that change color in the presence of acids or bases can be used as an acid-base indicator. Litmus, for example, is an acid-base indicator made from a dye extracted from certain species of lichens. The dye is applied to paper strips, which turn red in acidic solutions and blue in basic solutions. This is only a “ball park” indicator, however, since it only indicates on which side of pH 7 a solution lies. Other indicators are available that can be used to estimate the pH to about half a pH unit—for example, a pH of 3.5. There are also electrical instruments that measure pH by measuring the conductivity of a solution. In this experiment you will determine the characteristics of several commercial indicators, then test unknown solutions for acid and base properties. CAUTION: Acids and bases can damage skin and other materials. If spilled on your person or clothing, flush with running water for several minutes and inform the instructor. Dilute any tabletop spills with water, then use a sponge to wipe up the spill. Be sure to rinse the sponge with plenty of water. Figure 19.1 pH Ranges and Color Changes for Selected Indicators Procedure Obtain five clean test tubes and a test tube rack. Pour 5 mL of hydrochloric acid into each. Test the hydrochloric acid with litmus by placing a strip of red litmus paper and a strip of blue litmus paper in a clean watch glass. Dip a clean, dry stirring rod into the solution in one of the test tubes, then transfer a drop to each litmus paper. Record your observations in Data Table 19.1 on page 147. Save the litmus paper and watch glass for further testing. Add two drops of each indicator solution to a test tube of hydrochloric acid, recording your observations in Data Table 19.1. Include pH values or ranges where possible. Place a strip of universal indicator paper on the watch glass. Use the stirring rod to transfer a drop of hydrochloric acid to the indicator paper. The universal indicator paper container will have a color standard chart with a corresponding pH. Record the color and pH number in Data Table 19.1. If a color falls between two of the standards, estimate the pH to a fraction between the corresponding pH values. Repeat procedure steps 1 through 3 using sulfuric acid, acetic acid, sodium hydroxide, barium hydroxide, and dilute household ammonia solutions. Record the results for the acid solutions in Data Table 19.1 and the results for the base solutions in Data Table 19.2 on page 148. When all of the solutions have been tested, take a few minutes to analyze your findings. Compare the pH ranges of the indicators in Table 19.1 with the pH as shown by the universal indicator. Obtain an unknown solution and use the different indicator tests to determine if the solution is an acid or a base and to determine the pH of the solution. Record your test results and findings in Data Table 19.3 on page 149. Results A popular noncarbonated beverage turns thymol blue to yellow and methyl orange to red. What is the approximate pH of this soft drink? About 3. What is the approximate pH of a beer that turns methyl orange to a yellow color? About 4 How would the color changes of a mixture of methyl orange, methyl red, bromthymol blue, and phenolphthalein compare with those of the universal indicator? Results will be good for solutions with a pH between 3 and 10. However, the universal indicator is more convenient overall. Would you use a solution of phenolphthalein to find the strength of an acid? Explain. Yes. Use a calibrated basic solution, measure the amount of base needed to completely react with, and neutralize, a known amount of acid. This would be indicated by the phenolphthalein. The flowering shrub hydrangea produces blue flowers in a certain soil, but produces pink flowers if lime is added to the soil. Propose an explanation for this observation. Explain how you could do tests on the soil to check your explanation. (a) The flowers must have an organic material that responds to the pH in a way similar to litmus. (b). Run a pH test on a solution of soil from beneath a plant with blue flowers and another test from a solution derived from the soil beneath a plant with pink flowers. Was the purpose of this lab accomplished? Why or why not? (Your answer to this question should show thoughtful analysis and careful, thorough thinking.) (Students should be able to describe the difference between an acid and a base. The students should be familiar with the different acid-base indicators and have a good understanding of the pH scale.) Note: pH varies with concentration of solutions.
Data Table 19.3 Test Results for Unknown Solutions
Indicator Results
Red Litmus
Blue Litmus
Bromthymol Blue
Methyl Orange
Methyl Red
Phenolphthalein
Universal Indicator Color
pH of unknown solution: ______________
Is the unknown solution an acid or a base? _________________________
Name____________________________________________________Section________________Date___________ Experiment 20: Amount of Water Vapor in the Air Invitation to Inquiry Cobalt chloride is often used to test for the presence of water since it undergoes a reversible color change when exposed to moisture or humidity. For example, cobalt chloride is sometimes included with silica gel pellets to indicate when the gel has absorbed moisture. Experiment with cobalt chloride dried on filter paper strips. Find out if the color change is sensitive enough to water vapor and temperature to be used as an indicator of the relative humidity. Background The amount of water vapor in the air is referred to generally as humidity. A measurement of the amount of water vapor in the air at a particular time is called the absolute humidity. At room temperature, for example, humid air might contain 15 grams of water vapor in each cubic meter of air. At the same temperature air of low humidity might have an absolute humidity of only 2 grams per cubic meter. Absolute humidity can range from near zero up to a maximum that is determined by the temperature at a particular time, as shown in Figure 20.1. Since the temperature of the water vapor present in the air is the same as the temperature of the air, the maximum absolute humidity is usually said to be determined by the air temperature. What this really means is that the maximum absolute humidity is determined by the temperature of the water vapor; that is, the average kinetic energy of the water vapor. The relationship between the actual absolute humidity at a particular temperature and the maximum absolute humidity that can occur at that temperature is called the relative humidity. Relative humidity is a ratio between (1) the amount of water vapor in the air, and (2) the amount of water vapor needed to saturate the air at that temperature. The relationship is actual absolute humidity at present temperature ×100% = relative humidity maximum absolute humidity at present temperature For example, suppose a measurement of the water vapor in the air at 10˚C (50˚F) finds an absolute humidity of 5 g/m3. According to figure 20.1, the maximum amount of water vapor that can be in the air when the temperature is 10˚C is about 10 g/m3. The relative humidity is then 5 g / m3 ×100% = 50% 3 10 g / m If the absolute humidity were 10 g/m3, then the air would have all the water vapor it could hold and the relative humidity would be 100%. A relative humidity of 100% means that the air is saturated at the present temperature. Procedure Part A: Maximum Amount of Water Vapor Measure the present air temperature in your laboratory room and record it in Data Table 20.1 on page 156. Use the graph of maximum absolute humidity in figure 20.1 to estimate the maximum amount of water vapor a cubic meter of air can hold at this temperature. Since one gram of water has an approximate volume of one milliliter, find the maximum amount of water in liters per cubic meter that can be in the room at the present temperature. Record this maximum, in grams/cubic meter and in liters/cubic meter, in Data Table 20.1. Figure 20.1 Measure the length, width, and height of the laboratory room. Record these measurements in Data Table 20.1, then calculate the volume of the room in cubic meters. Record all data and calculations in the data table. Calculate the maximum amount of water vapor the air in the laboratory room can hold at the present temperature. This is found by multiplying the volume of the room (in cubic meters) by the maximum amount of water vapor that could be in the room at the present temperature (in liters per cubic meter). Record all data and calculations in Data Table 20.1. Part B: Actual Amount of Water Vapor Evaporation occurs at a rate that is inversely proportional to the relative humidity, ranging from a maximum rate when the air is driest to no net evaporation when the air is saturated. Since evaporation is a cooling process, it is possible to use a thermometer to measure humidity. An instrument called a psychrometer has two thermometers, one of which has a damp cloth wick around its bulb end. As air moves past the two thermometer bulbs, the ordinary thermometer (the dry bulb) will measure the present air temperature. Water will evaporate from the wet wick (the wet bulb) until an equilibrium is reached between water vapor leaving the wick and water vapor returning to the wick from the air. Since evaporation lowers the temperature, the depression of the temperature of the wet-bulb thermometer is an indirect measure of the water vapor present in the air. The relative humidity can be determined by obtaining the dry- and wet-bulb temperature readings and referring to a relative humidity chart, such as the one found inside the back cover. If the humidity is 100%, no net evaporation will take place from the wet bulb, and both wet- and dry-bulb temperatures will be the same. The lower the humidity, the greater the difference in the temperature reading of the two thermometers. Relative humidity is a ratio between the actual absolute humidity at a given temperature and the maximum absolute humidity that can occur at that temperature. Knowing the maximum absolute humidity and the relative humidity, you can find the amount of water vapor in the air at the present temperature. Wet the cotton wick on the wet bulb of a sling psychrometer. Whirl the thermometers in the air until the wet-bulb thermometer registers its lowest reading. Record the wet-bulb and dry-bulb temperatures, then use this data to find the relative humidity from the relative humidity chart inside the back cover. Record all data and calculations in the data table. Multiply the humidity as a fraction, times the maximum amount of water vapor that could be in your laboratory room at the present temperature. Record the amount of water vapor present in the room in the data table. Results What is the actual amount of water vapor present in the laboratory room air at the present temperature (in g/m3)? What is the maximum amount of water vapor that could be present? Answers will vary according to the relative humidity and temperature. Answers will vary but should be the absolute humidity for the present temperature asshown in Figure 20.1. Can the absolute humidity of the air in the room be increased? Explain. Yes, it can be increased by adding more water vapor to the air. Can the relative humidity of the air in the room be increased without adding more water vapor to the air? Explain. Yes, the relative humidity is increased by cooling the room, which reduces the capacity of air to hold water vapor. Suppose the room air has all the water vapor it will hold at 25˚ C, and the air is cooled to 15˚C. Considering the area of the floor from your measurements, how deep a layer of water will condense from the air? (Maximum of 23 g/m3) = (maximum of 13 g/m3) = 10 g/m3 that would condense out of the air. A room volume of 420 m3 ⇒ 4200 grams of water on the floor. 4200 grams of water ⇒ 4200 cm3 = 0.0042 m3 Thickness: 3 X 10-5 m, or not very much Was the purpose of this lab accomplished? Why or why not? (Your answer to this question should show thoughtful analysis and careful, thorough thinking.) (Students should know the difference between absolute and relative humidity. The Students should be able to explain why there is dew on the grass in the morning. The results of this investigation will vary drastically with the season and local climate.)
Data Table 20.1 Amount of Water Vapor in the Laboratory Room
1. Present temperature of laboratory room air ____________ ˚C22
2. Maximum absolute humidity at present temperature In grams per cubic meter In liters per cubic meter ____________ g/m17 3 ____________ L/m0.017 3
Room length ____________ m20
Room width ____________ m7
Room height ____________ m3
3. Volume of laboratory room (length × width × height) ____________ m420 3
4. Maximum amount of water vapor (row 2 × row 3) ____________ L7.2
5. Dry-bulb reading Wet-bulb reading Difference in wet- and dry-bulb readings Relative Humidity (from relative humidity chart) ____________ ˚C22 ____________ ˚C12 ____________ ˚C10 ____________ %28
6. Amount of water vapor in room (row 5 as decimal × row 4) ____________ L2.0
Name____________________________________________________Section________________Date___________ Experiment 21: Growing Crystals Invitation to Inquiry Investigate the crystal structure of some large samples of crystals. Consider, for example, measuring the angles between the faces, comparing how the crystal transmits light, and other measurements that would help you interpret the structure in terms of atoms or molecules. Does the structure provide any information about the conditions necessary for growing large and well-formed crystals? Or are the same conditions needed for all crystals? Are there any upper limits on the size of a crystal? Background A salt dissolves in water because the polar water molecules pull ions away from the crystal lattice of the salt. The oxygen end of a water molecule has a negative polar charge and the hydrogen ends are positive. The oxygen ends of water molecules tend to orient themselves toward the positive ions on the outside of the crystal lattice, as opposite charges attract one another. The hydrogen ends likewise orient themselves toward the negative ions of the lattice. If the attraction of water molecules is greater than the attraction between the ions in the lattice, the ions are pulled away and dissolving occurs. Saturation occurs as water molecules become “tied up” in their attraction for ions. Fewer water molecules means less attraction on the ionic solid, with more solute ions being pulled back to the surface of the crystal lattice. Each salt forms its own kind of crystal, which is one way to identify salts as well as other crystalline substances, such as minerals that occur in the earth’s crust. In this investigation a warm, saturated solution is cooled and this lowers the solubility. Rapid cooling—or rapid evaporation of water—results in many small crystals that can be seen with a hand lens. If the conditions are favorable, large beautiful crystals can be “grown” from a saturated solution. This investigation provides a procedure for growing large, beautiful crystals and will require an extended time period. CAUTION: Some salts can be poisonous if taken internally. Wash hands thoroughly after handling salts or solutions. Procedure l. Using Data Table 21.1 on page 161, select a salt to produce a large crystal. Note that several different chemicals may be sold as “alum” by drug and grocery stores. If alum is selected and purchased from a drug or grocery store, make sure it is potassium aluminum sulfate. Using a ring stand and ring, wire screen, and large Pyrex beaker, heat 1 L of distilled water. When the water is hot, begin stirring as you slowly add the approximate amount listed for the salt selected from Data Table 21.1. Stir while slowly adding the salt and continue stirring until no more will dissolve. Increase the solution temperature as necessary, stirring until you are confident that you have a saturated solution. If all the salt dissolves, continue adding small amounts until no more will dissolve. Remove the burner flame, allowing the solution to cool slightly. Decant (pour the solution from any undissolved salts in the bottom) the saturated solution while still warm into a large glass jar. Thread a length of fishing leader through a hole punched in the center of the lid, then tape it securely to the lid. When the lid is on the jar the leader should be long enough to extend halfway into the solution. If the leader floats, use the stirring rod to push it below the surface of the liquid. Place the jar where it can remain undisturbed overnight. Small crystals will form in the solution overnight. Carefully remove the lid and place the fishing leader across a paper towel. Remove all crystals from the leader, selecting the one largest crystal as your seed crystal. Epoxy this seed crystal to the end of the dry leader. Pour the solution from the jar back into the large beaker, being sure that all crystals (except the seed crystal) go with the solution. Heat the solution and again prepare a saturated solution. When you are confident that you have a saturated solution, cool it to room temperature, then return the seed crystal to the solution. Adjust the fishing leader and tape it to the lid so the seed crystal is about centered in the solution. If your seed crystal dissolves, you did not have a saturated solution or the solution was not cooled to room temperature. If you make this error, return to procedure step 3. The solution and seed crystal are now allowed to stand undisturbed, and at a constant temperature, for several days. To produce large, perfect crystals it is absolutely necessary for the solution to be isolated from physical disturbances and at a constant temperature. In addition, any small crystals that form on the surface of the seed crystal during a growth period must be completely removed by breaking, scraping, and sanding. The first growth period is complete when the crystal stops growing. When this occurs remove the crystal from the solution and provide more ions by repeating procedure steps 5 and 6. This process is repeated for more growth periods until the crystal reaches the desired size. Any imperfections that develop are the result of some physical or temperature disturbance. Results Explain why sudden temperature changes result in irregular growth. Sudden cooling causes growth of many small crystals because there are more nucleation sites possible when the molecules lose kinetic energy. Why is it necessary to remove other, smaller crystals from the solution? The small crystals compete with the seed crystal for the available and limited "growth" ions in solution. Why is it necessary to re-saturate the solution for each growth period? There is a need to resupply the solution with more ions as the forming crystal removes more and more for the crystal growth. A new supply also speeds the growth of the large crystals. Why does each salt form its own kind of crystal? The bonding arrangement and the most efficient packing is different for the ions of each salt. Was the purpose of this lab accomplished? Why or why not? (Your answer to this question should show thoughtful analysis and careful, thorough thinking.) (The students should be able to grow a reasonable size crystal. If they fail, they should be able to list some reasons that their crystal was small, things they would do differently next time. Be sure to give the students enough time to grow the crystals -- a few weeks should suffice.)
1. Name____________________________________________________Section________________Date___________ Experiment 22: Properties of Common Minerals Invitation to Inquiry Build or obtain a tumbler that gives minerals a smooth, polished surface. Investigate and predict which locally available minerals would be best for polishing. Learn how to use the tumbler and prepare some local specimens for a display. Background The earth’s crust is made up of rocks, which are solid aggregations of materials that have been cohesively brought together by rock-forming processes. The fundamental building blocks of rocks are minerals, which are naturally occurring, inorganic solid elements or chemical compounds with a crystalline structure. About 2,500 different minerals are known to exist, but only about 20 are common in the crust. Examples of these common minerals are quartz, calcite, and gypsum. Each rock-forming mineral has its own set of physical properties because each mineral has (1) a chemical composition, and (2) a crystal structure that is unlike any other mineral. The exact composition and crystal structure of an unknown mineral can be determined by using laboratory procedures. This type of analysis is necessary when the crystal structures are too small to be visible to the unaided eye. In this laboratory activity you will learn how to identify minerals by considering some identifying characteristics of large, well-developed mineral crystals. Procedure Part A: Developing a Table of Mineral Properties Examine each mineral specimen in the mineral collection set, one at a time. Consider each of the following properties and record all observations in Data Table 22.1 on pages 166 and 167. Color. The color of a mineral specimen is an obvious property, but the color of some minerals varies from one specimen to the next because of chemical impurities. Sometimes the property of color can be useful in identifying a mineral. Use combinations of words as necessary to describe mineral colors—for example, reddish-brown. Streak. Streak is the color of a mineral when it is finely powdered, a property that is more consistent from specimen to specimen than color. Streak is observed by rubbing the corner of a specimen across an unglazed streak plate. Determine the streak color of each mineral and record your observations in Data Table 22.1. Note that a mineral harder than the streak plate will not leave a streak. This is useful information, so record this observation if it occurs. Hardness. Hardness is the resistance of a mineral to being scratched. The Mohs hardness scale is used as a basis for measuring hardness. The hardness test is done by trying to scratch a mineral with a mineral (or some object with an equivalent hardness) from the Mohs scale. If the mineral being investigated scratches a Mohs hardness scale mineral, the mineral in question is harder than the test mineral. If the mineral being investigated is scratched by the test mineral, the mineral in question is not as hard as the test mineral. If both minerals are scratched by each other, they have the same hardness. The Mohs hardness scale is given in Table 22.1, along with the hardness of some common objects that can be used for hardness tests. A hardness of 1 on this scale is assigned to the softest mineral, and the hardest mineral has an assigned hardness of 10.
Table 22.1 Mohs Scale of Hardness for Minerals and Common Objects Tests
Hardness Number Mineral Example Common Object Test
1 Talc Can be scratched with fingernail.
2 Gypsum Can be scratched with fingernail, but not easily.
3 Calcite Can be scratched with copper penny.
4 Fluorite Can be scratched with pocketknife. Will not scratch glass.
5 Apatite Can be scratched with pocketknife, but not easily.
6 Feldspar Can be scratched with edge of steel file. Cannot be scratched with pocketknife.
7 Quartz Scratches glass.
8 Topaz Scratches quartz.
9 Corundum (ruby or sapphire, for example) Scratches all minerals but diamond.
10 Diamond Scratches all minerals. Can be scratched only by another diamond.
164 Cleavage. Cleavage is the tendency of a mineral to break along smooth planes. Cleavage occurs in parallel planes in some minerals and in one or more directions in other minerals. Sometimes cleavage is perfect, other minerals may have indistinct cleavage. Fracture. Minerals that do not have cleavage may show fracture, an irregular broken surface rather than the smooth planes of cleavage. Some minerals have a distinct way of fracturing, such as the conchoidal fracture of obsidian and quartz. Conchoidal fracture is the breaking along curved surfaces like a shell. Luster. Luster describes the surface sheen, the way a mineral reflects light. Minerals that have the surface sheen of a metal are described as being metallic. Nonmetallic luster is described by such terms as pearly (like a pearl), vitreous (like glass), glossy, dull, and so forth. Density. Density is a ratio of the mass of a mineral to its volume. Often the density of a mineral is expressed as specific gravity, which is a ratio of the mineral density to the density of water. To obtain an exact specific gravity, a mineral specimen must be pure and without cracks, bubbles, or substitutions of chemically similar elements. Other properties. There are a few other properties that are specific for one or more minerals. These properties are determined by tests such as a reaction to acid, reaction to a magnet, and others. Some of the specific properties of certain minerals, such as the double image seen through a calcite crystal, are unique enough to identify an unknown mineral in an instant. Part B: Unknown Minerals Once the Table of Mineral Properties (Data Table 22.1) is complete, you can use it to identify an unknown mineral. The procedure is to find out what the mineral is not. For example, suppose you start with the streak test and find the unknown mineral leaves a red streak. This test tells you the unknown mineral is not one of the minerals that leave some other color of streak. Then you look at the minerals that have a red streak and perform a second test. The second test eliminates still other possibilities. Eventually, you find what the unknown mineral is by finding out what it isn’t. Results: Was the purpose of this lab accomplished? Why or why not? (Your answer to this question should show thoughtful analysis and careful, thorough thinking.) (Students should be able to identify the unknown mineral.)
Data Table 22.1 Table of Mineral Properties
Mineral Color Luster Streak
Calcite Pink Vitreous, Dull White
Magnetite Metallic Black
Black
Pyrite Brass Yellow Metallic Greenish-Black
Hematite
Black/Red-Black Metallic, Dull Reddish
Limonite Black/Brown-Black Dull Yellow Brown
Quartz
White, Pink Vitreous White
Fluorite
Red, Violet Vitreous White
Apatite
Varies Vitreous, greasy White
Feldspar
Violet, Pink Vitreous White
Talc
White/Greenish Pearly White
Hornblende
Black, Green-Black Vitreous Gray, Gray-Green
Halite
White Vitreous White
Bauxite
Reddish Earthy, Dull Reddish
Biotite
Black, Brown Glassy White, Greyish
Muscovite
Clear, Light Brown Glassy White
Sphalerite
Brown-Yellowish Submetallic Light Brown
Galena
Lead Grey Metallic Gray-Black
Chrysotile
Green Vitreous Yellow-White
Gypsum White
White Vitreous, Silky
166
Data Table 22.1 Table of Mineral Properties, Continued
Mineral Hardness Cleavage/Fracture Other
Calcite
3 Rhombohral Brittle
Magnetite Indistinct
6 Magnetic
Pyrite "Fool's Gold"
6 Cubic
Hematite
6 None "Red Soil"
Limonite 5 None "Yellow-Brown Soil"
Quartz 7
Indistinct/Chon Often Transparent
Fluorite
4 Octahedral Brittle
Apatite Acid Soluble
5 Imperfect/Chon
Feldspar 6
Basal/Uneven Brittle
Talc
1 Basal/Uneven Soapy Feel
Hornblende
5.5 Prismatic Brittle
Halite
2 Cubic Brittle
Bauxite
1 None Clay Odor
Biotite
2.5-3 Sheets Flexible
Muscovite
2-3 Sheets Flexible
Sphalerite
3.5-4 Dodecahedral Brittle
Galena
2.5 Cubic Brittle
Chrysotile
6.5--7 Piracoidal Brittle
Gypsum 2
Clinopinacoidal Glassy
Name____________________________________________________Section________________Date___________ Experiment 23: Density of Igneous Rocks Invitation to Inquiry Survey the use of rocks used in building construction in your community. Compare the type of rocks that are used for building interiors and those that are used for building exteriors. Are any trends apparent for buildings constructed in the past and those built in more recent times? If so, are there reasons (cost, shipping, other limitations) underlying a trend or is it simply a matter of style? Background Igneous rocks are rocks that form from the cooling of a hot, molten mass of rock material. Igneous rocks, as other rocks, are made up of various combinations of minerals. Each mineral has its own temperature range at which it begins to crystallize, forming a solid material. Minerals that are rich in iron and magnesium tend to crystallize at high temperatures. Minerals that are rich in silicon and poor in iron and magnesium tend to crystallize at lower temperatures. Thus, minerals rich in iron and magnesium crystallize first in a deep molten mass of rock material, sinking to the bottom. The minerals that crystallize later will become progressively richer in silicon as more and more iron and magnesium are removed from the melt. Igneous rocks that are rich in silicon and poor in iron and magnesium are comparatively light in density and color. The most common igneous rock of this type is granite, which makes up most of the earth’s continents. Igneous rocks that are rich in iron and magnesium are dark in color and have a relatively high density. The most common example of these dark-colored, more dense rocks is basalt, which makes up the ocean basins and much of the earth’s interior. Basalt is also found on the earth’s surface as a result of volcanic activity. Procedure Use a balance to find the mass of a basalt rock. Record the mass in Data Table 23.1 on page 172. Tie a 20-cm length of nylon string around the rock so you can lift it with the string. Test your tying abilities to make sure you can lift the rock by lifting the string without the rock falling. Place an overflow can on a ring stand, adjusted so the overflow spout is directly over a graduated cylinder. Hold a finger over the overflow spout, then fill the can with water. Remove your finger from the spout, allowing the excess water to flow into the cylinder. Dump this water from the cylinder, then place it back under the overflow spout. Grasp the free end of the string tied around the basalt, then lower the rock completely beneath the water surface in the overflow can. The volume of water that flows into the graduated cylinder is the volume of the rock. Remembering that a volume of 1.0 mL is equivalent to a volume of 1.0 cm3, record the volume of the rock in cm3 in Data Table 23.1. Calculate the mass density of the basalt and record the value in the data table. Repeat procedure steps 1 through 5 with a sample of granite. Results In what ways do igneous rocks have different properties? Color, density, and grain size. Explain the theoretical process or processes responsible for producing the different properties of igneous rocks. Mineral mixtures of rock melt (magma) and different crystallization temperatures produce rocks of mixtures of minerals with different colors and densities. In general, more silicates produce rocks of lighter color and density and more ferromagnesian minerals produce rocks of greater density and darker colors. According to the experimental evidence of this investigation, propose an explanation for the observation that the bulk of the earth’s continents are granite, and that basalt is mostly found in the earth’s interior. Granite is less dense than basalt. Was the purpose of this lab accomplished? Why or why not? (Your answer to this question should show thoughtful analysis and careful, thorough thinking.) (Students should be able to differentiate between granite and basalt.)
Data Table 23.1 Density of Igneous Rocks
Mass Volume (g) (cm3) Density (g/cm3)
Basalt 192 ____________ g 64 ____________ cm3 3.0 ____________ g/cm3
Granite ____________ g173 ____________ cm64 3 ____________ g/cm2.7 3
Name____________________________________________________Section________________Date___________ Experiment 24: Latitude and Longitude Invitation to Inquiry There are several ways to find your latitude by measurement. First, determine your latitude by measuring the angle of the North Star above the horizon. Second, determine your latitude by measuring the angle between a vertical stick and a line to the noonday sun on the spring equinox (March 21) or the autumnal equinox (September 23). For the North Star, consider making two measurements 12 hours apart and averaging the two. Why do these two different methods tell you your latitude? Is one more in “agreement” with the stated latitude for your location? Background The continuous rotation and revolution of the earth establish an objective way to determine directions and locations on the earth. If the earth were an unmoving sphere there would be no side, end, or point to provide a referent for directions and locations. The earth’s rotation, however, defines an axis of rotation which serves as a reference point for determination of directions and locations on the entire surface. The reference point for a sphere is not as simple as on a flat, two-dimensional surface, because a sphere does not have a top or side edge. The earth’s axis provides the north-south reference point. The equator is a big circle around the earth that is exactly halfway between the two ends, or poles of the rotational axis. An infinite number of circles are imagined to run around the earth parallel to the equator. The east- and west-running parallel circles are called parallels. Each parallel is the same distance between the equator and one of the poles all the way around the earth. The distance from the equator to a point on a parallel is called the latitude of that point. Latitude tells you how far north or south a point is from the equator by telling you on which parallel the point is located. Since a parallel is a circle, a location of 40˚ N latitude could be anyplace on that circle around the earth. To identify a location you need another line, one that runs pole to pole and perpendicular to the parallels. North-south running arcs that intersect at both poles are called meridians. There is no naturally occurring, identifiable meridian that can be used as a point of reference such as the equator serves for parallels, so one is identified as the referent by international agreement. The reference meridian is the one that passes through the Greenwich Observatory near London, England, and is called the prime meridian. The distance from the prime meridian east or west is called the longitude. The degrees of longitude of a point on a parallel are measured to the east or to the west from the prime meridian up to 180˚. Locations identified with degrees of latitude north or south of the equator and degrees of longitude east or west of the prime meridian are more precisely identified by dividing each degree of latitude into subdivisions of 60 minutes (60’) per degree, and each minute into 60 seconds (60”). In this investigation you will do a hands-on activity that will help you understand how latitude and longitude are used to locate places on the earth’s surface. Procedure Obtain a lump of clay about the size of your fist. Knead the clay until it is soft and pliable, then form it into a smooth ball for a model of the earth. Obtain a sharpened pencil. Hold the clay ball in one hand and use a twisting motion to force the pencil all the way through the ball of clay. Reform the clay into a smooth ball as necessary. This pencil represents the earth’s axis, an imaginary line about which the earth rotates. Hold the clay ball so the eraser end of the pencil is at the top. The eraser end of the pencil represents the North Pole and the sharpened end represents the South Pole. With the North Pole at the top, the earth turns so the part facing you moves from left to right. Hold the clay ball with the pencil end at the top and turn the ball like this to visualize the turning earth. The earth’s axis provides a north-south reference point. The equator is a circle around the earth that is exactly halfway between the two poles. Use the end of a toothpick to make a line in the clay representing the equator. Hold the clay in one hand with the pencil between two fingers. Carefully remove the pencil from the clay with a back and forth twisting motion. Reform the clay into a smooth ball if necessary, being careful not to destroy the equator line. Use a knife to slowly and carefully cut halfway through the equator. Make a second cut down through the North Pole to cut away one-fourth of the ball as shown in Figure 24.1. Set the cut-away section aside for now. Figure 24.1 Place a protractor on the clay ball where the section was removed. As shown in Figure 24.2, the 0˚ of the protractor should be on the equator and the 90˚ line should be on the axis (the center of where the pencil was). You may have to force the protractor slightly into the clay so the 0˚ line is on the equator. Directly behind the protractor, stick toothpicks into the surface of the clay ball at 20˚, 40˚, and 60˚ above the equator on both sides. Remove the protractor from the clay and return the cut-away section to make a whole ball again. Use the end of a toothpick to make parallels at 20˚, 40˚, and 60˚ north of the equator, then remove the six toothpicks. Recall that parallels are east and west running circles that are the same distance from the equator all the way around the earth (thus the name parallels). The distance from the equator to a point on a parallel is called the latitude of that point. Latitude tells you how far north or south a point is from the equator. There can be 90 parallels between the equator and the North Pole, so a latitude can range from 0˚ North (on the equator) up to 90˚ North (at the North Pole). 90˚ Figure 24.2 Since a parallel is a circle that runs all the way around the earth, a second line is needed to identifya specific location. This second line runs from pole to pole and is called a meridian. To see how meridians identify specific locations, again remove the cut section from the ball of clay. This time place the protractor flat on the equator as shown in Figure 24.3, with the 90˚ line perpendicular to the axis. Stick toothpicks directly below the protractor at 0˚, 60˚, 120˚, and 180˚, then remove the protractor and return the cut-away section to make a whole ball again. Use a toothpick to draw lines in the clay that run from one pole, through the toothpicks, then through the other pole. By agreement, the 0˚ line runs through Greenwich near London, England and this meridian is called the prime meridian. The distance east or west of the prime meridian is called longitude. If you move right from the prime meridian you are moving east from 0˚ all the way to 180˚ East. If you move left from the prime meridian you are moving west from 0˚ all the way to 180˚ West. Use a map or a globe to locate some city that is on or near a whole number latitude and longitude. New Orleans, Louisiana, for example, has a latitude of about 30˚ N of the equator. It has a longitude of about 90˚ W of the prime meridian. The location of New Orleans on the earth is therefore described as 30˚ N, 90˚ W. Locate this position on your clay model of the earth and insert a toothpick. Compare your model to those of your classmates. Figure 24.3 Results What information does the latitude of a location tell you? How far north or south of the equator you are. What information does the longitude of a location tell you? How far east or west from the prime meridian you are. According to a map or a globe, what is the approximate latitude and longitude of the place where you live? Answers will vary. Explain how minutes and seconds are used to identify a location more precisely. There are 60 minutes in a degree and 60 seconds in a minute. Thus you can use these smaller divisions to describe a location more precisely. Was the purpose of this lab accomplished? Why or why not? (Your answer to this question should show thoughtful analysis and careful, thorough thinking.) (Students should know the difference between latitude and longitude. The students should know where the zero degree line is for both measurements. They should be able to locate places when given the latitude and longitude.)
Name____________________________________________________Section________________Date___________ Experiment 25: Telescopes Invitation to Inquiry Design an experiment to study the effect of the diameter of a lens on the image formed. Do the experiment. Use plane (flat), concave, and convex mirrors to find when you can see: an enlarged image. a reduced image. an image of the same size. an image that appears upright. an image that appears inverted. an image that appears upright, then inverted after some adjustment. What generalizations can you make to inform someone how to make the various images with mirrors? Background A convex lens can be used as a “burning glass” by moving the lens back and forth until the sunlight is focused into a small bright spot—an image of the sun. This image is hot enough to scorch the paper, perhaps setting it on fire. The lens is moved back and forth to refract the parallel rays of light from the sun, the point where the image is formed on the paper. The place where the image forms is called the focal point of the lens. The distance from the focal point to the lens is called the focal length (f). The focal length of a lens is determined by its index of refraction and the shape of the lens. The focal length is an indication of the refracting ability, or strength of a lens. A lens with a short focal length is considered to be a stronger lens than one with a longer focal length. There are three important measurements that are used to describe how lenses work as optical devices. These are (1) the focal length (f), (2) the image distance (di), the distance from the lens that an image is formed, and (3) the object distance (do), the distance from the object being imaged to the lens. The relationship between these measurements is given in the lens equation, which is 1 1 1 f = do + di The magnification produced by a lens is defined as the ratio of the height of the image (hi) to the height of the object (ho). This is also equal to the ratio of the image distance (di) to the object distance (do), or Magnification = ddoi In this investigation you will compare measuring the focal length of a lens directly with using the lens equation to calculate the focal length of a convex lens. Magnification of a lens will also be investigated by comparing direct measurement of magnification with theoretical magnification as calculated from the focal lengths of two lenses. Procedure Measure the focal length of a lens directly. First, place a lens in a lens holder and secure it at the 50 cm mark on a meter stick. Second, point the meter stick at some distant objects, such as a tree or a house about a block away. Move a cardboard screen in a holder back and forth until you obtain a sharp image on the screen. Finally, measure the distance between the sharp image and the lens, which is the focal length (f) of the lens. Record the focal length in Data Table 25.1 on page 184. Figure 25.1 Find the focal length of the lens used in procedure step 1 by use of the lens equation. Set up a meter stick in holders, a screen in a holder, and a luminous object in a holder as shown in Figure 25.1. The room should be darkened, then place the screen at the focal length distance found in procedure step 1. With the screen and object fixed in place, slowly move the lens along the meter stick to obtain the sharpest image possible. Note that the object and image should lie on a straight line along, and perpendicular to the principal axis of the lens. Measure and record in Data Table 25.1 the object distance (do) and the image distance (di) to the nearest 1 mm. Measure and record to the nearest 0.5 mm the height of the object (ho) and height of the image (hi). Record other observations here. Image is inverted when the object distance is greater than the focal length of the lens. The image is smaller than the object. Place the lens less than one focal length from the bulb, then move the screen back and forth to see if you can obtain an image on the screen. Look through the lens at the bulb. Record your observations here. No image is formed on the screen. Image is upright and on the same side of the lens as the object. The image is larger than the object. Repeat procedure steps 1 and 2 for three more lenses. Record all data and results of calculations in Data Table 25.1. Select one of the lenses with a short focal length (for eyepiece lens) and one with a longer focal length (for objective lens) for use in the next procedure step. i Figure 25.2 Make a telescope by mounting a short and longer focal length lenses on a meter stick as shown in Figure 25.2. Focus the telescope by adjusting the shorter focal length lens until it magnifies the image from the longer focal length lens. Calculate the theoretical magnification of your telescope by dividing the focal length of the objective lens by the focal length of the eyepiece lens. Record the theoretical magnification in Data Table 25.2 on page 185. Work with a partner to determine experimentally the magnification of your telescope. Focus your telescope on some object in front of your partner, who should hold a meter stick next to the object. Look at the object normally with one eye and look at the object through the telescope with the other eye. Direct your partner to position a pointer on the meter stick to indicate the apparent size of the enlarged image. Record the height of the image and the height of the object in Data Table 25.2, then calculate the magnification. Results What relationships did you find between do, di, and f? Students should note that, since the value of f calculated from 1 + 1 = 1 do di f and the value found by finding a focused image of a distant object were the same, the relationship 1 + 1 = 1 do di f is valid. What relationships did you find between do, di, hi, and ho? di = hi do ho Discuss the advantages, disadvantages, and possible sources of error involved in the two ways of finding the focal length of lenses. Main source of error occurs in determining where the image is in focus. A slight variation here can alter the distance measurements. When determining the focal length by finding a focused image of a distant object, the further the object is from the lens, the better. Discuss the advantages, disadvantages, and possible sources of error involved in the two ways of finding the magnification of lenses. Errors in measuring the height of the image and object. Was the purpose of this lab accomplished? Why or why not? (Your answer to this question should show thoughtful analysis and careful, thorough thinking.) Students should show an overall understanding about the relationship between f, do, and di, and the relationship between hi, ho, di, do and magnification. Students should also be able to describe the type of image formed based on the location of the object with respect to the focal length of the lens. (Note: A negative sign means an inverted image.)
Data Table 25.2 Lens Magnification
Objective lens focal length 25.0 cm ______________________
Eyepiece lens focal length 10.0 cm ______________________
Theoretical magnification (Focal length of objective ÷ focal length of eyepiece) 2.5 ______________________
Object height (ho) 3.2 ______________________
Image height (hi) 8.2 ______________________
Experimental magnification (hi ÷ ho) 2.6 ______________________
7. Name____________________________________________________Section________________Date___________ Experiment 26: Celestial Coordinates Background To an observer unencumbered by scientific models, the night sky appears to be an inverted bowl resting on a flat plane. The observer appears to be located at the center of the bowl. Sprinkled over the inner surface of the bowl are the stars arranged in fixed, identifiable patterns that do not change noticeably from day to day, year to year, or even century to century. Among what ancient observers called the “fixed stars” seven objects move: the sun, the moon, and five apparently star-like objects called planets. Other objects that moved in the sky but were transitory, such as meteors and comets, were considered by the ancient observers, Aristotle in particular, to be atmospheric phenomena and were not considered in modeling the heavens. The motion of the sun was readily apparent. The general direction of its rising is synonymous with what we call the east. The direction of its setting is the west. Each day the sun rises to its highest altitude above the horizon when it is due south. If we imagine a line extending from the north point on the horizon passing overhead and continuing to the south point on the horizon, then the sun reaches its highest altitude when it crosses this line called the meridian. When the sun, or any other object in the sky, crosses the meridian, the object is said to transit. During the course of a single night, the stars wheel around a fixed point as if the celestial bowl on which they are seemingly attached were spinning on an axis passing through this point and the observer’s position. On a time exposure photograph made with a camera pointed toward this fixed point, each star traces out an arc of a perfect circle. Careful observation shows that the bowl rotates from east to west at a uniform rate, completing one revolution in just under 24 hours (23h56m4.s091). That is, the time interval between successive meridian transits of a particular star, say, is a bit shorter than the average time between successive meridian transits of the sun, 24 hours exactly. (The word “average” is used because the time interval between successive solar transits varies throughout the year.) Since the sun rate and the star rate are close but not quite the same, it must be that the sun moves slowly with respect to the background stars, and in fact, the sun’s path among the stars can be delineated. By noticing the sun’s position among the stars at sunrise or sunset each day, we can plot its path among the stars. This path turns out to be a circle and is called the ecliptic, and the sun creeps from west to east at a nearly uniform rate, taking 365 days and almost 6 hours to complete a single circuit. Where is this circle on the celestial sphere? In order to answer this question, we will establish a coordinate system on the sky so that we can specify locations of points and circles. We will produce an analog of the system of latitude and longitude used to locate positions on the earth’s surface. Procedure Part A: The Celestial Pole and Equator Let us call the fixed point about which the sphere of stars turns the celestial pole. The direction on the ground toward this fixed point is called north if we are in the northern hemisphere of the earth. If we are in the southern hemisphere, a different fixed point is apparent, and the direction toward the point is south. In the northern sky a bright star, Polaris, is near the north celestial pole. There is no bright star near the south celestial pole. On the surface of the earth, the line that is everywhere 90° away from the pole is called the equator. It is a circle bisecting the earth halfway between each pole. In the sky, the line (circle) on the celestial sphere that is everywhere 90° from the celestial pole is called the celestial equator. Figure 26.1 shows the geometry of the celestial sphere relative to that of the earth. The angle α formed by the observer, the center of the earth, and a point on the equator is defined to be the observer’s latitude on the earth. Prove that the angle α, the observer’s latitude, is in fact the same as the angle (alpha) of the north celestial pole over the northern horizon. If the celestial pole is an angle α above the northern horizon, then we can establish the location of the celestial equator. From Figure 26.1, the angle along the meridian that the equator makes over the southern horizon is θ = 180° – 90° – α. So the celestial equator starts in the east, climbs up to an angle of 90°– α over the southern horizon, and then drops down to the west point of the horizon. If the celestial pole makes an angle of 40° over the northern horizon, what angle over the southern horizon does the equator make? 50° Suppose you are at Quito, Ecuador (latitude = 0°). Where is the celestial equator? 0 = 90° The ecliptic is a circle on the celestial sphere that is tipped at an angle of 23 1/2° to the celestial equator. Figure 26.2 illustrates the relationship between the celestial equator and the ecliptic. Since during the course of a year the sun traces out the path of the ecliptic, it appears at different places with respect to the celestial equator. At one time, the sun is maximally above (north of) the celestial equator, sometimes right on the equator, and at another time maximally below (south of) the equator. Figure 26.3 shows two different orientations of the sun with respect to the celestial equator. The figures are drawn for the case where the north celestial pole lies at an angle of 40° over the north point of the horizon (i.e., latitude = 40° north). The figures are also drawn for noon, when the sun is on the meridian. Figure 26.2 The ecliptic (dashed line) intersects the celestial equator at two points called equinoxes. The solstices mark the most northerly and southerly points. Part B: Celestial Coordinates We can now use the points and circles we have discussed thus far to establish a coordinate system to map the heavens. In analogy with the latitude-longitude system used to specify the location of a point on the earth’s surface, we will need two coordinates, called declination and right ascension to locate a particular point in the sky. The declination of a star is its angular distance measured in degrees between the celestial equator and the point or celestial object. The object’s declination is positive (+) if it is north of the celestial equator and negative (–) if it is south. Subdivisions are measured in the usual minutes and seconds of arc. Figure 26.4 illustrates this coordinate. Points on the celestial equator have a declination of 0°; the north celestial pole as dec = +90°; the south pole has dec = –90°. The right ascension coordinate is a bit more complicated. The right ascension of an object is the angular distance measured along the celestial equator between the vernal equinox and the point on the equator that intersects an hour circle passing through the object (see Figure 26.5). An hour circle is a great circle that passes through both celestial poles as well as through the object itself. Here is the peculiar aspect of the right ascension coordinate: it is measured in hours, with subdivisions of minutes and seconds of time. Instead of 360° all the way around the celestial equator, right ascension has 24h, that is, 24h is equivalent to 360°. Right ascension is measured eastward from the vernal equinox. How many degrees of arc are equivalent to 1h of right ascension? 360° = x 15° of arc = 1h of right ascension. 24 h 1 h How many minutes of right ascension are equivalent to 1° of arc? 360° = 1° 1m of right ascension15° = °1 of arc. 1440 min x How many minutes of arc are equivalent to 1m of right ascension? 0.25° of arc = 1 min of right ascension, but0.25° of arc =15 min of arc. So, 15 minutes of arc = 1 minute of right ascension. Part C: The Celestial Globe A celestial globe is a representation of the celestial sphere, the stars or other objects in the heavens, together with the coordinate system we have been discussing. Remember, the celestial sphere has you, the observer, at its center, so everything in the celestial globe is plotted from this perspective. As you look at the globe, you will have to visualize that you are at the center of the globe looking out. Locate on the globe the (1) north and south celestial poles, (2) the celestial equator, (3) the ecliptic, (4) the two equinoxes where the ecliptic and equator intersect, and (5) the solstices. Which stars are located at the following coordinates?
Right Ascension Declination Star Name
6h 43m –16° 39′
6h 22m –52° 40′
18h 35m +38° 44′
5h 12m –8° 15′
7h 37m +5° 39′
14h 36m –60° 38′
Which constellation is located at the approximate positions indicted? Right Ascension Declination Constellation 11h +50° 19h –25° 3h +20° 1h +60° 13h –50° 7h –40° The celestial globe is pivoted about the celestial poles and is held in place by a vertical circular ring of metal that is graduated in degrees. A large horizontal ring serves as a representation of the horizon. Adjust the vertical ring so that the celestial pole is at the proper altitude over the horizon for your latitude. (You may need to ask the laboratory instructor for your latitude to the nearest degree.) As you rotate the globe you will notice that the celestial equator always maintains a single orientation in the sky; it comes up out of the east, reaches a maximum altitude when it crosses the meridian (vertical metal circle), and sinks in the west. Part D: Relative Amounts of Daylight and Darkness On the ecliptic locate the position of the sun on June 21. Place a small piece of 3M self-stick paper to mark the location. Do not write on the globe or make any permanent markings. Note the sun’s declination for this day and record it in Table 26.1. Now rotate the globe until the sun is in its rising position on the eastern horizon. Does the sun rise exactly in the east on this day? Rotate the globe until the sun sets and note where on the horizon the sun disappears below the horizon. Bring the sun back to the rising position and by counting how many right ascension hour circles lie between the rising and setting points for the sun, estimate to the nearest 15 minutes the amount of time the sun is above the horizon on June 21. Enter your estimate in Table 1. By subtracting this number from 24h, calculate the length of darkness for this date. Repeat this for December 21 and March 21. Readjust the position of the celestial pole to correspond to an observer at Altengaard, Norway (latitude = 70° North). Fill in Table 26.2 as you did Table 26.1. Readjust the globe one more time to correspond to an observer in Quito, Ecuador (latitude = 0°). Fill in Table 26.3. Calculate the altitude of the sun (in degrees) over the southern horizon at noon on June 21 for an observer at latitude +40° N. Check your answer using the celestial globe. Repeat procedure 11 for December 21. Eratosthenes, who lived in the second century B.C., was a Renaissance man before his time. He was an astronomer, a geographer, a historian, a mathematician, and a poet. With such a diverse background, it is not surprising that he was the director of the Great Library of Alexandria. In one of the scrolls at the library, Eratosthenes read that at noon on June 21 in the southern frontier outpost of Syene, Egypt, near the first cataract of the Nile, obelisks cast no shadows. As noon approached, the shadows of temple columns grew shorter until at noon they were gone. Reportedly, a reflection of the sun could then be seen at the bottom of a well. On June 21 at Syene, the sun was directly overhead. Use the celestial globe to determine the latitude of Syene. Readjust the celestial globe for your latitude. What range of declinations can stars have so that they never set, that is, are always above the horizon? Your instructor will provide you with the coordinates of Mercury and Venus for the day of this lab session. Put marked pieces of 3M self-stick paper at the corresponding locations on the celestial globe. Locate the sun for today and place a piece of paper at the sun’s location on the globe. When is Mercury visible in a dark sky? (circle one) just after sunset just before sunrise When is Venus visible in a dark sky? just after sunset just before sunrise 16. Estimate how long each is visible in a dark sky. Mercury ____________________ Venus ______________________ Solution Manual Experiment for Integrated Science Bill W. Tillery, Eldon D. Enger , Frederick C. Ross 9780073512259
