P ART II Fundamental Science and Statistics Review L1681_book.fm Page 93 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 95 C HAPTER 5 Fundamental Chemistry and Hydraulics 5.1 INTRODUCTION It is not sufficient for a future working environmental engineer to understand the causes and effects of environmental problems in qualitative terms only. He or she must also be able to express the perceived problem and its potential solution in quantitative terms. To do this, environmental engineers must be able to draw on basic sciences such as chemistry, physics, and hydrology (as well as others) to predict the fate of pollutants in the environment and to design effective treatment systems to reduce impact. In this chapter, we discuss fundamental chemistry and basic hydraulics for environmental engineers. 5.2 FUNDAMENTAL CHEMISTRY The chemists are a strange class of mortals, impelled by an almost insane impulse to seek their pleasure among smoke and vapor, soot and flame, poisons and poverty; yet among all these evils I seem to live so sweetly that I may die if I would change places with the Persian King . Johann Joachim Becher All matter on Earth consists of chemicals. This simplified definition may shock those who think chemistry is what happens in a test tube or between men and women. Chemistry is much more; it is the science of materials that make up the physical world. Chemistry is so complex that no one person could expect to master all aspects of such a vast field; thus, it has been found convenient to divide the subject into specialty areas. For example: • Organic chemists study compounds of carbon. Atoms of this element can form stable chains and rings, giving rise to very large numbers of natural and synthetic compounds. • Inorganic chemists are interested in all elements, particularly in metals, and are often involved in the preparation of new catalysts. • Biochemists are concerned with the chemistry of the living world. • Physical chemists study the structures of materials, and rates and energies of chemical reactions. • Theoretical chemists use mathematics and computational techniques to derive unifying concepts to explain chemical behavior. • Analytical chemists develop test procedures to determine the identity, composition, and purity of chemicals and materials. New analytical procedures often discover the presence of previously unknown compounds. L1681_book.fm Page 95 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 96 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Why should we care about chemistry? Is it not enough to know that we do not want unnecessary chemicals in or on our food or harmful chemicals in our air, water, or soil? Chemicals are everywhere in our environment. The vast majority of these chemicals are natural. Chemists often copy from nature to create new substances superior to and cheaper than natural materials. Our human nature makes us work to make nature serve us. Without chemistry (and the other sciences), we are at nature’s mercy. To control nature, we must learn its laws and then use them. Environmental engineers must also learn to use the laws of chemistry; however, they must know even more. Environmental engineers must know the ramifications of chemistry out of control. Chemistry properly used can perform miracles, but, out of control, chemicals and their effects can be devastating. In fact, many current environmental regulations dealing with chemical safety and emergency response for chemical spills resulted because of catastrophic events involving chemicals. 5.2.1 Density and Specific Gravity When we say that iron is heavier than aluminum, we mean that iron has greater density than aluminum. In practice, what we are really saying is that a given volume of iron is heavier than the same volume of aluminum. Density ( p ) is the mass (weight) per unit volume of a substance at a particular temperature, although density generally varies with temperature. The weight may be expressed in terms of pounds, ounces, grams, kilograms, etc. The volume may be liters, milliliters, gallons, cubic feet, etc. Table 5.1 shows the relationship between the temperature, specific weight, and density of fresh water. Suppose we had a tub of lard and a large box of crackers, each with a mass of 600 g. The density of the crackers would be much less than the density of the lard because the crackers occupy a much larger volume than the lard occupies. The density of an object can be calculated by using the formula: (5.1) In water/wastewater operations, perhaps the most common measures of density are pounds per cubic foot (lb/ft 3 ) and pounds per gallon (lb/gal). •1 ft 3 of water weighs 62.4 lb — density = 62.4 lb/ft 3 •1 gal of water weighs 8.34 lb — density = 8.34 lb/gal Table 5.1 Water Properties (Temperature, Specific Weight and Density) Temperature (°F) Specific weight (lb/ft 3 ) Density (slugs/ft 3 ) Temperature (°F) Specific weight (lb/ft 3 ) Density (slugs/ft 3 ) 32 62.4 1.94 130 61.5 1.91 40 62.4 1.94 140 61.4 1.91 50 62.4 1.94 150 61.2 1.90 60 62.4 1.94 160 61.0 1.90 70 62.3 1.94 170 60.8 1.89 80 62.2 1.93 180 60.6 1.88 90 62.1 1.93 190 60.4 1.88 100 62.0 1.93 200 60.1 1.87 110 61.9 1.92 210 59.8 1.86 120 61.7 1.92 Source : Spellman, F.R., 2003, Handbook of Water and Wastewater Treatment Plant Operations . Boca Raton, FL: Lewis Publishers. Density Mass Volume = L1681_book.fm Page 96 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC FUNDAMENTAL CHEMISTRY AND HYDRAULICS 97 The density of a dry material (e.g., cereal, lime, soda, or sand) is usually expressed in pounds per cubic foot. The densities of plain and reinforced concrete are 144 and 150 lb/ft 3 , respectively. The density of a liquid (liquid alum, liquid chlorine, or water) can be expressed as pounds per cubic foot or as pounds per gallon. The density of a gas (chlorine gas, methane, carbon dioxide, or air) is usually expressed in pounds per cubic foot. As shown in Table 5.1, the density of a substance like water changes slightly as the temperature of the substance changes. This occurs because substances usually increase in volume (size — they expand) as they become warmer. Because of this expansion with warming, the same weight is spread over a larger volume, so the density is lower when a substance is warm than when it is cold. Specific gravity is defined as the weight (or density) of a substance compared to the weight (or density) of an equal volume of water. (The specific gravity of water is 1.) This relationship is easily seen when a cubic foot of water (62.4 lb) is compared to a cubic foot of aluminum (178 lb). Aluminum is 2.7 times as heavy as water. Finding the specific gravity of a piece of metal is not difficult. We weigh the metal in air, then weigh it under water. Its loss of weight is the weight of an equal volume of water. To find the specific gravity, divide the weight of the metal by its loss of weight in water. (5.2) Example 5.1 Problem : Suppose a piece of metal weighs 150 lb in air and 85 lb under water. What is the specific gravity? Solution : Step 1. 150 lb – 85 lb = 65 lb-loss of weight in water Step 2. Note that in a calculation of specific gravity, the densities must be expressed in the same units. As stated earlier, the specific gravity of water is one, which is the standard — the reference to which all other liquid or solid substances are compared. Specifically, any object that has a specific gravity greater than one will sink in water (rocks, steel, iron, grit, floc, sludge). Substances with a specific gravity of less than one will float (wood, scum, and gasoline). Because the total weight and volume of a ship is less than one, its specific gravity is less than one; therefore, it can float, The most common use of specific gravity in water/wastewater treatment operations is in gallons- to-pounds conversions. In many cases, the liquids handled have a specific gravity of 1.00 or very nearly 1.00 (between 0.98 and 1.02), so 1.00 may be used in the calculations without introducing significant error. However, in calculations involving a liquid with a specific gravity of less than 0.98 or greater than 1.02, the conversions from gallons to pounds must consider the exact specific gravity. The technique is illustrated in the following example. Example 5.2 Problem: A basin holds 1455 gal of a certain liquid. If the specific gravity of the liquid is 0.94, how many pounds of liquid are in the basin? Specific Gravity Weight of Substance Weight = of Equal Volume of Water Specific Gravity 150 65 ==23. L1681_book.fm Page 97 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 98 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Solution : If the substance’s specific gravity were between 0.98 and 1.02, we would use the factor 8.34 lb/gal (the density of water) for a conversion from gallons to pounds. However, in this instance the substance has a specific gravity outside this range, so the 8.34 factor must be adjusted. Step 1. Multiply 8.34 lb/gal by the specific gravity to obtain the adjusted factor: Step 2. Convert 1455 gal to pounds using the corrected factor: Example 5.3 Problem : The specific gravity of a liquid substance is 0.96 at 64°F. What is the weight of 1 gal of the substance? Solution : Example 5.4 Problem : A liquid has a specific gravity of 1.15. How many pounds is 66 gal of the liquid? Solution : Example 5.5 Problem : If a solid in water has a specific gravity of 1.30, what percent heavier is it than water? Solution : (8.34 lb/gal) (0.94) 7.84 lb/gal (rounde= dd) (1455 gal) (7.84 lb/gal) 11,407 lb (round= eed) Weight specific gravity weight of water=× =× 0.96 8.34 lb/gal = 8.01 lb Weight 66 gal 8.34 lb/gal 1.15=× × = 633 lb Percent heavier sp gr of solid sp gr of wa = − tter sp gr of water L1681_book.fm Page 98 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC FUNDAMENTAL CHEMISTRY AND HYDRAULICS 99 5.2.2 Water Chemistry Fundamentals Whenever we add a chemical substance to another chemical substance (adding sugar to tea or adding hypochlorite to water to make it safe to drink), we are performing the work of chemists because we are working with chemical substances; how they react is important to success. Envi- ronmental engineers involved with water treatment operations, for example, may be required to determine the amount of chemicals or chemical compounds to add (dosing) to various unit pro- cesses. Table 5.2 lists some of the chemicals and their common applications in water treatment operations. 5.2.2.1 The Water Molecule Just about everyone knows that water is a chemical compound of two simple and abundant elements: H 2 O. Yet scientists continue to argue the merits of rival theories on the structure of water. The fact is that we still know little about water — for example, we do not know how water works. The reality is that water is very complex, with many unique properties that are essential to life and determine its environmental chemical behavior. The water molecule is different. Two hydrogen atoms (the two in the H 2 part of the water formula) always come to rest at an angle of approximately 105° from each other. The hydrogens tend to be positively charged, and the oxygen tends to be negatively charged. This arrangement gives the water molecule an electrical polarity; that is, one end is positively charged and one end negatively charged. This 105° relationship makes water lopsided, peculiar, and eccentric; it breaks all the rules (Figure 5.1). In the laboratory, pure water contains no impurities, but in nature water contains a lot of materials besides water. The environmental professional tasked with maintaining the purest or cleanest water possible must always consider those extras that ride along in water’s flow. Water is often called the universal solvent , a fitting description when you consider that given enough time and contact, water will dissolve anything and everything on Earth. Table 5.2 Chemicals and Chemical Compounds Used in Water Treatment Name Common application Name Common application Activated carbon Taste and odor control Aluminum sulfate Coagulation Ammonia Chloramine disinfection Ammonium sulfate Coagulation Calcium hydroxide Softening Calcium hypochlorite Disinfection Calcium oxide Softening Carbon dioxide Recarbonation Copper sulfate Algae control Ferric chloride Coagulation Ferric sulfate Coagulation Magnesium hydroxide Defluoridation Oxygen Aeration Potassium permanganate Oxidation Sodium aluminate Coagulation Sodium bicarbonate pH adjustment Sodium carbonate Softening Sodium chloride Ion exchanger regeneration Sodium fluoride Fluoridation Sodium fluosilicate Fluoridation Sodium hexametaphosphate Corrosion control Sodium hydroxide pH adjustment Sodium hypochlorite Disinfection Sodium silicate Coagulation aid Sodium thiosulfate Dechlorination Sulfur dioxide Dechlorination Sulfuric acid pH adjustment Source : Spellman, F.R., 2003, Handbook of Water and Wastewater Treatment Plant Operations . Boca Raton, FL: Lewis Publishers. Percent heavier 1.0 = 130 10.–. = 30 L1681_book.fm Page 99 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 100 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK 5.2.2.2 Water Solutions A solution is a condition in which one or more substances are uniformly and evenly mixed or dissolved. In other words, a solution is a homogenous mixture of two or more substances. Solutions can be solids, liquids, or gases (drinking water, seawater, or air, for example). We focus primarily on liquid solutions. A solution has two components: a solvent and a solute (see Figure 5.2). The solvent is the component that does the dissolving. Typically, the solvent is the substance present in the greater quantity. The solute is the component that is dissolved. When water dissolves substances, it creates solutions with many impurities. Generally, a solution is usually transparent, not cloudy, and visible to longer wavelength ultraviolet light. Because water is colorless (hopefully), the light necessary for photosynthesis can travel to considerable depths. However, a solution may be colored when the solute remains uni- formly distributed throughout the solution and does not settle with time. When molecules dissolve in water, the atoms making up the molecules come apart (dissociate) in the water. This dissociation in water is called ionization . When the atoms in the molecules come apart, they do so as charged atoms (negatively charged and positively charged) called ions . The positively charged ions are called cations and the negatively charged ions are called anions . Figure 5.1 A molecule of water. (From Spellman, F.R., 2003, Handbook of Water and Wastewater Treatment Plant Operations . Boca Raton, FL: Lewis Publishers.) Figure 5.2 Solution with two components: solvent and solute. (From Spellman, F.R., 2003, Handbook of Water and Wastewater Treatment Plant Operations . Boca Raton, FL: Lewis Publishers.) O − − H + H + Solvent Solute L1681_book.fm Page 100 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC FUNDAMENTAL CHEMISTRY AND HYDRAULICS 101 Some of the common ions found in water are: Solutions serve as a vehicle to (1) allow chemical species to come into close proximity so that they can react; (2) provide a uniform matrix for solid materials, such as paints, inks, and other coatings, so that they can be applied to surfaces; and (3) dissolve oil and grease so that they can be rinsed away. Water dissolves polar substances better than nonpolar substances. Polar substances (mineral acids, bases and salts) are easily dissolved in water. Nonpolar substances (oils and fats and many organic compounds) do not dissolve easily in water. 5.2.2.3 Concentrations Because the properties of a solution depend largely on the relative amounts of solvent and solute, the concentrations of each must be specified. Key point : Chemists use relative terms such as saturated and unsaturated, as well as more exact concentration terms, such as weight percentages, molarity, and normality. Although polar substances dissolve better than nonpolar substances in water, polar substances dissolve in water only to a point; that is, only so much solute will dissolve at a given temperature. When that limit is reached, the resulting solution is saturated. At that point, the solution is in equilibrium — no more solute can be dissolved. A liquid/solids solution is supersaturated when the solvent actually dissolves more than an equilibrium concentration of solute (usually when heated). Specifying the relative amounts of solvent and solute, or specifying the amount of one com- ponent relative to the whole, usually gives the exact concentrations of solutions. Solution concen- trations are sometimes specified as weight percentages. (5.3) To understand the concepts of molarity , molality , and normality , we must first understand the concept of a mole . The mole is the amount of a substance that contains exactly the same number of items (i.e., atoms, molecules, or ions) as 12 g of Carbon-12. By experiment, Avagodro determined this number to be 6.02 × 10 23 (to three significant figures). Examples of Ionization CaCO 3 ↔ Calcium carbonate Ca ++ Calcium ion (cation) + CO 3 –2 Carbonate ion (anion) NaCl ↔ Sodium chloride Na + Sodium ion (cation) + Cl – Chloride ion (anion) Ion Symbol Hydrogen H + Sodium Na + Potassium K + Chloride Cl – Bromide Br – Iodide I – Bicarbonate HCO 3 – %ofSolute Mass of Solute Mass of Solute =×1000 L1681_book.fm Page 101 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 102 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK If 1 mol of C atoms equals 12 g, how much is the mass of 1 mol of H atoms? • Note that carbon is 12 times as heavy as hydrogen. • Therefore, we need only 1/12 the weight of H to equal the same number of atoms of C. Key point : 1 mol of H equals 1 g. By the same principle: •1 mol of CO 2 = 12 + 2(16) = 44 g •1 mol of Cl – = 35.5 g •1 mol of Ra = 226 g In other words, we can calculate the mass of a mole if we know the formula of the “item.” Molarity ( M ) is defined as the number of moles of solute per liter of solution. The volume of a solution is easier to measure in the lab than its mass . (5.4) Molality ( m ) is defined as the number of moles of solute per kilogram of solvent. (5.5) Key point : Molality is not as frequently used as molarity, except in theoretical calculations. Especially for acids and bases, the normality ( N ) rather than the molarity of a solution is often reported — the number of equivalents of solute per liter of solution (1 equivalent of a substance reacts with 1 equivalent of another substance). (5.6) In acid/base terms, an equivalent (or gram equivalent weight) is the amount that will react with 1 mol of H + or OH – . For example, 1 mol of HCl will generate 1 mol of H + Therefore, 1 mol HCl = 1 equivalent 1 mol of Mg(OH) 2 will generate 2 mol of OH – Therefore, 1 mol of Mg(OH) 2 = 2 equivalents By the same principle: A 1- M solution of H 3 PO 4 =3 N A 2- N solution of H 2 SO 4 =1 M A 0.5- N solution of NaOH = 0.5 M A 2- M solution of HNO 3 = 2 N M No. of moles of solute No.ofliters of = ssolution M No. of moles of solute No. of kilograms = oof solution N No. of equivalents of solute No.oflite = rrs of solution HCl H Cl - ⇒ + + Mg(OH) Mg 2OH 22-++ ⇒ + L1681_book.fm Page 102 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC FUNDAMENTAL CHEMISTRY AND HYDRAULICS 103 Chemists titrate acid/base solutions to determine their normality. An endpoint indicator is used to identify the point at which the titrated solution is neutralized. Key point: If 100 mL of 1 N HCl neutralizes 100 mL of NaOH, then the NaOH solution must also be 1 N. 5.2.2.4 Predicting Solubility Predicting solubility is difficult, but as a general rule of thumb: like dissolves like. Liquid–Liquid Solubility Liquids with similar structure and hence similar intermolecular forces are completely miscible. For example, we would correctly predict that methanol and water are completely soluble in any proportion. Liquid–Solid Solubility Solids always have limited solubilities in liquids because of the difference in magnitude of their intermolecular forces. Therefore, the closer the temperature comes to the melting point of a particular solid, the better the match is between a solid and a liquid. Key point: At a given temperature, lower melting solids are more soluble than higher melting solids. Structure is also important; for example, nonpolar solids are more soluble in nonpolar solvents. Liquid–Gas Solubility As with solids, the more similar the intermolecular forces, the higher the solubility. Therefore, the closer the match is between the temperature of the solvent and the boiling point of the gas, the higher the solubility is. When water is the solvent, an additional hydration factor promotes solubility of charged species. Other factors that can significantly affect solubility are temperature and pressure. In general, raising the temperature typically increases the solubility of solids in liquids. Key point: Dissolving a solid in a liquid is usually an endothermic process (i.e., heat is absorbed), so raising the temperature “fuels” this process. In contrast, dissolving a gas in a liquid is usually an exothermic process (it emits heat). Therefore lowering the temperature generally increases the solubility of gases in liquids. Interesting point: “Thermal” pollution is a problem because of the decreased solubility of O 2 in water at higher temperatures. Pressure has an appreciable effect only on the solubility of gases in liquids. For example, carbonated beverages like soda water are typically bottled at significantly higher atmospheres. When the beverage is opened, the decrease in the pressure above the liquid causes the gas to bubble out of solution. When shaving cream is used, dissolved gas comes out of solution, bringing the liquid with it as foam. 5.2.2.5 Colligative Properties Some properties of a solution depend on the concentrations of the solute species rather than their identity: L1681_book.fm Page 103 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC [...]... 0.03 = 0.33 M 0.090 Example 5. 9 Problem: Find the normality of 0.3 mol CaCl2 in 300 mL of solution Solution: 2 × 0.3 = 2.0 N 0.300 Example 5. 10 Problem: Find the molality of 5. 0 g of NaOH in 50 0 g of water Solution: 5. 0 = 0.2 m 50 × 0 .50 0 © 20 05 by CRC Press LLC 111 L1681_book.fm Page 112 Tuesday, October 5, 2004 10 :51 AM 112 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK 5. 2.2.8 Simple Solutions and... FW (5. 10) Example 5. 14 Problem: Convert 6 .5% solution of a chemical with FW = 351 to molarity Solution: [6.5g/100 ml) (10)] = 0.1 852 M 351 g/L To convert from molarity to percent solution, multiply the molarity by the FW and divide by 10 % Solution = (molarity)(FW) 10 Example 5. 15 Problem: Convert a 0.00 4 5- M solution of a chemical having FW 176 .5 to percent solution Solution: [0.00 45 mol/L) (176 .5 g/mol)]... Example 5. 12 Problem: Given the following data, determine how many grams of reagent to use Chemical FW = 1 95 g/mol; to make 0. 15 M solution Solution: (1 95 g/mol)(0. 15 mol/L) = 29. 25 g/L Example 5. 13 Problem: A chemical has a FW 190 g/mol and we need 25 mL (0.0 25 L) of 0. 15 M (M = mol/L) solution How many grams of the chemical must be dissolved in 25 mL of water to make this solution? © 20 05 by CRC... (H) is one of the elements The relative strengths of acids in water, listed in descending order of strength, are classified in Table 5. 5 © 20 05 by CRC Press LLC L1681_book.fm Page 108 Tuesday, October 5, 2004 10 :51 AM 108 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Table 5. 5 Relative Strengths of Acids in Water Acid Chemical symbol Perchloric acid Sulfuric acid Hydrochloric acid Nitric acid Phosphoric... solution © 20 05 by CRC Press LLC (5. 9) L1681_book.fm Page 113 Tuesday, October 5, 2004 10 :51 AM FUNDAMENTAL CHEMISTRY AND HYDRAULICS 113 Cf = concentration of final solution Vf = volume of final solution Example 5. 11 Problem: How much water must be added to 60 mL of 1.3-M HCl solution to produce a 0 . 5- M HCl solution? Solution: Ci × Vi = Cf × Vf 1.3 × 60 = 0 .5 × X X= 1.3 × 60 = 156 ml 0 .5 The volume of... (pounds per day), we must first determine chlorine dose To do this, we must use Equation 5. 15 © 20 05 by CRC Press LLC L1681_book.fm Page 122 Tuesday, October 5, 2004 10 :51 AM 122 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Solution: Cl Dose, mg/L = Cl Demand, mg/L + Cl Residual, mg/L = 12 mg/L + 2 mg/L = 14 mg/L (5. 15) Then we can make the milligrams per liter to pounds per day calculation: (12 mg/L)(3... × 10 6 gal/day × 8.34 lb/gal × 1.0 Dosage = wt of F/wt of water = 66.2 × 8.34 × 1.191 12 x 10 6 × 8.34 lb/gal × 1.0 = 6 .58 10 6 = 6 .58 mg/L © 20 05 by CRC Press LLC L1681_book.fm Page 126 Tuesday, October 5, 2004 10 :51 AM 126 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Example 5. 25 Problem: A raw water flow that averages 8.6 MGD is continuously fed 10 mg/L of liquid alum with 60% strength How much liquid... Hypochlorite 15% Available Cl Step 3 Calculate the gallons per day sodium hypochlorite: 933 lb/day = 112 gal/day Sodium Hypochlorite 8.34 lb/gal Example 5. 20 Problem: How many pounds of chlorine gas are necessary to treat 5, 000,000 gal of wastewater at a dosage of 2 mg/L? © 20 05 by CRC Press LLC L1681_book.fm Page 124 Tuesday, October 5, 2004 10 :51 AM 124 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Solution:... Substitute 5 × 10 6 gal × 2 mg/L × 8.34 = 83 lb Cl Additional Dosage Calculations Example 5. 21 Problem: Chlorine dosage at a treatment plant averages 112 .5 lb/day Its average flow is 11 .5 MGD What is the chlorine dosage in milligrams per liter? Solution: Dosage = 112 .5 lb/day 11 .5 × 10 6 gal = 112 .5 lb/day 11 .5 × 10 6 gal × 8.34 lb/gal = 1.2 10 6 = 1.2 ppm = 1.2 mg/L Example 5. 22 Problem: Twenty-five pounds... CRC Press LLC L1681_book.fm Page 114 Tuesday, October 5, 2004 10 :51 AM 114 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Solution: Number grams/desired volume (L) = desired molarity (mol/L) × FW (g/mol) rearranging Number grams = desired volume (L) × desired molarity (mol/L) × FW (g/mol) Number grams = (0.0 25 L)(0. 15 mol/L) (190 g/mol) = 0.71 25 g/ 25 mL Percent Solutions Many reagents are mixed as percent . of Water Specific Gravity 150 65 ==23. L1681_book.fm Page 97 Tuesday, October 5, 2004 10 :51 AM © 20 05 by CRC Press LLC 98 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Solution : If the. water. (5. 2) Example 5. 1 Problem : Suppose a piece of metal weighs 150 lb in air and 85 lb under water. What is the specific gravity? Solution : Step 1. 150 lb – 85 lb = 65 lb-loss of. previously unknown compounds. L1681_book.fm Page 95 Tuesday, October 5, 2004 10 :51 AM © 20 05 by CRC Press LLC 96 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Why should we care about chemistry?