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Keys to the Study of Chemistry. Chapter 1. Chapter 1 : Keys to the Study of Chemistry. 1.1 Some Fundamental Definitions 1.2 The Scientific Approach: Developing a Model 1.3 Chemical Problem Solving 1.4 Measurement in Scientific Study
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Keys to the Study of Chemistry Chapter 1
Chapter 1 : Keys to the Study of Chemistry 1.1 Some Fundamental Definitions 1.2 The Scientific Approach: Developing a Model 1.3 Chemical Problem Solving 1.4 Measurement in Scientific Study 1.5 Uncertainty in Measurement: Significant Figures
CHEMISTRY Is the study of matter, its properties, the changes that matter undergoes, and the energy associated with these changes.
Matter anything that has mass and volume -the “stuff” of the universe: books, planets, trees, professors, students Composition the types and amounts of simpler substances that make up a sample of matter Properties the characteristics that give each substance a unique identity Definitions Physical Properties those which the substance shows by itself without interacting with another substance such as color, melting point, boiling point, density Chemical Properties those which the substance shows as it interacts with, or transforms into, other substances such as flammability, corrosiveness
A Physical change B Chemical change The distinction between physical and chemical change. Figure 1.1
Figure 1.2 The physical states of matter.
PROBLEM: Decide whether each of the following process is primarily a physical or a chemical change, and explain briefly: Sample Problem 1.1 Distinguishing Between Physical and Chemical Change (a) Frost forms as the temperature drops on a humid winter night. (b) A cornstalk grows from a seed that is watered and fertilized. (c) Dynamite explodes to form a mixture of gases. (d) Perspiration evaporates when you relax after jogging. (e) A silver fork tarnishes slowly in air. PLAN: “Does the substance change composition or just change form?” SOLUTION: (a) physical change (b) chemical change (c) chemical change (d) physical change (e) chemical change
Energy is the capacity to do work. energy due to the position of the object or energy from a chemical reaction Potential Energy Kinetic Energy energy due to the motion of the object Potential and kinetic energy can be interconverted.
Energy is the capacity to do work. Figure 1.3A less stable change in potential energyEQUALS kinetic energy more stable A gravitational system. The potential energy gained when a lifted weight is converted to kinetic energy as the weight falls.
Energy is the capacity to do work. Figure 1.3B less stable change in potential energyEQUALS kinetic energy more stable A system of two balls attached by a spring. The potential energy gained by a stretched spring is converted to kinetic energy when the moving balls are released.
Energy is the capacity to do work. Figure 1.3C less stable change in potential energyEQUALS kinetic energy more stable A system of oppositely charged particles. The potential energy gained when the charges are separated is converted to kinetic energy as the attraction pulls these charges together.
Energy is the capacity to do work. Figure 1.3D less stable change in potential energyEQUALS kinetic energy more stable A system of fuel and exhaust. A fuel is higher in chemical potential energy than the exhaust. As the fuel burns, some of its potential energy is converted to the kinetic energy of the moving car.
Natural phenomena and measured events; universally consistent ones can be stated as a natural law. Observations : Hypothesis: Tentative proposal that explains observations. revised if experiments do not support it Procedure to test hypothesis; measures one variable at a time. Experiment: Set of conceptual assumptions that explains data from accumulated experiments; predicts related phenomena. Model (Theory): altered if predictions do not support it Further Experiment: Tests predictions based on model. Scientific Approach: Developing a Model
A Systematic Approach to Solving Chemistry Problems • Problem statement Clarify the known and unknown. • Plan Suggest steps from known to unknown. Prepare a visual summary of steps. • Solution • Check Comment and Follow-up Problem
PROBLEM: To wire your stereo equipment, you need 325 centimeters (cm) of speaker wire that sells for $0.15/ft. What is the price of the wire? ft in 12 in 2.54 cm $0.15 ft Sample Problem 1.2 Converting Units of Length PLAN: Known - length (in cm) of wire and cost per length ($/ft) We have to convert cm to inches and inches to ft followed by finding the cost for the length in ft. SOLUTION: length (cm) of wire Length (in) = length (cm) x conversion factor 2.54 cm = 1 in = 325 cm x = 128 in length (in) of wire 12 in = 1 ft Length (ft) = length (in) x conversion factor length (ft) of wire = 128 in x = 10.7 ft 1 ft = $0.15 Price ($) = length (ft) x conversion factor Price ($) of wire = 10.7 ft x = $1.60
SI Base Units Table 1. 1 time second s temperature kelvin K electric current ampere A amount of substance mole mol luminous intensity candela cd Physical Quantity (Dimension) Unit Abbreviation Unit Name mass kilogram kg length meter m
Table 1.2 Common Decimal Prefixes Used with SI Units
Common SI-English Equivalent Quantities Table 1.3 1 kilometer(km) 1000(103)m 0.62miles(mi) 1 mi = 1.61km 1 meter(m) 100(102)cm 1000(103)mm 1.094yards(yd) 39.37inches(in) 1 yd = 0.9144m 1 foot (ft) = 0.3048m 1 centimeter(cm) 0.01(10-2)m 0.3937in 1 in = 2.54cm (exactly!) 1 kilometer(km) 1000(103)m 0.62mi Quantity SI Unit SI Equivalent English Equivalent Length English to SI Equivalent
Common SI-English Equivalent Quantities Table 1.3 1 cubic meter(m3) 1,000,000(106) cubic centimeters 35.2cubic feet (ft3) 1 ft3 = 0.0283m3 1 cubic decimeter(dm3) 1000cm3 0.2642 gallon (gal) 1.057 quarts (qt) 1 gal = 3.785 dm3 1 qt = 0.9464 dm3 1 cubic centimeter (cm3) 0.001 dm3 0.0338 fluid ounce 1 qt = 946.4 cm3 1 fluid ounce = 29.6 cm3 Quantity SI Unit SI Equivalent English Equivalent Volume English to SI Equivalent
Common SI-English Equivalent Quantities Table 1.3 1 kilogram (kg) 1000 grams 2,205 pounds (lb) 1 (lb) = 0.4536 kg 1 gram (g) 1000 milligrams 0.03527 ounce(oz) 1 lb = 453.6 g 1 ounce = 28.35 g Quantity SI Unit SI Equivalent English Equivalent Mass English to SI Equivalent
PROBLEM: The volume of an irregularly shaped solid can be determined from the volume of water it displaces. A graduated cylinder contains 19.9mL of water. When a small piece of galena, an ore of lead, is submerged in the water, the volume increases to 24.5mL. What is the volume of the piece of galena in cm3 and in L? subtract 1 cm3 10-3 L mL mL 1 mL = 1 cm3 1 mL = 10-3 L Sample Problem 1.3 Converting Units of Volume PLAN: The volume of galena is equal to the change in the water volume before and after submerging the solid. SOLUTION: volume (mL) before and after addition (24.5 - 19.9)mL = volume of galena volume (mL) of galena 4.6 mL x = 4.6 cm3 volume (cm3) of galena volume (L) of galena 4.6 mL x = 4.6x10-3 L
PROBLEM: International computer communications are often carried by optical fibers in cables laid along the ocean floor. If one strand of optical fiber weighs 1.19 x 10-3lbs/m, what is the total mass (in kg) of a cable made of six strands of optical fiber, each long enough to link New York and Paris (8.84 x 103km)? 8.84 x 103km x 1 km = 103 m 8.84 x 106m x 103m 1.19 x 10 -3lbs 1 m = 1.19x10-3 lb 6.30x 104lb 6 fibers km m = 1.05 x 104lb x cable cable 6 fibers = 1 cable 6.30x 104lb 1kg 2.86x104 kg x = cable cable 2.205 lb 2.205 lb = 1 kg Sample Problem 1.4 Converting Units of Mass PLAN: The sequence of steps may vary but essentially you have to find the length of the entire cable and convert it to mass. SOLUTION: length (km) of fiber = 8.84 x 106m length (m) of fiber = 1.05 x 104lb mass (lb) of fiber mass (lb) of cable mass (kg) of cable
PROBLEM: Lithium (Li) is a soft, gray solid that has the lowest density of any metal. If a slab of Li weighs 1.49 x 103 mg and has sides that measure 20.9 mm by 11.1 mm by 11.9 mm, what is the density of Li in g/cm3 ? 10 mm = 1 cm 1cm 10mm 103 mg = 1 g multiply lengths 10-3g 1mg 1.49g 2.76 cm3 Sample Problem 1.5 Calculating Density from Mass and Length PLAN: Density is expressed in g/cm3 so we need the mass in grams and the volume in cm3. SOLUTION: lengths (mm) of sides 1.49x103mg x = 1.49g mass (mg) of Li lengths (cm) of sides 20.9mm x = 2.09cm Similarly the other sides will be 1.11 cm and 1.19 cm, respectively. mass (g) of Li volume (cm3) 2.09 x 1.11 x 1.20 = 2.76cm3 density (g/cm3) of Li density of Li = = 0.540 g/cm3
The freezing and boiling points of water. Figure 1.8
Temperature Scales and Interconversions Kelvin ( K ) - The “Absolute temperature scale” begins at absolute zero and only has positive values. Celsius ( oC ) - The temperature scale used by science, formally called centigrade, most commonly used scale around the world; water freezes at 0oC, and boils at 100oC. Fahrenheit ( oF ) -Commonly used scale in the U.S. for our weather reports; water freezes at 32oF and boils at 212oF. T (in K) = T (in oC) + 273.15 T (in oC) = T (in K) - 273.15 T (in oF) = 9/5 T (in oC) + 32 T (in oC) = [ T (in oF) - 32 ] 5/9
PROBLEM: A child has a body temperature of 38.70C. 9 (38.70C) + 32 = 101.70F 5 Sample Problem 1.6 Converting Units of Temperature (a) If normal body temperature is 98.60F, does the child have a fever? (b) What is the child’s temperature in kelvins? PLAN: We have to convert 0C to 0F to find out if the child has a fever and we use the 0C to kelvin relationship to find the temperature in kelvins. SOLUTION: (a) Converting from 0C to 0F (b) Converting from 0C to K 38.70C + 273.15 = 311.8K
Figure 1.9A The number of significant figures in a measurement depends upon the measuring device. 32.330C 32.30C
Rules for Determining Which Digits are Significant except zeros that are used only to position the decimal point. All digits are significant • Make sure that the measured quantity has a decimal point. • Start at the left of the number and move right until you reach the first nonzero digit. • Count that digit and every digit to it’s right as significant. Zeros that end a number and lie either after or before the decimal point are significant; thus 1.030 ml has four significant figures, and 5300. L has four significant figures also. Numbers such as 5300 L are assumed to only have 2 significant figures. A terminal decimal point is often used to clarify the situation, but scientific notation is the best!
PROBLEM: For each of the following quantities, underline the zeros that are significant figures(sf), and determine the number of significant figures in each quantity. For (d) to (f), express each in exponential notation first. (a) 0.0030 L (b) 0.1044 g (c) 53.069 mL (d) 0.00004715 m (e) 57,600. s (f) 0.0000007160 cm3 Sample Problem 1.7 Determining the Number of Significant Figures (a) 0.0030 L (b) 0.1044 g (c) 53,069 mL (d) 0.00004715 m (e) 57,600. s (f) 0.0000007160 cm3 PLAN: Determine the number of sf by counting digits and paying attention to the placement of zeros. SOLUTION: 2sf 4sf 5sf (f) 7.160x10-7 cm3 (d) 4.715x10-5 m 4sf (e) 5.7600x104 s 5sf 4sf
83.5 mL + 23.28 mL 865.9 mL - 2.8121 mL Rules for Significant Figures in Answers 1. For addition and subtraction. The answer has the same number of decimal places as there are in the measurement with the fewest decimal places. Example: adding two volumes 106.78 mL = 106.8 mL Example: subtracting two volumes 863.0879 mL = 863.1 mL
Rules for Significant Figures in Answers 2. For multiplication and division. The number with the least certainty limits the certainty of the result.Therefore, the answer contains the same number of significant figures as there are in the measurement with the fewest significant figures. Multiply the following numbers: 9.2 cm x 6.8 cm x 0.3744 cm = 23.4225 cm3 =23cm3
Rules for Rounding Off Numbers 1. If the digit removed is more than 5, the preceding number increases by 1. 5.379 rounds to 5.38 if three significant figures are retained and to 5.4 if two significant figures are retained. 2. If the digit removed is less than 5, the preceding number is unchanged. 0.2413 rounds to 0.241 if three significant figures are retained and to 0.24 if two significant figures are retained. 3.If the digit removed is 5, the preceding number increases by 1 if it is odd and remains unchanged if it is even. 17.75 rounds to 17.8, but 17.65 rounds to 17.6. If the 5 is followed only by zeros, rule 3 is followed; if the 5 is followed by nonzeros, rule 1 is followed: 17.6500 rounds to 17.6, but 17.6513 rounds to 17.7 4. Be sure to carry two or more additional significant figures through a multistep calculation and round off only the finalanswer.
Electronic Calculators Choice of Measuring Device Exact Numbers Issues Concerning Significant Figures be sure to correlate with the problem FIX function on some calculators graduated cylinder < buret ≤ pipet 60 min = 1 hr numbers with no uncertainty 1000 mg = 1 g These have as many significant digits as the calculation requires.
PROBLEM: Perform the following calculations and round the answer to the correct number of significant figures: 1 g 1 g 4.80x104 mg 4.80x104 mg 1000 mg 1000 mg 16.3521 cm2 - 1.448 cm2 16.3521 cm2 - 1.448 cm2 (b) (b) (a) (a) 11.55 cm3 11.55 cm3 7.085 cm 7.085 cm 14.904 cm2 = 7.085 cm 48.0 g = 11.55 cm3 Sample Problem 1.8 Significant Figures and Rounding PLAN: In (a) we subtract before we divide; for (b) we are using an exact number. SOLUTION: = 2.104 cm = 4.16 g/ cm3
Precision and Accuracy Errors in Scientific Measurements Precision - Refers to reproducibility orhow close the measurements are to each other. Accuracy - Refers to how close a measurement is to the real value. Systematic Error - Values that are either all higher or all lower than the actual value. Random Error - In the absence of systematic error, some values that are higher and some that are lower than the actual value.
precise and accurate precise but not accurate Figure 1.10 Precision and accuracy in the laboratory.
random error systematic error Precision and accuracy in the laboratory. Figure 1.10 continued