1 - Chemical Foundations Lecture modified

1 - Chemical Foundations Lecture modified - Chemical...

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Unformatted text preview: Chemical Foundations Chapter 1 General Chemistry 1 Scientific Method Make an observation Look for patterns in the observation Formulate theory Design experiments to test the theory 1. 2. 3. 4. • • Hypothesis – tentative explanation or idea about Hypothesis how things work how Theory – explanation of general principles of certain Theory phenomena with considerable facts to support it phenomena 2 Hypothesis vs Theory vs Fact Hypothesis • • “The moon is made of cheese” How do we test this? Theory • Remains valid only if every new piece of Remains information supports it information Fact • An indisputable truth 3 Development of Simple Theory Development Observation: Observation: Every swan I have ever seen is white. is 4 Development of Simple Theory Development Observation: Observation: Every swan I have ever seen is white. is Hypothesis: All swans must be Hypothesis: white. white. 5 Development of Simple Theory Development Observation: Observation: Every swan I have ever seen is white. is Hypothesis: All swans must be Hypothesis: white. Test: Random sampling of swans 6 Development of Simple Theory Development Observation Hypothesis Test Publication: Publication: “My global research has indicated that swans are always white, wherever they are observed.” wherever 7 Development of Simple Theory Development Observation Hypothesis Test Publication: Publication: “My global research has indicated that swans are always white , wherever they are observed.” wherever Verification: Scientists observe white Verification: swans swans 8 Development of Simple Theory Development Observation Hypothesis Test Publication Verification Theory: All swans are white. 9 Development of Simple Theory Development Observation Hypothesis Test Publication Verification Theory: All swans are white. Prediction: The next swan I see will be Prediction: white. white. 10 Units of Measurement Mass kilogram kg meter m Time second s Temperature Kelvin K ampere A Amt of substance mole mol Volume liter L candella cd Length Electric Current Luminous Intensity 11 11 SI System Prefixes SI Prefix Prefix Symbol Meaning Exponential Exponential notation notation giga G 1,000,000,000 109 mega M 1,000,000 106 kilo k 1,000 103 hecto h 100 102 deka da 10 101 -- -- 1 100 12 12 SI System Prefixes Prefix Symbol Meaning Exponential Exponential notation notation -- -- 1 100 deci d 0.1 10-1 centi c 0.01 10-2 milli m 0.001 10-3 micro µ 0.000001 10-6 nano n 0.000000001 10-9 13 13 Exponential Scale Powers of 10 perspective http://micro.magnet.fsu.edu/primer/java/scienceo 14 14 Mass vs Weight Mass • • Amount of matter Measured by force necessary to give an Measured object a given acceleration object Weight • • • • Gravitational effect on mass On earth weigh 200 lbs. On moon weigh 33.2 lbs On Jupiter weigh 472.8 lbs 15 15 Uncertainty in Measurement Accuracy Accuracy : agreement of a value with the true value true Precision: agreement among a set of Precision: measurements measurements http://www.mhhe.com/physsci/chemistry/chang7/esp/folder_structure/ch/m2/s2/index.htm 16 16 Errors Random: equal probability of being high or low Systematic: occurs in the same direction each time 17 17 Significant Figures Example: Postage Scale 3g Two-pan balance 2.73 g Analytical How balance 2.731 g many significant figures? 18 18 Significant Figures Significant 19 Significant Figures Significant Ruler A Ruler B 20 Significant Figures Significant Ruler A Ruler B 4.40 cm 21 Significant Figures Significant Ruler A 4.40 cm Ruler B 4.5 cm 22 Significant Figures Significant 23 Sig. Figs. In Measurements A C B D 24 24 Sig. Figs. In Measurements A B C D E 25 25 Counting Significant Figures All numbers significant except zero 18.693 Zeros 5 sig. figs. within a number are significant 70.4006 6 sig. figs. For For numbers < 1 count digits starting with first nonzero digit on left and continue right first 0.004289 4 sig. figs. 26 26 Counting Significant Figures Leading Leading zeros left of decimal are not significant significant 0.0000915 3 sig. figs. Trailing Trailing zeros to the right of decimal point are significant are 0.002700 4 sig. figs. Trailing Trailing zeros to the left of decimal point are ambiguous are 473,000 3 – 6 sig. figs. 27 27 Counting Significant Figures Trailing Trailing zeros to the left of decimal point are ambiguous are 230,000 230,000 230,000 230,000 2-6 sig. figs. 3 sig. figs. 4 sig. figs. 6 sig. figs. Trailing zeros are not significant unless otherwise specified by the line above the zero. 28 28 Class Problem #1 Number Significant Figures .0012 6500 408 408.02 408.020 0.0107 6500 29 29 Class Problem #1 Number Significant Figures .0012 2 6500 408 408.02 408.020 0.0107 6500 30 30 Class Problem #1 Number Significant Figures .0012 6500 2-4? 408 408.02 408.020 0.0107 6500 31 31 Class Problem #1 Number Significant Figures .0012 6500 408 3 408.02 408.020 0.0107 6500 32 32 Class Problem #1 Number Significant Figures .0012 6500 408 408.02 5 408.020 0.0107 6500 33 33 Class Problem #1 Number Significant Figures .0012 6500 408 408.02 408.020 6 0.0107 6500 34 34 Class Problem #1 Number Significant Figures .0012 6500 408 408.02 408.020 0.0107 3 6500 35 35 Class Problem #1 Number Significant Figures .0012 6500 408 408.02 408.020 0.0107 6500 4 36 36 Exact Numbers Unlimited number of significant figures Sources: Counting objects Defined quantities Numbers that are part of an equation Exact Numbers Determined by counting Never limit the number of significant Never figures in a calculation figures Examples: 7 beakers 3 feet/yard 15 batches 38 38 Exact Numbers Examples Within English system: 16 oz. = 1 lb. 12 in. = 1 ft. Within 10 Metric system: mm = 1 cm Between English and Metric system: 1 in = 2.54 cm 454 g = 1 lb or 454 g/1 lb Significant Figures in Calculations Addition and Subtraction 14.105 100.9__ 100.9__ 115.005 115.005 Multiplication 115.0 and Division 36 * 1075 = 2580 15 15 2.6 x 103 40 40 Significant Figures in Calculations 6.404 x 2.91 18.7 – 17.1 18.7 41 41 Significant Figures in Calculations 6.404 x 2.91 18.7 – 17.1 18.7 18.6356 1.6 1.6 42 42 Significant Figures in Calculations 6.404 x 2.91 18.7 – 17.1 18.7 18.6356 1.6 1.6 11.64725 43 43 Significant Figures in Calculations 6.404 x 2.91 18.7 – 17.1 18.7 18.6356 1.6 1.6 11.64725 12 44 44 Rounding Off When When working problems, carry the extra digits through to the final result and then round. round. Less than 5 round down Equal to or more than 5 round up 45 45 Rounding Examples Round to the nearest tens position 1356 1353 1355 1345 46 46 Rounding Examples Round 1356 to the nearest tens position 1360 1353 1355 1345 47 47 Rounding Examples Round 1356 1353 to the nearest tens position 1360 1350 1355 1345 48 48 Rounding Examples Round 1356 1353 1355 to the nearest tens position 1360 1350 1360 1345 49 49 Rounding Examples Round 1356 1353 1355 1345 to the nearest tens position 1360 1350 1360 1350 50 50 Exponential Notation Simplifies numbers by getting rid of zeros Nonscientific Scientific Notation 10000 1 x 104 7354 7.354 x 103 0.01 1 x 10-2 0.00044 4.4 x 10-4 51 51 Class Problem #2 0.053 7025 (4.07*102)(6.39*10-4) 4.36*104 - 2.796*102 52 52 Class Problem #2 0.053 5.3 * 10-2 7025 (4.07*102)(6.39*10-4) 4.36*104 - 2.796*102 53 53 Class Problem #2 0.053 7025 7.025 * 103 (4.07*102)(6.39*10-4) 4.36*104 - 2.796*102 54 54 Class Problem #2 0.053 7025 (4.07*102)(6.39*10-4) 2.60 * 10-1 4.36*104 - 2.796*102 55 55 Class Problem #2 0.053 7025 (4.07*102)(6.39*10-4) 4.36*104 - 2.796*102 4.33 * 104 56 56 Dimensional Analysis Use Use the equivalence statement that relates the two units two Cancel the unwanted units Multiply the quantity to be Multiply converted by the unit factor to give quantity with desired units desired Industry Week, 1981 November 30 57 57 Dimensional Analysis Example #1 You are planning a party Friday night and You expect 30 people. How much pizza and beer should you get? beer Estimate that one person eats a third of a Estimate pizza and drinks 4 beers pizza 58 58 Dimensional Analysis Example #1 0.333 pizza 30humans * ÷= 10 pizza 1human 4beers 30humans * ÷= 120beers 1human 59 59 Dimensional Analysis Example #1 0.333 pizza 30humans * ÷= 10 pizza 1human 4beers 30humans * ÷= 120beers 1human 60 60 Dimensional Analysis Example #1 Should you buy the beer in six-packs or in cases? 1sixpack 120beers * ÷ = 20 sixpacks 6beers 1case 120beers * ÷ = 5cases 24beers 61 61 Dimensional Analysis Example #1 How do we know the equation was set up How correctly? correctly? 6beers 2 120beers * ÷ = 720beers / sixpack 1sixpack 62 62 Class Problem #3 The bike distance in an Olympic length The triathlon is 40 km. How many miles is this? triathlon 63 63 Class Problem #3 The bike distance in an Olympic length The triathlon is 40 km. How many miles is this? triathlon Need to go from kilometers to miles kilometers meters yards miles 64 64 Class Problem #3 The bike distance in an Olympic length The triathlon is 40 km. How many miles is this? triathlon Need the following equivalence statements: 1 km 1000 m 1m 1.094 yd 1.094 1760 yd 1 mii 1760 m 65 65 Class Problem #3 The bike distance in an Olympic length The triathlon is 40 km. How many miles is this? triathlon 40 km * 1000 m = 40 1000 1 km 66 66 Class Problem #3 The bike distance in an Olympic length The triathlon is 40 km. How many miles is this? triathlon 40 km * 1000 m = 40 1000 1 km 40000 m 67 67 Class Problem #3 The bike distance in an Olympic length The triathlon is 40 km. How many miles is this? triathlon 40 km * 1000 m = 40000 m 40 1000 1 km 40000 m * 1.094 yd = 1.094 1m 68 68 Class Problem #3 The bike distance in an Olympic length The triathlon is 40 km. How many miles is this? triathlon 40000 m * 1.094 yd = 43760 yd 40000 1.094 1m 43760 yd * 1 mi = 1760 yd 1760 69 69 Class Problem #3 The bike distance in an Olympic length The triathlon is 40 km. How many miles is this? triathlon 40000 m * 1.094 yd = 43760 yd 40000 1.094 1m 43760 yd * 1 mi = 24.864 mi 1760 yd 1760 70 70 Class Problem #3 The bike distance in an Olympic length The triathlon is 40 km. How many miles is this? triathlon 43760 yd * 1 mi 1760 yd 1760 = 24.864 mi 24.864 mi rounds to 25 mi 24.864 71 71 Dimensional Analysis Example #3 You have a 2 in x 2 in x 2 in cube. What is You the volume of the cube? the 72 72 Dimensional Analysis Example #3 You have a 2 in x 2 in x 2 in cube. What is You the volume of the cube? the Need the following equivalence statements: 1 inch 2.54 cm 100 cm 1m 73 73 Dimensional Analysis Example #3 You have a 2 in x 2 in x 2 in cube. What is You the volume of the cube? the 2 in x 2 in x 2 in = 8 in3 in 74 74 Dimensional Analysis Example #3 You have a 2 in x 2 in x 2 in cube. What is You the volume of the cube? the 2 in x 2 in x 2 in = 8 in3 in 8 in3 x 2.54 cm 3 = in 2.54 1 in 75 75 Dimensional Analysis Example #3 You have a 2 in x 2 in x 2 in cube. What is You the volume of the cube? the 2 in x 2 in x 2 in = 8 in3 in 8 in3 x 2.54 cm 3 = 1.310965 * 102 cm3 in 2.54 1 in 76 76 Dimensional Analysis Example #3 You have a 2 in x 2 in x 2 in cube. What is You the volume of the cube? the 8 in3 x 2.54 cm 3 = 1.310965 * 102 cm3 in 2.54 1 in 1.310965 * 102 cm3 x 1 m 3 = 1.310965 100 cm 100 77 77 Dimensional Analysis Example #3 You have a 2 in x 2 in x 2 in cube. What is the You volume of the cube? volume 8 in3 x 2.54 cm 3 = 1.310965 * 102 cm3 in 2.54 1 in 1.310965 * 102 cm3 x 1 m 3 = 1.310965 100 cm 78 78 Dimensional Analysis Example #3 You have a 2 in x 2 in x 2 in cube. What is the You volume of the cube? volume 1.3110 * 102 cm3 x 1.3110 * 10-4 m3 1m 100 cm 100 3 = 1.3110*10-4 m3 1.3110*10 1 * 10-4 m3 10 79 79 Dimensional Analysis Summary Used in numerical calculations Used in converting units Can help identify if an equation is set up Can correctly correctly If a variable in an equation is squared, If then the associated dimensions are squared squared 80 80 Temperature http://www.chem.ufl.edu/~itl/2045/matter/FG01_018.GIF 81 81 Temperature Conversions TK = TC + 273.15 TC = TK - 273.15 (TF – 32°F)(5/9) = TC TF = TC x (9/5) + 32°F Where do these equations come from? 82 82 Temperature Conversions 212 – 32 = 180°F 100 – 0 = 100°C 180°F = 9°F 9°F 100°C 5°C 83 83 Temperature Conversions Different zero points: 32°F = 0°C Apply unit factor of 5/9 (TF – 32°F)(5°C/9°F) = TC 84 84 Temperature Conversions Different zero points: 32°F = 0°C Apply unit factor of 5/9 (TF – 32°F)(5°C/9°F) = TC 85 85 Temperature Problem 432°C = ? K 432°C 86 86 Temperature Problem 432°C = ? K 432°C TK = TC + 273.15 87 87 Temperature Problem 432°C = ? K 432°C TK = TC + 273.15 432 + 273.15 = 705.15 K 705 K 88 88 Temperature Problem 2835 K = ? °C 2835 89 89 Temperature Problem 2835 K = ? °C 2835 TC = TK - 273.15 90 90 Temperature Problem 2835 K = ? °C 2835 TC = TK - 273.15 2835 – 273.15 = 2561.85°C 2562°C 91 91 Temperature Problem 92°F = ? °C 92 92 Temperature Problem 92°F = ? °C (TF – 32°F)(5°C/9°F) = TC 93 93 Temperature Problem 92°F = ? °C (TF – 32°F)(5°C/9°F) = TC (92 – 32)(5/9) = (92 94 94 Temperature Problem 92°F = ? °C (TF – 32°F)(5°C/9°F) = TC (92 – 32)(5/9) = 33.3°C 33°C 95 95 Temperature Problem 3°C = ? °F 96 96 Temperature Problem 3°C = ? °F TF = TC x (9/5) + 32°F 97 97 Temperature Problem 3°C = ? °F TF = TC x (9/5) + 32°F TF = 3 x (9/5) + 32°F 98 98 Temperature Problem 3°C = ? °F TF = TC x (9/5) + 32°F TF = 3 x (9/5) + 32°F TF = 37.4°F rounds to 37°F 99 99 Volume Represent Represent amounts in three dimensional space space 1 mL = 1 cm3 1 L = 1000 cm3 100 100 Density Mass per unit volume Used as an “identification Used tag” tag” _mass_ Density = volume http://www.chem.com.au/products/lstheavyliquid/ancillary/hydrometer.jpeg 101 101 Density Problem #1 Stirring Stirring bar with a mass of 7.68 g is put in a graduated cylinder containing 6.6 mL of water. The level of water was raised such that the new volume was 8.5 mL. that Calculate the density of the stirring rod Initial 10 5 After addition of stir bar 10 5 102 102 Density Problem #1 Volume 8.5 – 6.6 = 1.9 mL Density 7.68 g / 1.9 mL = 4.042 g/mL round to 4.0 g/mL 103 103 Density Problem #2 Diamonds Diamonds are measured in carats, and 1 carat = 0.200 g. The density of diamond is 3.51 g/cm3. 3.51 What is the volume of a 5.0 carat What diamond? diamond? 104 104 Density Problem #2 Diamonds Diamonds are measured in carats, and 1 carat = 0.200 g. The density of diamond is 3.51 g/cm3. 3.51 What is the volume of a 5.0 carat What diamond? diamond? First need mass of 5.0 carat diamond (5.0 carat)(0.200 g/1 carat) = 1.000 g 105 105 Density Problem #2 Diamonds Diamonds are measured in carats, and 1 carat = 0.200 g. The density of diamond is 3.51 g/cm3. 3.51 What is the volume of a 5.0 carat What diamond? diamond? Density = mass / volume 3.51 g/cm3 = (1.000 g / V) 106 106 Density Problem #2 Diamonds Diamonds are measured in carats, and 1 carat = 0.200 g. The density of diamond is 3.51 g/cm3. 3.51 What is the volume of a 5.0 carat What diamond? diamond? 3.51 g/cm3 = (1.000 g / V) V = 0.2849 cm3 0.2849 107 107 Density Problem #2 Diamonds Diamonds are measured in carats, and 1 carat = 0.200 g. The density of diamond is 3.51 g/cm3. 3.51 What is the volume of a 5.0 carat What diamond? diamond? V V = 0.2849 cm3 = 0.28 cm3 round to 108 108 State of Matter Solid: Definite volume Fixed shape Liquid: Definite volume Flexible shape Gas: No definite volume or shape Occupies entire container http://library.tedankara.k12.tr/chemistry/vol1/matter/trans7.jpg 109 109 Matter Can be separated by a physical process? No Mixture Pure Substance Uniform material of definite composition w/ characteristic properties Can it be decomposed by a chemical process? No Element Cannot be broken down into simpler substances hydrogen, zinc, sulfur Yes Yes Compound Can be broken down into two or more elements Water, zinc sulfide Variable composition whose parts can be separated by physical means Is it uniform throughout? Yes Homogeneou s Mixture (Solution) Uniform mixture of different substances seawater, air No Heterogeneous Mixture Nonuniform mixture of different substances that remain distinct wood, concrete 110 110 Matter Examples Cookie Yogurt Lead Sinker 111 111 Physical Changes Change Change in the form of a substance, not in its chemical composition its Water Water Steam Liquid Ice 112 112 Physical Changes Piece of paper Cut Tear Folded Written on Painted 113 113 Chemical Change Substance Substance becomes a new substance with different properties and different composition composition Electrolysis of water 114 114 Chemical Change New matter formed Burning Rusting Rusting Cooking Dehydration of sugar 115 115 Separation Methods Distillation 116 116 Separation Methods Filtration: 117 117 Chromatography: 118 118 Separation Methods Paper Chromatography: 119 119 Learning Objectives Scientific Method Units of Measurement Significant Figures Dimensional Analysis Temperature Density Matter Classification Classification Separation 120 120 ...
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