Chemistry Unit 2 pp.pptx [repaired]

November 8, 2017 | Author: UsmanShahzad1977 | Category: Education
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1. This is a unit of study called “Matter and Energy”… …in which basic properties of matter and energy transformations are discussed. Slides that utilize media are: …Slide 52 has video on demand that can be viewed online at any time for students. …slide 69 has sound to correspond to blocks falling in water. Slide is animated so students can visual what is happening. A transparent overlay covers the block of wood and ice. As a teacher I point out that the wood is 25% submerged as it has a specific gravity of 0.25 while the ice is submerged 90% as it has a specific gravity of 0.90. I then show this actual demonstration to students using a document camera. …slide 71 has sound to correspond to cloud of CO2 falling in down stairs because it is denser than air. A voice explanation is also included to give a brief explanation. … slide 202-207 demonstrate the theory and show chromatography by animated clicks on PowerPoint I would give a verbal explanation in class to explain the slides. … slide 227-231 concern electrolysis slides 230 and 231 have an embedded video I that plays when clicked. Video came from PBS tv show. A voice explanation is also included to give a brief explanation …slide 238-261 explain properties of matter. A slide 238 and 241 have video to give additional information …Slide 255 are a series of jpeg images I captured from video I shot to sequence as a slide that looks like video. Slide 256 has the actual video of class demonstration I recorded. It plays regular speed and slow motion. Each was created using Windows Movie Maker. …slide 261 has an external link to an animation by a textbook author (bottom right corner) …Many slides have additional links to additional information (e.g. slide 8 (mouse over the word “matches “to get more information from Wikipedia) 2. www.unit5.org/chemistry Energy and Matter Unit 2 3. Guiding Questions Why do substances boil or freeze at different temperatures? Why do we put salt on the roads in the winter? Why does sweating cool us? What is energy? How do we measure energy? 4. Table of Contents ‘Matter and Energy’ (13) Introduction - Bonding (14) Temperature vs. Heat (11) Density (6) Carbon Dioxide & Monoxide (4) Archimede’s Principle (3) Galilean Thermometer (11) Golf Ball Lab (15) Solid, Liquid, and Gas (3) Heating Curve (13) Classification of Matter (6) Crystalline Structure (10) Allotropes (9) Alloys (4) Separation Techniques (11) Distillation (2) Centrifugation (3) Electrolysis (5) Properties of Matter (6) Energy (11) Exothermic vs. Endothermic (29) Calorimetry (12) Nuclear Energy 5. Lecture Outline – Energy and Matter Keys Lecture Outline – Energy and Matter Lecture Outline – Energy and Matter student notes outline textbook questions http://www.unit5.org/chemistry/Matter.html textbook questions text 6. Chemistry of Matches P4S3 + KClO3 P2O5 + KCl + SO2 tetraphosphorus trisulfide potassium chlorate diphosphorus pentaoxide potassium chloride sulfur dioxide D The substances P4S3 and KClO3 are both present on the tip of a strike anywhere match. When the match is struck on a rough surface, the two chemicals (reactants) ignite and produce a flame. Charles H.Corwin, Introductory Chemistry 2005, page 182 Safety matches The products from this reaction are P2O5, KCl, and SO2,the last of which is responsible for the characteristic sulfur smell. Strike anywhere matches The substances P4S3 and KClO3 are separated. The P4S3 is on the matchbox cover. Only when the chemicals combine do they react and produce a flame. 7. block of wood: length = 2.0 m width = 0.9 m height = 0.5 m block of wood: force = 45 N 2.205 pounds = 1 kilogram 10 Newton (9.8 N) 8. Force versus Pressure Area = 0.9 m x 2.0 m = 1.8 m2 Area = 0.5 m x 2.0 m = 1.0 m2 Area = 0.5 m x 0.9 m = 0.45 m2 area force Pressure  2 m1.8 N45.0 Pressure  2 m1.00 N45.0 Pressure  2 m0.45 N45.0 Pressure  block of wood: length = 2.0 m width = 0.9 m height = 0.5 m 25 N/m2 45 N/m2 100 N/m2 Herron, Frank, Sarquis, Sarquis, Schrader, Kulka, Chemistry, Heath Publishing,1996, page Section 6.1 9. Pressure area force pressure  Which shoes create the most pressure? 10. During a “physical change” a substance changes some physical property… H2O 11. …but it is still the same material with the same chemical composition. H2O gas solid liquid 12. Chemical Property: The tendency of a substance to change into another substance. caused by iron (Fe) reacting with oxygen (O2) to produce rust (Fe2O3) Steel rusting: 4 Fe + 3 O2 2 Fe2O3 13. Chemical Change: Any change involving a rearrangement of atoms. 14. Chemical Reaction: The process of a chemical change... 15. During a “chemical reaction” new materials are formed by a change in the way atoms are bonded together. 16. Physical and Chemical Properties Examples of Physical Properties Boiling point Color Slipperiness Electrical conductivity Melting point Taste Odor Dissolves in water Shininess (luster) Softness Ductility Viscosity (resistance to flow) Volatility Hardness Malleability Density (mass / volume ratio) Examples of Chemical Properties Burns in air Reacts with certain acids Decomposes when heated Explodes Reacts with certain metals Reacts with certain nonmetals Tarnishes Reacts with water Is toxic Ralph A. Burns, Fundamentals of Chemistry 1999, page 23 Chemical properties can ONLY be observed during a chemical reaction! 17. The formation of a mixture The formation of a compound 18. Physical & Chemical Changes Limestone, CaCO3 crushing PHYSICAL CHANGE Crushed limestone, CaCO3 heating CHEMICAL CHANGE Pyrex CO2 CaO Lime and carbon dioxide, CaO + CO2 19. Pyrex O2 H2O Pyrex H2O2 Light hastens the decomposition of hydrogen peroxide, H2O2. The dark bottle in which hydrogen peroxide is usually stored keeps out the light, thus protecting the H2O2 from decomposition. Sunlight energy H H O O 20. Three Possible Types of Bonds + - d+ d- Covalent e.g. H2 Polar Covalent e.g. HCl Ionic e.g. NaCl 21. Metallic Bonding Metallic bonding is the attraction between positive ions and surrounding freely mobile electrons. Most metals contribute more than one mobile electron per atom. “electron sea” e1- e1- e1- e1- e1- e1- e1- e1- e1- e1- e1- e1- e1- e1- e1- e1- Free electrons + + + + + ++ + + + + + + + Cations Bailar, Jr, Moeller, Kleinberg, Guss, Castellion, Metz, Chemistry, 1984, page 245 22. Shattering an Ionic Crystal; Bending a Metal Bailar, Jr, Moeller, Kleinberg, Guss, Castellion, Metz, Chemistry, 1984, page 248 + + ++ + ++ +++ + + + + + + + ++ + ++ +++ + + + + + ++ + + + ++ + + + + + + + + ++ + + + + + ++ + ++ +++ + + + + + + + ++ + ++ +++ + + + + + ++ + + + ++ + + + + + + + + ++ + + + +- + -- - + + - -+ - ++- + - - + + -+-- - -++ - + + - + - -+ + - +- + -- - + + - -+ - ++- + - - + + -+-- - -++ - + + - + - -+ + - An ionic crystal A metal No electrostatic forces of repulsion – metal is deformed (malleable) Electrostatic forces of repulsion Force Force broken crystal 23. Properties of Ionic Compounds • Crystalline solids • Hard and brittle • High melting points • High boiling points • High heats of vaporization • High heats of fusion • Good conductors of electricity when molten • Poor conductors of heat and electricity when solid • Many are soluble in water 24. Chemical Bonds Increasing ionic character Covalent bonding Electrons are shared equally Cl Cl Polar covalent bonding Electrons are shared unequally ClH Ionic bonding Electrons are transferred Cl1-Na1+ Ralph A. Burns, Fundamentals of Chemistry 1999, page 229 • between two identical nonmetal atoms are non-polar covalent. • between two different nonmetal atoms are polar covalent. • between nonmetals and reactive metals are primarily ionic. 25. Chemical Bonds Increasing ionic character Nonpolar covalent Electrons are shared equally Cl Cl Polar covalent Electrons are shared unequally ClH Ionic bonding Electrons are transferred Cl1-Na1+ Ralph A. Burns, Fundamentals of Chemistry 1999, page 229 • between two identical nonmetal atoms are nonpolar covalent. • between two different nonmetal atoms are polar covalent. • between nonmetals and reactive metals are primarily ionic. 26. Covalent vs. Ionic Covalent Transfer electrons (ions formed) + / - Between Metal and Nonmetal Strong Bonds (high melting point) Share electrons (polar vs. nonpolar) Between Two Nonmetals Weak Bonds (low melting point) Alike Different Electrons are involved Chemical Bonds Ionic Different Topic Topic 27. Photoelectric Generator Solar Calculator Radiant energy Evacuated chamber Metal surface Current indicator Positive terminal Voltage source cathode anode Symbolic representation of a photoelectric cell cathode anode evacuated glass envelope Photoelectric Cell 28. Celsius & Kelvin Temperature Scales Boiling point of water Freezing point of water Absolute zero Celsius 100 Celsius degrees 100oC 0oC -273oC Kelvin 100 Kelvins 373 K 273 K 0 K 29. Temperature is Average Kinetic Energy Fast Slow “HOT” “COLD” Kinetic Energy (KE) = ½ m v2 *Vector = gives direction and magnitude 30. Temperature Scales Fahrenheit 212 oF 180 oF 32 oF Celcius 100 oC 100 oC 0 oC Kelvin 373 K 100 K 273 K Boiling point of water Freezing point of water Notice that 1 kelvin degree = 1 degree Celcius 31. Kelvin Scale blue white yellow 4300 K PIAA HID Bulb 5000 K PIAA Plasma Blue 5250 K Sunlight 4150 K PIAA Xtreme White 3800 K PIAA Super White 3200 K Halogen Bulb 2600 K Incandescent Bulb 32. Temperature Scales Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 136 33. Compare Celsius to Fahrenheit oF – 32 = 1.8 oC Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 139 34. Converting 70 degrees Celsius to Kelvin units. oC + 273 = K Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 137 35. Temperature Scales • Temperature can be subjective and so fixed scales had to be introduced. • The boiling point and freezing point of water are two such points. • Celsius scale (oC) – The Celsius scale divides the range from freezing to boiling into 100 divisions. – Original scale had freezing as 100 and boiling as 0. – Today freezing is 0 oC and boiling is 100 oC. • Fahrenheit scale (oF) • Mercury and alcohol thermometers rely on thermal expansion 36. Thermal Expansion • Most objects e-x-p-a-n-d when heated • Large structures such as bridges must be built to leave room for thermal expansion • All features expand together COLD HOT Cracks in sidewalk. 37. Equal Masses of Hot and Cold Water Thin metal wall Insulated box Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291 38. Water Molecules in Hot and Cold Water Hot water Cold Water 90 oC 10 oC Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291 39. Water Molecules in the same temperature water Water (50 oC) Water (50 oC)Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291 40. Heat versus Temperature Kinetic energy Fractionsofparticles lower temperature higher temperature TOTAL Kinetic ENERGY = Heat 41. Molecular Velocities speed Fractionsofparticles many different molecular speeds molecules sorted by speed the Maxwell speed distribution http://antoine.frostburg.edu/chem/senese/101/gases/slides/sld016.htm 42. Temperature vs. Heat Measured with a Calorimeter Total Kinetic Energy Joules (calories) Measured with a Thermometer Average Kinetic Energy oCelcius (or Kelvin) Alike Different A Property of Matter Have Kinetic Energy Heat Different Topic Topic Temperature 43. Conservation of Matter Reactants yield Products 44. Heavy Metal Poisoning Exposure to mercury made the Hatter “mad”. Eating chips of lead paint causes brain damage. TREATMENT: Chelation therapy EDTA (ethylenediamine tetra acetic acid) Arsenic treated lumber. ‘Green-treated’ wood will not rot outdoors for 50 years. 45. Density • Density is an INTENSIVE property of matter. - does NOT depend on quantity of matter. - color, melting point, boiling point, odor, density • Contrast with EXTENSIVE - depends on quantity of matter. - mass, volume, heat content (calories) Styrofoam Brick 46. Properties of Matter http://antoine.frostburg.edu/chem/senese/101/matter/slides/sld001.htm Pyrex Pyrex Extensive Properties Intensive Properties volume: mass: density: temperature: 100 mL 99.9347 g 0.999 g/mL 20oC 15 mL 14.9902 g 0.999 g/mL 20oC 47. Styrofoam Brick ? It appears that the brick is ~40x more dense than the Styrofoam. 48. MM V = =D V D BrickStyrofoam Styrofoam Brick 49. Which liquid has the highest density? 52 3 1 4 Coussement, DeSchepper, et al. , Brain Strains Power Puzzles 2002, page 16 least dense 1 < 3 < 5 < 2 < 4 most dense 50. Cube Representations 1 m3 = 1 000 000 cm3 Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 119 51. Volume and Density Relationship Between Volume and Density for Identical Masses of Common Substances Cube of substance Mass Volume Density Substance (face shown actual size) (g) (cm3) (g/cm3) Lithium Water Aluminum Lead 10 19 0.53 10 10 1.0 10 3.7 2.7 10 0.58 11.4 52. Density D M Vensity ass olume D = M V M = D x V V = M D 53. Volume Dorin, Demmin, Gabel, Chemistry The Study of Matter 3rd Edition, page 41 6 cm 3 cm 2 cm 1 cm 4 cm4 cm 54. 2 cm 2 cm 2 cm Volume Dorin, Demmin, Gabel, Chemistry The Study of Matter 3rd Edition, page 41 V = length x width x heightV = 2 cm x 2 cm x 2 cmV = 8 cm3 Volume = length x width x heightVolume = 6 cm x 2 cm x 3 cm 6 cm 3 cm 2 cm 1 cm 4 cm Volume = 36 cm3 Volume = Volume = 28 cm3 36 cm3 8 cm3 - 55. Density of Some Common Substances Density of Some Common Substance Substance Density (g / cm3) Air 0.0013* Lithium 0.53 Ice 0.917 Water 1.00 Aluminum 2.70 Iron 7.86 Lead 11.4 Gold 19.3 *at 0oC and 1 atm pressure 56. Consider Equal Volumes The more massive object (the gold cube) has the _________ density. Equal volumes… …but unequal masses aluminum gold GREATER Density = Mass Volume Dorin, Demmin, Gabel, Chemistry The Study of Matter , 3rd Edition, 1990, page 71 57. Consider Equal Masses Equal masses… …but unequal volumes. The object with the larger volume (aluminum cube) has the density. aluminum gold Dorin, Demmin, Gabel, Chemistry The Study of Matter , 3rd Edition, 1990, page 71 smaller Christopherson Scales Made in Normal, Illinois USA 58. Two ways of viewing density Dorin, Demmin, Gabel, Chemistry The Study of Matter , 3rd Edition, 1990, page 71 Equal volumes… …but unequal masses The more massive object (the gold cube) has the greater density. aluminum gold (A) Equal masses… …but unequal volumes. (B) gold aluminum The object with the larger volume (aluminum cube) has the smaller density. 59. Carbon Dioxide Detector • Where is the best location to place a CO2 detector in your home? Recall: Density Air = 1.29 g/L Density CO2 = 1.96 g/L A. Top floor of home B. Basement (near ceiling) C. Basement (near floor) D. It doesn’t matter, if your batteries are dead in the detector C. Basement (near floor) Carbon dioxide is denser than air and sinks. 60. Symptoms of CO Poisoning Concentration of CO in air (ppm)* Hemoglobin molecules as HbCO Visible effects 100 for 1 hour or less 10% or less no visible symptoms 500 for 1 hour or less 20% mild to throbbing headache, some dizziness, impaired perception 500 for an extended period of time 30 - 50% headache, confusion, nausea, dizziness, muscular weakness, fainting 1000 for 1 hour or less 50 - 80% coma, convulsions, respiratory failure, death *ppm is parts per million Davis, Metcalfe, Williams, Castka, Modern Chemistry, 1999, page 760 61. Carbon Monoxide Poisoning ‘The Silent Killer’ Poisoning: Hb + CO  HbCO Hemoglobin (Hb) binds with carbon monoxide (CO) in the capillaries of the lungs. If caught in time, giving pure oxygen (O2) revives victim of CO poisoning. Treatment causes carboxyhemoglobin (HbCO) to be converted slowly to oxyhemoglobin (HbO2). Treatment: O2 + HbCO  CO + HbO2 Carbon monoxide, CO, has almost 200 times the affinity to bind with hemoglobin, Hb, in the blood as does oxygen, O2. Davis, Metcalfe, Williams, Castka, Modern Chemistry, 1999, page 760 62. Exchange of Blood Gases 63. Tank of Water Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 143 64. Person Submerged in Water Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 143 65. Galilean Thermometer • Density = Mass / Volume • Mass is constant • Volume changes with temperature – Increase temperature  larger volume In the Galilean thermometer, the small glass bulbs are partly filled with a different (colored) liquid. Each is filled with a slightly different amount, ranging from lightest at the uppermost bulb to heaviest at the lowermost bulb. The clear liquid in which the bulbs are submerged is not water, but some inert hydrocarbon (probably chosen because its density varies with temperature more than that of water does). Temp = 68 oC 66. Galilean Thermometer In the Galilean thermometer, the small glass bulbs are partly filled with a different (colored) liquid. Each is filled with a slightly different amount, ranging from lightest at the uppermost bulb to heaviest at the lowermost bulb. The clear liquid in which the bulbs are submerged is not water, but some inert hydrocarbon (probably chosen because its density varies with temperature more than that of water does). The correct temperature is the lowest floating bulb. As temperature increases, density of the clear medium decreases (and bulbs sink). RECALL: Density equals mass / volume. 80o 76o 80o 76o 72o 64o 68o 64o 68o 72o 76oF 68 oF 67. Dissolving of Salt in Water NaCl(s) + H2O  Na+(aq) + Cl-(aq) Cl- ions Na+ ions Water molecules 68. Determine the minimum amount of salt needed to make a golf ball float in 100 mL water. Weigh out 50.0 g of NaCl Trial Salt (g) Total Float /Sink 1 5.0 g 5.0 g Sink 2 5.0 g 10.0 g Sink 3 5.0 g 15.0 g Sink 4 5.0 g 20.0 g Sink 5 5.0 g 25.0 g Float Add 5 g additions of salt to the water, dissolve, check to see if ball floats. Continue with this method of successive additions until ball floats. Re-weigh remaining salt and subtract this amount from 50.0 g to determine the amount of salt needed. Finally, repeat…begin 5 g less salt and add 1 g increments to narrow range. 69. Theorize, but Verify Jaffe, New World of Chemistry, 1955, page 1 …We must trust in nothing but facts. These are presented to us by nature and cannot deceive. We ought in every instance to submit our reasoning to the test of experiment. It is especially necessary to guard against the extravagances of imagination which incline to step beyond the bounds of truth. Antoine Laurent Lavoisier, 1743 - 1794 70. Theory Guides, Laboratory Decides! Density of water = 1.0 g/mL Need to determine density of a golf ball. mass =______ g (electronic balance) volume = ______ mL (water displacement method) or formula? Density of golf ball cannot be made to decrease. Therefore, you need to increase the density of the water by dissolving salt into the water. Limiting Factor: accurate determination of volume of golf ball Solubility Curve of salt in water. Water has a limit to how much salt can be dissolved. Saturation – point at which the solution is full and cannot hold anymore solute. 71. Packing of NaCl Ions Electron Microscope Photograph of NaCl 72. Dissolving of Salt in Water NaCl(s) + H2O  Na+(aq) + Cl-(aq) Cl- ions Na+ ions Water molecules 73. Dissolving of Salt in Water NaCl(s) + H2O  Na+(aq) + Cl-(aq) Cl- ions Na+ ions Water molecules 74. Dissolving of Salt in Water NaCl(s) + H2O  Na+(aq) + Cl-(aq) Cl- ions Na+ ions Water molecules 75. Copyright 2007 Pearson Benjamin Cummings. All rights reserved. 76. Dissolving of NaCl Timberlake, Chemistry 7th Edition, page 287 HH O Na+ + - - + - + + - Cl- + - + hydrated ions 77. 100 mL Interstitial Spaces and Particle Size Interstitial spaces (holes in water where substances dissolve) Parking at school if you arrive at 7:00 AM = _____ Parking at school if you arrive at 7:45 AM = _____ More available spaces if you arrive early. Salt dissolves quicker when you begin because there are more available spaces to 'park'. Analogy: Compact car is easier to park than SUV. STIR Easy Hard Theory: Crush salt to make particles smaller (increase surface area) …it will dissolve more rapidly. 78. 100 mL of water = 100 g Add 3.0 mL water,stir…float You determine the density of golf ball to be 1.18 g/mL density of water= 1.00 g/mL Add 19 g salt to 100 g water = 119 g salt + water Volume remains100 mL (saltwater) Density = Mass volume 119 g 100 mLor Density (saltwater) = 1.19 g/mL If golf ball doesn’t float, add 2 mL additions of salt until it floats. Add 3.0 mL water,stir…float Add 3.0 mL water,stir…sink mL100 saltgx mL6mL100 g119   79. Goals and Objectives: a. Given materials and problem, formulate and test a hypothesis to determine if a golf ball can float in salt water. b. Collect accurate data and compare own data to other class data. Evaluate own results. Investigation Procedure: a. Design an experiment to accurately determine how dense salt water must be in order for a golf ball to float. Use metric units. Be sure to control as many variables as possible. b. Write down the procedure that you and your partner(s) are going to use prior to lab day. Record any researched facts that may be useful in knowing before conducting your experiment. c. Carefully run your experiment, make observations and record your measurements in a data table. Use grams and milliliters in your measurements. Include a calculation column in your data table. d. Critique your own procedure, discuss and compare your process with another group, then modify your own steps as needed. e. Repeat your experiment to check for accuracy, if time allows. 80. Discussion Questions for Understanding: a. How did you determine the density of your golf ball? b. Why does a golf ball normally sink to the bottom of a pond at the golf course? c. What variables were difficult or impossible for you to control during this experiment? How much salt can be dissolved in 100 mL of water? (saturated) effect of temperature on solubility Surface area of salt may affect rate of dissolving (may need to crush salt finely) d. What variables may have changed as time went on that could have affected the outcome of your results? e. Did you improve the accuracy of your results after conferring with another group? f. Describe your sources of error. (Human error and faulty equipment are unacceptable answers) 81. Materials: electronic balance 100 mL & 500 mL graduated cylinder mortar / pestle glass stirring rod golf ball salt (Kosher, iodized table salt, table salt) 250 mL beaker Extension: a. Research the manufacturing of golf balls to determine why they sink in pond water. b. Research to determine which body of salt water in the world would float a golf ball the highest. Lab Report : (10 - 12 point font two page maximum length) Background / problem Hypothesis (if...then) Procedure (protocol) Data (table, graph) Analysis Conclusions / Future directions (limitations) Sample calculations - Appendix Do not use references to yourself or others in your writing of a lab report (except for citing past research). OR Poster (25 words or less) A picture is worth 1000 words! 82. Solid, Liquid, Gas (a) Particles in solid (b) Particles in liquid (c) Particles in gas 83. Solid H2O(s) Ice Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 31 84. Ice H2O(s) Ice Photograph of ice model Photograph of snowflakes Copyright © 2007 Pearson Benjamin Cummings. All rights reserved. 85. Liquid H2O(l) Water Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 31 In a liquid • molecules are in constant motion • there are appreciable intermolecular forces • molecules are close together • Liquids are almost incompressible • Liquids do not fill the container 86. Gas H2O(g) Steam Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 31 87. Liquids The two key properties we need to describe are EVAPORATION and its opposite CONDENSATION add energy and break intermolecular bonds EVAPORATION release energy and form intermolecular bonds CONDENSATION 88. States of Matter 89. Gas, Liquid, and Solid Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 441 Gas Liquid Solid 90. States of Matter Solid Liquid Gas Holds Shape Fixed Volume Shape of Container Free Surface Fixed Volume Shape of Container Volume of Container heat heat 91. Some Properties of Solids, Liquids, and Gases Property Solid Liquid Gas Shape Has definite shape Takes the shape of Takes the shape the container of its container Volume Has a definite volume Has a definite volume Fills the volume of the container Arrangement of Fixed, very close Random, close Random, far apart Particles Interactions between Very strong Strong Essentially none particles 92. • To evaporate, molecules must have sufficient energy to break IM forces. • Molecules at the surface break away and become gas. • Only those with enough KE escape. • Breaking IM forces requires energy. The process of evaporation is endothermic. • Evaporation is a cooling process. • It requires heat. Evaporation 93. Change from gas to liquid Achieves a dynamic equilibrium with vaporization in a closed system. What is a closed system? A closed system means matter can’t go in or out. (put a cork in it) What the heck is a “dynamic equilibrium?” Condensation 94. When first sealed, the molecules gradually escape the surface of the liquid. As the molecules build up above the liquid - some condense back to a liquid. The rate at which the molecules evaporate and condense are equal. Dynamic Equilibrium 95. As time goes by the rate of vaporization remains constant but the rate of condensation increases because there are more molecules to condense. Equilibrium is reached when: Rate of Vaporization = Rate of Condensation Molecules are constantly changing phase “dynamic” The total amount of liquid and vapor remains constant “equilibrium” Dynamic Equilibrium 96. • Vaporization is an endothermic process - it requires heat. • Energy is required to overcome intermolecular forces • Responsible for cool earth • Why we sweat Vaporization 97. Energy Changes Accompanying Phase Changes Solid Liquid Gas Melting Freezing Deposition CondensationVaporization Sublimation Energyofsystem Brown, LeMay, Bursten, Chemistry 2000, page 405 98. solid liquid gas Heat added Temperature(oC) A B C D E Heating Curve for Water 0 100 LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 487 99. solid liquid gas vaporization condensation melting freezing Heat added Temperature(oC) A B C D E Heating Curve for Water 0 100 LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 487 100. Latent Heat • Take 1 kg of water from –10 oC up to 150 oC we can plot temperature rise against absorbed heat water steam (water vapor) -10 C 0 C 100 C ice Lf = 80 cal/g Lv = 540 cal/g Lf is the latent heat of fusion Lv is the latent heat of vaporization Q heat absorbed 101. MATTER Can it be physically separated? Homogeneous Mixture (solution) Heterogeneous Mixture Compound Element MIXTURE PURE SUBSTANCE yes no Can it be chemically decomposed? noyesIs the composition uniform? noyes Colloids Suspensions Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 102. Elements only one kind of atom; atoms are bonded it the element is diatomic or polyatomic Compounds two or more kinds of atoms that are bonded substance with definite makeup and properties Mixtures two or more substances that are physically mixed two or more kinds of and Both elements and compounds have a definite makeup and definite properties. Packard, Jacobs, Marshall, Chemistry Pearson AGS Globe, page (Figure 2.4.1) 103. Matter Flowchart Examples: – graphite – pepper – sugar (sucrose) – paint – soda Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem element hetero. mixture compound solution homo. mixture hetero. mixture 104. Pure Substances Element – composed of identical atoms – EX: copper wire, aluminum foil Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 105. Pure Substances Compound – composed of 2 or more elements in a fixed ratio – properties differ from those of individual elements – EX: table salt (NaCl) Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 106. Pure Substances Law of Definite Composition – A given compound always contains the same, fixed ratio of elements. Law of Multiple Proportions – Elements can combine in different ratios to form different compounds. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 107. Pure Substances For example… Two different compounds, each has a definite composition. Carbon, C Oxygen, O Carbon monoxide, CO Carbon, C Oxygen, O Oxygen, O Carbon dioxide, CO2 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 108. Mixtures Variable combination of two or more pure substances. Heterogeneous Homogeneous Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 109. Mixtures Solution – homogeneous – very small particles – no Tyndall effect Tyndall Effect – particles don’t settle – EX: rubbing alcohol Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 110. Mixtures Colloid – heterogeneous – medium-sized particles – Tyndall effect – particles don’t settle – EX: milk Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 111. Mixtures Suspension – heterogeneous – large particles – Tyndall effect – particles settle – EX: fresh-squeezed lemonade Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 112. Mixtures Examples: – mayonnaise – muddy water – fog – saltwater – Italian salad dressing Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem colloid suspension colloid solution suspension 113. Classification of Matter Materials Homogeneous Heterogeneous Heterogeneous mixture Homogeneous mixture Substance Element Compound Solution Mixture Order / Disorder Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 43 114. Classification of Matter MATTER (gas. Liquid, solid, plasma) PURE SUBSTANCES MIXTURES HETEROGENEOUS MIXTURE HOMOGENEOUS MIXTURES ELEMENTSCOMPOUNDS Separated by physical means into Separated by chemical means into Kotz & Treichel, Chemistry & Chemical Reactivity, 3rd Edition , 1996, page 31 115. Classification of Matter uniform properties? fixed composition? chemically decomposable? no no no yes hetero- geneous mixture solution element compound http://antoine.frostburg.edu/chem/senese/101/matter/slides/sld003.htm 116. Elements, Compounds, and Mixtures (a) an element (hydrogen) (b) a compound (water) (c) a mixture (hydrogen and oxygen) (d) a mixture (hydrogen and oxygen) Dorin, Demmin, Gabel, Chemistry The Study of Matter , 3rd Edition, 1990, page 68 hydrogen atoms hydrogen atoms oxygen atoms 117. Elements, Compounds, and Mixtures (a) an element (hydrogen) (b) a compound (water) (c) a mixture (hydrogen and oxygen) (d) a mixture (hydrogen and oxygen) Dorin, Demmin, Gabel, Chemistry The Study of Matter , 3rd Edition, 1990, page 68 hydrogen atoms hydrogen atoms oxygen atoms 118. Mixture vs. Compound Mixture Fixed Composition Bonds between components Can ONLY be separated by chemical means Variable Composition No bonds between components Can be separated by physical means Alike Different Contain two or more elements Can be separated into elements Involve substances Compound Different Topic Topic 119. Compounds vs. Mixtures • Compounds have properties that are uniquely different from the elements from which they are made. – A formula can always be written for a compound – e.g. NaCl  Na + Cl2 • Mixtures retain their individual properties. – e.g. Salt water is salty and wet 120. Diatomic Elements, 1 and 7 H2 N2 O2 F2 Cl2 Br2 F2 121. Products made from Sulfur Magazines and printing papers Writing and fine papers Wrapping and bag papers Sanitary and tissue papers Absorbent papers Rayon Cellophane Carbon Tetrachloride Ruber processing chemicals Containers and boxes Newsprint Pulp for rayon and film PULP 3% NONACID 12% Insecticides Fungicides Rubber vulcanizing Soil sulfur Specialty steels Magnessium Leather processing Photography Dyestuffs Bleaching Soybean extraction Aluminum reduction Paper sizing Water treatment Pharmaceuticals Insecticides Antifreeze Superphosphates Ammonium phosphate Ammonium sulfate Mixed fertilizers Autos Appliances Tin and other containers Galvanized products Explosives Nonferrous metals Synthetic rubber Storage batteries Textile finishing Tire cords Viscose textiles Acetate textiles Blended fabrics Cellophane Photographic film Paints and enamels Linoleum and coated fabrics Paper Printing inks Aviation Gasoline Lubricants Other Refinery products SULFURIC ACID 88% CARBON DISULFIDE 3% GROUND & DEFINED 3% IRON & STEEL 1% PETROLEUM 2% CHEMICAL 17% Synthetic detergents Feed additives Anti-knock gasoline Synthetic resins Protective coating Dyestuffs Oil well acidizing Petroleum catalysts 122. • Rhombic sulfur – “Brimstone” (when molten) – Polyatomic (S8) – Forms SO2 Amorphous sulfur – (without shape) Sulfur The sudden cooling of m-sulfur produces amorphous sulfur. 123. Amorphous (Glass) Crystalline 124. The Haber Process 125. Matter Substance Definite composition (homogeneous) Element (Examples: iron, sulfur, carbon, hydrogen, oxygen, silver) Mixture of Substances Variable composition Compound (Examples: water. iron (II) sulfide, methane, Aluminum silicate) Homogeneous mixture Uniform throughout, also called a solution (Examples: air, tap water, gold alloy) Heterogeneous mixture Nonuniform distinct phases (Examples: soup, concrete, granite) Chemically separable Physically separable 126. The Organization of Matter MATTER PURE SUBSTANCES HETEROGENEOUS MIXTURE HOMOGENEOUS MIXTURES ELEMENTS COMPOUNDS Physical methods Chemical methods Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 41 127. Top Ten Elements in the Universe Percent Element (by atoms) 1. Hydrogen 73.9 2. Helium 24.0 3. Oxygen 1.1 4. Carbon 0.46 5. Neon 0.13 6. Iron 0.11 7. Nitrogen 0.097 8. Silicon 0.065 9. Magnesium 0.058 10.Sulfur 0.044 A typical spiral galaxy (Milky Way is a spiral galaxy) Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 26 128. The Composition of Air Air Nitrogen OxygenHelium Water vapor Neon Carbon dioxide Argon Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 34 129. Chart Examining Some Components of Air Nitrogen consists of molecules consisting of two atoms of nitrogen: Oxygen consists of molecules consisting of two atoms of oxygen: Water consists of molecules consisting of two hydrogen atoms and one oxygen atom: Argon consists of individual argon atoms: Carbon dioxide consists of molecules consisting of two oxygen atoms and one carbon atom: Neon consists of individual neon atoms: Helium consists of individual helium atoms: N2 O2 H2O Ar CO2 Ne He Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 35 130. Reviewing Concepts Classifying Matter • Why does every sample of a given substance have the same properties? • Explain why the composition of an element is fixed. • Describe the composition of a compound. • Why can the properties of a mixture vary? • On what basis can mixtures be classified as solutions, suspensions, or colloids? 131. Unit Cells The kind of symmetry found throughout a crystalline substance is determined by the type of unit cell which generates the lattice structure. Simple cubic Body-centered cubic Face-centered cubic Monoclinic HexagonalTetragonal 132. Simple cubic Body-centered cubic Face-centered cubic 133. Phosphorous (P4) TWO ALLOTROPIC FORMS White phosphorous spontaneously ignites Red phosphorous used for matches Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 457 134. Sodium Chloride Crystal Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 455 = Cl- = Na+ 135. Packing of NaCl Ions Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 456 = Na1+ = Cl1- 136. Packing of NaCl Ions Electron Microscope Photograph of NaCl 137. Molecular Structure of Ice Hydrogen bonding Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 455 138. Dry Ice – Carbon Dioxide 139. Allotropes of Carbon Graphite BuckminsterfullereneDiamond Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 27 140. Allotropes of Carbon Graphite 141. Copyright © 2007 Pearson Benjamin Cummings. All rights reserved. Graphite 142. Allotropes of Carbon Diamond Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 455 143. Diamond 144. Diamonds in Garage • Made gem quality diamonds by burning wood • Scholarship Molecular structure of diamond LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 476 145. Molecular structure of Diamond 146. Allotropes of Carbon C60 & C70 “Buckyballs” “Buckytubes” Buckminsterfullerene 147. Credit: Baughman et al., Science 297, 787 (2002) 148. Trojan Horse Can use ‘camouflage’ to hide things. Be careful what’s in the Trojan! Buckyballs can hide medicine to treat the human body. 149. Gold 24 karat gold 18 karat gold 14 karat gold Gold Copper Silver 18/24 atoms Au24/24 atoms Au 14/24 atoms Au 150. Solid Brass An alloy is a mixture of metals. • Brass = Copper + Zinc • Solid brass • homogeneous mixture • a substitutional alloy Copper Zinc 151. Brass Plated • Brass = Copper + Zinc • Brass plated • heterogeneous mixture • Only brass on outside Copper Zinc 152. Steel Alloys • Stainless steel • Tungsten hardened steel • Vanadium steel • We can engineer properties – Add carbon to increase strength – Too much carbon  too brittle and snaps – Too little carbon  too ductile and iron bends Tensilestrength Force is added 153. Galvanized Nails and Screws • Zinc coating prevents rust – Use deck screws for any outdoor project • Iron will rust if untreated – Weaken and break 154. Nitinol Wire • Alloy of nickel and titanium • Remembers shape when heated Applications: surgery, shirts that do not need to be ironed. 155. Properties of Matter • Electrical Conductivity • Heat Conductivity • Density • Melting Point • Boiling Point • Malleability • Ductility 156. Methods of Separating Mixtures • Magnet • Filter • Decant • Evaporation • Centrifuge • Chromatography • Distillation 157. Filtration separates a liquid from a solid Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 40 Mixture of solid and liquid Stirring rod Filtrate (liquid component of the mixture) Filter paper traps solid Funnel 158. Chromatography • Tie-dye t-shirt • Black pen ink • DNA testing – Tomb of Unknown Soldiers – Crime scene – Paternity testing 159. Paper Chromatography 160. Paper Chromatography of Water-Soluble Dyes orange red yellow Initial spots of dyes Direction of Water (mobile phase) movement Filter paper (stationary phase) Orange mixture of red and yellow Suggested red dye is not homogeneous 161. Separation by Chromatography sample mixture a chromatographic column stationary phase selectively absorbs components mobile phase sweeps sample down column detector http://antoine.frostburg.edu/chem/senese/101/matter/slides/sld006.htm 162. Separation by Chromatography sample mixture a chromatographic column stationary phase selectively absorbs components mobile phase sweeps sample down column detector http://antoine.frostburg.edu/chem/senese/101/matter/slides/sld006.htm 163. Ion chromatogram of orange juice time (minutes) detector response 0 5 10 15 20 25 Na+ K+ Mg2+ Fe3+ Ca2+ 164. Setup to heat a solution Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 42 Ring stand Beaker Wire gauze Ring Bunsen burner 165. long spout helps vapors to condense mixture for distillation placed in here Furnace Glass retort Glass Retort Eyewitness Science “Chemistry” , Dr. Ann Newmark, DK Publishing, Inc., 1993, pg 13 166. A Distillation Apparatus liquid with a solid dissolved in it thermometer condenser tube distilling flask pure liquid receiving flaskhose connected to cold water faucetDorin, Demmin, Gabel, Chemistry The Study of Matter , 3rd Edition, 1990, page 282 167. The solution is boiled and steam is driven off. Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 39 168. Salt remains after all water is boiled off. Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 39 169. No chemical change occurs when salt water is distilled. Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 40 Saltwater solution (homogeneous mixture) Distillation (physical method) Salt Pure water 170. Separation of a sand-saltwater mixture. Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 40 171. Separation of Sand from Salt 1. Gently break up your salt-crusted sand with a plastic spoon. Follow this flowchart to make a complete separation. Salt- crusted sand. Dry sand. Wet sand. Weigh the mixture. Decant clear liquid. Evaporate to dryness. Pour into heat-resistant container. Fill with water. Stir and let settle 1 minute. Weigh sand. Calculate weight of salt. Repeat 3 times? Yes No 2. How does this flow chart insure a complete separation? 172. Four-stroke Internal Combustion Engine 173. Different Types of Fuel Combustion 2 C8H18 + 25 O2  16 CO2 + 18 H2O __CH3OH +__O2 __CO2 +__H2O Methanol (in racing fuel) Gasoline (octane) 174. Combustion Chamber -The combustion chamber is the area where compression and combustion take place. -Gasoline and air must be mixed in the correct ratio. 175. •Methanol can run at much higher compression ratios, meaning that you can get more power from the engine on each piston stroke. •Methanol provides significant cooling when it evaporates in the cylinder, helping to keep the high-revving, high- compression engine from overheating. •Methanol, unlike gasoline, can be extinguished with water if there is a fire. This is an important safety feature. •The ignition temperature for methanol (the temperature at which it starts burning) is much higher than that for gasoline, so the risk of an accidental fire is lower. The Advantages of Methanol - Burning Engines 176. •At 900 hp, it has about two to three times the horsepower of a "high- performance" automotive engine. For example, Corvettes or Vipers might have 350- to 400-horsepower engines. •At 15,000 rpm, it runs at about twice the rpm of a normal automotive engine. Compared to a normal engine, an methanol engine has larger pistons and the pistons travel a shorter distance up and down on each stroke. •The motor is lighter. This lowers their inertia and is another factor in the high rpm. A Race Car - Basic Information 177. Centrifugation • Spin sample very rapidly: denser materials go to bottom (outside) • Separate blood into serum and plasma – Serum (clear) – Plasma (contains red blood cells ‘RBCs’) • Check for anemia (lack of iron) Blood RBC’s Serum A B C AFTER Before 178. Water Molecules Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 8 179. The decomposition of two water molecules. 2 H2O  O2 + 2 H2 Electric current Water molecules Diatomic Diatomic oxygen molecule hydrogen molecules+ 180. Electrolysis *Must add acid catalyst to conduct electricity *H1+ water oxygen hydrogen “electro” = electricity “lysis” = to split Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 32 Water Hydrogen gas forms Oxygen gas forms ElectrodeSource of direct current H2O(l) O2 (g) + 2 H2 (g) 181. Electrolysis of Water Half reaction at the cathode (reduction): 4 H2O + 4 e -  2 H2 + 4 OH 1- Half reaction at the anode (oxidation): 2 H2O  O2 + 4 H 1+ + 4 e - hydrogen gas cathode oxygen gas anode D.C. power source water 182. Reviewing Concepts Physical Properties • List seven examples of physical properties. • Describe three uses of physical properties. • Name two processes that are used to separate mixtures. • When you describe a liquid as thick, are you saying that it has a high or low viscosity? 183. Reviewing Concepts Physical Properties • Explain why sharpening a pencil is an example of a physical change. • What allows a mixture to be separated by distillation? 184. Reviewing Concepts Chemical Properties • Under what conditions can chemical properties be observed? • List three common types of evidence for a chemical change. • How do chemical changes differ from physical changes? 185. Reviewing Concepts Chemical Properties • Explain why the rusting of an iron bar decreases the strength of the bar. • A pat of butter melts and then burns in a hot frying pan. Which of these changes is physical and which is chemical? 186. 2 H2 O2 2 H2O+ + E Copyright © 2007 Pearson Benjamin Cummings. All rights reserved. 187. The Zeppelin LZ 129 Hindenburg catching fire on May 6, 1937 at Lakehurst Naval Air Station in New Jersey. 188. S.S. Hindenburg 35 people died when the Hindenburg exploded. May 1937 at Lakehurst, New Jersey • German zeppelin luxury liner • Exploded on maiden voyage • Filled with hydrogen gas 189. Hydrogen is the most effective buoyant gas, but is it highly flammable. The disastrous fire in the Hindenburg, a hydrogen-filled dirigible, in 1937 led to the replacement of hydrogen by nonflammable helium. 190. Erosion Takes a Powder TreatedUntreated 191. Sodium Polyacrylate • Absorbent Material – Absorbs 700x volume of water • Magicians – Pour water in hat and it “disappears” • Diapers • Farmers – Anti-erosion powder • Add to Soils – hold moisture between watering 192. Specific Heats of Some Substances Specific Heat Substance (cal/ g oC) (J/g oC) Water 1.00 4.18 Alcohol 0.58 2.4 Wood 0.42 1.8 Aluminum 0.22 0.90 Sand 0.19 0.79 Iron 0.11 0.46 Copper 0.093 0.39 Silver 0.057 0.24 Gold 0.031 0.13 193. Copyright © 2007 Pearson Benjamin Cummings. All rights reserved. (a) Radiant energy (b) Thermal energy (c) Chemical energy (d) Nuclear energy (e) Electrical energy 194. The energy something possesses due to its motion, depending on mass and velocity. Potential energy Energy in Energy out kinetic energy kinetic energy 195. School Bus or Bullet? Which has more kinetic energy; a slow moving school bus or a fast moving bullet? Recall: KE = ½ m v2 KE = ½ m v2 KE = ½ m v2 BUS BULLET KE(bus) = ½ (10,000 lbs) (0.5 mph)2 KE(bullet) = ½ (0.002 lbs) (240 mph)2 Either may have more KE, it depends on the mass of the bus and the velocity of the bullet. Which is a more important factor: mass or velocity? Why? (Velocity)2 196. Kinetic Energy and Reaction Rate Kinetic energy Fractionsofparticles lower temperature higher temperature minimum energy for reaction 197. Kinetic Energy and Reaction Rate Kinetic energy Fractionsofparticles lower temperature higher temperature minimum energy for reaction 198. Hot vs. Cold Tea Kinetic energy Many molecules have an intermediate kinetic energy Few molecules have a very high kinetic energy Low temperature (iced tea) High temperature (hot tea) Percentofmolecules 199. Decomposition of Nitrogen Triiodide 200. Decomposition of Nitrogen Triiodide 2 NI3(s) N2(g) + 3 I2(g) NI3 I2 N2 201. Exothermic Reaction Reactants  Products + Energy 10 energy = 8 energy + 2 energy Reactants Products -DH Energy Energy of reactants Energy of products Reaction Progress 202. Endothermic Reaction Energy + Reactants  Products +DH Endothermic Reaction progress Energy Reactants ProductsActivation Energy 203. Effect of Catalyst on Reaction Rate reactants products Energy activation energy for catalyzed reaction Reaction Progress No catalyst Catalyst lowers the activation energy for the reaction.What is a catalyst? What does it do during a chemical reaction? 204. An Energy Diagram activated complex activation energyEa reactants products course of reaction energy Animation by Raymond Chang All rights reserved 205. Energy Sources in the United States Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 307 Wood Coal Petroleum / natural gas Hydro and nuclear 1850 1900 1940 1980 1990 100 80 60 40 20 0 Percent 9 91 21 71 5 3 10 50 40 20 70 10 26 58 16 206. Energy Sources in the United States Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 307 Wood Coal Petroleum / natural gas Hydro and nuclear 1850 100 80 60 40 20 0 Percent 9 91 1900 21 71 5 3 1940 10 50 40 1980 20 70 10 1990 26 58 16 207. Energy Sources in the United States Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 307 Wood Coal Petroleum / natural gas Hydro and nuclear 1850 100 80 60 40 20 0 Percent 9 91 1900 21 71 5 3 1940 10 50 40 1980 20 70 10 1990 26 58 16 2005 50 21 26 208. Energy Conversion Timberlake, Chemistry 7th Edition, page 202 fan electrical energy to mechanical energy light bulb electrical energy to light energy to thermal and radiant energy coffee maker electrical energy to thermal energy pencil sharpener electrical energy to mechanical energy 209. Conservation of Energy in a Chemical Reaction Surroundings System Surroundings System Energy Before reaction After reaction In this example, the energy of the reactants and products increases, while the energy of the surroundings decreases. In every case, however, the total energy does not change. Myers, Oldham, Tocci, Chemistry, 2004, page 41 Endothermic Reaction Reactant + Energy Product 210. Conservation of Energy in a Chemical Reaction Surroundings System Surroundings System Energy Before reaction After reaction In this example, the energy of the reactants and products decreases, while the energy of the surroundings increases. In every case, however, the total energy does not change. Myers, Oldham, Tocci, Chemistry, 2004, page 41 Exothermic Reaction Reactant Product + Energy 211. Direction of Heat Flow Surroundings ENDOthermic qsys > 0 EXOthermic qsys < 0 System Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 207 System H2O(s) + heat  H2O(l) melting H2O(l)  H2O(s) + heat freezing 212. Caloric Values Food joules/grams calories/gram Calories/gram Protein 17 000 4000 4 Fat 38 000 9000 9 Carbohydrates 17 000 4000 4 Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 51 1000 calories = 1 Calorie "science" "food" 1calories = 4.184 joules 213. Units of energy Most common units of energy 1. S unit of energy is the joule (J), defined as 1 (kilogram•meter2)/second2, energy is also expressed in kilojoules (1 kJ = 103J). 2. Non-S unit of energy is the calorie where 1 calorie (cal) is the amount of energy needed to raise the temperature of 1 g of water by 1°C. One cal = 4.184 J or 1J = 0.2390 cal. Units of energy are the same, regardless of the form of energy 214. Typical apparatus used in this activity include a boiler (such as large glass beaker), a heat source (Bunsen burner or hot plate), a stand or tripod for the boiler, a calorimeter, thermometers, samples (typically samples of copper, aluminum, zinc, tin, or lead), tongs (or forceps or string) to handle samples, and a balance. Experimental Determination of Specific Heat of a Metal 215. A Coffee Cup Calorimeter Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 302 Thermometer Styrofoam cover Styrofoam cups Stirrer Thermometer Glass stirrer Cork stopper Two Styrofoam ® cups nested together containing reactants in solution 216. A Bomb Calorimeter 217. Heating Curves Melting - PE  Solid - KE  Liquid - KE  Boiling - PE  Gas - KE  Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 218. Heating Curves Temperature(oC) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time Melting - PE  Solid - KE  Liquid - KE  Boiling - PE  Gas - KE  219. Heating Curves • Temperature Change – change in KE (molecular motion) – depends on heat capacity • Heat Capacity – energy required to raise the temp of 1 gram of a substance by 1°C – “Volcano” clip - – water has a very high heat capacity Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 220. Heating Curves • Phase Change – change in PE (molecular arrangement) – temp remains constant • Heat of Fusion (DHfus) – energy required to melt 1 gram of a substance at its m.p. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 221. Heating Curves • Heat of Vaporization (DHvap) – energy required to boil 1 gram of a substance at its b.p. – usually larger than DHfus…why? • EX: sweating, steam burns, the drinking bird Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 222. Phase Diagrams • Show the phases of a substance at different temps and pressures. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 223. Calculating Energy Changes - Heating Curve for Water Temperature(oC) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time DH = mol x DHfus DH = mol x DHvap Heat = mass x Dt x Cp, liquid Heat = mass x Dt x Cp, gas Heat = mass x Dt x Cp, solid 224. Equal Masses of Hot and Cold Water Thin metal wall Insulated box Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291 225. Water Molecules in Hot and Cold Water Hot water Cold Water 90 oC 10 oC Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291 226. Water Molecules in the same temperature water Water (50 oC) Water (50 oC)Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291 227. Heat Transfer Al Al m = 20 g T = 40oC SYSTEM Surroundings m = 20 g T = 20oC 20 g (40oC) 20 g (20oC) 30oC Block “A” Block “B” Final Temperature Assume NO heat energy is “lost” to the surroundings from the system.       C30 g)20g(20 C20g20C40g20 o oo    What will be the final temperature of the system ? a) 60oC b) 30oC c) 20oC d) ? 228. Heat Transfer Al Al m = 20 g T = 40oC SYSTEM Surroundings m = 10 g T = 20oC 20 g (40oC) 20 g (20oC) 30.0oC Block “A” Block “B” Final Temperature Assume NO heat energy is “lost” to the surroundings from the system. 20 g (40oC) 10 g (20oC) 33.3oC       C3.33 g)10g(20 C20g10C40g20 o oo    What will be the final temperature of the system ? a) 60oC b) 30oC c) 20oC d) ? ? 229. Heat Transfer Al Al m = 20 g T = 20oC SYSTEM Surroundings m = 10 g T = 40oC 20 g (40oC) 20 g (20oC) 30.0oC Block “A” Block “B” Final Temperature Assume NO heat energy is “lost” to the surroundings from the system. 20 g (40oC) 10 g (20oC) 33.3oC       C7.26 g)10g(20 C40g10C20g20 o oo    20 g (20oC) 10 g (40oC) 26.7oC 230. Heat Transfer m = 75 g T = 25oC SYSTEM Surroundings m = 30 g T = 100oC 20 g (40oC) 20 g (20oC) 30.0oC Block “A” Block “B” Final Temperature 20 g (40oC) 10 g (20oC) 33.3oC       C46 g)30g(75 C100g30C25g75 o oo    20 g (20oC) 10 g (40oC) 26.7oC AgH2O Real Final Temperature = 26.6oC Why? We’ve been assuming ALL materials transfer heat equally well. 231. Specific Heat • Water and silver do not transfer heat equally well. Water has a specific heat Cp = 4.184 J/goC Silver has a specific heat Cp = 0.235 J/goC • What does that mean? It requires 4.184 Joules of energy to heat 1 gram of water 1oC and only 0.235 Joules of energy to heat 1 gram of silver 1oC. • Law of Conservation of Energy… In our situation (silver is “hot” and water is “cold”)… this means water heats up slowly and requires a lot of energy whereas silver will cool off quickly and not release much energy. • Lets look at the math! 232. “loses” heat Calorimetry                    C26.6x 320.8x8550 7845313.8xx05.7705 algebra.thesolveandunitsDrop C25-xg75CgJ184.4C100-xg30CgJ235.0 equation.intovaluesSubstitute TTmCTTmC TmCTmC qq o oooo ifpinitialfinalp pp OHAg 2      DD  m = 75 g T = 25oC SYSTEM Surroundings m = 30 g T = 100oC AgH2O Tfinal = 26.6oC 233. Calorimetry                    C26.6x 8550320.8x 7845313.8xx05.7705 algebra.thesolveandunitsDrop C25-xg75CgJ184.4C100-xg30CgJ235.0 equation.intovaluesSubstitute TTmCTTmC TmCTmC qq o oooo ifpinitialfinalp pp OHAg 2      DD  m = 75 g T = 25oC SYSTEM Surroundings m = 30 g T = 100oC AgH2O 234. 1 Calorie = 1000 calories “food” = “science” Candy bar 300 Calories = 300,000 calories English Metric = _______Joules 1 calorie - amount of heat needed to raise 1 gram of water 1oC 1 calorie = 4.184 Joules 235. Cp(ice) = 2.077 J/g oC It takes 2.077 Joules to raise 1 gram ice 1oC. X Joules to raise 10 gram ice 1oC. (10 g)(2.077 J/g oC) = 20.77 Joules X Joules to raise 10 gram ice 10oC. (10oC)(10 g)(2.077 J/g oC) = 207.7 Joules Heat = (specific heat) (mass) (change in temperature) q = Cp . m . DT Temperature(oC) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time DH = mol x DHfus DH = mol x DHvap Heat = mass x Dt x Cp, liquid Heat = mass x Dt x Cp, gas Heat = mass x Dt x Cp, solid 236. Heat = (specific heat) (mass) (change in temperature) q = Cp . m . DT TmCq p(ice) D  initialfinalp(ice) TTmCq   C)30(C20-g10 Cg J2.077 q oo o         Given Ti = -30oC Tf = -20oC q = 207.7 Joules Temperature(oC) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time DH = mol x DHfus DH = mol x DHvap Heat = mass x Dt x Cp, liquid Heat = mass x Dt x Cp, gas Heat = mass x Dt x Cp, solid 237. 240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC). When thermal equilibrium is reached, the system has a temperature of 42oC. Find the mass of the iron. Calorimetry Problems 2 question #5 Fe T = 500oC mass = ? grams T = 20oC mass = 240 g LOSE heat = GAIN heat- - [(Cp,Fe) (mass) (DT)] = (Cp,H2O) (mass) (DT) - [(0.4495 J/goC) (X g) (42oC - 500oC)] = (4.184 J/goC) (240 g) (42oC - 20oC)] Drop Units: - [(0.4495) (X) (-458)] = (4.184) (240 g) (22) 205.9 X = 22091 X = 107.3 g Fe 238. A 97 g sample of gold at 785oC is dropped into 323 g of water, which has an initial temperature of 15oC. If gold has a specific heat of 0.129 J/goC, what is the final temperature of the mixture? Assume that the gold experiences no change in state of matter. Calorimetry Problems 2 question #8 Au T = 785oC mass = 97 g T = 15oC mass = 323 g LOSE heat = GAIN heat- - [(Cp,Au) (mass) (DT)] = (Cp,H2O) (mass) (DT) - [(0.129 J/goC) (97 g) (Tf - 785oC)] = (4.184 J/goC) (323 g) (Tf - 15oC)]Drop Units: - [(12.5) (Tf - 785oC)] = (1.35x 103) (Tf - 15oC)] -12.5 Tf + 9.82 x 103 = 1.35 x 103 Tf - 2.02 x 104 3 x 104 = 1.36 x 103 Tf Tf = 22.1oC 239. If 59 g of water at 13oC are mixed with 87 g of water at 72oC, find the final temperature of the system. Calorimetry Problems 2 question #9 T = 13oC mass = 59 g LOSE heat = GAIN heat- - [(Cp,H2O) (mass) (DT)] = (Cp,H2O) (mass) (DT) - [(4.184 J/goC) (59 g) (Tf - 13oC)] = (4.184 J/goC) (87 g) (Tf - 72oC)]Drop Units: - [(246.8) (Tf - 13oC)] = (364.0) (Tf - 72oC)] -246.8 Tf + 3208 = 364 Tf - 26208 29416 = 610.8 Tf Tf = 48.2oC T = 72oC mass = 87 g 240. A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC. Find the system's final temperature. Calorimetry Problems 2 question #10 ice T = -11oC mass = 38 g T = 56oC mass = 214 g LOSE heat = GAIN heat- - [(Cp,H2O) (mass) (DT)] = (Cp,H2O) (mass) (DT) + (Cf) (mass) + (Cp,H2O) (mass) (DT) - [(4.184 J/goC)(214 g)(Tf - 56oC)] = (2.077 J/goC)(38 g)(11oC) + (333 J/g)(38 g) + (4.184 J/goC)(38 g)(Tf - 0oC) - [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)] - 895 Tf + 50141 = 868 + 12654 + 159 Tf - 895 Tf + 50141 = 13522 + 159 Tf Tf = 34.7oC 36619 = 1054 Tf Temperature(oC) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time DH = mol x DHfus DH = mol x DHvap Heat = mass x Dt x Cp, liquid Heat = mass x Dt x Cp, gas Heat = mass x Dt x Cp, solid A B C D warm icemelt ice warm water water cools 241. 25 g of 116oC steam are bubbled into 0.2384 kg of water at 8oC. Find the final temperature of the system. Calorimetry Problems 2 question #11 - [(Cp,H2O) (mass) (DT)] + (Cv,H2O) (mass) + (Cp,H2O) (mass) (DT) = [(Cp,H2O) (mass) (DT)] - [ - 816.8 - 56400 + 104.5Tf - 10450] = 997Tf - 7972 - [qA + qB + qC] = qD qA = [(Cp,H2O) (mass) (DT)] qA = [(2.042 J/goC) (25 g) (100o - 116oC)] qA = - 816.8 J qB = (Cv,H2O) (mass) qA = (2256 J/g) (25 g) qA = - 56400 J qC = [(Cp,H2O) (mass) (DT)] qC = [(4.184 J/goC) (25 g) (Tf - 100oC)] qA = 104.5Tf - 10450 qD = (4.184 J/goC) (238.4 g) (Tf - 8oC) qD = - 997Tf - 7972 - [qA + qB + qC] = qD 816.8 + 56400 - 104.5Tf + 10450 = 997Tf - 7972 67667 - 104.5Tf = 997Tf - 7979 75646 = 1102Tf 1102 1102 Tf = 68.6oC Temperature(oC) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time DH = mol x DHfus DH = mol x DHvap Heat = mass x Dt x Cp, liquid Heat = mass x Dt x Cp, gas Heat = mass x Dt x Cp, solid A B C D (1000 g = 1 kg) 238.4 g 242. A 322 g sample of lead (specific heat = 0.138 J/goC) is placed into 264 g of water at 25oC. If the system's final temperature is 46oC, what was the initial temperature of the lead? Calorimetry Problems 2 question #12 Pb T = ? oC mass = 322 g Ti = 25oC mass = 264 g LOSE heat = GAIN heat- - [(Cp,Pb) (mass) (DT)] = (Cp,H2O) (mass) (DT) - [(0.138 J/goC) (322 g) (46oC - Ti)] = (4.184 J/goC) (264 g) (46oC- 25oC)]Drop Units: - [(44.44) (46oC - Ti)] = (1104.6) (21oC)] - 2044 + 44.44 Ti = 23197 44.44 Ti = 25241 Ti = 568oC Pb Tf = 46oC 243. A sample of ice at –12oC is placed into 68 g of water at 85oC. If the final temperature of the system is 24oC, what was the mass of the ice? Calorimetry Problems 2 question #13 H2O T = -12oC mass = ? g Ti = 85oC mass = 68 g GAIN heat = - LOSE heat [ qA + qB + qC ] = - [(Cp,H2O) (mass) (DT)] 458.2 m = - 17339 m = 37.8 g ice Tf = 24oC qA = [(Cp,H2O) (mass) (DT)] qC = [(Cp,H2O) (mass) (DT)] qB = (Cf,H2O) (mass) qA = [(2.077 J/goC) (mass) (12oC)] qB = (333 J/g) (mass) qC = [(4.184 J/goC) (mass) (24oC)] [ qA + qB + qC ] = - [(4.184 J/goC) (68 g) (-61oC)] 24.9 m 333 m 100.3 m 458.2 mqTotal = qA + qB + qC 458.2 458.2 244. Endothermic Reaction Energy + Reactants  Products +DH Endothermic Reaction progress Energy Reactants Products Activation Energy 245. Catalytic Converter Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 454 N O N O N O N O N O N O N O N O N O N O N O O O N N O O O O N NN N One of the reactions that takes place in the catalytic converter is the decomposition of nitrogen (II) oxide (NO) to nitrogen and oxygen gas. N O 246. O Catalytic Converter C O N O C O O CO N N One of the reactions that takes place in the catalytic converter is the decomposition of carbon monoxide (CO) to carbon dioxide and nitrogen (II) oxide (NO) to nitrogen gas. C O N N N OO O C OCO 2 CO(g) + 2 NO(g) N2(g) + 2 CO2(g)catalyst 247. Enthalpy Diagram H2O(g) H2O(l) H2(g) + ½ O2(g) -44 kJ Exothermic +44 kJ Endothermic DH = +242 kJ Endothermic 242 kJ Exothermic 286 kJ Endothermic DH = -286 kJ Exothermic Energy H2(g) + 1/2O2(g)  H2O(g) + 242 kJ DH = -242 kJ Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 211 248. Hess’s Law Calculate the enthalpy of formation of carbon dioxide from its elements. C(g) + 2O(g)  CO2(g) Use the following data: 2O(g)  O2(g) DH = - 250 kJ C(s)  C(g) DH = +720 kJ CO2(g)  C(s) + O2(g) DH = +390 kJ Smith, Smoot, Himes, pg 141 2O(g)  O2(g) DH = - 250 kJ C(g) + 2O(g)  CO2(g) DH = -1360 kJ C(g)  C(s) DH = - 720 kJ C(s) + O2(g)  CO2(g) DH = - 390 kJ 249. In football, as in Hess's law, only the initial and final conditions matter. A team that gains 10 yards on a pass play but has a five-yard penalty, has the same net gain as the team that gained only 5 yards. initial position of ball final position of ball 10 yard pass 5 yard penalty 5 yard net gain 250. Fission vs. Fusion Fuse small atoms 2H2 He NO Radioactive waste Very High Temperatures ~5,000,000 oC (SUN) Split large atoms U-235 Radioactive waste (long half-life) Nuclear Power Plants Alike Different Create Large Amounts of Energy E = mc2 Transmutation of Elements Occurs Change Nucleus of Atoms Fusion Different Topic Topic Fission 251. • Use fear and selective facts to promote an agenda • Eating animals? • Radiation = Bad Look who is funding research; it may bias the results. 252. Shielding Radiation 253. Nuclear Fission 254. Nuclear Fission First stage: 1 fission Second stage: 2 fission Third stage: 4 fission 255. Nuclear Fission 256. Nuclear Power Plants map: Nuclear Energy Institute 257. Fermi Approximations FERMI APPROXIMATIONS An educated guess – based on a series of calculations of known facts – to arrive at a reasonable answer to a question. How many piano tuners are there in New York City? ANSWER: Enrico Fermi 400 piano tuners 258. Nuclear Fusion Sun + + Four hydrogen nuclei (protons) Two beta particles (electrons) One helium nucleus Hee2H4 4 2 0 1- 1 1  + Energy 259. Conservation of Mass …mass is converted into energy Hydrogen (H2) H = 1.008 amu Helium (He) He = 4.004 amu FUSION 2 H2  1 He + ENERGY 1.008 amu x 4 4.0032 amu = 4.004 amu + 0.028 amu This relationship was discovered by Albert Einstein E = mc2 Energy= (mass) (speed of light)2 260. Time Travel? …Albert Einstein also discovered the Geometry of Space Near a Black Hole Einstein’s theory of general relativity maybe interpreted in terms of curvature of space in the presence of a gravitational field. Here we see how this curvature varies near a black hole. 261. Time Travel? …Albert Einstein also discovered the Geometry of Space Near a Black Hole Einstein’s theory of general relativity maybe interpreted in terms of curvature of space in the presence of a gravitational field. Here we see how this curvature varies near a black hole. 262. Time Travel? …Albert Einstein also discovered the Geometry of Space Near a Black Hole Einstein’s theory of general relativity maybe interpreted in terms of curvature of space in the presence of a gravitational field. Here we see how this curvature varies near a black hole. 263. Tokamak Reactor • Fusion reactor • 10,000,000 o Celcius • Russian for torroidial (doughnut shaped) ring • Magnetic field contains plasma 264. Cold Fusion? • Fraud? • Experiments must be repeatable to be valid 265. 0 1 2 3 4 Number of half-lives Radioisotoperemaining(%) 100 50 25 12.5 Half-life of Radiation Initial amount of radioisotope t1/2 t1/2 t1/2 After 1 half-life After 2 half-lives After 3 half-lives 266. Triple Point Plot LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 488 solid liquid gas melting freezing sublimation deposition vaporization condensation Temperature (oC) Pressure(atm) 0.6 2.6 267. Liquid VaporSolid Normal melting point Normal boiling point 101.3 0.61 0.016 1000 Temperature (oC) Pressure(KPa) Triple point Triple Point Critical pressure Critical point Critical temperature 373.99 22,058 268. Copyright © 2007 Pearson Benjamin Cummings. All rights reserved. 269. Objectives - Matter • Explain why mass is used as a measure of the quantity of matter. • Describe the characteristics of elements, compounds, and mixtures. • Solve density problems by applying an understanding of the concepts of density. • Distinguish between physical and chemical properties and physical and chemical changes. • Demonstrate an understanding of the law of conservation of mass by applying it to a chemical reaction. 270. Objectives - Energy • Identify various forms of energy. • Describe changes in energy that take place during a chemical reaction. • Distinguish between heat and temperature. • Solve calorimetry problems. • Describe the interactions that occur between electrostatic charges. 271. Law of Conservation of Energy Eafter = Ebefore 2 H2 + O2  2 H2O + energy +  + WOOF! 272. Law of Conservation of Energy ENERGY CO2 + H2OC2H2 + O2 PEreactants PEproducts KEstopper heat, light, sound Eafter = Ebefore 2 H2 + O2  2 H2O + energy +  + WOOF! 273. Law of Conservation of Energy ENERGY C2H2 + O2 C2H2 + O2 PEreactants PEproducts KEstopper heat, light, sound Eafter = Ebefore 2C2H2 + 5O2  4 CO2 + 2H2O + energy Energy Changes 274. First experimental image showing internal atomic structures © 2005 University of Augsburg, Experimental Physics VI, http://www.physik.uni-augs 275. Heating CurvesTemperature(oC) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140 Time Melting - PE  Solid - KE  Liquid - KE  Boiling - PE  Gas - KE  276. Heating Curves  Temperature Change • change in KE (molecular motion) • depends on heat capacity  Heat Capacity • energy required to raise the temp of 1 gram of a substance by 1°C • “Volcano” clip - water has a very high heat capacity Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 277. Heating Curves  Phase Change • change in PE (molecular arrangement) • temp remains constant  Heat of Fusion (DHfus) • energy required to melt 1 gram of a substance at its m.p. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 278. Heating Curves  Heat of Vaporization (DHvap) • energy required to boil 1 gram of a substance at its b.p. • usually larger than DHfus…why?  EX: sweating, steam burns, the drinking bird Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 279. Phase Diagrams  Show the phases of a substance at different temps and pressures. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem 280. Resources - Matter and Energy Objectives - matter and energy Objectives - measurement Objectives - phases of matter Worksheet - vocabulary Worksheet II - percentage composition Worksheet - properties Worksheet - density problems Activity - density blocks & Part 2 Lab - golf ball lab Worksheet - classifying matter Outline (general) Activity - chromatography Outline - causes of change - calorimetry Worksheet - calorimetry problems 1 Worksheet - calorimetry problems 2 Worksheet - heat energy problems Worksheet - conversion factors Worksheet - atoms, mass, and the mole activity - mole pattern Article - buried in ice (questions) Lab - beverage density (PowerPoint) Textbook - questions Article - buckyball (pics) & (video) questions Episode 5 - A Matter of State General Chemistry PP


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