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Chapter 3

Chapter 3. The Properties of Matter and Energy. Setting the Stage – Matter and Energy. Understanding the properties of matter allows us to draw conclusions about compounds and elements found on other planets

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Chapter 3

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  1. Chapter 3 The Properties of Matter and Energy

  2. Setting the Stage – Matter and Energy • Understanding the properties of matter allows us to draw conclusions about compounds and elements found on other planets • Energy, which has no mass, is a second component of the universe, in addition to matter • We can quantify energy and how it interacts with matter Malone and Dolter - Basic Concepts of Chemistry 9e

  3. Setting a Goal - Part AThe Properties of Matter • You will learn how a sample of matter can be described by its properties and how they can be quantitatively expressed Malone and Dolter - Basic Concepts of Chemistry 9e

  4. Objective for Section 3-1 • List and define several properties of matter and distinguish them as physical or chemical Malone and Dolter - Basic Concepts of Chemistry 9e

  5. 3-1 The Physical and Chemical Properties of Matter • Properties describe the particular characteristics of a substance • Pure substances have definite composition and definite, unchanging properties • Physical properties - can be observed without changing the substance • Chemical properties - require that the substance changes into another Malone and Dolter - Basic Concepts of Chemistry 9e

  6. The Physical States of Matter • The three fundamental physical states are solid, liquid and gas • solids have a definite shape and volume • liquids have a definite volume but not a definite shape • Gases have neither a definite volume nor shape • A substance exists in a particular physical state under defined conditions Malone and Dolter - Basic Concepts of Chemistry 9e

  7. The Physical States of Matter Malone and Dolter - Basic Concepts of Chemistry 9e

  8. Changes in Physical State • Melting point (reverse: freezing point) • temperature at which a substance changes from solid to liquid (reverse: liquid to solid) • Boiling point (reverse: condensation point • temperature at which a substance changes from liquid to gas (reverse: gas to liquid) Malone and Dolter - Basic Concepts of Chemistry 9e

  9. Types of Physical Properties • Intensive properties – those properties that depend on the type or identity (but not the amount) of material present • Examples: color, density, melting point • Extensive properties – those properties that depend on the amount of material present • Examples: mass, volume Malone and Dolter - Basic Concepts of Chemistry 9e

  10. Chemical Changes and Properties • Chemical properties – tendency of a pure substance to undergo chemical changes • Sometimes quite difficult to determine • Some examples of chemical changes are burning (as opposed to boiling) and color changes • Law of the Conservation of Mass - matter is neither created nor destroyed in chemical reactions Malone and Dolter - Basic Concepts of Chemistry 9e

  11. Chemical Changes Fast reaction: “Thermite Reaction” Medium fast reaction: Zn and HCl Slow reaction: rusting of iron Malone and Dolter - Basic Concepts of Chemistry 9e

  12. Objective for Section 3-2 • Perform calculations involving the density of liquids and solids Malone and Dolter - Basic Concepts of Chemistry 9e

  13. 3-2 Density – A Physical Property • Density is the ratio of the mass of a substance to the volume of that mass – it is usually measured in g/mL for solids and liquids; g/L for gases • Specific gravity is the ratio of the mass of a substance to the mass of an equal volume of water Malone and Dolter - Basic Concepts of Chemistry 9e

  14. Density as a Conversion Factor • Density can also be used to convert between mass and volume • Typically g → mL or mL→ g Malone and Dolter - Basic Concepts of Chemistry 9e

  15. Determination of Density This example shows just one way of determining the density of a solid substance Malone and Dolter - Basic Concepts of Chemistry 9e

  16. Objective for Section 3-3 • Describe the differences in properties between a pure substance and a mixture • Perform calculations involving percent as applied to mixtures Malone and Dolter - Basic Concepts of Chemistry 9e

  17. 3-3 The Properties of Mixtures • A mixture is an aggregation of two or more pure substances • Can be separated by physical means (filtration, distillation, crystallization, chromatography) • Have chemical and physical properties that are different from the substances that make them up • The percentages by mass of the components of a mixture can be varied continuously Malone and Dolter - Basic Concepts of Chemistry 9e

  18. Types of Mixtures • Heterogeneous mixture - nonuniform mixture containing two or more phases with definite boundaries between the phases (e.g. ice and water; sand and water) • Homogeneous mixture - same throughout and contains only one phase (substances are mixed at the atomic or molecular level) (e.g. air; aqueous solution of glucose) • A phase isone physical state with distinct boundaries and uniform properties Malone and Dolter - Basic Concepts of Chemistry 9e

  19. Separation of Mixtures • A heterogeneous mixture of a solid and liquid can be separated by filtration • A heterogeneous mixture of two liquids can be separated using a separating funnel • Several substances in solution can be separated by chromatography • A solution of a solid in a liquid or of two liquids can be separated by distillation Malone and Dolter - Basic Concepts of Chemistry 9e

  20. Basic Distillation Kit Malone and Dolter - Basic Concepts of Chemistry 9e

  21. Distillation is Important in Industry Malone and Dolter - Basic Concepts of Chemistry 9e

  22. Solutions • A type of homogeneous mixture • Usually involves a liquid phase, but can be solid-solid, gas-gas, solid-liquid, etc. • The pure substances can be in different phases but form a homogeneous mixture (table salt or glucose and water, for example) Malone and Dolter - Basic Concepts of Chemistry 9e

  23. Alloys • Alloys are homogeneous mixtures of metallic elements existing in one phase • Important solid solutions of two or more metals include: • brass (copper and zinc) • dental fillings (silver and mercury) • stainless steel (iron, chromium and nickel) • solder (tin and lead) Malone and Dolter - Basic Concepts of Chemistry 9e

  24. Analysis • Solder is an alloy made from 60.0 % tin and 40.0 % lead. What mass of lead is present in 72 g of solder? Malone and Dolter - Basic Concepts of Chemistry 9e

  25. Setting a Goal - Part BThe Properties of Energy • You will be able to qualitatively and quantitatively describe processes in terms of the forms and types of energy associated with them Malone and Dolter - Basic Concepts of Chemistry 9e

  26. Objectives for Section 3-4 • Distinguish among the forms and types of energy • Define the terms endothermic and exothermic, providing several examples of each type of process Malone and Dolter - Basic Concepts of Chemistry 9e

  27. 3-4 The Forms and Types of Energy • Energy is the capacity to do work • There are many forms of energy • heat • light • chemical (stored energy) • electrical energy • mechanical • nuclear Malone and Dolter - Basic Concepts of Chemistry 9e

  28. Energy • Law of the Conservation of Energy - energy can neither be created nor destroyed, but can only transformed from one form to another • The transformation from one type to another may not be efficient (the efficiency of transforming chemical energy to electricity energy (Figure 3-8) is only about 35% efficient). The other 65% is lost as heat but the total amount of energy is constant Malone and Dolter - Basic Concepts of Chemistry 9e

  29. Energy Flow • Exothermic reactions - produce energy (release energy to the surroundings) • Endothermic reactions - require energy input (store energy) • Potential energy is that energy available due to position or composition • Kinetic energy is that energy resulting from motion Malone and Dolter - Basic Concepts of Chemistry 9e

  30. Objective for Section 3-5 • Perform calculations involving the specific heat of a substance, and use it to identify a substance Malone and Dolter - Basic Concepts of Chemistry 9e

  31. 3-5 Energy Measurement and Specific Heat • Specific heat - the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius (or Kelvin) • Reflects how some substances heat up faster than others Malone and Dolter - Basic Concepts of Chemistry 9e

  32. Temperature and Specific Heat • Recall that we measure temperature in °C or K • Energy units • calorie (cal) - amount of heat required to raise the temperature of one gram of water from 14.5 °C to 15.5 °C • *joule: 1 cal = 4.184 J (exactly) • Nutritional ‘Calorie’ is actually 1000 cal (indicated as 1 C) Malone and Dolter - Basic Concepts of Chemistry 9e

  33. Heat Flow • When two substances at different temperatures are put in contact with each other, or mixed, heat flows spontaneously from the substance at higher temperature to the substance at lower temperature • This heat flow continues until the temperatures are the same Malone and Dolter - Basic Concepts of Chemistry 9e

  34. Brewing beer Distilling rice wine Making mist from “dri ice” Dyeing clothes Frying chicken C P P C C Worked Example 1 (Ch. 3) Label the following as physical (P) or chemical (C) changes Malone and Dolter - Basic Concepts of Chemistry 9e

  35. Pewter Cloudy apple juice Sugar (for coffee) Clear apple juice Distilled water Tap water Ho M He M PS Ho M PS Ho M Worked Example 2 (Ch. 3) Classify the following as pure substances, (PS) homogeneous (Ho M) or heterogeneous (He M) mixtures Malone and Dolter - Basic Concepts of Chemistry 9e

  36. Worked Example 3 (Ch. 3) • An unknown metal with a mass of 23.6 g is placed in a graduated cylinder that had an initial water volume of 22.0 mL. After the sample is submerged in the water, the level of water in the cylinder was observed to be 24.1 mL. Determine the identity of the metal Malone and Dolter - Basic Concepts of Chemistry 9e

  37. Worked Example 3 (Ch. 3) • Solution 23.6 g occupies 2.1 mL, hence the density of the metal is 23.6 g/2.1 mL = 11.2 g/mL The metal is likely to be Pb (density = 11.3 g/mL) Malone and Dolter - Basic Concepts of Chemistry 9e

  38. Worked Example 4 (Ch. 3) • A gas can (438 g) contains 33.5 mL of kerosene. If the mass of the can plus the kerosene is 465 g, what is the density of kerosene? • Solution: The 33.5 mL of kerosene weighs 465 g – 438 g = 27.0 g, hence its density is 27.0/33.5 = 0.806 g/mL Malone and Dolter - Basic Concepts of Chemistry 9e

  39. Worked Example 5 (Ch. 3) • At room temperature, the specific heat of benzene is 1.060 J/goC. If a 30.-g sample of benzene releases 450 J of energy, what is the change in temperature? • Solution: Energy (heat) transferred = m x sp ht x Change in temp (DT) 450 (J) = 30 (g) x 1.060 (J/goC) x DT (oC) Change in temp = 14 oC Malone and Dolter - Basic Concepts of Chemistry 9e

  40. Worked Example 6 (Ch. 3) • A 440 g piece of metal at 100.0 oC is placed in 258 g of water initially at 25.0 oC. If the final temperature is 36.5 oC, what is the specific heat of the metal? • Solution: Heat lost (metal) = heat gained (water) -m(m) x sp ht (m) x Dt (m) = m(w) x sp ht (w) x Dt (w) Malone and Dolter - Basic Concepts of Chemistry 9e

  41. Worked Example 6 (Ch. 3)…contd. Hence, -440 g x sp ht(m) x (36.5 - 100.0 ) oC = 258 g x 4.184 J/g oC x (36.5 – 25.0) oC Therefore, sp ht (m) = 258 g x 4.184 J/g oC x 11.5 oC 440 g x 63.5 oC = 0.444 J /g oC (iron) Malone and Dolter - Basic Concepts of Chemistry 9e

  42. Objective for Special Topic – Units of Energy • Distinguish amongst different units of energy and carry out interconversions between them Malone and Dolter - Basic Concepts of Chemistry 9e

  43. Special Topic – Units of Energy • The SI (m-kg-s or MKS system) unit of energy is the joule(J) (= kg m2 s-2) • The cgs (cm-g-s) system unit of energy is the erg (= g cm2 s-2) • The conversion factor is 1 J = 107 ergs • The joule is a much bigger unit of energy than the erg Malone and Dolter - Basic Concepts of Chemistry 9e

  44. An Energy Unit for Atomic Systems • The potential energy associated with a proton and an electron separated by 1 Å (= 10-10 m), a typical atomic distance, is only –2.307 x 10-18 J (or –2.307 x 10-11 ergs) • Hence a more convenient unit is needed to express energies relating to atoms. • This unit is the electron-volt (eV) Malone and Dolter - Basic Concepts of Chemistry 9e

  45. The Electron-Volt Unit • The kinetic energy (KE) gained by an electron falling through an electrical potential difference (voltage) of 1 volt (V) is defined as 1 eV • 1 V is defined as the voltage that gives 1 joule of energy to 1 Coulomb (C) of charge; that is, 1 V = 1 J C-1 Malone and Dolter - Basic Concepts of Chemistry 9e

  46. The Electron-Volt Unit • KE = electronic charge x potential difference • Hence, 1 eV = 1.602 x 10-19 C x 1 V = 1.602 x 10-19 C x 1 J C-1 = 1.602 x 10-19 J (1 J = 6.173 x 1018 eV) Malone and Dolter - Basic Concepts of Chemistry 9e

  47. The Electron-Volt Unit Malone and Dolter - Basic Concepts of Chemistry 9e

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