Transcript Chapter 3
Matter and Energy Chapter 3 1 Properties • Characteristics of the substance under observation • Properties can be either directly observable or the manner something interacts with other substances in the universe 2 Universe Classified • Matter is the part of the universe that has mass and volume • Energy is the part of the universe that has the ability to do work • Chemistry is the study of matter – The properties of different types of matter – The way matter behaves when influenced by other matter and/or energy 3 Properties of Matter • Physical Properties are the characteristics of matter that can be changed without changing its composition – Characteristics that are directly observable • Chemical Properties are the characteristics that determine how the composition of matter changes as a result of contact with other matter or the influence of energy • Characteristics that describe the behavior of matter 4 Classify Each of the following as Physical or Chemical Properties The boiling point of ethyl alcohol is 78°C. – Physical property – describes inherent characteristic of alcohol – boiling point Diamond is very hard. – Physical property – describes inherent characteristic of diamond – hardness Sugar ferments to form ethyl alcohol. – Chemical property – describes behavior of sugar – forming a new substance (ethyl alcohol) 5 States of Matter • solid, liquid, gas State Solid Liquid Gas Shape Keeps Shape Takes Shape of Container Takes Shape of Container Volume Compress Flow Keeps Volume Keeps Volume No No No Yes Takes Volume of Container Yes Yes 6 Figure 3.2: The three states of water 7 Changes in Matter • Physical Changes are changes to matter that do not result in a change of the fundamental components that make that substance – State Changes – boiling, melting, condensing • Chemical Changes involve a change in the fundamental components of the substance – Produce a new substance – Chemical reaction – Reactants Products 8 Figure 3.3: Electrolysis 9 Classify Each of the following as Physical or Chemical Changes Iron is melted. – Physical change – describes a state change, but the material is still iron Iron combines with oxygen to form rust. – Chemical change – describes how iron and oxygen react to make a new substance, rust Sugar ferments to form ethyl alcohol. – Chemical change – describes how sugar forms a new substance (ethyl alcohol) 10 Elements and Compounds • Substances which can not be broken down into simpler substances by chemical reactions are called elements • Most substances are chemical combinations of elements. These are called compounds. – Compounds are made of elements – Compounds can be broken down into elements – Properties of the compound not related to the properties of the elements that compose it – Same chemical composition at all times 11 Classification of Matter Matter Pure Substance Constant Composition Homogeneous Mixture Variable Composition Heterogeneous • Homogeneous = uniform throughout, appears to be one thing – pure substances – solutions (homogeneous mixtures) • Heterogeneous = non-uniform, contains regions with different properties than other regions 12 Pure Substances vs. Mixtures • Pure Substances – All samples have the same physical and chemical properties – Constant Composition all samples have the same composition – Homogeneous – Separate into components based on chemical properties • Mixtures – Different samples may show different properties – Variable composition – Homogeneous or Heterogeneous – Separate into components based on physical properties • All mixtures are made of pure substances 13 Figure 3.4: Table salt is stirred into water (left), forming a homogeneous mixture called a solution (right) 14 Figure 3.5: Sand and water do not mix to form a uniform mixture 15 Identity Each of the following as a Pure Substance, Homogeneous Mixture or Heterogeneous Mixture Gasoline – a homogenous mixture A stream with gravel on the bottom – a heterogeneous mixture Copper metal – A pure substance (all elements are pure substances) 16 Separation of Mixtures • Separate mixtures based on different physical properties of the components – Physical change Different Physical Property Technique Boiling Point Distillation State of Matter (solid/liquid/gas) Adherence to a Surface Filtration Chromatography Volatility Evaporation 17 Figure 3.6: Distillation of a solution consisting of salt dissolved in water 18 Figure 3.7: No chemical change occurs when salt water is distilled 19 Figure 3.8: Filtration separates a liquid from a solid 20 Figure 3.9: Separation of a sand-saltwater mixture 21 Figure 3.10: The organization of matter 22 Energy and Energy Changes • Capacity to do work – chemical, mechanical, thermal, electrical, radiant, sound, nuclear • Energy may affect matter – e.g. raise its temperature, eventually causing a state change – All physical changes and chemical changes involve energy changes 23 Units of Energy • One calorie is the amount of energy needed to raise the temperature of one gram of water by 1°C – kcal = energy needed to raise the temperature of 1000 g of water 1°C • joule – 4.184 J = 1 cal • In nutrition, calories are capitalized – 1 Cal = 1 kcal 24 Example - Converting Calories to Joules Convert 60.1 cal to joules 1 cal 4.184 joules 4.184 J 60.1cal 251J 1 cal 25 Energy and the Temperature of Matter • The amount the temperature of an object increases depends on the amount of heat added (Q). – If you double the added heat energy the temperature will increase twice as much. • The amount the temperature of an object increases depends on its mass – If you double the mass it will take twice as much heat energy to raise the temperature the same amount. 26 Specific Heat Capacity • Specific Heat (s) is the amount of energy required to raise the temperature of one gram of a substance by one Celsius degree J By definition , the specific heat of water is 4.184 g C Amount of Heat = Specific Heat x Mass x Temperature Change Q = s x m x T 27 Example – Calculate the amount of heat energy (in joules) needed to raise the temperature of 7.40 g of water from 29.0°C to 46.0°C JJ Specific Heat of Water = 4.184 gg-CC Mass = 7.40 g Temperature Change = 46.0°C – 29.0°C = 17.0°C Q = s x m x T J Heat 4.184 7.40g 17.0C 526 J g C 28 Example – A 1.6 g sample of metal that appears to be gold requires 5.8 J to raise the temperature from 23°C to 41°C. Is the metal pure gold? Q s m T Q s m T T 41C - 23C 18C 5.8 J J s 0.20 1.6 g x 18C g C Table 3.2 lists the specific heat of gold as 0.13 Therefore the metal cannot be pure gold. J g C 29