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Matter and Energy An Introduction Still Another View. What is Matter?. Matter Anything that occupies space and has mass Atoms Tiny particles too small to see Molecules Atoms bonded together- also too small to see. Classifying Matter. States of Matter Solid Fixed Volume Rigid Shape
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What is Matter? • Matter • Anything that occupies space and has mass • Atoms • Tiny particles too small to see • Molecules • Atoms bonded together- also too small to see
Classifying Matter • States of Matter • Solid • Fixed Volume • Rigid Shape • Liquid • Fixed volume • Assume the shape of the container • Gas • Atoms separated by large distances • Assume the volume and shape of the container
Classifying MatterBased on Composition Pure substance • Matter that has a uniform and unchanging composition Mixture • Two or more substances physically placed together in any proportion; each retains its characteristic properties
Classifying Matter • Mixtures • Heterogeneous Mixture • Two or more regions with different composition • Homogeneous Mixture • Uniform composition • Elements • Cannot be broken into simpler substances • Compounds • Composed of two or more elements
Classifying Matter • Matter • Pure substance • Element • Compound • Mixture • Heterogeneous • Homogeneous
Physical and Chemical Properties • Physical properties • No change of composition • Chemical properties • Change of composition
Physical and Chemical Properties • Physical properties • Boiling point • Odor • Taste • Color • Melting point • Density
Chemical properties • Oxidation Rust • Flammability • Chemical activity
Matter and EnergyConservation of Mass • Antoine Lavoisier • Law of Conservation of Mass • “Matter is neither created nor destroyed”
Conservation of Mass • Total amount of matter remains constant in a chemical reaction • 58 grams of butane burns in 208 grams of oxygen to form 176 grams of carbon dioxide and 90 grams of water. 58 grams + 208 grams = 176 grams + 90 grams 266 grams = 266 grams
Energy • Energy • Law of Conservation of Energy • “Energy is neither created nor destroyed”
Energy • Energy • Capacity to do work- what does this mean? • Types of Energy • Kinetic Energy – motion • Potential Energy – position • Electrical Energy – flow of electrons • Chemical Energy – chemical changes • Units of Energy • Joule (J) • Calorie (cal)
Conversion of Energy Units A candy bar contains 225 Cal of nutritional energy. How many joules does it contain? Given: 225 Cal Find: J Conversion Factors: 1000 calories = 1 Cal 4.184 J = 1 calorie Solution Map: Cal cal J 225 Cal X 1000 cal X 4.184 J = 9.41 X 105 J 1 Cal 1 cal
“flow” ceases at thermal equilibrium EOS Heat (q) Heat is a common form of energy associated with the energy transfer that results from thermal differences between objects or between the system and its surroundings Heat gained q and change in T are positive Heat lost q and change in T are negative “flows” spontaneously from higher Tlower T
Temperature Changes • Heat Capacity • Quantity of heat energy required to change the temperature of a given amount of substance by a standard increment. • Specific Heat Capacity • Amount of substance = exactly 1 gram • T = exactly 1 degree • Units of heat in joules or calories • Can be used to identify materials
Many metals have low specific heats. The specific heat of water is higher than that of almost any other substance.
Calculations • Specific Heat • Used to quantify relationship between heat added and the temperature change • Equation Heat = Mass X Heat Capacity X Temperature Change q = m X C X T
Suppose that you are making a cup of tea. How much heat energy will be needed to warm 236 grams of water (about 8 ounces) from 25 0C to 100 0C ? Given: 235 grams of water m 25 0C initial temperature Ti 100.0 0C final temperature Tf Find: Amount of Heat needed q Equation: q = m X C X T Solution Map: C, m, T q
Equation: q = m X C X T Solution Map: C, m, T q C = 4.18 J/g 0C m = 235 grams T = 100.0 0C – 25 0C = 75 0C q = m X C X T = 235 g X 4.18 J/g 0C X 75 0C = 7.4 X 104 J
A chemistry student finds a shiny rock that she suspects is gold. She weighs the rock on a balance and obtains the mass, 14.3 g. She then finds that the temperature of the rock rises from 25 0C to 52 0C upon absorption of 174 J of heat. Find the heat capacity of the rock and determine if the value is consistent with the heat capacity of gold. Given: m = 14.3 g q = 174 J T = 52 0C – 25 0C = 27 0C Find: C Equation: q = m X C X T Solution Map: m, q, T C
q = m X C X T C = q m X T = 174 J 14.3 g X 27 0C = 0.45J g 0C
Conceptualizing an Exothermic Reaction Surroundings are at 25 °C Typical situation: some heat is released to the surroundings, some heat is retained by the solution. T increases 35.4 °C 32.2 °C 25 °C In an isolated system, all heat remains in the system. Maximum temperature rise. Hypothetical situation: all heat is instantly released to the surroundings. Heat = qrxn
Heat Capacity of Water • Water • High Heat Capacity • Changing water temperature requires a lot of heat energy • Important property for living organisms
Matter and EnergyChapter in Review • Matter • Classification of Matter • Properties and Changes in Matter • Conservation of Matter • Energy • Heat Capacity • Specific Heat Capacity
Matter and EnergyChemical Skills • Classify Matter • Physical and Chemical Properties • Chemical and Physical Changes • Conservation of Mass • Conservation of Energy Units • Energy, Temperature Change, and Specific Heat Capacity Calculations