Copyright©2004 by Houghton Mifflin Company. All rights reserved. 1 Matter and Energy Chapter 3

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 2 Common Elements Iodine Manganese Chromium Gold Calcium Strontium Phosphorous Silver Nitrogen Platinum Magnesium Titanium Tungsten Arsenic Aluminum Copper Barium Tin Uranium Potassium Hydrogen Cobalt Bismuth Zinc Chlorine Lead Nickel Mercury Sodium Oxygen Sulfur Fluorine Iron Bromine Antimony Carbon

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 3 Matter: Anything that has mass and takes up space. Sec 3.1 Properties of Matter

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 4

5 solid, liquid, gas States of Matter

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 6 Physical Properties are the characteristics of matter that can be observed without changing its composition –Characteristics that are directly observable –Examples: color, melting point, density, volume 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 –Examples: burning, color change, decomposing

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 7 Intensive Properties: Independent of the amount of the substance that is present. Density, boiling point, color, etc. Extensive Properties: Dependent upon the amount of the substance present. Mass, volume, energy, etc.

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 8 Classify Each of the following as Physical or Chemical Properties ¬The boiling point of ethyl alcohol is 78°C. ­Diamond is very hard. ®Sugar ferments to form ethyl alcohol.

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 9 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)

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 10 Changes in Matter Physical Changes are changes to matter that do not result in a change the fundamental components that make that substance –Example: state changes – boiling, melting, condensing –Grinding, crushing, bending, filtering Chemical Changes involve a change in the fundamental components of the substance –Produce a new substance –Chemical reaction –Reactants  Products

Copyright©2004 by Houghton Mifflin Company. All rights reserved Indicators of a Chemical Reaction Produces a gas (bubbles) Precipitate forms (solid) Color change (not due to blending) Changes in temperature (hot or cold)

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 12 Classify Each of the following as Physical or Chemical Changes ¬Iron metal is melted. ­Iron combines with oxygen to form rust. ®Sugar ferments to form ethyl alcohol.

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 13 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)

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 14 Sec 3.3 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

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 15 Sec 3.4 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

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 16 Classification of Matter Homogeneous = uniform throughout, appears to be one thing –pure substances –solutions (homogeneous mixtures) Heterogeneous = non-uniform, contains regions with different properties than other regions

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 17 Identity Each of the following as a Pure Substance, Homogeneous Mixture or Heterogeneous Mixture ¬Gasoline ­Rocky Road ice cream ®Copper metal

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 18 Identity Each of the following as a Pure Substance, Homogeneous Mixture or Heterogeneous Mixture ¬Gasoline –a homogenous mixture ­Rock Road ice cream –a heterogeneous mixture ®Copper metal –A pure substance (all elements are pure substances)

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 19 Sec 3.5 Separation of Mixtures Separate mixtures based on different physical properties of the components –Physical change EvaporationVolatility ChromatographySolubility FiltrationState of Matter (solid/liquid/gas) DistillationBoiling Point TechniqueDifferent Physical Property

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 20 Separates homogeneous mixture on the basis of differences in boiling point.

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 21 Filtration: Separates solid substances from liquids and solutions.

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 22 Chromatography: Separates substances on the basis of differences in solubility in a solvent.

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 23 Sec 3.6 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

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 24 Heat Heat: a flow of energy due to a temperature difference 1.Exothermic = A process that results in the evolution of heat. Example: when a match is struck, it is an exothermic process because energy is produced as heat. 2.Endothermic = A process that absorbs energy. Example: melting ice to form liquid water is an endothermic process.

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 25 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

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 26 Example - Converting Calories to Joules Convert 60.1 cal to joules

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 27 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.

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 28 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 Amount of Heat = Specific Heat x Mass x Temperature Change Q = s x m x  T

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 29 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 Mass = 7.40 g Temperature Change = 46.0°C – 29.0°C = 17.0°C Specific Heat of Water = Q = s x m x  T

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 30 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? Table 3.2 lists the specific heat of gold as 0.13 Therefore the metal cannot be pure gold.