Matter and Energy

Matter and Energy

Matter and Energy I. Matter Matter is anything that has mass and volume Mass o Amount of matter o Measured in grams (g) Volume o Space matter occupies o Measured in milliliters (mL), liters (L) or cubic centimeters (cm3 )

Two types of matter o Pure substances o Mixtures A. Pure Substances Uniform composition The same throughout the sample Two Types

Elements Compounds 1. Elements Simplest form of matter Smallest part called atom o Cannot breakdown Represented

using a capital letter or capital letter and lower case letter o Carbon (C) o Sodium (Na) 2. Compound Two or more elements chemically joined in a specific ratio o Properties of the compound are different than the elements that make it up

Compounds can be broken down o Decomposed Examples o Water (H2O) o Hydrogen peroxide (H2O2) B. Mixture Two

or more substances physically joined in any ratio o Keep the properties of their components (parts) Can be separated by physical means Exist in two forms o Heterogeneous o Homogeneous Heterogeneous

Visible difference between components (parts) Homogeneous No visible differences

between components (parts) Called a solution Represented using (aq) C. Properties of matter Physical Properties Properties that can be observed without

changing the substance Chemical Properties Properties that show how a substance reacts (changes) Densitya physical property used to identify a substance Density = mass volume

Determine the density of an object with a mass of 59.3g and a volume of 84.2 mL. Density = Mass Volume d = 59.3g 84.2mL d = .704g/mL Determine the mass of an object with a density of 0.94g/mL and a volume of 24.8 mL. Density = Mass Volume Solve for mass Mass = Density x Volume m = 0.94g/mL x 24.8mL

m = 23.3g Determine the volume of an object with a density of 1.47g/mL and a mass of 28.6g. Density = Mass Volume Solve for volume Volume = Mass Density v = 28.6g 1.47g/mL v = 19.5mL II. Energy Energy

is the driving force behind change Cannot be created or destroyed It can change its form Two types of energy Kinetic Energy of motion Potential Stored energy A. Measurements involving energy

1. Temperature Average kinetic energy of particles Measured using a thermometer (unit: degrees) Fahrenheit Celsius Kelvin To convert F to C -- use C = 5/9( F - 32) C to F -- use F = 9/5 C + 32 C to K -- use K = C + 273 K to C use K = C + 273 2. Calorimetry

Measures the actual energy (q) in a system Related to mass (m), specific heat capacity (C) and temperature change (T) Measured using a calorimeter (unit: joules) To calculate energy use Reference Table T How many joules are required to heat 40g water at 30C to 80C? q = m C T q = 40g x 4.18J/gC x 50C q = 8360J

5000J were added to 30g water at 25C. What is the new temperature? q = m C T 5000J = 30g x 4.18J/gC x T 5000 = 125.4 x T T = 39.9 ~ 40 T new = 25 + 40 T new = 65C How

many joules are needed to melt 100g ice at 0C q = m Hfus q = 100g x 334J/g q = 33400J III. Phases of Matter Solids Liquids Gases A. Solids

Matter that has specific shape and specific volume Atoms closely packed together o Cannot be compressed B. Liquids Matter that has a specific volume but takes the shape of the container

Atoms are close but have some space between them o Cannot be compressed o Can be poured C. Gases Matter that takes the shape and volume of the container Atoms have free space between them o Compressible

o Can be poured D. Phase Changes If energy is added Solid to liquid Melting Liquid to gas Boiling Solid to gas Sublimation

If energy is removed Liquid to solid Freezing Gas to liquid Condensing Gas to solid Deposition Temperature E. Phase Diagram

Boil Melt Liquid Solid Time Gas Temperature Heat

of Fusion Heat of Vaporization Liquid Solid Time Gas

Temperature Kinetic Potential Energy Energy Potential Energy Kinetic Energy

Kinetic Energy Time F. Gas Laws Gas volume is controlled by pressure and temperature 1. Units Volume mL L cm3 Pressure atm kPa

Temperature C K is the only unit to be used F. Gas Laws 2. Boyle Law As pressure increases, volume decreases P1V1 = P2V2 3. Charles Law As temperature increases, volume increases V1 = V2 T1

T2 Converting pressure units Convert 1.9 atm to kPa 101.3 kPa = x kPa 1 atm 1.9 atm (101.3)(1.9) = (1)(x) x = 192.7 kPa Convert 180 kPa to atm 101.3 kPa = 180 kPa 1 atm

x atm (101.3) (x) = (1)(180) x = 1.8 atm 20 mL of CO2(g) at 1.2 atm is compressed to 15 mL. What is the new pressure of the gas? P1V1 = P2V2 20 mL = volume (V) 1.2 atm = pressure (P) 15 mL = volume (V) (1.2 atm)(20 ml) = P2(15 ml) P2 = 1.6 atm

100 mL of NH3(g) at 25C is heated to 50C. What is the new volume of the gas? V1 = V2 T 1 T2 25C + 273 = 298 K 50C + 273 = 323 K 100 ml = V2 298 K 323 K (100)(323) = (298) V 32300 = 298V 298 298

V2 = 108 ml When 500 mL of NH3(g) at 25C was cooled, its new volume became 250 mL. What is the Celsius temperature of the gas? V1 = V2 T1 T2 25C + 273 = 298 K 500 ml = 250 mL 298 K T2

(298)(250) = (500) T 74500 = 500 T 500 500 T2 = 149 K 149 K = C + 273 -124C

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