Matter and Energy I. Matter Matter is anything that has mass and volume Mass Amount of matter Measured in grams (g) Volume Space matter occupies Measured in milliliters (ml), liters (L) or cubic centimeters (cm 3 ) Two types of matter Pure substances Mixtures A. Pure Substances Uniform composition The same throughout the sample Two Types Elements Compounds 1. Elements Simplest form of matter Cannot breakdown Smallest part called atom Represented using a capital letter or capital letter and lower case letter Carbon (C) Sodium (Na) 2. Compound Two or more elements chemically joined in a specific ratio Properties of the compound are different than the elements that make it up Compounds can be broken down Decomposed Examples Water (H 2 O) Hydrogen peroxide (H 2 O 2 ) B. Mixture Two or more substances physically joined in any ratio Keep the properties of their components (parts) Can be separated by physical means Filtration Distillation Exist in two forms Heterogeneous Homogenous 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) Density a physical property used to identify a substance D = mass Volume II. Energy Energy is the driving force behind change Cannot be created or destroyed Does change its form SCREAM T Sound Chemical Radiant (light) Electrical Atomic (nuclear) Mechanical Thermal (heat) 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 o q = m C T (used for heating or cooling) o q = m H f (used for melting or freezing) o q = m H v (used for boiling or condensing) Constants found on Table B C water = 4.18J/g C H f = 334J/g H v = 2260J/g How many joules are required to heat 40g water at 30 C to 80 C? q = m C T q = 40g x 4.18J/g C x 50 C q = 8360J 5000J were added to 30g water at 25 C. What is the new temperature? q = m C T 5000J = 30g x 4.18J/g C x T 5000 = 125.4 x T T = 39.9 ~ 40 Since energy was added, temp increases T new = 25 + 40 T new = 65 C How many joules are needed to melt 100g ice at 0 C q = m H f q = 100g x 334J/g q = 33400J III. Phases of Matter Solids Liquids Gases A. Solids o Matter that has specific shape and specific volume o Atoms closely packed together o Cannot be compressed B. Liquids o Matter that has a specific volume but takes the shape of the container o Atoms are close but have some space between them o Cannot be compressed o Can be poured
C. Gases o Matter that takes the shape and volume of the container o Atoms have free space between them o Compressible o Can be poured D. Phase Changes If energy is added o Solid to liquid Melting o Liquid to gas Boiling o Solid to gas Sublimation If energy is removed o Liquid to solid Freezing o Gas to liquid Condensing o Gas to solid Deposition E. Phase Diagram
F. Gas Laws Gas volume is controlled by pressure and temperature 1. Units o Volume ml L cm 3 o Pressure atm kpa o Temperature C K is the only unit to be used 2. Boyle Law As pressure increases, volume decreases P 1 V 1 = P 2 V 2 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 CO 2 (g) at 1.2 atm is compressed to 15 ml. What is the new pressure of the gas? P 1 V 1 = P 2 V 2 (1.2 atm)(20 ml) = P 2 (15 ml) P 2 = 1.6 atm
3. Charles Law As temperature increases, volume increases V 1 = V 2 T 1 T 2 100 ml of NH 3 (g) at 25 C is heated to 50 C. What is the new volume of the gas? V 1 = V 2 T 1 T 2 25 C + 273 = 298 K 50 C + 273 = 323 K 100 ml = V 2 298 K 323 K (100)(323) = (298) V 32300 = 298V 298 298 V 2 = 108 ml When 500 ml of NH 3 (g) at 25 C was cooled, its new volume became 250 ml. What is the Celsius temperature of the gas? V 1 = V 2 T 1 T 2 25 C + 273 = 298 K 500 ml = 250 ml 298 K T 2 (298)(250) = (500) T 74500 = 500 T 500 500 T 2 = 149 K 149 K = C + 273-124 C