UNIT 6 PHYSICAL BEHAVIOR OF MATTER I CLASSIFICATION

UNIT 6: PHYSICAL BEHAVIOR OF MATTER

I. CLASSIFICATION OF MATTER Substance - Definite Composition (Homogenous) Physically Separable Mixture of Substances

I. CLASSIFICATION OF MATTER Element Compound (Fe, K, Ca, Two or more Ne) different elements Chemicallybonded Separable Individual atoms Ionic (Metal and Nonmetal) Molecular (covalent bond) (Nonmetals)

CHECKS FOR UNDERSTANDING 1. A) B) C) D) A compound differs from an element in that a compound Is homogeneous Has a definite composition Has a definite melting point Can be decomposed by a chemical reaction 2. Which of the following substances cannot be separated by chemical change? A) Nitrogen (g) B) Sodium chloride (s) C) Carbon dioxide (g) D) Magnesium Sulfate (aq)

I. CLASSIFICATION OF MATTER Homogenous Uniform throughout (air, tap water, solutions) Heterogeneous Nonuniform; distinct phases

CHECK FOR UNDERSTANDING 1. A pure substance that is composed only of identical atoms is classified as a A) A compound B) An element C) A heterogeneous mixture D) A homogeneous mixture 3. A heterogeneous material may be A) An element B) A compound C) A mixture D) Pure substance

II. SEPARATING MATTER • Certain types of matter can be separated using various methods. • Monatomic Elements - _______ be CANNOT decomposed (broken apart) using PHYSICAL CHEMICAL _______ or _______ means. • Diatomic Elements and Compounds (ie – O 2 CHEMICAL MEANS and H 2 O) – can be decomposed using _________ only

II. SEPARATING MATTER Mixtures – can be separated using PHYSICAL MEANS __________ Separation by particle size Filtration – Separation by boiling point Evaporation – Separation by polarity Chromatography – Separation by boiling point Distillation –

CHECK FOR UNDERSTANDING 1. Which of the substances could be decomposed by a chemical change? A) sodium B) aluminum C) magnesium D) ammonia 2. A sample of a material is passed through a filter paper. A white deposit remains on the paper, and It is a heterogeneous a clear liquid passes through. The clear liquid is mixture then evaporated, leaving a white residue. What can you determine about the nature of the sample? In a mixture the elements are not bonded with each other and can be physically separated. In 3. What are some of the differences between a a compound the elements are bonded and can mixture of iron and oxygen and compound

THINK ABOUT THIS What happens to the spacing and speed of particles at each of the phases? SOLID LIQUID GAS

III. FORMS OF MECHANICAL ENERGY Kinetic Energy of movement (similar to temperature) (how fast atoms are moving) Potential Energy Stored energy (energy of position) More spread out (gas) = High PE Closer together (solid) = Low PE

IV. HEATING AND COOLING CURVES (ANIMATION) Heating Curve: ______ - Energy is being ENDOTHERMIC ABSORBED ____ l g s l liquid gas s g solid Sublimation (video)- Solid changes directly to a gas Heating Curve Animation

IV. HEATING AND COOLING CURVES A B Kinetic ↑ Energy Potentia Conl Energy stant Phase solid B C C D Constant ↑ ↑ Constant s l liquid melting D E Constant ↑ l g boiling E F ↑ Constant gas

CHECK FOR UNDERSTANDING 1. A substance begins to a melt. What happens to the potential and kinetic energy? PE increase, KE stays the same 2. The temperature of a substance refers to what type of energy? Kinetic energy 3. How does the speed and space of water molecules compare when in a liquid phase to a gas phase Molecules move faster and more spread out in gas phase

IV. HEATING AND COOLING CURVES Cooling Curve: ______ - Energy is being EXOTHERMIC RELEASED ____ gas g l liquid l s g s solid Deposition - Gas changes directly to a solid

IV. HEATING AND COOLING CURVES A B Kinetic ↓ Energy Potentia Conl Energy stant Phase gas B C C D Constant ↓ ↓ Constant g l condensing D E Constant ↓ liquid l s Freezing E F ↓ Constant solid

CHECK FOR UNDERSTANDING 1. As a substance condenses, what happens to its potential and kinetic energy? PE decreases, KE stays the same 2. What phase is a substance in when it has its highest kinetic energy? gas 3. How does the speed and space of water molecules compare when in a liquid phase to a solid phase Molecules move slower and are closer together in solid phase

V. TEMPERATURE VS. HEAT Amount of energy Average kinetic energy transferred from one of its particles (how substance to fast they’re moving) another Joules (J) or Celsius (o. C) or Kelvin (K) Calories (cal) (K = o. C + 273) 1 cal = 4. 18 J T q

V. TEMPERATURE VS. HEAT Temperature Scales (See Ref. Tabs. ): K = o. C + 273 K = Kelvin Celsius o. C = degrees Convert: 200 degrees Celsius to Kelvin K = o. C + 273 K = 200 o. C + 273 = 473 K Law of Conservation of Energy: Energy (heat) cannot be created or destroyed. Energy (heat) can be TRANSFERRED. Heat Transfer: HEAT ALWAYS MOVES FROM WARMER OBJECTS TO COLDER OBJECTS

VI. MEASUREMENT OF HEAT ENERGY • The amount of heat given off or absorbed in a reaction can be calculated using the following equation: (See Ref. T Tabs. on Table ______ ) q = heat (J) q = m c ∆T m = mass (g) c = specific heat (J/g*o. C) ∆T = change in temperature (o. C) • Specific Heat: The amount of heat it takes to raise the temperature of 1 g of a substance 1 o. C 4. 18 B • Specific Heat for water: _____ (Found on Table J/g*K _______ in Ref. Tabs) Specific heat for concrete is 0. 84 J/(g. K) – why concrete is much hotter than water on a sunny

CHECK FOR UNDERSTANDING 1. You wake up in the morning and your barefoot touches the ceramic floor and it feels cold. Explain which way heat is being transferred. Heat moves from your body (warm) to the floor (cold) 2. You are cooking pasta in a boiling metal pot of water. You grab the metal handles with your bare hands (ouch!). Explain which way heat is being transferred. Heat moves from metal handles (warm) to your hands (cold). 3. Why do you feel cold after you get out of a hot shower. (link)

VI. MEASUREMENT OF HEAT ENERGY Example: How many joules are absorbed when 50. 0 g of water are hater from 30. 2 o. C to 58. 6 o. C? m = 50. 0 g q = m c ∆T Ti = 30. 2 o. C ) (58. 6 o. C – 30. 2 q = (50. 0 g) (4. 18 J/g o Tf = 58. 6 C q = ? q = 5935. 6 J 5940 J

VI. MEASUREMENT OF HEAT ENERGY Example: How many joules of heat energy are released when 50. 0 g of water are cooled from 70. 0 o. C to 60. 0 o. C? q = m c ∆T m = 50. 0 g q = (50. 0 g) (4. 18 J/go. C ) (60. 0 o. C Ti = 30. 2 o. C Tf = 58. 6 o. C – 70. 0 o. C) q = - 2090 J ( - means heat is q = ? released) Example: 50. 0 g of water goes from 289. 6 K to 309. 6 K. A) Is heat energy released or absorbed? B) Calculate the amount energy. m = 50. 0 g q = m c ∆T Ti = 289. 6 K o. C ) (36. 6 o. C q = (50. 0 g) (4. 18 J/g o 16. 6 C – 16. 6 o. C ) Tf = 309. 6 K q = 4180 J o 36. 6 C

VII. HEAT OF FUSION • Heat of Fusion: Amount of heat absorbed (endothermic) to change a substance from s to l at its melting point 334 J/g B • Heat of Fusion for water: ______ (Found on Table ____ in Ref. Tabs) T • Equation: (Found on Table _______ in Ref. Tabs) q = m. Hf

VII. HEAT OF FUSION Example: How many joules are required to melt 255 g of ice at 0. 00 o. C? q = m. Hf m = 255 g q = (255 g) (334 J/g) H = 334 J/g f q = ? q = 85, 170 J 85, 200 J Example: What is the total number of joules of heat needed to change 150 g of ice to water at 0. 00 o. C? q = 50, 100 J 5. 0 x 10 4 J or 50. k. J

VIII. HEAT OF VAPORIZATION • Heat of Vaporization: Amount of heat absorbed (endothermic) to change a substance from l to g at its boiling point 2260 J/g B • Heat of Vaporization for water: ______ (Found on Table ____ in Ref. Tabs) T • Equation: (Found on Table _______ in Ref. Tabs) q = m. Hv

VIII. HEAT OF VAPORIZATION Example: How many joules of energy are required to vaporize 423 g water at 100 o. C and 1 atm? q = m. Hv m = 423 g q = (423 g) (2260 J/g) Hv = 2260 J/g q = ? q = 955, 980 J 956, 000 J Example: What is the total number of joules required to completely boil 125 g of water at 100 o. C at 1 atmosphere? q = 282, 500 J 283, 000 J

IX. CALORIMETRY • Used to: - Measure the amount of heat given off in a reaction. - Use q = m c ∆T to find the amount of heat lost or gained in a sample