Unit 6: Physical Behavior of matter
I. Classification of Matter Substance - Definite Composition (Homogenous) Mixture of Substances Physically Separable
I. Classification of Matter Element (Fe, K, Ca, Ne) Compound Two or more different elements bonded Chemically Separable Ionic (Metal and Nonmetal) Molecular (covalent bond) (Nonmetals) Individual atoms
Checks for Understanding 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 Which of the following substances cannot be separated by chemical change? Nitrogen (g) Sodium chloride (s) Carbon dioxide (g) Magnesium Sulfate (aq)
I. Classification of Matter Heterogeneous Nonuniform; distinct phases Homogenous Uniform throughout (air, tap water, solutions)
Check for Understanding A pure substance that is composed only of identical atoms is classified as a A compound An element A heterogeneous mixture A homogeneous mixture 3. A heterogeneous material may be A mixture Pure substance
II. Separating Matter Certain types of matter can be separated using various methods. Monatomic Elements - _____________ be decomposed (broken apart) using _____________ or ______________ means. Diatomic Elements and Compounds (ie – O2 and H2O) – can be decomposed using __________________ only CANNOT PHYSICAL CHEMICAL CHEMICAL MEANS
II. Separating Matter Mixtures – can be separated using ___________________ Filtration – Evaporation – Chromatography – Distillation – PHYSICAL MEANS Separation by particle size Separation by boiling point Separation by polarity Separation by boiling point
Check for Understanding Which of the substances could be decomposed by a chemical change? A) sodium B) aluminum C) magnesium D) ammonia A sample of a material is passed through a filter paper. A white deposit remains on the paper, and a clear liquid passes through. The clear liquid is then evaporated, leaving a white residue. What can you determine about the nature of the sample? What are some of the differences between a mixture of iron and oxygen and compound composed of iron and oxygen? It is a heterogeneous mixture In a mixture the elements are not bonded with each other and can be physically separated. In a compound the elements are bonded and can only be separated through a chemical reaction.
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 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) ENDOTHERMIC ABSORBED Heating Curve: ___________ - Energy is being ________ gas l g liquid s l s g solid Sublimation (video)- Solid changes directly to a gas Heating Curve Animation
IV. Heating and Cooling Curves AB BC CD DE EF Kinetic Energy Potential Energy Phase Con-stant Con-stant ↑ ↑ ↑ Con-stant Con-stant Con-stant ↑ ↑ gas solid l g boiling s l melting liquid
Check for Understanding 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 EXOTHERMIC RELEASED Cooling Curve: ___________ - Energy is being ________ gas g s g l liquid l s solid Deposition - Gas changes directly to a solid
IV. Heating and Cooling Curves AB BC CD DE EF Kinetic Energy Potential Energy Phase Con-stant Con-stant ↓ ↓ ↓ Con-stant Con-stant Con-stant ↓ ↓ g l condensing solid gas liquid l s Freezing
Check for Understanding 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 transferred from one substance to another Average kinetic energy of its particles (how fast they’re moving) Joules (J) or Calories (cal) 1 cal = 4.18 J Celsius (oC) or Kelvin (K) (K = oC + 273) T q
V. Temperature vs. Heat K = oC + 273 K = Kelvin oC = degrees Celsius Temperature Scales (See Ref. Tabs.): K = oC + 273 K = Kelvin oC = degrees Celsius Convert: 200 degrees Celsius to Kelvin Law of Conservation of Energy: Heat Transfer: K = oC + 273 K = 200oC + 273 = 473 K Energy (heat) cannot be created or destroyed. Energy (heat) can be TRANSFERRED. 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. Tabs. on Table ______ ) Specific Heat: Specific Heat for water: __________ (Found on Table _______ in Ref. Tabs) T q = heat (J) m = mass (g) c = specific heat (J/g*oC) ∆T = change in temperature (oC) q = m c ∆T The amount of heat it takes to raise the temperature of 1g of a substance 1oC 4.18 J/g*K B Specific heat for concrete is 0.84 J/(gK) – why concrete is much hotter than water on a sunny summer day
Check for Understanding 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) 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. Why do you feel cold after you get out of a hot shower. (link) Heat moves from metal handles (warm) to your hands (cold).
VI. Measurement of Heat Energy Example: How many joules are absorbed when 50.0 g of water are hater from 30.2 oC to 58.6 oC? m = 50.0 g Ti = 30.2 oC Tf = 58.6 oC q = ? q = m c ∆T q = (50.0 g) (4.18 J/goC ) (58.6 oC – 30.2 oC) 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 oC to 60.0 oC? 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. q = m c ∆T m = 50.0 g Ti = 30.2 oC Tf = 58.6 oC q = ? q = (50.0 g) (4.18 J/goC ) (60.0 oC – 70.0 oC) q = - 2090 J ( - means heat is released) m = 50.0 g Ti = 289.6 K 16.6 oC Tf = 309.6 K 36.6 oC q = ? q = m c ∆T q = (50.0 g) (4.18 J/goC ) (36.6oC – 16.6oC ) q = 4180 J
VII. Heat Of FUSION q = mHf Heat of Fusion for water: ____________ (Found on Table ________ in Ref. Tabs) Equation: (Found on Table _______ in Ref. Tabs) Amount of heat absorbed (endothermic) to change a substance from s to l at its melting point 334 J/g B T q = mHf
VII. Heat Of FUSION q = mHf m = 255 g Hf = 334 J/g q = ? Example: How many joules are required to melt 255 g of ice at 0.00oC? Example: What is the total number of joules of heat needed to change 150 g of ice to water at 0.00oC? q = mHf m = 255 g Hf = 334 J/g q = ? q = (255 g) (334 J/g) q = 85,170 J 85,200 J q = 50,100 J 5.0 x 10 4 J or 50. kJ
VIII. Heat Of Vaporization Heat of Vaporization for water: ____________ (Found on Table ________ in Ref. Tabs) Equation: (Found on Table _______ in Ref. Tabs) Amount of heat absorbed (endothermic) to change a substance from l to g at its boiling point 2260 J/g B T q = mHv
VIII. Heat Of Vaporization Example: How many joules of energy are required to vaporize 423 g water at 100 oC and 1 atm? Example: What is the total number of joules required to completely boil 125 g of water at 100 oC at 1 atmosphere? q = mHv m = 423 g Hv = 2260 J/g q = ? q = (423g) (2260 J/g) q = 955,980 J 956,000 J q = 282,500 J 283,000 J
IX. Calorimetry - Measure the amount of heat given off in a reaction. 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