Unit 6: Physical Behavior of matter

Unit 6: Physical Behavior of matter

I. Classification of Matter
Substance - Definite Composition (Homogenous) Mixture of Substances Physically Separable

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)

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 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

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
AB BC CD DE EF Kinetic Energy Potential Energy Phase Con-stant Con-stant ↑ ↑ ↑ Con-stant Con-stant Con-stant ↑ ↑ gas solid l  g boiling s  l melting liquid

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

EXOTHERMIC RELEASED Cooling Curve: ___________ - Energy is being ________  gas g  s g  l  liquid l  s  solid Deposition - Gas changes directly to a solid

AB BC CD DE EF Kinetic Energy Potential Energy Phase Con-stant Con-stant ↓ ↓ ↓ Con-stant Con-stant Con-stant ↓ ↓ g  l condensing solid gas liquid l  s Freezing

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 K = Kelvin oC = degrees Celsius Convert: 200 degrees Celsius to Kelvin Law of Conservation of Energy: Heat Transfer: K = oC + 273 K = 200oC = 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

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).

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 = J  5940 J

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 K to 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 = J ( - means heat is released) m = 50.0 g Ti = K  16.6 oC Tf = K  36.6 oC q = ? q = m c ∆T q = (50.0 g) (4.18 J/goC ) (36.6oC – 16.6oC ) q = 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

Unit 6: Part B Behavior of Gases

X. Endothermic and exothermic (revisited)
Energy is either absorbed or released in chemical reactions Remember: - Breaking bonds is ____________________ - Heat is ___________ the reaction from the surroundings - Ex) heat + Br2  Br + Br - Creating bonds is _________________ - Ex) N+ N N2+ energy ENDOTHERMIC ENTERING EXOTHERMIC EXITING

Where does the heat come from (or go to)? ______________________ For exothermic reactions, heat (energy) leaves the reaction and moves __________________________. Therefore, making the surrounding temperature _________________ Exothermic Chemical Equation: The surroundings into the surroundings warmer Reactant(s)  Product(s) + HEAT

For endothermic reactions, heat (energy) leaves the surroundings and moves __________________________. Therefore, making the surrounding temperature _________________ Endothermic Chemical Equation: Into the reaction colder Reactant(s) + HEAT  Product(s) ENDOTHERMIC REACTION VIDEO

DISORDER Entropy: Degree of ___________________ in a system Entropy Increases, if 1) _________________________________________________ 2) __________________________________________________ 3) __________________________________________________ Substance goes from a s  l  g More moles of gas are produced Large molecule breaks up

Enthalpy: It is a measure of the heat released or absorbed in a reaction. If heat is released then the reaction is an exothermic reaction, heat will be on the products side and the ΔH will be negative. If heat is absorbed then the reaction is an endothermic reaction, heat will appear on the reactants side and the ΔH will be positive. (see Table ____ on Reference Tables). Examples: Tell whether each of the following reactions are endothermic or exothermic (you may have to use table I). Then tell whether the temperature of the surroundings increases or decreases as a result. Entropy Endo/Exothermic _______Surroundings _____________________C6H12O O2  6CO H2O + heat _________ I warmer ↑ Exo (-∆H)

Entropy Endo/Exothermic _______Surroundings 1. ___________________C6H12O6 (s) + 6O2(g)  6CO2 (g) + 6H2O (l)+ heat ____________ 2. _________________ 2H2O kJ  2H2 (g) + O2 (g) __________ 3. ____________________ N2 (g) + 3H2 (g)  2NH3 (g) ________ 4. ______________________ NaOH(s)  Na+ (aq) + OH- (aq) _________ 5. _____________________ C2H5OH(l) + 3O2 (g)  2CO2 (g) H2O(l) _____ 6. ________________________2KClO3(s)  2KCl(s) + O2(g) + heat ________ 7. ______________ _______H+(aq) + C2H3O2(aq) + heat  HC2H3O2(l) ______ ↑ exo(-∆H) warmer ↑ colder endo(+∆H) ↓ exo (-∆H) warmer ↑ exo(-∆H) warmer SKIP exo(-∆H) warmer ↑ exo(-∆H) warmer ↓ colder endo(+∆H)

XI. Kinetic Molecular Theory
DO NOT ACTUALLY EXIST Ideal Gas: A set of gases that ________________________________, but represents a standard that we can use to predict the behavior of gases. Kinetic Molecular Theory of Ideal Gases tell how ideal gases behave: 1 . 2. 3. 4. 5. However, Real gases behave ideally under ________________ temperature and ____________ pressure Particles are small and spread far apart Travel in constant, random, straight line motion Can not happen in real-world Have no attractive forces (IMF’s) Don’t lose energy when they collide, but they can transfer energy in the collision Move faster when hotter HIGH LOW

XII. Vapor Pressure ______ VP = _________________________ Example:
______ VP = _________________________ Example: Factors for Vapor Pressure 1. 2. Pressure a liquid “feels” pushing it to evaporate (turn to gas) HIGH Evaporates easily Ethanol has higher VP than H2O, so it evaporates more easily Strength of intermolecular force: molecules held together by dipole-dipole (polar) have lower VP than Van der Waals (nonpolar) (H2O does not evaporate as fast as methane (CH4)) Temperature/Pressure: increase temp., increase VP

XII. Vapor Pressure Vapor Pressure of Four Liquids (See Table _____)
Questions: 1. What is the vapor pressure of propanone at 35 oC? _______ 2. What temperature does water boil at if the pressure is 70 kPa? ______ 3. What is the normal boiling point of ethanoic acid? __________ Sublimation: Example: “Dry Ice” CO2 (s)  CO2 (g) H Measured in: kPa (101.3 kPa = 1 atm = 760 mmHg = 760 torr.) 48 kPa 90 degrees C 117 degrees C Occurs because solids have very weak IMF (usually Van de Waals). They go directly from s  g and have HIGH VP

How could you change the boiling point temperature of a substance without adding anything to the substance? You spill a glass of water on the floor. How could you get the water to evaporate faster, without using a mop or something to soak it up? Change your altitude (higher altitude (lower pressure), lower boiling point) Increase the temperature of the room. Spread out the puddle (increase surface area).

XIII. Evaporation vs. Boiling Point
Converting l  g at the surface of a liquid that is NOT BOILING (leaving clothing out to dry) When the vapor pressure of liquid = air pressure Temperature when a liquid boils at standard pressure (101.3 kPa or 1 atm = sea level) Water = 100oC 1. Substance (strong IMF, slower evaporation) 2. Temperature (high temp., faster evaporation) 3. Surface area (increase, faster evaporation) Water Boiling at 70.0oC Video

You move to Denver, CO, where the altitude is 5,183 feet above sea level (mile high city – approximately). What would this high altitude due to the spacing of air particles? Based on your answer to question 1, what would this do to the vapor pressure of water? Based on your answer to question 2, what would this do to the boiling point temperature of water in Denver? The air pressure would be lower, so the particles would be more spread out. The vapor pressure would be less (less pressure pushing on the water – approximately 90 kPa in Denver) Boiling temperature would be lower (95 degrees Celsius)

Expanding Marshmallows
Think about this Expanding Marshmallows Why did this happen to the marshmallow? Decreased the pressure inside and the marshmallow expanded 2. What is the relationship between pressure and volume? As pressure decreases, volume increases 3. What remained constant in this video? temperature

XIV. Gas Laws A. Pressure and Volume (Boyle’s Law)
A. Pressure and Volume (Boyle’s Law) As pressure _______________, volume ________________, and _____________ is constant Volume Relationship: Pressure increases decreases temperature inverse Equation: P1V1 = P2V2 Question: If the pressure on a gas is halved, under constant temperature, what will happen to the volume of the gas? The volume will double

Think about this You blow up a balloon and it expands each time you exhale into the balloon. What happens to the number of gas particles you put in the balloon each exhale? increases 2. What happens to the pressure on the balloon? It increases (more gas particles pushing against the wall of the balloon) 3. What is the relationship between gas particles and pressure More gas particles, more pressure

XIV. Gas Laws B. Pressure and Number of Gas Particles
B. Pressure and Number of Gas Particles As number of gas particles _______________, pressure ________________ Pressure Relationship: # of Gas Particles increases increases Direct Question: If the number of particles of a gas is tripled, under constant volume, what will happen to the pressure on the gas? The pressure will triple

Think about this You blow up a balloon inside the school and bring it outside (it is – 10oC out). What would you see happen to the balloon as you stand outside? The balloon would get smaller 2. Why would this occur? The cold air would cool the gas inside, bringing the molecules closer together 3. What would you see happen once you brought it back inside? Why? It would expand. Warm air would warm the gas inside, causes the molecules to spread out (volume increases).

XIV. Gas Laws C. Temperature and Volume (Charles’s Law)
C. Temperature and Volume (Charles’s Law) As temperature_______________, volume ________________ and ______________ is constant Volume Relationship: Temperature increases increases pressure Direct Equation: V 1 = V 2 T1 T2 Question: If the temperature of a gas is halved, under constant pressure, what will happen to the volume of the gas? The volume will be halved

Think about this: Pressure and Temperature of A Gas
On cold days the sensor in my car says that my tires have lower than normal pressure. Why would this happen? The cold air makes the gas inside the tire slow down causing the tire pressure to be less 2. What two properties and being compared? Temperature and pressure 3. What is the relationship between these two properties As temperature decreases, pressure decreases Video Demonstration

XIV. Gas Laws D. Temperature and Pressure (Gay-Lussac’s Law)
D. Temperature and Pressure (Gay-Lussac’s Law) As temperature_______________, pressure________________ and ______________ is constant Pressure Relationship: Temperature increases increases volume Direct Equation: P 1 = P 2 T1 T2 Question: If the temperature of a gas is tripled, under constant volume, what will happen to the pressure of the gas? The pressure will triple

XV. Combined Gas Law Equation (See Table _____)
The relationships among pressure, temperature, and volume can be mathematically represented by an equation known as the combined gas law. Equation (See Table _____) Important notes when using the equation: T P 1 V1 = P 2 V2 T T2 1. Temperature must be in Kelvin, NOT Celsius 2. For pressure and volume, make sure the units of the initial and final are the same 3. If a variable remains constant, you can take it out of the equation

XV. Combined Gas Law 1. The pressure of a gas at 200. K is increased from 200. kPa to 305. kPa at a constant volume. What is the new temperature? 2. If a gas at 8.00 atm is cooled from 600. K to 150. K in a rigid container, what is the final pressure? 305 K 2.00 atm

XV. Combined Gas Law 3. If I have 2.90 L of gas at a pressure of 5.00 atm and a temperature of 320. K, what will be the temperature of the gas if I decrease the volume of the gas to 2.40 L and decrease the pressure to 3.00 atm? 4. If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 250 K, and then I raise the pressure to 14 atm and increase the temperature to 350 K, what is the new volume of the gas? 159 K 28 L

Unit 6: Physical Behavior of matter

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