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SOL REMEDIATION 2016. SESSIONS AND TOPICS Date & TimeSubject Monday 4/18/16 8:00 – 9:00 Am Foundation Wednesday 4/20/16 8:00- 9:00 Am Matter Friday 4/22/16.

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Presentation on theme: "SOL REMEDIATION 2016. SESSIONS AND TOPICS Date & TimeSubject Monday 4/18/16 8:00 – 9:00 Am Foundation Wednesday 4/20/16 8:00- 9:00 Am Matter Friday 4/22/16."— Presentation transcript:

1 SOL REMEDIATION 2016

2 SESSIONS AND TOPICS Date & TimeSubject Monday 4/18/16 8:00 – 9:00 Am Foundation Wednesday 4/20/16 8:00- 9:00 Am Matter Friday 4/22/16 8:00 – 9:00 Am Atomic Theory Monday 4/25/16 8:00 – 9:00 Am Modern Atomic Theory Wednesday 4/27/16 8:00- 9:00 Am Periodic Table/ Period Trends Friday 4/29/16 8:00 – 9:00 Am Bonding Monday 5/2/16 8:00 – 9:00 Am Naming Wednesday 5/4/16 8:00 – 9:00 Pm Reactions Friday 5/6/16 8:00 – 9:00 Am Moles Monday 5/9/16 8:00 – 9:00 Am Stoichiometry Wednesday 5/11/16 8:00 – 9:00 Am Gas Laws Friday 5/13/16 8:00 – 9:00 Am Energy Monday 5/18/16 8:00 – 9:00 Am Solutions Wednesday 5/20/16 8:00 – 9:00 Am Equilibrium Friday 5/22/16 8:00 – 9:00 Am Acid and Base

3 SESSION 1: FOUNDATIONS Topics 1.Metric conversion 2.Scientific Notation 3.Significant Figures 4.Percent Error 5.SI units

4 METRIC CONVERSIONS When converting any unit follow the steps below 1.Start with the information given 2.Set up a conversion bar  On the top place the new unit on the top  Place the old unit on the bottom  Smaller unit hold the non- one factor 3.Do the math! 4.Write the answer WITH units http://www.kentchemistry.com/links/Measurements/metricconversions.htm

5 SAMPLE PROBLEMS

6 SCIENTIFIC NOTATION A system of writing big and small numbers 4.67 x 10 6 5.23 x10 -3 4670000.00523 Base Number (argument) – tells the value Times and Exponent – tells the scale of the number If exponent is negative Number is smaller than one If exponent is positive Number is greater than one

7 SIGNIFICANT FIGURES A system of rounding to account for the precision of the measurement Measurements that do not apply to Significant Figures Counted numbers Conversion factors

8 SIGNIFICANT DIGITS 1.All non-zero numbers are significant 2.Zeros that are sandwiched between non-zero numbers are significant 3.The base of a number in scientific notation

9 NON-SIGNIFICANT NUMBERS Place holders Zeros to the right of a number WITHOUT decimal 100000000 ONLY 1 sig fig Zeros to the left of decimal point.0000025 ONLY 2 sig figs

10 SAMPLE PROBLEMS 1)2804 m 2) 2.84 km 3) 5.029 m 4) 0.003068 m 5) 4.6 x 10 5 m 6) 4.06 x 10 -5 m 1) 4 2) 3 3) 4 4) 4 5) 2 6) 3

11 PERCENT ERROR

12 SCIENTIFIC INTERNATIONAL UNITS For every measurable amount there is a standard unit in which it is to be reported. http://www.learnalberta.ca/content/memg/division03/International%20System%20of%20Units/index.html

13 MATTER- SESSION 2 Topics 1.Basic Atomic Structure A.Isotope B.Ion 2.Nuclear Decay 3.Half life 4.Average Atomic Mass 5.Physical and Chemical A.Properties B.Changes

14 ATOMIC STRUCTURE Nucleus is the center of an atom Outside nucleus are electrons E P P N N Neutron Proton E

15 ISOTOPE AND ION Isotope: The same atom with different number of neutrons Ion: The same atom with different numbers of electrons Atoms from ion to become more stable  achieve a full shell Metals: lose electron(s) to form cation (positive) Non-Metals gain electron(s) to for anion (negative) http://mrbloch516.edublogs.org/2015/01/page/2/

16 NUCLEAR DECAY Stability When the number of protons and neutrons are one to one the nucleus is stable When it is more than one to one nucleus is unstable and decay/transmutation can occur ParticlesSymbols Alpha Beta Beta (positron) Neutron Gamma

17 TRANSMUTATION EXAMPLES http://www.chemteam.info/ChemTeamIndex.html

18 HALF LIFE The amount of time it takes for half the nuclear material within a substance to transmute/ decay Calculated using a time and mass chart Always start with time zero

19 HALF LIFE Sample Problem The half-life of chromium-51 is 28 days. If the sample contained 510 grams, how much chromium would remain after 56 days? MassTime 510 g0 days 255 g28 days 127.5 g56 days Divide by 2 Add ½ life

20 AVERAGE ATOMIC MASS The mass of an atom factors the abundance of the all naturally occurring isotopes Weighted average How to solve 1.Make a table of values that include A.Isotope B.Mass C.Percent 2.Covert the percent to decimals (divide by 100) 3.Multiple by the mass of the isotope 4.Add the answers together

21 SAMPLE PROBLEM The element copper has naturally occurring isotopes with mass numbers of 63 and 65. The relative abundance and atomic masses are 69.2% for a mass of 62.93amu and 30.8% for a mass of 64.93amu. Calculate the average atomic mass of copper. IsotopePercentMass Cu - 6369.2 %0.69262.93 amu Cu – 6530.8 %0.30864.93 amu (0.692) (62.93) = 43.55 (0.308) (64.93) = 24.67 68.22 amu

22 NOW YOU TRY! Magnesium has three naturally occurring isotopes. 78.70% of Magnesium atoms exist as Magnesium-24 (23.9850 amu), 10.03% exist as Magnesium-25 (24.9858 amu) and 11.17% exist as Magnesium-26 (25.9826 amu). What is the average atomic mass of Magnesium?

23 PHYSICAL AND CHEMICAL PROPERTIES Intrinsic Is dependent on the amount of matter Extrinsic Does not depend on the amount of matter Density Describes the sample Think 5 senses Physical Chemical A description of a change Examples Flammability Reactivity

24 PHYSICAL AND CHEMICAL CHANGES Physical Does not change the identity of the substance State changes Chemical Changes the identity of the substance Color change Precipitate Production of a gas Change in energy

25 ATOMIC THEORY –SESSION THREE Topics 1.Dalton 2.Thompson A.Cathode Ray experiment B.Electrons 3.Rutherford A.Gold Sheet experiment B.Nucleus C.Atomic structure 4.Bohr A.Energy level B.Emission spectra

26 DALTON Five principles All matter was made up of atoms Atoms were the smallest particle Atoms of the same elements had the same properties Atoms of different elements had different properties Atoms can rearrange and combine to form other substances Two Principles where incorrect atoms can only combine with other elements Diatomic molecules Atoms are NOT the smallest particle

27 THOMPSON Plum Pudding model Atoms were positive sphere with negative specs in them to be neutral in charge Cathode Ray experiment Beam was run through a vacuum and when introduced to magnetic fields the beam bent away from negative and toward positive Proved that there was a negative subatomic particle

28 RUTHERFORD Fired an alpha particle through a thin sheet of gold Found three things 1.Most went through foil A.Atoms are mostly empty space 2.Some reflected at an high angle A.There are positive particles in an atom 3.Some bounced back A.There is a dense mass in the center of an atom B.Nucleus  Neutrons  Protons http://www.daviddarling.info/encyclopedia/R/Rutherfords_experiment_and_atomic_model.html

29 BOHR Previously scientists could not explain why electrons did not crash into the positively charged nucleus Bohr determined that electron could only exist in specific energy level Quanta: specific amount of energy Each element has its own set of energy levels Each element produces an light emission spectra that corresponds to its energy level

30 EMISSION SPECTRA An electron absorbs a photon of energy and jumps to a higher energy level. Becomes excited moves to an excited state An electron releases a photon in the form of wavelength De-excites back down to it normal energy state Ground state The wavelength is seen as color http://imagine.gsfc.nasa.gov/educators/lessons/xray_spectra/background-atoms.html

31 MODERN ATOMIC THEORY-SESSION 4 1.Modern Model 2.Electron Configurations

32 MODERN MODEL Atoms is made up of three main subatomic particles Proton (positively charged) Neutron (no charge) Electron (negatively charged) Proton and neutrons make up the nucleus in the center of an atom Electrons exist in the electron cloud around the nucleus Atoms do not have definite edges Electron only exist within energy levels NEVER between

33 ELECTRON CONFIGURATIONS A system that details where the electrons can most likely be found Principle number (n) corresponds to the row of the periodic table Orbital type (s,p,d,f) corresponds to the region of the periodic table Aufbau Energy levels are filled lowest to highest energy Exception are transition metal Hund’s rule Every orbital must be occupied with one electron before electrons can double up Pauli Exclusion Electrons of opposite spins can exist within a single orbital

34 ORBITAL TYPE 4 types of orbitals Orbital can hold a total of two electrons Orbital typeWhere on Periodic Table Number of orbitalTotal number of electrons sGroups 1 – 2 and Helium 12 dGroups 3-12510 pGroups 13-1836 fRows under the main periodic table 714

35 HOW TO WRITE AN ELECTRON CONFIGURATION 1.Find the element on the periodic table 2.Start at the first row on the periodic table 3.Write the row number 4.Write the orbital block (s,p,d,f) 5.Write the number of electrons

36 SAMPLE PROBLEMS Li 1s 2 2s 1 V 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 3 Al 1s 2 2s 2 2p 6 3s 2 3p 1 Br 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 5

37 AUFBAU ORDER slideplayer.com chemistry.tutorcircle.com

38 USING HUND’S RULE AND AUFBAU ORDER Carbon 2p 2s 1s 3d 3p 2s 2p 3s 4s Iron

39 PERIODIC TABLE AND TRENDS –SESSION 5 1.Layout 2.Groups 3.Valence A.Ion formation 4.Trends A.Atomic radii B.Ionization energy C.Electronegativity

40 LAYOUT Metals Non- Metals metalloids Alkali Metals Alkaline Earth Metals Transition Metals Halogens Noble Gases H Lanthanide Series Actinide Series

41 DEFINITIONS Period: Row/ energy level Groups: Column / Valence Valence: Electrons in highest energy level

42 GROUPS/ REGIONS Regions Metals make up most of the periodic table. Left of the stair step. Metal lose electron to have a full outer shell Non-metals are to the right of the stair step. Non-Metal gain electron to have a full outer shell Metalloids: can act as either metal or non metal depending on conditions Groups Alkali Metals: Group 1, most reactive metals. Alkaline Earth Metals: Group 2, reactive metals. Transition Metals: Charge and valence change depending on what they are bonded with. Halogens: Group 17, most reactive non- metals Noble Gases: Group 18, inert gases. Non- reactive due to their full outer shell.

43 VALENCE GroupValence numberChargeGains or loses electrons Alkali Metals11+Loses 1 Alkaline Earth Metals22+Loses 2 Transition MetalsChangesDependent on bonding pair Loses Halogen71-Gains 1 Noble Gas80N/A

44 TRENDS Atomic Radii: The distance between two bonded nuclei. The size of a atom. Ionization Energy: The energy it takes to remove a valence electron. Electronegativity: The amount of relative attraction of electrons by the nucleus.

45 ATOMIC RADII Increases: Down a group because of shielding and the addition of energy levels. Decreases: Across a period due to increased nuclear pull www.dayasriojm.top www.shodor.org

46 IONIZATION ENERGY Decreases: Down a groups because inner electrons shield the pull of the nucleus which increases as energy levels are added. Increases: Across a period due to increase nuclear charge of the same energy level. www.shodor.org http://2012books.lardbucket.org/

47 ELECTRONEGATIVITY Decreases: Down a groups because inner electrons shield the attractive force of the nucleus which increase as energy levels are added. Increases: Across a period due to increase nuclear charge of the same energy level. www.shodor.org www.webelements.com Electronegativity

48 www.dayasriojf.top chemistry.about.com High and tight to the right Low, Left, and Loose

49 BONDING – SESSION 6 Topics 1.Ionic bonding A.Polyatoms ions 2.Covalent bonding A.Lewis dot structures B.Geometric shapes

50 IONIC BONDING Metal and a non-metal Ions form attract to oppositely charged ions Crystalline structures Metals form cations Lose electrons Positively charged Non-metals form anions Gain electrons Negatively charged www.geo.arizona.edu

51 PROPERTIES OF IONIC SALTS All salts have a unique bonding ratio Dependent on the size of the ions Causes different crystalline patterns High melting points and boiling points Rigid structures Fracture: the amount of pressure needed to break a salt along a line Can conduct electricity when dissolved Not when in their solid state www.science.uwaterloo.ca

52 POLYATOMIC IONS When two or more elements are covalently bonded together and carry a charge

53 COVALENT COMPOUNDS When electrons are shared Can share up to 6 electrons Single bond (2 electrons) Double bond (4 electrons) Triple Bonds (6 electrons Between two non-metals www.bbc.co.uk www.chemistryforkids.net

54 LEWIS DOT STRUCTURES Steps to draw a LDS 1.Calculate valence electrons for molecule A.That is the maximum number 2.Determine the central atom A.Most electro-positive 3.Place the electrons to make octet for EACH atom A.Use multiple bonds if needed 4.Double check the electron number and octet A.Duet rule for H and He

55 SAMPLE PROBLEMS H2OH2O CO 2 Valence calculation H: 1 x 2 = 2 O: 6 x 1 = 6 Total: 8 Valence calculation C: 4 x1 = 4 O: 6 x 2 = 12 Total: 16 socratic.org chemwiki.ucdavis.edu

56 MOLECULAR GEOMETRY How to predict the shape 1.Draw the Lewis Dot Structure 2.Count the number of lone pairs and bonded pairs around the CENTRAL atom A.Predict the shape www.studyblue.com

57 PROPERTIES OF COVALENT COMPOUNDS Low melting and boiling points Have different bond natures Polar Non-polar Exhibit intermolecular forces London Dispersion/Van Der Walls Dipole- dipole Hydrogen bonding

58 SHAPE DETERMINES PROPERTY Polar Regions of partial positive (electron deficient) and partial negative (electron rich) at the poles of a molecule Due to differences in electronegativity – uneven sharing of electrons Intermolecular forces Dipole-dipole Hydrogen bonding N,F,O and H Non-Polar Electronegativity differences at the poles of the molecules are small or zero Intermolecular forces London Dispersion

59 NAMING- SESSION 7 1.Ionic Naming A.Monatomic B.Polyatomic C.Transition metals 2.Covalent Naming A.Number prefixes

60 IONIC NAMING Main group metals Cation (metal) keeps it’s name Anion (non-metal) Take root of name and end it in “ide” Transition metal Cation (metal) keeps it’s name Followed by a roman numeral Indicates charge on the metal Is determined by the anion Anion (non-metal) Take root of name and end it in “ide” Polyatomic Polyatomic ion keeps its name no matter if it is the anion or cation

61 EXAMPLES/ NOW YOU TRY! 1.Na 2 CO 3 2.MgBr 2 3.FeCl 2 4.FeCl 3 5.Zn(OH) 2 6.Al 2 S 3 Sodium Carbonate Magnesium Bromide Iron (II) Chloride Iron (III) Chloride Zinc Hydroxide Aluminum Sulfide

62 FORMULAS TO NAME Ionic compounds have an overall charge of zero Cations and anions charges are equal and opposite If they are not then balance for charge by using subscript sodium phosphate Na 3 (PO4) 1 iron (II) bromide Fe 1 Br 2 PO 4 3 - Na 1+ ?1+ 3+ ? 3 - 2+ Fe 2+ Br 1- ?2+ 1- ?2-

63 NOW YOU TRY! 1.vanadium (V) phosphate 2.calcium oxide 3.magnesium acetate 4.aluminum sulfate 5.copper (I) carbonate V 3 (PO 4 ) 5 CaO Mg(C 2 H 3 O 2 ) 2 Al 2 (SO 4 ) 3 Cu 2 CO 3

64 COVALENT NAMING Binary Covalent compounds First element in formula keeps it’s name If there is more than one of the first element must include number prefix Second element Take the root and end it in “ide” Always include number prefix www.uzinggo.com

65 EXAMPLE/ NOW YOU TRY 1.BrO 3 2.BN 3.N 2 O 3 4.NI 3 5.SF 6 6.XeF 4 7.PCl 3 1.bromine trioxide 2.boron mononitride 3.dinitrogen trioxide 4.nitrogen triiodide 5.sulfur hexafluoride 6.xenon tetrafluoride 7.phosphorous trichloride

66 REACTIONS- SESSION 8 Types of Reactions Balancing equations

67 REACTION TYPES Synthesis Decomposition Single Replacement Double Replacement Neutralization Combustion Incomplete Complete

68 SYNTHESIS/ DECOMPOSITION Synthesis Two reactants form one product X +Y  XY 2H 2 + O 2  2H 2 O Decomposition One reactant break into two or more products Binary- two elements Breaks into the elements XY  X + Y 2H 2 O  2H 2 + O 2 Ternary- three or more elements Breaks in to simpler compounds CaCO 3  CaO + CO 2

69 SINGLE/ DOUBLE REPLACEMENT Single When an element replaces an another element within a compound Determined by an activity series Like replaces like Metal replaced metal Non-metal replaces non-metal A + BX  AX + B Double Occurs between ions in aqueous solution. A reaction will occur when pair of ions come together to produce at least one of the following A precipitate (solid) A gas Water or some other non-ionized substance AX + BY  AY + BX

70 COMBUSTION Incomplete When a hydrocarbon burns in oxygen gas and produces Carbon Monoxide, Carbon Dioxide, and Water 4CH 4 + 7O 2  2CO + 8H 2 O Complete When a hydrocarbon burns in oxygen gas and produces Carbon Dioxide and Water CH 4 + 2O 2  CO 2 + 2H 2 O

71 BALANCING EQUATIONS Law of Conservation of Mass Mass of Reactants must equal mass of the products Same number and type of atoms must be on both sides of the equation ____ Na 3 PO 4 + ____ KOH  ____ NaOH + ____ K 3 PO 4 Na PO 4 OH K 3 1 1 1 11 1 3 3 3 3 33

72 NOW YOU TRY! ____ MgF 2 + ____ Li 2 CO 3  ____ MgCO 3 + ____ LiF ____ P 4 + ____ O 2  ____ P 2 O 3 ____ RbNO 3 + ____ BeF 2  ____ Be(NO 3 ) 2 + ____ RbF

73 MOLES- SESSION 9 1.Percent composition 2.Mole A.Avogadro's number  Simple conversions 3.Empirical Formula 4.Molecular Formula

74 PERCENT COMPOSITION The percent of different elements/ions within a compound How to find percent composition 1.Find the molar mass of the compound A.Add up all the elements masses from the periodic table If there are subscripts multiply mass by the subscripts 2.Take the mass of the element an divide by the total mass of the compound 3.Multiple by 100

75 SAMPLE PROBLEMS Sample What is the percent of Carbon in C 2 H 6. C : 12.01(2) = 24.02 g/mol H: 1.009 (6) = 6.054 g/mol 30.074 g/mol 24.02 g/mol 30.074 g/mol Now you try! NaHSO 4 100 = 79.86 %

76 THE MOLE S.I. Unit for amount In one mole of ANY substance there are 6.022 x 10 23 particles Particles can be Atoms Ions Molecules Formula units The amount of mass in one mole is equal to the molar mass Calculated from periodic table The subscripts of a compound are the relative number of mole C 6 H 12 Carbon has 6 moles and Hydrogen has 12 moles in that compound

77 THE MOLE CONVERSION Mass (grams) Mole Particles Ions Atoms Molecules Formula units Volume (Liters) Gases at STP X by 6.022 x 10 23 ÷ by molar mass X by Molar mass ÷ by 6.022 x 10 23 ÷ by 22.4 L x by 22.4 L

78 PARTICLE TO MOLE/ MOLES TO PARTICLES 1.How many moles are 1.20 x 10 25 atoms of phosphorous? 1.20 x 10 25 atoms X 1 mole 6.022 x10 23 atoms 2.How many atoms are in 0.750 moles of zinc? 0.750 moles X 6.022 x10 23 atoms 1 mole = 20.0 moles = 4.52 x 10 23 atoms

79 MASS TO MOLE/ MOLES TO MASS 1.Find the number of moles of argon in 452 g of argon. 452 g Ar x 1 mol 39.94 g 2.Find the mass in 2.6 mol of lithium bromide. 2.6 moles x [(6.941 g Li) +(79.904 g Br)] 1 mole = 11.3 moles = 225.7 gram = 230 grams (sig fig)

80 NOW YOU TRY 1.How many molecules are in 0.400 moles of N 2 O 5 2.Find the grams in 1.26 x 10 -4 mol of HC 2 H 3 O 2

81 EMPIRICAL FORMULA Lowest ratio of atoms in a compound How to solve 1.Take the percent's and “convert” to grams Just switch the sign because we assume there are 100 grams 2.Divide each mass amount by the elements molar mass 3.Re-divide by the lowest answer 4.Write the compound Your answers become the subscripts of the elements within the compound

82 SAMPLE PROBLEM Find the empirical formula for a compound which contains 32.8% chromium and 67.2% chlorine. ElementPercentDivide by molar Mass MolesRe-dividesubscripts Cr32.8%32.8 g 51.99 g/mol 0.631mol 0.631 1 Cl67.2 %67.2 g 35.45 g/mol 1.90 mol 0.631 3 CrCl 3

83 NOW YOU TRY! What is the empirical formula for a compound which contains 67.1% zinc and 32.9 % oxygen?

84 MOLECULAR FORMULA This is a multiple of the empirical formula There is a common multiple How to solve 1.Find the mass of the empirical formula 2.Divide the molecular formula mass by the empirical formula 3.Multiple the subscripts of the empirical formula by the multiple

85 SAMPLE PROBLEM You and your lab partner determined that the mass of the sample was 475.02 grams using the empirical formula from the previous sample. What is the molecular formula of the compound. Empirical formula CrCl 3 Cr: 51.99475.02 Cl: 35.45 x 3 = 106.35158.34 158.34 g/mol = 3 Cr 3 Cl 9

86 STOICHIOMETRY- SESSION 10 1.Topics A.Mole ratio  How to determine B.Mass to mass problems

87 DEFINITION Stoichiometry using molar relationships between reactants and/or products in a chemical reaction to determine desired quantitative data

88 MOLE RATIO Comes from the balance equation The coefficients are the relative number of moles in a balanced equation 2 Na 3 PO 4 + 3 CaCl 2  6 NaCl + 1 Ca 3 (PO 4 ) 2 2:1 Mole ratio between Na 3 PO 4 and Ca 3 (PO 4 ) 2 is 2:1

89 MASS TO MASS How to solve: 1.Identify the knowns and unknowns 2.Find the mole ratio 3.Setup the problem A.Write what you are given B.Convert to Moles C.Switch substances using mole ratio D.Convert to the unit the problem calls for 4.Solve!

90 SAMPLE PROBLEM 25.O g X g 2:2

91 NOW YOU TRY! How many grams of AgCl, silver chloride, are produced from 5.0 g of AgNO 3, silver nitrate? 2AgNO 3 + BaCl 2 → 2AgCl + Ba(NO 3 ) 2

92 GAS LAWS- SESSION 11 Topics 1.Properties of Gases 2.Boyle’s Law 3.Charles’ Law 4.Combine Gas Law 5.Dalton’s Law of Partial Pressure 6.Avogadro's relationship 7.Ideal Gas Law A.Density B.Molecular mass

93 PROPERTIES OF GASES No definite volume Must be enclosed in a container No definite shape Move Independently of each other At high speeds Behave ideally under most conditions No appreciable volume No attractive forces Pressure The amount of force, per unit area, in which the molecules hit the container

94 STANDARD TEMPERATURE AND PRESSURE Pressure 3 main units of pressure 1.Atmospheres (atm) 2.Millimeters of Mercury (mm Hg) 3.KiloPascals (kPa) Atmospheric pressure is standard pressure (sea level) 1 atm = 760 mm Hg = 101.325 kPa Temperature 2 temperature units 1.Celsius ( o C) 2.Kelvin (K) For gas law problem MUST convert to Kelvin T K = T C + 273 Cannot divide by zero Standard temperature 0 o C 273 K

95 BOYLE’S LAW Relationship between pressure and volume Indirect relationship Pressure increases Volume decreases Pressure decrease Volume increases P 1 V 1 = P 2 V 2 www.kentchemistry.com

96 BOYLE’S LAW PROBLEM A gas occupies 12.3 liters at a pressure of 40.0 mm Hg. What is the volume when the pressure is increased to 60.0 mm Hg? P 1 V 1 = P 2 V 2 P 1 = 40.0 mm Hg P 2 = 60.0 mm Hg V 1 = 12.3 L V 2 = X (40.0 mm Hg)(12.3 L) = (60.0 mm Hg) (V 2 ) V 2 = 8.2 L

97 NOW YOU TRY! If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 2.50 atm?

98 CHARLES LAW www.kentchemistry.com

99 CHARLES' LAW PROBLEM A sample of oxygen occupies a volume of 350 mL at 35 o C. What volume will it occupy at 85 o C? V 1 : 350 mL V 2 : X T 1 : 35 o C + 273 = 308 K T2: 85 o C + 273 = 358 K V 2 = 406.82 mL

100 NOW YOU TRY! Neon gas was heated from 50 o C to 150 o C. Its new volume is 175 mL. What was the original volume?

101 COMBINED GAS LAW

102 SAMPLE PROBLEM V 1 = 400 mL P 1 = 300 mm Hg T 1 = O o C + 273 = 273K V 2 = ? P 2 = 140 mm Hg T 2 = 10 o C + 273 = 283K V 2 = 888.54 mL

103 NOW YOU TRY! A gas has a volume of 39 liters at STP. What will its volume be at 4 atm and 25 o C?

104 DALTON’S PARTIAL PRESSURE Each gas in a system exerts it’s own pressure All the partial pressures of the gas can be added up to find the total pressure. P 1 + P 2 +…..P n = P total A metal tank contains three gases: oxygen, helium, and nitrogen. If the partial pressures of the three gases in the tank are 35 atm of O 2, 5 atm of N 2, and 25 atm of He, what is the total pressure exerted inside the tank? P N2 + P O2 +P He = P total 5 atm + 35 atm + 25 atm = P total 65 atm = P total

105 NOW YOU TRY! Blast furnaces give off many unpleasant and unhealthy gases. If the total air pressure is 0.99 atm, the partial pressure of carbon dioxide is 0.05 atm, and the partial pressure of hydrogen sulfide is 0.02 atm, what is the partial pressure of the remaining air?

106 AVOGADRO'S RELATIONSHIP ONLY AT Standard Temperature and Pressure (STP) does this relationship apply Standard Pressure: 1 atm Standard Temperature: 0 o C A mole of a gas is equal to 22.4 L Used for conversions

107 NOW YOU TRY! 100 g of oxygen(O2) is added to the gas in Question 16. What is the volume of the combined gases at STP.

108 IDEAL GAS LAW

109 ALTERNATE FORM To Find Density PM = DRT P: Pressure M: Molecular Mass (periodic table) D: Density R: Gas Constant T: Temperature (K) To Find Molecular Mass

110 SAMPLE PROBLEM

111 NOW YOU TRY! At what pressure would 0.150 mole of nitrogen gas at 23.0 °C occupy 8.90 L?

112 ENERGY- SESSION 12 Topics 1.Energy 1.Heat A.Temperature 2.Endothermic 3.Exothermic 4.Phase changes A.Phase change diagram 5.Specific heat A.Latent heat B.Heating and cooling curves 6.Potential energy diagram

113 ENERGY SI unit for energy is Joules (J) Energy can be in different forms such as Heat Light Sound Chemical Mechanical

114 HEAT / TEMPERATURE Heat is a form of energy The transfer of energy between objects of different temperatures Flows from Hot to cold More energy to less energy Temperature Average Kinetic Energy The average speed that the molecules are moving

115 ENDOTHERMIC / EXOTHERMIC Endothermic Energy is added into the system Heat of reaction ( △ H) is negative Reactant have less energy than the products Exothermic Energy is released from the system Heat of reaction ( △ H) is positive Reactants have more energy than the products

116 PHASE CHANGES Four phases of matter 1.Solid 2.Liquid 3.Gas 4.Plasma There are six main phase changes 1.Freezing (exo) 2.Melting (endo) 3.Evaporation/ Vaporization (endo) 4.Condensation (exo) 5.Sublimation (endo) 6.Deposition (exo) www.elementalmatter.info

117 PHASE CHANGE DIAGRAM Triple point: where all three phases are at equilibrium Critical point: the point at which the gas and liquid phase can co- exist Normal Melting point: The point where a substance melts at normal atmospheric pressure (standard pressure) Normal boiling point: The temperature where a substance will boil at standard pressure Normal freezing point Normal boiling point 1atm www.course-notes.org

118 SPECIFIC HEAT/ LATENT HEAT Specific Heat The amount of energy it takes to raise one gram of a substance one degree Celsius Q = m X Cp X ΔT Q: heat/energy (J) M: mass (g) Cp: specific heat J/gC ΔT: change in temperature Final temperature- initial temperature Latent heat/ Heat of Fusion/heat of Vaporization The amount of energy it take for a substance to phase change. Q = moles X ΔH vap/fus Q: heat/energy (J) Moles ΔH vap or ΔH fus : Heat of vaporization/ Heat of fusion

119 SAMPLE PROBLEMS Specific heat Latent Heat

120 HEAT AND COOLING CURVES The areas of the graph that are sloped indicate and temperature Increasing KE Use the specific heat equations The flat segments of the graph indicate a phase change No temperature occurs All energy is being used to over come intermolecular forces Use latent heat equation www.slideshare.net www.teachinghighschoolchemistry.com

121 SAMPLE PROBLEM A fluorescent light contains 0.1 g of mercury which needs to be vaporized to allow the light to work. How much energy does it take to boil away the mercury if it starts at 20 °C? Boiling point356.73 °C Specific heat0.14 J/g C Heat of vaporization59.11 KJ/mol Part 1: 20 °C  356.73 °C Q = m X Cp X ΔT Q: ? m: 0.1 g Cp: 0.14 J/g C ΔT: 356.73°C – 20 °C = 336.73°C Q = (0.1g)(0.14 J/g C)(336.73°C) Q = 4.714 J Total Energy = 4.714 J + 29.46 J = 34.18 J

122 NOW YOU TRY What is the total amount of energy needed to melt 54 grams of ice starting at -5 o C? Assume that the molar heat of fusion of ice is 6 kJ/mol. Specific Heat of Ice is 2.03 J/g o C.

123 POTENTIAL ENERGY DIAGRAMS chemwiki.ucdavis.edu 1: Potential Energy (PE) of reactants 2: Activation Energy (Ae) of forward reaction 3: Potential Energy of activation complex (transition state) 4: Activation Energy of the reverse reaction 5: Heat of Reaction ( △ H) - Different of energy between products and reactants 6: Potential Energy of products

124 SOLUTIONS – SESSION 13 Topics 1.Properties 2.Solubility A.Rules B.Curves  Calculations C.Net ionic equations 3.Colligative properties A.Boiling point elevation B.Freezing point depression

125 PROPERTIES OF SOLUTIONS They are homogeneous mixture Light can pass through them The particles of the solution will pass through a filter The particles will not separate on their own

126 SOLUBILITY The ability for a compound to dissolve Solute: the substance being dissolved Solvent: the medium of the solution The substance the solute is being dissolved into

127 FACTORS OF AFFECTING SOLUBILITY AND RATE OF SOLUBILITY Solubility Volume /pressure Gases ONLY pressure solubility Nature of Solute/Solvent Like dissolves like Temperature temperature (solids/liquids) solubility Gases in temperature in solubility Rate Amount of solute already in solution More of solute already in solution rate Surface area S.A. increase rate Stirring rate Temperature temperature (solids/liquids) rate Gases in temperature rate

128 SOLUBILITY RULES How to read: 1.Find the ion on the chart 2.Check if it is 1.Soluble 2.Insoluble 3.Are there any exceptions movies-in-theaters.net

129 IS THE COMPOUND SOLUBLE 1.KBr 2.PbCO 3 3.BSO 3 4.Mg 3 (PO 4 ) 2 5.KOH 6.NiCl 2 7.NH 4 OH 8.Hg 2 SO 4 9.PbI 2 1.Soluble 2.Insoluble 3.Soluble 4.Soluble 5.Soluble 6.Soluble 7.Soluble 8.Soluble 9.Insoluble

130 SOLUBILITY CURVE Graph showing the relationship between amount of solute and temperature. How to read: Find temperature Find the line for the compound Read the amount of solute www.sciencegeek.net

131 INFORMATION FROM THE CURVE Unsaturated: More solute can be added into the solution (under the curve) Saturated: the maximum amount of solute is in solution (on the curve) Supersaturated: there is more solute in solution than at its saturation point (above the curve)

132 SAMPLE PROBLEMS Amount change Temperature Change A saturated solution of KClO 3 is formed from one hundred grams of water. If the saturated solution is cooled from 90°C to 50°C, how many grams of precipitate are formed? 90°C: 47 g KClO 3 50°C: 18 g KClO 3 29 g KClO 3

133 NET IONIC EQUATIONS Phase indicators Solids (s) Liquids (l) Gases (g) Aqueous (aq) Solutions which are dissolved in water  determine if the compound breaks into ions

134 PROCESS TO SOLVE 1.Write a balanced equation with phase indicators A.Balance for charge (subscripts) B.Balance for mass (coefficients) 2.Determine if there is an insoluble product 3.Separate all soluble compounds A.Do not separate A.Solids B.Pure liquids C.Gases D.Insoluble compounds 4.Cross out spectator ions A.Ions that do not directly contribute to the reaction 5.Re-write the equations with ions/cpmpound that contribute to the reactions

135 SAMPLE PROBLEM ____Ni(NO 3 ) 2 + ____NaOH  _____ Ni(OH) 2 + _____ NaNO 3 Ni 2+ (aq) +2(NO 3 ) 1- (aq) +2Na 1+ (aq) +2(OH 1- ) (aq)  Ni(OH) 2 + 2Na 1+ (aq) + 2(NO 3 ) 1- (aq) Ni 2+ (aq) +2(OH 1- ) (aq)  Ni(OH) 2 (s) 2 2 (aq) (s)(aq) (s)

136 NOW YOU TRY! ____ NaCl + ____ AgC 2 H 3 O 2  ____ NaC 2 H 3 O 2 + _____ AgCl

137 COLLIGATIVE PROPERTIES Might have to do side calculations

138 SAMPLE PROBLEM Ca x 1: 40.08g Cl X 2: 70.9 110.98 g/mol

139 STEP 2  T = (0.90 mol/Kg)(3)(0.512 o C/m)  T =1.38 o C 100+1.38 = 101.38 o C

140 NOW YOU TRY! Find the boiling point of a solution containing 6.0 g benzene, C 6 H 6, in 35 g of napthalene. (Kb of naphthalene = 5.65 o C/m)

141 EQUILIBRIUM/KINETICS- SESSION 13 Topics 1.Kinetics A.Collison theory B.Reaction rates 2.Equilibrium A.Constants B.Le Chatelier Shifts

142 COLLISION THEORY In order to react molecules must Impact each other with the Sufficient amount of Kinetic energy Correct orientation iverson.cm.utexas.edu

143 KINETICS Factors that affect reaction rates Concentration Particle size Pressure/volume Temperature Activation energy

144 WHAT WILL HAPPEN WHEN? ScenarioIncrease or Decrease Adding heat. Removing heat Adding a catalyst Diluting a solution Removing an enzyme Lowering the temperature Decreasing the surface area Increasing concentration of a solution Breaking reactants into smaller pieces Increases Decreases Increases Decreases

145 EQUILIBRIUM mx1.ibchem.com

146 EQUILIBRIUM CONSTANTS Keq > 1 Products are favored Keq < 1 Reactants are favored Keq = 1 Neither products nor reactants are favored

147 SAMPLE PROBLEM A reaction vessel with a capacity of 1.0 L, in which the following reaction: SO 2 (g) + NO 2 (g) ↔ SO 3 (g) + NO(g) Had reached a state of equilibrium, was found to contain 0.40 mol of SO3, 0.30 mol of NO, 0.15 mol of NO2, and 0.20 mol of SO2. Write and calculate the Keq for this reaction. Which side of the reaction is favored? Keq= 4 Products are favored

148 NOW YOU TRY! Consider the following equation: H 2 (g) + I 2 (g) ↔ 2HI(g) Write and calculate the Keq if at 300 K the concentrations are [H 2 ] = 0.40 M, [I 2 ] = 0.45 M and [HI] = 0.30 M. Which side of the reaction is favored?

149 LE CHATELIER’S PRINCIPLE How an equilibrium system reacts when a stress is placed on the system The system will shift in a direction that relieves the stress The equilibrium position changes but NOT the Keq Exception is when temperature changes Temperature is the ONLY variable that affects the Keq

150 FACTORS THAT AFFECT LE CHATELIER Temperature: treat as a product/reactant Exothermic: temp shifts to left, Keq increases Endothermic: temp. shifts right, Keq decreases Concentration Reactants concentration shifts left When removed shifts right Products concentration shift right When removed shifts left Volume/Pressure volume Dec. Pressure Shifts toward the side with more moles volume Inc. Pressure Shift towards the side with less moles

151 WHAT WILL HAPPEN WHEN? ReactionStressorShift 2 SO 2 (g) + O 2 (g) ⇄ 2 SO 3 (g) + energy decrease temperature C (s) + CO 2 (g) + energy ⇄ 2 CO (g) increase temperature N 2 O 4 (g) ⇄ 2 NO 2 (g) increase total pressure CO (g) + H 2 O (g) ⇄ CO 2 (g) + H 2 (g) decrease total pressure 3 Fe (s) + 4 H 2 O (g) ⇄ Fe 3 O 4 (s) + 4 H 2 (g) add Fe (s) 2 SO 2 (g) + O 2 (g) ⇄ 2 SO 3 (g) add catalyst CaCO 3 (s) +170 KJ ⇄ CaO (s) + CO 2 (g) [CO 2 ] is decreased 4 NH 3 (g) + 5 O 2 (g) ⇄ 4 NO (g) + 6 H 2 O (g) add H 2 O Left Right No shift Left Right

152 StressEquilibrium shift[N 2 ][H 2 ][NH 3 ] Add N 2 Add H 2 Add NH 3 Remove N 2 Remove H 2 Remove NH 3 Increase Pressure Decrease pressure N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g)

153 ACID AND BASE- SESSION 14 Topics 1.Properties 2.Categorizing A.Arrhenius B.Bronsted Lowry C.Lewis 3.Conjugate pairs 4.Water constant (Kw) 5.Calculations A.pH/pOH B.Concentration 6.Titrations

154 PROPERTIES Acid Are electrolytes Sour pH under 7 Reacts with metal to produce H 2 Colorless in phenolphthalein Turns blue litmus paper red Base Are electrolytes Tastes bitter pH over 7 Feels slippery to the skin Pink in phenolphthalein Turns red litmus paper blue

155 CLASSIFICATION Arrhenius Theory Acid: any substance which releases H + ion in a water solution Base: Any substance which releases OH - ions in water solution Bronsted-Lowry Theory Acid: Any substance which donates a proton Base: Any substance which can accepts a proton Lewis Theory Acid: Any substance which accepts an electron pair Base: Any substance which can donate an electron pair.

156 DEFINITIONS/CONCEPTS Amphoteric: any substance that can act as both an acid or a base Dissociation: ability to dissolve Monoprotic: gives off one proton Polyprotic: can give off more than one proton

157 CONJUGATE ACID/BASE PAIRS In an equilibrium system acid/base conjugate pairs are formed Acid forms conjugate base Base forms conjugate acid H 2 SO 4 + H 2 O  HSO 4 - + H 3 O + AcidConjugate Base Base Conjugate Acid

158 IDENTIFY THE FOLLOWING HClO 4 (aq) + H 2 O(l) ⇄ H 3 O + (aq) + ClO 4 – (aq) NH 3 (g) + H 2 O(l) ⇄ NH 4 +(aq) + OH–(aq) HC 2 H 3 O 2 (aq) + H 2 O(l) ⇄ H 3 O + (aq) + C 2 H 3 O 2 – (aq) H 2 S(g) + H 2 O(l) ⇄ H 3 O + (aq) + HS – (aq)

159 WATER CONSTANT Water self ionized into OH - and H 3 O + For pure water the [OH - ] = [H 3 O + ] Kw is the water constant  1.0 x 10 -14 [OH - ] x [H3O + ] = 1.0 x 10 -14 [OH - ] = 1.0 x 10 -7 [H3O + ] = 1.0 x 10 -7 Basis of the pH scale Neutral = 7 -log [1.0 x 10 -7 ] top of the pH scale is 14 -log[1.0 x 10 -14 ]

160 CALCULATIONS pH = -log [H + ] pOH = -log [OH - ] [H + ]= 10 ^ -pH [OH - ] = 10 ^ -pOH pH + pOH = 14

161 SAMPLE PROBLEM/NOW YOU TRY! pH[ H 3 O 1+ ]pOH[ OH 1– ] ACID or BASE? 3.78 10 ^-3.78 = 1.65 x10 -4 M 14- 3.78 = 10.22 10 ^-10.22 = 6.02 x 10 -11 M Acid 3.89 x 10 –4 M 5.19 4.88 x 10 –6 M 8.46 Base 10 ^-8.81 = 1.5 x 10 -9 M 14-5.19 = 8.81 10 ^-5.19 = 6.4 x 10 -6 M Base - log [3.89 x 10 –4 ] = 3.41 14-3.41 = 10.59 10 ^-10.59 = 2.57 x 10 -11 M Acid 14-5.31 = 8.68 10 ^-8.68 = 2.04 x 10 -9 M -log [4.88 x 10 –6 ] = 5.31 Base 10 ^-8.46 = 3.47 x 10 -9 M 14-8.46= 5.54 10 ^-5.54 = 2.88 x 10 -6 M

162 TITRATIONS Purpose is to determine an unknown concentration of an acid or base solution This is done by slowly adding a solution of a known concentration (titrant) into a reaction vessel, with solution of unknown concentration and indicator, until the equivalence point is reached Equivalence point: moles of acid = moles of base Indicators: chemical change color according to the pH range of the solution

163 TITRATION SETUP www.scimath.org intranet.tdmu.edu.ua

164 CALCULATIONS M A V A =M B V B M A : molarity of acid V A : Volume of acid M B : molarity of base V B : volume of base

165 NOW YOU TRY! How many milliliters of 0.360 M H 2 SO 4 are required to neutralize 25.0 mL of 0.100 M Ba(OH) 2 ?


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