1. CP Chemistry Period 5

2. Physical and Chemical Properties Physical PropertiesChemical Properties Size Color Texture Smell Mass Taste Density Volume Area Melting Point Malleability Elasticity Solubility Flammability Digestibility Decomposable Ability to oxidize Reactivity inert (not reactive) Physical Property- a trait or characteristic that you can observe without changing the identity of the substance Chemical Property - a trait you can observe by changing the identity of the substance

3. Physical and Chemical Changes Physical ChangesChemical Changes Examples in terms of a piece of paper Crumpled paper Ripped paper Drawn on paper Stomping on paper Examples in terms of a piece of paper Eating it Digesting it Burning it Physical Change- a change that affects only physical properties and does not alter the identity of the substance Chemical Change- a change that alters the identity of our substance

4. Elements, Compounds, and Mixtures Element- a substance that cannot be separated by chemical or physical means Compound- a substance made up of two or more elements only separated by chemical means Atom- smallest unit of an element Molecule- smallest unit of a compound

5. Elements, Compounds and Mixtures cont. Mixture- a combination of substances thhat are not chemically combined. These can be separated physically – Homogeneous: looks same throughout – Heterogeneous: composed of different parts HomogeneousHeterogeneous

6. Accuracy and Precision True Data is accurate. Repeatable data is precise. Accurate & Precise

7.  The mass number of the element eqauls the number of protons in an element  The number of protons is also the number of electrons unless its an ion  To find the neutrons you subtract the atomic number from the mass number

8.  Are atoms that have lost or gained neutrons, same element but different number of neutrons

9.  Alpha- 4 2 H- stopped by clothing or skin  Beta- 0 -1 e- stopped by a sheet of lead  Gamma- stopped by several inches of lead, most dangerous  Nuclear reactions happened when there is an unstable particle and eventually gives off a particle of radiation

10.  Atoms that have lost or gained electrons  atoms turn into ions when electrons move  Ions have a charge  There are negative electrons and positive electrons  How Ions are formed:  Positive ions have lost electrons  Negative ions have gained electrons  Positive ions are called cations  Negative ions are called anions  When atoms are most stable they have an octet  Octet- 8 electrons in the outer most energy level

11.  Share electrons between two atoms  Properties  Low melting point  Molecule structure  Gases, liquids, soft solids  Poor conductors of heat  Poor conductors of electricity  Typically 2 non-metal atoms

12.  They trade electrons between the two atoms  Ions must form from the atom  Properties  High melting point  Crystal lattice structure  Hard solids  Brittle  Good conductors of heat  Good conductors of electricity  Typically 1 metal and 1 non metal

13. Wavelength Waves of Light: electromagnetic radiation (light) moves as a wave Crest Trough Wave

14. Wavelength/Frequency λ= Wavelength: Distance from crest to crest on a wave v= Frequency: How often a wave passes by in a second (s-1) Wavelength and Frequency are inversely related – Wavelength increases, Frequency decreases – Wavelength decreases, Frequency increases

15. Calculations

16. Answer – Frequency

17. Answer- Energy E=hv E= ? h= 6.626 x10 -34 J(s) v= 5.0 x 10 14 s-1 E= (6.626 x10 -34 ) 5.0 x 10 14 – 3.31 x 10 -19 J

18. Ionization Energy The amount of energy needed to remove one electron from an atom Size of atom determines how easily electrons are removed Big atoms lose e- with minimal effort Little atoms lose e- with a huge amount of energy needed - Increases up and to the right on the Periodic Table Noble gases all have elect negativities equal to zero If an atom needs a lot of energy to remove an electron its because it really wants the e-

19. Periodic Trends Properties of elements can be predicted using the location on the periodic table – Electron Configuration – Family and Periods – Densities – Reactivity – Atomic radius – Ionization Energy – Electronegativity

20. Atomic Radius The distance from the nucleus to the outer- most elections (in the highest energy orbital filled) As electrons fill into higher energy orbitals, the radius of the atoms gets bigger! Na K Rb Na 3s1 K 4s1 Rb 5s1

21. Atomic Radius Within an energy level adding more protons makes the radius of atoms smaller because the protons can hold the electrons in closer P S Cl P 3P 3 S 3 P 4 Cl 3P 5

22. Question Put in order smallest to largest: – Rb, P, Na Answer on next slide

23. Answer Na, Rb, P

24. Periodic Table Families and Periods Groups (families) = The columns on the Periodic Table Periods = Rows on the Periodic Table Elements arranged by atomic number – Column 1: #1 Silvery White – Column 2: React with water to form a base – Column 3: 5,8,9,10,11,12 All metals Form colorful solutions Hard Brittle Metallic Versatile in bonding ability Charges: +2, +3

25. Periodic Table Families and Periods Column 14 – Charge +4 -4 – Non-metallic – Can bond with positive or negative ions – Solid at room temperature – Relatively low reactivity Column 16: Oxygen Family – Charge: -2 – Non-metallic – Bonds with itself – Shares electrons with other elements Column 17: Halogens -Non-metallic -Charge -1 -Very reactive with positive ions -Not solid at room temperature -Colored Gas Column 18: Noble Gases -Non-Metallic -Non Reactive -Charge: 0

26.  VSEPR is a model of molecular structures based on the idea that ideal structures minimize electron pair repulsions  Used to draw and evaluate Lewis Structures  Bare electrons are the most repulsive!

27.  Looking at the molecular geometry of a single atom, not of an entire molecule  3D Figures to represent Lewis Structures  Constituent groups are the things bonded to the atom under scrutiny  Dashed lines represent a bond behind the plane of the paper; wedged lines represent a bond coming toward you

28. Linear 1-2 Constituents 0 Lone Pairs Bond Angle: 180 Trigonal Planar 3 Constituents 0 Lone Pair Bond Angle: 120 Bent 2 Constituents 1 Lone Pair Bond Angle: <120

29. Tetrahedral 4 Constituents 0 Lone Pair Bond Angle: 109.5 Trigonal Pyramidal 3 Constituents 1 Lone Pair Bond Angle: 107.3 Bent 2 Constituents 2 Lone Pairs Bond Angle: 104.5

30.  Lewis dot structure is a drawing of how the atoms are bonded together covalently using valence electrons.  You need to know  Shared pair= 2 electrons shared by 2 atoms (bond)  Lone pair= 2 electrons not shared by atoms (unshared pair)

31. 1. Count the total valence electrons for the molecule Ex: SCl2=20 valence electrons 2. Select a central atom. look for= *the only one of its kind. *less electronegative Ex: SCl2= S is central atom because it’s alone

32. 3. Set up the elements as symmetrical as possible. Ex: SCl2= Cl S Cl 4. Draw in shared pairs by drawing a line. Ex: SCl2= Cl-S-Cl

33. 5. Account for electrons used from total you started with.(Shared pairs=2 electrons) Ex: 20 valence electrons -4 shared electrons ___ 16 unshared electrons

34. 6. Fill in unshared pairs around the outside of elements Ex:

35. 7. When there isn’t enough electrons for every element to have an octet, we share more pairs. Ex:

36.  When finding the polarity in molecules you need to find out if the bonds are polar or non- polar first.  Polar bond- when 2 atoms share electrons unequally  Non-Polar bond- share electrons completly even.

37.  Polar Molecules- Molecules where one side of the molecule has more electrons than the other.  1. if there is a lone pair on the center atom, it is polar  2. If bonds are unequal polarity, than the molecule is polar

38.  Non-Polar Molecules  If there is no lone pairs on the center atom  If the bonds are equal polarity

39. Synthesis: Ex: Cu +3 + O 2  Cu 2 O 3 Decomposition Ex: MgCl 2  Mg +2 + Cl Single Replacement Ex: MgCl 2 + Cu +2  Mg + CuCl 2 Double Replacement Ex: 3MgCl 2 + Cu 2 O 3  3MgO + 2CuCl 3 Combustion Hydrocarbon + oxygen  CO 2 +H 2 O CH 3 OH + O 2  CO 2 + H 2 O

40. Synthesis all you have to do is combine the two reactants to get your products Decomposition you break up your reactants and get two products Single replacement take either the positive or negative ion (by itself) and replace it with the positive or negative ion from a formula in the equation Double replacement you take the positive ion from one formula and put the negative ion from the other formula to create a new formula, do this again with the left over positive and negative ions. (positive come first) Combustion always ends up with carbon dioxide and water

41. Matter can neither be created nor destroyed In chemical equation it’s crucial to make sure its balanced because, if its not balances it goes against the law of conservation of matter because it creates (or destroys) matter

42. According to the law of conservation of matter you have to balance all of your equations so that you don’t create or destroy matter. NaOH + Cl 2  NaCl + OH This is not balance because you have two chlorines in the reactants and only one on the product side So, all you have to do is add a coefficient in order to balance it: 2NaOH + Cl 2  2NaCl + 2OH

43. Aqueous substance dissociate A complete ionic equation shows all the ions and molecules in a reaction Zn (s) + CuSO 4 (aq)  ZnSO 4 (aq) Complete ionic equation: Zn (s) + Cu +2 (aq) + SO4 -2 (aq)  Zn +2 (aq) + SO4 -2 (aq) +Cu (s) NET IONIC: Anything that’s AQUEOUS that’s the same on both sides you can cancel out So you can get rid of SO4 -2 (reactant side) AND SO4 -2 (product side) Final net ionic equation: Zn (s) + Cu (aq)  Zn (aq) + Cu (s)

44. 1. What type of reaction is this? Mg + KCl  MgCl + K 2. Balance the following equations: ___ Al + ___ O 2  _____ Al 2 O 3 ___CuS + ____ O 2  ____CuO + _____ SO 2 ____ Ca 3 P 2 + ____ H 2 O  ____ Ca(OH) 2 + ___ PH 3 3. Write the net ionic equation for: AgNO 3 (aq) + NaCl (aq)  AgCl (s) + NaNO 3 (aq)

45. 1. single replacement 2. 4 Al + 3 O 2  2 Al 2 O 3 3. 2 CuS + 3 O 2  2 CuO + 2 SO 2 4. (1) Ca 3 P 2 + 6 H 2 O  3 Ca(OH) 2 + 2 PH 3 5. AgNO 3 (aq) + NaCl (aq)  AgCl (s) + NaNO 3 (aq) Complete ionic = Ag +1 (aq) + NO 3 -1 (aq) + Na +1 (aq) + Cl -1 (aq)  Ag +1 (aq) + Cl -1 (s) + Na +1 (aq) + NO 3 -1 (aq) Net Ionic = Ag +1 (aq) +Cl -1 (aq)  AgCl (s)

46. The Factor Label Method A ratio used to convert the unit you have into the desired unit Example: If you are given one day, how do you convert it into the amount of seconds in a day? Answer: 1 day* 24 hours*60 minutes*60 sec 1 day 1 hour 1 min

47. The Factor Label Method Cont. The factor label method is useful in converting metric prefixes The Metric Prefixes are: – TGMKHDBDCMMNP – The Great Mister King Henry Died By Drinking Chocolate Milk Monday Night Partying – Tetra Giga Mega Kilo Hecto Deca BASE Deci Centi Mili Micro Nano Pico You can use Metric conversions to change from prefix to prefix!

48. Converting Moles, Liters, Grams & Particles 1 Mole = 6.02 X 10 23 “things” – 6.02 X 10 23 = Avogadro’s Number Moles to Particles/Liters/Grams : mole of element X 6.02*10 23 = # with desired unit 1 Mole

49. Molar Mass To find the molar mass you must refer to the periodic table Look up each atomic mass of the element and add them all together to find the molar mass If there is a subscript then you must multiply the atomic mass by the subscript

50. Empirical Formula Empirical formula= the lowest terms ratio of elements in a formula (Not the true formula) To calculate the empirical formula you must find 1.Percent to Mass 2.Mass to Mole 3.Divide by small 4.Multiply until whole

51. Molecular Formula Molecular Formula= The true formulas for a compound (Not lowest terms) To find the molecular formula you must divide the actual molecule mass by the empirical formula mass – Example : OH is the empirical formula, the actual formula has a mass of 34 g/mol, what is the molecular formula? – 34 g/mol = 2 so OH becomes O 2 H 2 17 g/mol

52. Percent Composition To calculate the % composition you must use the following equation: Mass of particle * 100 Mass of whole

53. States of Matter Gas  Forms to shape of container  Spread apart atoms Solid  Atoms tightly compacted  Definite shape and volume Liquid  Definite volume  Moderate spacing of atoms

54. Parts of Solutions Solute  Dissolved by the solvent in the solution  Ex: Salt in salt water Solvent  Substance that does the dissolving  Ex: Water in salt water

55. Phase Diagrams/ Heating Curves Phase Diagrams  Shows what temperature and pressure combinations can create each state of matter for a particular chemical. Heating Curves  Shows the temperatures at which changes in states of matter occur and describe how a substance uses heat.

56. Molarity Calculations Molarity= moles of substance/ volume(L)  Ex: The chemical Carbon Dioxide has a volume of 2L. Find the concentration of Carbon Dioxide if it has a mass of 24.02g  Grams to Moles using Molar Mass conversions.  24.02g*1mol/12.01g/mol= 2mol  Then use the formula M(Molarity)=mol(Moles)/Volume(L)  M=2mol/2L  M=1M

57. Intermolecular Forces  The forces of attraction between molecules. Vander Waal's(London Dispersion) Hydrogen Bonding Dipole-Dipole

58. SPECIFIC HEAT  Specific Heat Capacity- The amount of heat needed to raise the temperature of one gram of substance one degree Celsius or one degree Kelvin.  Molar Heat Capacity- The amount of heat needed to raise the temperature of one mole of substance one degree Celsius or one degree Kelvin.

59. SPECIFIC HEAT CALCULATIONS  q = m C T Heat (Joules) Mass (g) or moles (mols) Specific or molar heat capacity Change in temperature (kelvin or celsius) T f – T i

60. EXAMPLE PROBLEM  12g of water is heated from 15 degrees Celsius to 35 degrees Celsius. How much heat was absorbed by the water? (The specific heat capacity for water is 4.184 J/g degrees Celsius).  Q = ?  m = 12 g  C = 4.184 J/g  T = 20 degrees Celsius q = 12 * 4.184 * 20 q = 1004.16 J

61. PROPERTIES OF ACIDS  Sour  Burn/ sting  React with metal  Electrolyte (conducts electricity)  pH less than 7  Releases hydrogen ions in water  Accept an electron pair

62. PROPERTIES OF BASES  Bitter  Slippery  Non-reactive with metals  Electrolyte  Releases hydroxide ions in water  Donate an electron pair

63. P H CALCULATIONS  The pH scale is a numerical system that expresses the acidity of a solution What is the pH of a solution that has [H + ] of 1.38 * 10 -11 ? (plug into your calculator) Ans: pH = 10.86

64. P OH CALCULATIONS  A numerical scale that measures solutions by basicity. pOH = -log [OH - ] What is the pOH of the solution you used in the last slide? (plug into your calculator) Ans: 3.14 Hint: pH + pOH = 14

65. H + CALCULATIONS  [H + ] = 10 -pH  Example: Determine the concentration of [H+] in the solution. pH = 3.0  Plug into your into your calculator by clicking “2 nd ” and log  Ans: 1 x 10 -3 M

66. [OH-] CALCULATIONS  [OH-] = 10 -pOH  Example: Determine the [OH-] in the solution given the pOH is 4.0. Ans: 1 x 10 -4 M

67.  Definition: A theory concerning the thermodynamic behavior of matter, especially the relationships among pressure, volume, and temperature in gases.

68.  STP is used for measuring gas temperature and volume. STP means standard temperature pressure.  Absolute zero is used for Absolute zero is the point where no more heat can be removed from a system, according to the absolute or thermodynamic temperature scale. This corresponds to 0 K or -273.15°C.

69.  Pressure is measured using a barometer. The glass tube on the barometer contains a vacuum that allows mercury flow up it when pressure is excreted on the surface of the mercury  Pressure is usually measured in:  Atmospheres (atm)  Bar (bar)  Pascals (pa)  Millimeter of Mercury (mmHg)  Torr (torr)

70. PRESSURE CONVERSION PROBLEM  1 atm= 101.325 Pa  1 bar= 100.025 Pa  1 Torr= 133.32 Pa  1 MMHg= 133.32 Pa  1 MMhg= 1 Torr  A radio station announcer reports the atmospheric pressure to be 99.6 kPa. What is the pressure in atmospheres? In millimeters of mercury?

71. 99.6 kPa x 1 atm/101.3 kPa = 0.983 atm 0.983 atm x 760 mm Hg/1 atm = 747 mm Hg Answer 0.983 atm; 747 mm Hg

74. 1) In a thermonuclear device, the pressure of 0.050 liters of gas within the bomb casing reaches 4.0 x 106 atm. When the bomb casing is destroyed by the explosion, the gas is released into the atmosphere where it reaches a pressure of 1.00 atm. What is the volume of the gas after the explosion? 2) On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250 mL bag at a temperature of 19 0 C, and I leave it in my car which has a temperature of 60 0 C, what will the new volume of the bag be?

75. 1) 2.0 x 10­ 5 L 2) 285 mL