1) THE NUCLEUS IS SMALL AND MASSIVE (HEAVY…MOST OF ATOMS MASS) AND IS POSITIVE FROM THE PROTONS THERE. THE ELECTRON CLOUD IS DIFFUSE AND NEGATIVE. THE ANSWER IS #2.

RELATED FACTS:A) THE ATOMIC NUMBER INDICATES: THE NUMBER OF PROTONS, NUCLEAR CHARGE, NUMBER OF e- IN A NEUTRAL ATOM, THE SPECIFIC ELEMENT…EACH ELEMENT HAS ITS OWN ATOMIC #.

2) METALLIC BEHAVIOR INCREASES AS WE PROCEED TWARD THE LOWER LEFT. IN THIS QUESTION TOP TO BOTTOM DOES NT APPLY, SO LOOK FOR THE LEFT MOST ELEMENT (IN OR CLOSEST TO GROUP 1). #4Sc (SCANDIUM) IS THE ANSWER.

3) THE SEPARATION OF COMPONENTS FROM A LIQUID CAN EASILY BE PERFORMED BY DISTILLATION, SEPARATION BASED ON BOILING POINTS…THE LOWEST BOILING POINT WILL EVAPORATE FIRST.

RELATED FACTS: THE LOWEST BOILING POINTS CORRELATE WITH THE WEAKEST INTER PARTICULATE ATTRACTIONS . Ans =CHOICE 1

4) THESE ARE TWO DIFFERENT COMPOUNDS AS INDICATED BY THE SUBSCRIPTS, O3 IS OZONE AND O2 IS MOLECULAR ELEMENTAL OXYGEN. THEY WILL HAVE VERY DIFFERENT CHEMICAL BEHAVIOR BASED ON DIFFERENTIAL STRUCTURE. #2

5)IN ANY ELECTROCHEMICAL CELL, VOLTAIC(SPONTANEOUS) OR ELECTROLYTIC(NON-SPONTANEOUS) THE FOLLOWING IS TRUE:

ANODE…OXIDATION (AN OX)CATHODE…REDUCTION (RED CAT) …CHOICE 3

6) FACTS: THE OXIDE ION IS O2- AND IRON CAN BE 2+ OR 3+ FROM PERIODIC TABLE. USE THE ALGEBRA REDOX RULE TO SOLVE

2X + 3(-2) = 0X = OXIDATION STATE OF Fe = 3+

Fe HAS MULTIPLE OXIDATION STATES AND HAS NO RULE, SOLVE FOR IT AS THE VARIABLE X

EACH O is 2-

OXYGEN RULE THE CHARGE OF THIS IONIC COMPOUND IS IS 0, ALL THE SUM OF THE OXIDATION STATES MUST EQUAL 0. NO CHARGE WAS GIVEN IN QUESTION, ASSUME NEUTRAL

Fe2O3 0

7) EXAMPLES OF CHEMICAL CHANGES INCLUSE REACTIONS TYPES: SYNTHESIS, DECOMPOSITION, COMBUSTION, NEUTRALIZATION, REDOX ETC. PHYSICAL CHANGES (PHASE CHANGES) ARE EVAPORIZATION/CONDENSATION, MELTING /FREEZING, SUBLIMATION/DEPOSITION. DECOMPOSITON IS A REACTION AND IS THE ANSWER, #4.

8) THE TRANSFER OF ELECTRONS (FROM A METAL TO A NON METAL) DEFINES IONIC SUBSTANCES.

LOOK FOR AN ACTIVE METAL WITH AN ACTIVE NONMETAL CALCULATE IONIC CHARACTER, IF IT IS ABOVE 1.7 THE COMPOUND IS IONIC.KBr IS THE ANSWER (#3) BECAUSE:

1) K IS IN GROUP 1, GROUP ONE BONDS IONIC ONLY 2) K IS AN ACTIVE METAL AND Br IS AN ACTIVE NONMETAL3) Br (3.0 ELECTRONEGATIVITY) AND K (0.8 ELECTRONEGATIVITY) HAVE AN IONIC CHARACTER OF 2.2, VERY IONIC

9) FOR THE MOLECULE N2 THE LEWIS STRUCTURE IS

CONISTANT WITH A TRIPLE BOND…6 SHARED e-.

N N N

10) TWO OF THE SAME ATOM BONDIED (A2) WILL HAVE AN IONIC CHARACTER OF 0, THEREFORE THE BOND IS NON POLAR COVALENT, HERE THIS IS EXEPLIFIED BY H2 ANSWER #4.

RELATED FACTS:H2, O2, N2, Cl2, Br2, I2, F2 ARE ALL IN THE 0 OXIDATION STATE,

ELEMENTAL DIATOMIC LINEAR A2 MOLECULES AND NONPOLAR.

CCl4 IS NONPOLAR(SYMMETRICAL WITH POLAR BONDS ANSWER IS 2

11) FOR A MOLECULE TO BE POLAR TWO FACTS MUST APPLY SIMULTANEOUSLY:A) THE IONIC CHARACTER MUST BE ABOVE 0.4B) THE MOLECULE MUST BE ASYMMETRICAL (NOT SYMMETRICAL).

* ASYMETRICAL GEOMETRY IS PYRAMIDAL AND BENT.

IN THIS QUESTION THE BEST ANSWER IS PRESENTS THE ELECTRONEGATIVITY DIFFERENCE AS THE CRITERIA IN ANSWER # 3. SEE NEXT SLIDE.

IONIC CHARACTER• The IONIC CHARACTER is the difference of electro

negativity that exists between 2 atoms in a compound.• The ionic character is used to classify the type of

bonding and can be thought of as a measurement of a “tug of war” on the electrons by the atoms.

IONIC:IONIC CHARACTER IS 1.7 AND ABOVE

POLAR COVALENT:IONIC CHARACTER IS 0.4 TO ABOUT 1.7

NON POLAR COVALENT:IONIC CHARACTER IS 0 TO ABOUT 0.4

INCREASING IONIC CHARACTER

12) AN ALUMINUM CYLINDER IS COMPOSED OF Al(S) METAL. A SOLID ALWAYS HAS ITS OWN SHAPE AND VOLUME, IT DOES NOT ASSUME THE SHAPE AND VOLUME OF ITS CONTAINER. THE ANSWER IS #1, SOLIDS HAVE A DEFINITE SHAPE AND VOLUME.

13) ELEMENTS (ONE CAPITOL LETTER) CANNOT BE BROKEN DOWN IN A CHEMICAL REACTION (BUT CAN IN A NUCLEAR REACTION). COMPOUNDS (TWO OR MORE DIFFERENT ELEMENTS…2 OR MORE CAPITOL LETTERS) CAN BE BROKEN DOWN. THE ANSWER IS #2, C AND Cu ARE TWO ELEMENTS ON THE PERIODIC TABLE AND AS WITH ALL ELEMENTS ON THE TABLE…CANNOT BE DECOMPOSED BY CHEMICAL MEANS.

14) GROUP 1 METALS ARE SOLUBLE ELIMINATING 3 AND 4 CHOICES. SULFATES ARE SOLUBLE ELIMINATING CHOICE 1. THEREFORE CHOICE 2 IS INSOLUBLE.

15) SOME FACTS WE NEED TO REMEMBER:A) A CLOSED CONTAINER PREVENTS MOLES FROM CHANGING…NO

CHANGE IN PARTICLES.B) A RIDGID CYLINDER IS AT CONSTANT VOLUME…UNLESS IT IS A

PISTON.C) WHEN YOU HEAT A GAS, VELOCITY INCREASES… ANSWER #3

16) IMPORTANT CATALYST FACTS:1) INCREASES RATE2) DECREASES ACTIVATION ENERGY3) LOWERS ACTIVATED COMPLEX ENERGY LEVEL.4) PROVIDES ALTERNATE PATHWAY FOR RX5) DOES NOT CHANGE HEAT OF REACTION6) DOES NOT CHANGE MOLE RATIOS7) DOES NOT CHANGE YIELD OF PRODUCT

THE ANSWER IS # 2---CHANGE PATH AND LOWER ACTIVATION ENERGY

17) 2-BUTANOL IS AN ALCOHOL (-OH FUNCTIONAL GROUP) ANDE 2-BUTENE IS AN ALKENE (DOUBLE BOND). THE ONLY THING IN COMMON IS CARBON ATOMS.

18) GIVEN IS TEMP AND PRESSURE OF 4 GASES, ALL AT 2L. KNOW: EQUAL MOLES OF ANY GASES AT THE SAME T AND P WILL HAVE THE SAME # OF PARTICLES (EQUIMOLAR). THE ANSWER IS Ne ANS CH4, BOTH ARE 2L (1 SF)AT 300K (3 SF, A) AND 1.00 (3 SF,) ATM, CHOICE 2.

19) THE DEFINITION OF EQUILIBRIUM IS THE RATE OF THE FORWARD AND RATE OF REVERSE ARE EQUAL. IN DYNAMIC EQUILIBRIUM THE REACTION CONTINUES IN BOTH FORWARD AND REVERSE INDEFINITLEY. CHOICE 4.

20) FOR ORGANIC SUBSTANCES SATURATED IS DEFINED AS :1) NO MULTIPLE BONDS.2) BONDED TO AS MUCH H AS THE COMPOUND CAN .3) ALKANES ARE SATURATED.4) ALKENES(DOUBLE BOND) AND ALKYNES (TRIPLE BOND) ARE

UNSATURATED. THE ANSWER IS ETHANE, THE ALKANE. CHOICE 4.

21) CONDUCTIVITY REQUIRES MOBILE CHARGE:1) METALS CARRY CHARGE IN THE SEA OF ELECTRONS.2)IN AQUEOUS SOLUTION IONS CARRY THE CHARGE, THE IONS ARE

FROM AN ELECTROLYTE (ACID, BASE, SALT). NOTE ELECTRONS DO NOT LOW THROUGH WATER. CHOICE 3.

22) THE 2 HALF REACTIONS ARE REDUCTION AND OXIDATION., CHOICE 4.

23) ISOMERS MUST HAVE THE SAME MOLECULAR FORMULA AND DIFFERENT STRUCTURES. FORMULAS A AND B BOTH HAVE C2H6O AS THE MOLECULAR FORMULA, A IS AN ALCOHOL (ETHANOL) AND B IS AN ETHER (DI METHYL ETHER). CHOICE 1. NOTE A AND B ARE 2 CARBONS, C AND D ARE 3.

24) FOR THE REACTION

2Al0 + 3Cu2+ 2 Al3+ + 3Cu0

1) EACH ALUMINUM GOES FROM 0 UP TO 3: 3 LOST e- EACH.2) EACH COPPER GOES FROM 2 DOWN TO 0: 2 MOLE ELECTRONS

GAINED FOR EACH MOLE COPPER. CHOICE 2.

25) BY DEFINITION A VOLTAIC CELL SPONTANEOUSLY CONVERTS CHEMICAL TO ELECTRICAL ENERGY. REMEMBER THE OXIDATION MUST BE THE METAL ON TABLE J THAT IS ON TOP OF THE REDUCTION METAL ON THAT TABLE. CHOICE 4

26) NaOH IS A METAL HYDROXIDE, A STRONG BASE (AN ELECTROLYTE ALSO).BY DEFINITION A BASE IS A HYDROXIDE DONOR WITH OH- AS THA ONLY NEGATIVE ION. CHOICE 1.

27) BY DEFINITION A NEUTRALIZATION IS : ACID + BASE SALT + WATER BY DEFINITION.

CHOICE 3. ALSO NO OTHER CHOICE IS EVEN CLOSE!

28) USE TABLE N AND O TO LOOK THIS UP!3H EMITS BETA, AN ELECTRON.

29) BY DEFINITION TRANSMUTATION CONVERTS ONE ELEMENT TO ANNOTHER.CHOICE 4. 30) BOTH 37-K 9(18 N) AND 42-K (23 N) HAVE AN ATOMIC NUMBER OF 19, THEREFORE THEY DO NOT DIFFER IN P OR e-. THEY DO DIFFER IN NEUTRONS CHOICE 2. POSITRONS ARE RADIATION PARTICLES AND DO NOT LINGER IN ATOMS.

31) BY DEFINITION, THE MASS # IS THE SUM OF NEUTRONS AND PROTONS, HERE IT IS 6 + 8 = 14, CHOICE 3.

32) 27-Al HAS AN ATOMIC # OF 13, THEREFORE THE NEUTRONS MUST BE 14 TO ADD TO 27.CHOICE 2.

34) BY DEFINITION, NON METALS ARE RELATIVLEY ELECTRONEGATIVE AND THUS HAVE RELATIVLEY GREAT ATTRACTION FOR ELECTRONS IN BONDS, CHOICE 3

33) BY DEFINITION THE ELECTRON IS APX 2000 (1837 ACTUALLY) TIMES SMALLER THAT A PROTON. SO THE PROTON SHOULD BE ABOUT 1837 GRAMS, 2000 GRAMS IS CLOSEST, CHOICE 4.

35) AS YOU READ DOWN ANY GROUP…RADIUS INCREASES Be < Mg<CA. CHOICE 1

36) CHOICE 2 +2 + -2 (2e-) = 0

37) % COMP = PART/WHOLE * 100 1) THE WHOLE IS THE MOLAR MASS (GIVEN AS 184 G/MOL)

2) THE PART IS I MOL Mg, 24.3 G/MOL FROM MASS # ON P TABLE.

% COMP = PART/WHOLE * 100 = 24.3/184 * 100 = 13.2%

THIS IS CONSISTANT WITH THE SETUP IN CHOICE 1

38) IN THE REACTION THE COEFFICIENT MOLE RATIO IS 2 (HCl) TO 1 CARBON DIOXIDE.

20 MOL HCl * 1 MOLE CO2 = 10 MOLE CO2 CHOICE 2

2 MOL HCl

39) THIS IS MELTING, A PHASE CHANGE (AT A CONSTANT 0 C).

THE EQUATION IS q = Hfm : WHERE Hf = HEAT OF FUSION 334 J/g m = MASS, GIVEN AS 29.95 g

THERFORE q = 334 * 29.95 = 10003.3 = 1 X 104 CHOICE 4

40) THE DIAGRAM MUST HAVE 2 DIFFERENT SYMBOLS NOT CONNECTED (NO BONDING INTO MOLECULES). CHOICE 4

41) IF YOU LOOK AT TABLE “G” AT 90 c ABOUT 10.0 G DISSOLVES, CHOICE 2.

42) MOLARITY = MOLES/LITER VOLUME.

2.0 M = X/1.5 L = 3.0 MOL CHOICE 3

43) IN ORDER TO LIQUIFY A GAS, YOU MUST BRING THE PARTICLES CLOSE IN ORDER FOR ATTRACTIONS TO FORM…ALSO LOW TEMP IS NEEDED TO ALLOW WEAK ATTRACTIONS TO FORM. CHLORINE Cl2 MOLECULES ARE NONPOLAR AND THERFORE HASE VERY WEAK VAN DER WAALS INTERMOLECULAR ATTRACTIONS…. CHOICE 3

45) ACCORDING TO LECHATELIER, IF YOU ADD SOMETHING IT WILL BE CONSUMED. ALL SUBSTANCES ON THE SIDE YOU ADD TO WILL DECREASE, ALL SUBSTANCES ON THE OTHER SIDE WILL INCREASE.

N2 + 3H2 2NH3

WHEN HYDROGEN IS ADDED, MORE COLLISIONS BETWEEN THE REACTANTS DRIVE THE REACTION FORWARD, AND NITROGEN IS CONSUMED, CHOICE 1

46) ARROW 4, THE HEAT OF REACTION. CHOICE 4

47) CaCl2 IS IONIC BASED ON IONIC CHARACTER ( 3.2 – 1 = 2.2, VERY IONIC…ANBOVE 1.4)) AND METAL WITH NONMETAL FROM P TABLE. IONIC SUBSTANCES HAVE VERY STRONG ATTRACTIONS IN A CRYSTAL…CHOICE 3

48) MULTIPLY THE FRACTION ( PERCENT /100) BY THE MASS FOR EACH ISOTOPE AND ADD THOSE PRODUCTS, CHOICE 1.

1)THE MASS NUMBER ON THE P TABLE IS THE WEIGHTED AVERAGE OF THE MOST ABUNDANT ISOTOPES ONLY. 2)HERE THE FRACTIONAL ABUNDANCE IS THE GRAMS /TOTAL:

80.22g/100g = .8022

49) ENTROPY…RANDOMNESS…FREEDOM OF MOTION IS HIGHEST IN THE GAS PHASE AND LOWEST IN THE SOLID PHASE. IT ALSO INCREASES WITH TEMP. CHOICE 2 SHOWS THIS.

50) EACH pH UNIT IS A 10X CHANGE, HERE THE pH IS GOING FROM 6 DOWN TO 3. THAT’S 10 X 10 x10 = 1000 TIMES MORE HYDRONIUM. IT WOULD BE 1000 TIMES LESS HYDRONIUM FROM 3 TO 6.

51) Sulfur (S) HAS A CONFIGURATION OF 2-6, THEREFORE SIX VALENCE ELECTRONS, SO:

52) S AND Se (SELENIUM) ARE IN THE SAME GROUP, THEREFORE:

1)SAME VALENCE e- NUMBER2)SAME VALENCE CONFIGURATION AND THUS SAME CHEMICAL PROPERTIED.3) SIMILAR ELECTRONEGARTIVITY AND

IONIZATION ENERGY.

S1 2

65

4

3

53) HETEROGENOUS (NOT UNIFORM)

54) COMPOUND (2 OR MORE DIFFERENT ELEMENTS BONDED, WRITTEN AS ONE WORD FORMULA AND MAY HAVE SUBSCRIPTS)

53) THE PARTICLES ARE EVENLY DISPERSED, NOT SEGREGATED.

4) FILTRATON, USED TO SEPARATE A SOLID (OR PRECIPITATE) FROM A LIQUID. THIS WILL NOT WORK WITH HOMOGENOUS (aq) SOLUTIONS.

59) AS THE CONCENTRATION OF EITHER REACTANT OR BOTH INCREASE:

1) COLLISIONS INCREASE CAUSING THE NUMBER OF EFFECTIVE COLLISIONS (THE ONES THAT CAUSE CONVERSION OF PRODUCT TO REACTANT) TO INCREASE.

60)TEMPERATURE (AVERAGE KINETIC E) CAUSES PARTICLE VELOCITY (SPEED) TO INCREASE …THEREFORE COLLISIONS INCREASE, THEREFORE EFFECTIVE COLLISIONS INCREASE AND THE RATE OF CONVERSION OF PRODUCTS TO REACTANTS INCREASE. Also increases the energy level…sufficient energy is needed to make collisions effective.

59PRACTICE EXAMPLE, NOT THE ANSWER

H H H I I I H-C-C-C -H I I I Br Br H

1 2 3

1)THE LONGEST CHAIN OF CARBONS GIVES THE ROOT OF THE NAME…TABLE P GIVES THE SUFFIX…PROP FOR THREE CARBONS. 2) ALL SINGLE BONDS IN LONGEST CHAIN…ALKANE…NAME ENDS IN ANE

THERE IS A BROMINE ON CARBON 1THERE IS A SECOND BROMINE ON CARBON 2

1,2 DIBROMO PROPANE

59) ACTUAL ANSWER

H H H I I I H-C=C-C -H I H

1 2 3

1)THE LONGEST CHAIN OF CARBONS GIVES THE ROOT OF THE NAME…TABLE P GIVES THE SUFFIX…PROP FOR THREE CARBONS. 2) A DOUBLE BOND IN LONGEST CHAIN …ALKENE…NAME ENDS IN ENE

PROPENE

62) ADDITION (TO A MULTIPLE BOND): A HALOGEN MOLECULE ADDS ITS ATOMS TO THE DOUBLE BOND, THE NUMBER OF H ATOMS IS CONSTANT.

63) THE MOLECULAR MASS IS THE SUM OF ALL OF THE ATOMIC MASSES OF ALL ATOMS IN THE FORMULA.

202 g/mol

ELEMENT MOLES OF ELEMENT

ATOMIC MASS

SUB

H 6 X 1 = 6

Br 2 X

80 = 160

C 3 X 12 = 36

60) Polonium 218 decays to Pb 214…a loss of 4 from the atomic mass, Polonium has an atomic number of 84, and Lead has 82…a loss of 2 from Polonium to Lead. Therefore the particle is alpha

4 mass and 2 atomic number He.

218 214 4

Po Pb + X X = He 84 82 2

65. Lead 206 is the most stable element listed.

4

2

66)

24 0

Na e + X 11 -1

67. Divide by the number of half life intervals(T) within elapsed time(t) 60 (elapsed)/15 (half life) = 4 half lives

24

12

From reference table This is Mg -- 24

100 % One T 50 % 25 % 12.5 %two T three T

Four T

6.25 %

68) Solid…most periodic elements are.

69) It is a metal, lets remember metal properties:1)Good conductors of heat and electricity.2)Mobile electrons in solid state – ANS.3) Malleable, ductable and lusterous

70) When an atom loses electrons (becomes positive ion), the shells (electron cloud) contracts to a smaller radius.

71) M is +2 due to it being in group 2, so use cris cross to get formula. NOTE…halogens are -2 in binary compounds.

M+2 I-1 thus MI2