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
EACH O is 2OXYGEN RULE
Fe HAS MULTIPLE
OXIDATION STATES
AND HAS NO RULE,
SOLVE FOR IT AS
THE VARIABLE X
Fe2O3 0
THE CHARGE OF THIS IONIC
COMPOUND IS 0, ALL THE
SUM OF THE OXIDATION
STATES MUST EQUAL 0. NO
CHARGE WAS GIVEN IN
QUESTION, ASSUME
NEUTRAL
2X + 3(-2) = 0
X = OXIDATION STATE OF Fe = 3+
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 NONMETAL
3) 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
Connect only single
electrons for
covalent (molecular)
substances.
STEP 1
N N
STEP 2
N N
The electron configuration
of nonmetals will resemble
the noble gas of their own
period
Each nitrogen has
an octet,
isoelectronic to
Ne
Each nitrogen
shares 6 electrons,
triple bond.
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(TETRAHEDRAL
SYMMETRICAL WITH POLAR BONDS
ANSWER IS 2
3.2(Cl) -2.6(C)= 0.6
IONIC CHARACTER-WEAKLY POLAR
11) FOR A MOLECULE TO BE POLAR TWO FACTS MUST APPLY SIMULTANEOUSLY:
A) THE IONIC CHARACTER MUST BE ABOVE 0.4
B) THE MOLECULE MUST BE ASYMMETRICAL (NOT SYMMETRICAL).
* ASYMETRICAL GEOMETRY IS PYRAMIDAL AND BENT AND
OTHERS.
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.
INCREASING IONIC CHARACTER
NON POLAR
COVALENT:
IONIC
CHARACTER IS
0 TO ABOUT 0.4
POLAR COVALENT:
IONIC CHARACTER
IS
0.4 TO ABOUT 1.7
IONIC:
IONIC
CHARACTER
IS
1.7 AND ABOVE
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 RATE
2) DECREASES ACTIVATION ENERGY
3) LOWERS ACTIVATED COMPLEX ENERGY LEVEL.
4) PROVIDES ALTERNATE PATHWAY FOR RX
5) DOES NOT CHANGE HEAT OF REACTION
6) DOES NOT CHANGE MOLE RATIOS
7) 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 THE 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 OF NUCLEAR ORIGIN,
NOT A SHELL ELECTRON, PENETRATING AND VERY
DEADLY.
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
2 MOL HCl
= 10 MOLE CO2 CHOICE 2
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:
1
5
2
S
3
6
4
52) S AND Se (SELENIUM) ARE IN THE SAME GROUP,
THEREFORE:
1)SAME VALENCE e- NUMBER
2)SAME VALENCE CONFIGURATION AND THUS
SAME CHEMICAL PROPERTIED.
3) SIMILAR ELECTRONEGARTIVITY AND
IONIZATION ENERGY.
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
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
I I I
H-C-C-C -H
1
THERE IS A BROMINE
ON CARBON 1
THERE IS A SECOND
BROMINE ON
CARBON 2
2
3
I I I
Br Br H
1,2 DIBROMO PROPANE
59) ACTUAL ANSWER
H H H
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
I I I
H-C=C-C -H
1
2
3
I
H
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.
ELEMENT
MOLES OF
ELEMENT
ATOMIC
MASS
SUB
H
6
X
1
=
6
Br
2
X
80
=
160
C
3
X
12
=
36
202 g/mol
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
Po  Pb
84
82
4
4
+ X
X = He
2
2
65. Lead 206 is the most stable element listed.
66)
24
0
24
Na  e + X
11
-1
From reference
table
12
This is Mg -- 24
67. Divide by the number of half life intervals(T) within
elapsed time(t) 60 (elapsed)/15 (half life) = 4 half lives
100 %
One T
50 %
two T
25 %
three
T
12.5 %
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.
+2
M
-1
I
thus MI2