Honors ChemistryHonors ChemistryFinal Assessment ReviewFinal Assessment Review

Your TurnYour Turn

Let’s see if you can do the last 4 orbital filling diagrams and the 4 quantum numbers for the last electron of #112, Copernicium

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2

5d1 4f14 5d10 6p1-6 7s2 6d1 5f14 6d10

The Electrons of the 4 Configurations The Electrons of the 4 Configurations for Coperniciumfor Copernicium

7s = ___ 6p = ___ ___ ___ 0 -1 0 +1

6d = ___ ___ ___ ___ ___ -2 -1 0 1 2

5f = ___ ___ ___ ___ ___ ___ ___ -3 -2 -1 0 1 2 3

The 4 Quantum Numbers for the The 4 Quantum Numbers for the Circled Electron (112Circled Electron (112thth) of Copernicium) of Copernicium

n

6

l2

m

2

s-1/2

Ionization Energy TrendsIonization Energy Trends

Ionization Energy TrendsIonization Energy Trends

Top to Top to BottomBottom Left to RightLeft to Right

DecreasesDecreases IncreasesIncreases

Electronegativity TrendsElectronegativity Trends

Electronegativity TrendsElectronegativity Trends

Top to Top to BottomBottom Left to RightLeft to Right

DecreasesDecreases IncreasesIncreases

SAMPLE PROBLEM: Predicting Molecular Shapes with Two, Three, or Four Electron Groups

PROBLEM: Draw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) PF3 and (b) COCl2.

SOLUTION: (a) For PF3 - there are 26 valence electrons, 1 nonbonding pair

The shape is based upon the tetrahedral arrangement.

The F-P-F bond angles should be <109.50 due to the repulsion of the nonbonding

electron pair.

The final shape is trigonal pyramidal.<109.50

The type of shape is

AX3E

SAMPLE PROBLEM: Predicting Molecular Shapes with Two, Three, or Four Electron Groups

(b) For COCl2, C has the lowest EN and will be the center atom.

There are 24 valence e-, 3 atoms attached to the center atom.

CCl O

Cl

C does not have an octet; a pair of nonbonding electrons will move in from the

O to make a double bond.

The shape for an atom with three atom attachments and no nonbonding pairs on the

central atom is trigonal planar.CCl

O

Cl The Cl-C-Cl bond angle will be less than 1200 due to the electron density of

the C=O.

CCl

O

Cl

124.50

1110

Type AX3

SAMPLE PROBLEM: Predicting Molecular Shapes with Five or Six Electron Groups

PROBLEM: Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) SbF5 and (b)

BrF5.SOLUTION: (a) SbF5 - 40 valence e-; all electrons around central atom will be in bonding pairs; shape is AX5 - trigonal

bipyramidal.F

SbF

F F

FF Sb

F

F

F

F

(b) BrF5 - 42 valence e-; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX5E, square pyramidal.

BrF

F F

F

F

Exit QuizExit Quiz

Al3+ combines with sulfate (SO4)2– to make aluminum sulfate.

Write the chemical formula for aluminum sulfate.

Al3+(SO4)

2-

32

Exit Quiz Answer

Naming Organics: The IUPAC RulesNaming Organics: The IUPAC Rules

Find longest carbon chain Number the chain so the substituent groups

have the lowest total number Give alkyl groups attached to the longest chain a

name and a number Multiple alkyl groups named alphabetically.

ex. 3-ethyl-2-methylpentane Multiple groups that are the same: di(2), tri(3),

tetra(4), penta(5), hexa(6), ex. 2,4-dimethylpentane

Halogens are named “halo” groups – fluoro, chloro, bromo, iodo

The IUPAC RulesThe IUPAC Rules

Double bonds = ene, triple bonds - yne Multiple double/triple bonds have a prefix just in

front of the ending. ex. 2,3,4,5-octatetraene

Cyclics = cyclo- ex. cyclobutane Benzene = 3 resonating double bonds

The IUPAC RulesThe IUPAC Rules

Alcohols – OH, -ol, many = -diol, must show the H of OH

Acids - C=OOH, -oic acid, on the end, must show the H of OH

Aldehyde - C=O on the end, -al Ketone - C=O in the middle, -one,

The IUPAC RulesThe IUPAC Rules

Ethers - O in the middle, R1-R2 ether, R1 = first in alphabet, ex. methylpropyl ether.

Esters - C=O-O in the middle. Acid part is named last with -oate ending, other part is named as radical. ex. methylpropanoate

First priority - functional groups above Second priority - double/triple bonds Third priority - side chains (radicals)

Convert 15.2 m/s to km/hrConvert 15.2 m/s to km/hr

1. 54.7 km/hr

2. 0.912 km/hr

3. 4.22 km/hr

4. 5.47 x 107

5. not listed

Converting Word Equations into Converting Word Equations into Chemical Equations #10Chemical Equations #10

Converting Word Equations into Converting Word Equations into Chemical Equations #10Chemical Equations #10

Strontium iodide + Lead (II) phosphate Strontium phosphate + lead (II) iodide

SrI2 + Pb3(PO4)2 ---->

Sr3(PO4)2 + PbI2

3

3

Practice Quiz Practice Quiz Net Ionic EquationsNet Ionic Equations

Write the molecular, complete ionic, and net ionic equations for this reaction: Silver nitrate reacts with calcium chloride

Molecular:

2AgNO3 + CaCl2 2AgCl + Ca(NO3)2

Complete Ionic:

2Ag+ + 2NO3-+ Ca2+ + 2Cl- 2AgCl + Ca2+ +

2NO3-

Net Ionic:

2Ag+ + 2Cl- 2AgCl

Mixed PracticeMixed Practice

• State the type, predict the products, and balance the following reactions:

1. BaCl2 + H2SO4

2. C6H12 + O2

3. Zn + CuSO4

4. Cs + Br2

5. NaCl

Double Displacement

Combustion

Single Displacement

Synthesis

Decomposition

AnswersAnswers

1. BaCl2 + H2SO4 BaSO4(s) + 2 HCl

2. C6H12 + 9 O2 6 CO2 + 6 H2O

3. Zn + CuSO4 Cu + ZnSO4

4. 2 Cs + Br2 2 CsBr

5. 2 NaCl 2 Na + Cl2

Try This…Try This…

• Predict the products, balance the following reactions and show the change in oxidation numbers :

• Zinc reacts with aqueous copper (II) sulfate• Zinc is higher relative activity so…

Zn + CuSO4 Cu + ZnSO4

Each Zn loses 2e- oxidation, reducing agent

Each Cu(II) gains 2e- reduction, oxidizing agent

Learning CheckLearning Check

What radioactive isotope is produced in the following bombardment of boron?

10B + 4He ? + 1n

5 2 0

13N7

Write Nuclear Equations!Write Nuclear Equations!

Write the nuclear equation for the beta decay of Co-60.

6060Co Co 00e + e + 6060Ni Ni 27 -1 2827 -1 28

Write Nuclear Equations!Write Nuclear Equations!

In the following reaction, what is being emitted and what is the daughter nuclide?

5959Fe Fe 00e + e + 5959Co Co 26 -1 2726 -1 27

Beta Particle

Review Mass-Mole-Molecules: Review Mass-Mole-Molecules: Determine the number of molecules in Determine the number of molecules in

73 g of water73 g of water

73 g H2O

# H2O molecules =

x 1 mol H2O

18.02 g H2O

= 2.4 x 1024 molecules H2O

x 6.02x1023 molecules

1 mol H2O

Try this one:Try this one:

Calculate the mass in grams of iodine required to react completely with 0.50 moles of aluminum.

Al + I2 AlI32 Al + 3 I2 2 AlI3

= 190 g I2

0.50 mol Al x 3 mol I2 2 mol Al

x 253.80 g I2

1 mol I2

0.50 g Al

Try this one:Try this one:

Calculate the mass in grams of iodine required to react completely with 0.50 g of aluminum.

Al + I2 AlI32 Al + 3 I2 2 AlI3

= 7.1 g I2

x 3 mol I2 2 mol Al

x 253.80 g I2

1 mol I2

x 1 mol Al

26.98 g Al

x 3 mol Zn

2 mol MoO3

What mass of ZnO is formed when 20.0 g of MoO3 is reacted with 10.0 g of Zn? ( you must first determine which one is the limiting reactant!)

20.0 g MoO3 x 1 mol MoO3

143.94 g MoO3

3 Zn + 2 MoO3 Mo2O3 + 3 ZnO

x 65.39 g Zn

1 mol Zn

= 13.6 g Zn needed, have 10.0g. Zn is the limiting

reactant

Use 10.0g of Zn to calculate the amount of ZnO produced.

=12.4g ZnO produced

253.80 g I2

x 1 mol I298.60 g I2

How many grams of antimony(III) iodide would be produced using 98.60g of iodine?

Sb + I2 SbI32 Sb + 3 I2 2 SbI3

= 130.1 g SbI3

x 2 mol SbI3

3 mol I2

x 502.46 g SbI3

1 mol SbI3

x 118.00 g SbI3

130.1 g SbI3

Using the previous problem, determine the percent yield if 118.00 g of antimony (III) iodide is produced.

(130.1 g of Sbl3 should have been produced). What is the percent yield?

2 Sb + 3 I2 2 SbI3

= 90.70 % Yield

x 100

Empirical Formulas:

72% iron and 28% oxygenDetermine the formula for this substance.

moles of Fe 72 g Fe

55.85 g/mole

= 1.29 moles Fe

moles of O = 28 g H

16.00 g/mole

= 1.75 moles O

Formula:

1.29 1.75 Fe O 1.29 1.79 1.36

1.29 1.29

Fe FeOO

Multiply subscripts by 3 to get

3 4 Fe O

1.36FeO

How much of 3.0 M HCl do I need to use How much of 3.0 M HCl do I need to use (and diluet to 1 L) to make 1 L of 1.0 M (and diluet to 1 L) to make 1 L of 1.0 M

HCl?HCl?

How much of 3.0 M HCl do I need to use How much of 3.0 M HCl do I need to use (and diluet to 1 L) to make 1 L of 1.0 M (and diluet to 1 L) to make 1 L of 1.0 M

HCl?HCl?

X= .33L or 330mL

3.0 M HCl x X

= 1.0 M HCl x 1 L

Name HCl.Name HCl.

1. Hydrogen chloride

2. Hydrochloric acid

3. Chloric acid

4. Perchloric acid

5. Chlorous acid

6. Hypochlorus acid

7. I have no idea!!

8. Who cares??1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

Name HClOName HClO44..

1. Hydrogen chloride

2. Hydrochloric acid

3. Chloric acid

4. Perchloric acid

5. Chlorous acid

6. Hypochlorus acid

7. I have no idea!!

8. Who cares??1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

Name Fe(OH)Name Fe(OH)33

1. Iron (III) Hydroxide

2. Iron Hydroxide

3. Ironic Acid

4. Iron (I) Hydroxide

5. Iron Oxyhydride

6. Not listed

Which of the following definitions of Which of the following definitions of an acid includes conjugate acids?an acid includes conjugate acids?

1. Arrhenius

2. Bronsted-Lowry

3. Lewis

4. Kenzig

5. Woods

6. Toburen

7. Sabol

8. Sanson

Identify the conjugate base in the Identify the conjugate base in the following equation.following equation.

1. NH3

2. H2O

3. NH4+

4. OH-

HH22O + COO + CO332-2- OH OH-- + HCO + HCO33

According to Bronsted-Lowry theory, in According to Bronsted-Lowry theory, in the above reaction, Hthe above reaction, H22O is a(n)O is a(n)

1. Acid

2. Base

3. Conjugate acid

4. Conjugate base

According to Lewis theory,PClAccording to Lewis theory,PCl33 is a(n) is a(n)

1. Acid

2. Base

3. Salt

4. Conjugate Acid

5. Conjugate Base

The pOH of a 0.0030M solution of The pOH of a 0.0030M solution of HH22SOSO44 is: is:

1. 2.52

2. 11.48

3. 2.22

4. 11.78

5. 0.99

6. 13.01

100.0 mL of 3.000 M nitric acid 100.0 mL of 3.000 M nitric acid neutralizes 3.000 M of aluminum neutralizes 3.000 M of aluminum

hydroxide. How many mL of the base hydroxide. How many mL of the base did you use?did you use?

1. 100.0 mL

2. 50.00 mL

3. 33.33 mL

4. 16.67 mL

5. 8.333 mL

6. Not listed

Learning CheckLearning Check

A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

CalculationCalculation

P1 = 0.850 atm V1 = 675 mL T1 = 308 K

P2 = 1.06 atm V2 = 315 mL T2 = ??

P1 V1 P2 V2

= P1 V1 T2 = P2 V2 T1

T1 T2

T2 = 1.06 atm x 315 mL x 308 K

0.850 atm x 675 mL

T2 = 179 K - 273 = -94 °C

= 179 K

Zinc will react with hydrochloric acid. Zinc will react with hydrochloric acid. What are the 2 products for this What are the 2 products for this

reaction? reaction?

1. ZnCl + H

2. ZnCl + H2

3. Zn2Cl + H2

4. ZnCl2 + H2

5. Not listed

Zinc will react with hydrochloric acid. Zinc will react with hydrochloric acid. What kind of reaction is this? What kind of reaction is this?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

1. Double replacement

2. Single replacement

3. Synthesis

4. Decomposition

5. Not listed

Zinc will react with hydrochloric acid. This Zinc will react with hydrochloric acid. This reaction will form reaction will form ZnCl ZnCl22 + H + H22. What are the . What are the

4 coefficients for the balanced chemical 4 coefficients for the balanced chemical equation?equation?

1. 1,2,1,2

2. 2,1,2,1

3. 1,1,1,1

4. 2,2,1,2

5. Not listed

Zinc will react with hydrochloric acid. The Zinc will react with hydrochloric acid. The hydrogen gas is collected through water at hydrogen gas is collected through water at

30.030.0ooC and 782 mm Hg. The vapor C and 782 mm Hg. The vapor

pressure of water at 30.0pressure of water at 30.0ooC is 32.0 mm Hg. C is 32.0 mm Hg. What is the partial pressure of H What is the partial pressure of H22??

1. 250 atm

2. 314 atm

3. 0.329 atm

4. 0.987 atm

5. Not listed

Zinc ( 65.39 g/mole) will react with Zinc ( 65.39 g/mole) will react with hydrochloric acid. Determine the grams of hydrochloric acid. Determine the grams of zinc that must be reacted to produce this zinc that must be reacted to produce this quantity of hydrogen if the volume is 142 quantity of hydrogen if the volume is 142 mL . (P = 0.987 atm, T = 30.0mL . (P = 0.987 atm, T = 30.0ooC) C)

1. 0.112 g

2. 1.18 g

3. 628 g

4. 0.000 628 g

5. 0.369 g

When a gas forms a liquid, which process is taking place?

1. freezing

2. condensation

3. boiling

4. evaporation

Chemical or Physical

Based on the melting points shown in the table, which

material would still be a solid at 400°C?

1. beeswax

2. gold

3. lead

4. oxygen

Based on the melting points shown in the table, which

material would still be a solid at 400°C?

1. beeswax

2. gold

3. lead

4. oxygen

Substance Melting Point (°C)

Beeswax 62

Gold 1,063

Lead 327

Oxygen –218

Substance Melting Point (°C)

Beeswax 62

Gold 1,063

Lead 327

Oxygen –218

Chemical or Physical

A chemical change for a piece of metal would be

1. being bent in half.

2. getting cut into two pieces.

3. being painted.

4. getting rusty.

States of Matter States of Matter

Solid

Liquid

Gas

Melting

Deposition

Freezing

Sublimation

Condensation

Boiling

Enthapy Used to Change StateEnthapy Used to Change State

Solid

Liquid

Gas

Heat of Fusion

Heat of Vaporization

Determine the energy released in Determine the energy released in joules as a 152.00 g sample of metal joules as a 152.00 g sample of metal

(0.0335 cal/g(0.0335 cal/gooC cools 51.5C cools 51.5ooCC

q = cp x m x t:

q = x Jcp = 0.0335 cal/goC (4.184 J/cal)

m = 152.00 g t = -51.5oC

x = (0.140164 J/goC )(152.00 g)(-51.5 oC)

= -1.10 x 103 J

The End!!!