honors chemistry final assessment review
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Honors Chemistry Final Assessment Review. Your 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 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 - PowerPoint PPT PresentationTRANSCRIPT
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Honors ChemistryHonors ChemistryFinal Assessment ReviewFinal Assessment Review
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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
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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
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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
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Ionization Energy TrendsIonization Energy Trends
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Ionization Energy TrendsIonization Energy Trends
Top to Top to BottomBottom Left to RightLeft to Right
DecreasesDecreases IncreasesIncreases
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Electronegativity TrendsElectronegativity Trends
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Electronegativity TrendsElectronegativity Trends
Top to Top to BottomBottom Left to RightLeft to Right
DecreasesDecreases IncreasesIncreases
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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
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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
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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
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Exit QuizExit Quiz
Al3+ combines with sulfate (SO4)2– to make aluminum sulfate.
Write the chemical formula for aluminum sulfate.
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Al3+(SO4)
2-
32
Exit Quiz Answer
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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
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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
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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,
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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)
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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
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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
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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
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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
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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
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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
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Learning CheckLearning Check
What radioactive isotope is produced in the following bombardment of boron?
10B + 4He ? + 1n
5 2 0
13N7
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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
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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
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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
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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
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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
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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
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Use 10.0g of Zn to calculate the amount of ZnO produced.
=12.4g ZnO produced
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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
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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
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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
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Multiply subscripts by 3 to get
3 4 Fe O
1.36FeO
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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
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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
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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
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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
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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
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Identify the conjugate base in the Identify the conjugate base in the following equation.following equation.
1. NH3
2. H2O
3. NH4+
4. OH-
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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
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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
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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
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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
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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?
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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
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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
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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
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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
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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
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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
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When a gas forms a liquid, which process is taking place?
1. freezing
2. condensation
3. boiling
4. evaporation
Chemical or Physical
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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
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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.
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States of Matter States of Matter
Solid
Liquid
Gas
Melting
Deposition
Freezing
Sublimation
Condensation
Boiling
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Enthapy Used to Change StateEnthapy Used to Change State
Solid
Liquid
Gas
Heat of Fusion
Heat of Vaporization
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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
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The End!!!