CHEM125 – CHAPTER 09 – REGULAR HOMEWORK

CHAPTER 09 REGULAR HOMEWORK

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Question 1

Arrange the following series of compounds in order of increasing lattice energies.
Select one:
 Correct

Explanation

Smaller the radius of element = higher lattice energy smaller the radius the stronger the bond.
CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 1

Question 2

Arrange the following series of compounds in order of increasing lattice energies.
Select one:
 Correct

Explanation

Smaller the radius of element = higher lattice energy smaller the radius the stronger the bond.
CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 1

Question 3

Arrange the following series of compounds in order of increasing lattice energies.
Select one:
 Correct

Explanation

Smaller the radius of element = higher lattice energy smaller the radius the stronger the bond.
CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 1

Question 4

Write Lewis structures for the following compounds. Match compounds with the same Lewis Structure. All must be correct to get credit.
LiF Correct
Cl2 Correct
NO+ Correct
NO Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 4
CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 5
CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 6
CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 7

Question 5

Write the Lewis structure for BF4 (each F is bonded to B). How many single bonds are there?
Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 8

Question 6

Write the Lewis structure for CH2CHCHO . How many single bonds are there? Help–skeleton structure: CH2-CH-CHO, meaning the first C is bonded to 2 H’s and a 2nd C. The 2nd C is also bonded to a H and the 3rd C. The 3rd C is also bonded to a H and an O.
Correct

Explanation

Your content goes here…
CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 9

Question 7

How many double bonds are there in CH2CHCHO?
Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 9

Question 8

How many triple bonds are there in CH2CHCHO?
Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 9

Question 9

Write the Lewis structure for CF2CF2 . How many single bonds are there? Help–skeleton structure: CF2-CF2 meaning, The first C is bonded to two F’s and the second C. In addition, the second C is bonded to two F’s.
Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 12

Question 10

Write the Lewis structure for CH2CHCCH How many single bonds are there? Help–skeleton structure CH2-CH-C-CH.
Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 13

Question 11

How many double bonds are there in CH2CHCCH?
Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 13

Question 12

How many triple bonds are there in CH2CHCCH?
Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 13

Question 13

Is the following Lewis structure correct for CH2CHOCOOH? Check all that apply. Help given in feedback.
CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 16
Select one or more:
 Correct
 Correct
 Correct

Explanation

Formal charge = number of valence electrons – (number of nonbonding electrons + number of bonding electrons/2) C : 4 – 0 – 5 = -1 O: 6 – 4 -3 = -1

Question 14

Is the following Lewis structure correct for CH2CHOCOOH?
CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 17
Select one:
 Correct

Explanation

Formal charge = number of valence electrons – (number of nonbonding electrons + number of bonding electrons/2) Formal charge on elements are equal to zero similar to the formula.

Question 15

Is the following Lewis structure correct for CH2CHOCOOH? Check all that apply.
CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 18
Select one or more:
 Correct
 Correct

Explanation

Formal charge = number of valence electrons – (number of nonbonding electrons + number of bonding electrons/2) O : 6 – 6 – 1 = -1 C: 4 – 3 = +1

Question 16

Is the following Lewis structure correct for CH2CHOCOOH? Check all that apply.
CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 19
Select one or more:
 Correct
 Correct

Explanation

O : 6 – 6 – 1 = -1 Charge of Lewis structure (-1) not equal to formula’s charge (0)

Question 17

Which of the following molecules has the most polar bond: O2, BrCl, or ICl?
Select one:
 Correct

Explanation

The greater the difference in electronegativity, the greater the bond polarity.
CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 20

Question 18

There are two equivalent resonance structures for CHOO (all atoms bonded to C). How many valence electrons are there? Remember all C and O atoms obey the octet rule.
Correct

Explanation

CHOO : 4 + 1 + 6 + 6 +1 = 18 valence electrons

Question 19

Describe the carbon-oxygen bonds in CHOO.
Select one:
 Correct

Explanation

The resonance structures obey the octet rule (except H which has 2 electrons like He).
CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 21

Question 20

With respect to either resonnace structure, which of the atoms have a formal charge of 0 in CHOO?
Select one or more:
 Correct
 Correct
 Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 21

Question 21

What is the sum of the formal charges on the H, C, and 2 O atoms in CHOO?
Correct

Explanation

The sum of charge is equal to the formula’s overall charge which in this case is -1.

Question 22

How many valence electrons are in HONO (H-O-N-O).
Correct

Explanation

H-O-N-O : 1 + 6 + 5 + 6 = 18 valence electrons

Question 23

Draw two resonance structures for HONO. In the more important structure, which of the atoms have a formal charge of 0? Remember all N and O atoms obey the octet rule.
Select one or more:
 Correct
 Correct
 Correct
 Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 23
The best structure is the one that corresponds to the overall charge of the compound.

Question 24

In the less important resonance structure for HONO, which of the atoms have a formal charge of 0?
Select one or more:
 Correct
 Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 23

Question 25

How many valence electrons are there in ClO3OH (Cl is the central atom–each O is bonded to it. The H is bonded to one of the O’s)?
Correct

Explanation

ClO3OH : 7 + (3 x 6) + 6+1 = 42 valence electrons

Question 26

For ClO3OH, write a resonance structure with the octet rule satisfied for the central atom and a second resonance structure that minimizes formal charges. In the one that satisfies the octet rule, which atoms have a formal charge of zero?
Select one or more:
 Correct
 Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 25
———————O ———————-l —————–O-Cl-O ———————-l ———————OH

Question 27

In the one that satisfies the octet rule, what is the formal charge of the chlorine?
Correct

Explanation

Look Above at question 26

Question 28

In the one that minimizes formal charges, which atoms have a formal charge of zero?
Select one or more:
 Correct
 Correct
 Correct
 Correct

Explanation

By the way, the one that minimizes formal charge is expected to be the more important resonance structure. Look Above at question 26

Question 29

In the one that minimizes the formal charge, how many double bonds does the chlorine atom make?
Correct

Explanation

Your content goes here…

Question 30

In the gas phase, the oxide N2O5 has a structure with an N-O-N core,and each N bonds to two additional O atoms. In contrast, in the solid phase the stable form is an ionic solid, (NO2+)(NO3 ). How many valence electrons does N2O5 have?
Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 26

Question 31

Draw one of the Lewis structure of the N2O5. In each case one oxygen bridges the nitrogens (N-O-N single bond) and 2 other oxygens are bonded to each N. How many equivalent resonance structures are there that satisfy the octet rule and where O makes at most 2 bonds?
Correct

Explanation

Move the double for each oxygen and nitrogen connection for resonance strucuture
CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 27

Question 32

How many double bonds does each N2O5 resonance structures have?
Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 27

Question 33

Draw the dominant Lewis structure for NO2+. Note, that this ion is isoelectronic with CO2. How many double bonds does this resonance structures have?
Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 29

Question 34

How many valence electrons does NO3 have?
Correct

Explanation

NO3 : 5+ (3 x 6) + 1 = 24 valence electrons

Question 35

How many equivalent resonance structures does NO3 have?
Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 30

Question 36

The reaction of XeF6 + H2O → XeOF4 + HF. Write Lewis dot structures for all the molecules. In the case of Xe containing molecules, Xe is the central atom. How many of the F atoms in XeF6 satisfy the octet rule and form one bond?

Explanation

Question 37

How many lone pairs of electrons does Xe have in XeF6?
Correct

Explanation

Question 38

How many valence electrons does XeOF4?
Correct

Explanation

XeOF4: 8 + 6 + (4 x 7) = 42 valence electrons

Question 39

How many lone pairs of electrons does Xe have in XeOF4? (Use the structure where all the atoms have a formal charge of 0.)
Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 31

Question 40

How many lone pairs of electrons does O have in XeOF4? (Use the structure where all the atoms have a formal charge of 0.)
Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 32

Question 41

Using Table 9.4 (see below), calculate the energy (in kJ) required to break all of the bonds in 1 mole(s) of methanol (CH3OH). Hint given in feedback
Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 33
First calculate the per mole energy. C- H : 414 kJ/mol x 3 = 1242 C- O : 351 kJ/ mol x 1 = 351 O – H : 463 kJ/ mol x 1 = 463 —————————————— 2056 kJ/mol

Question 42

Using Table 9.4, calculate an approximate enthalpy (in kJ) for the reaction of 3.67 g gaseous methanol (CH3OH) in excess molecular oxygen to form gaseous carbon dioxide and gaseous water. (Hint, remember to first write the balanced equation.)
CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 34
Correct

Explanation

2CH3OH +3O2 –> 2CO2 +4H2O sum of bonds reactants: 2*(3CH + CO + OH) + 3(O2) Sum of bond products: 2*(2CO) + 4*(2OH) Now use your table: 2*(3*414 + 351 + 463) + (3*498) = 5606 kJ/mol reactant 4*799 + 8*463 = 6900 kJ /mol product Now use the equation = rect – prod = -1294 kJ -1294 kJ/ 2 mol CH3OH -647 kJ / 1 mol CH3OH so no. of moles of methanol = 3.67/32 = 0.1147 So Enthalpy = -647 *0.1147 = -74.3

Question 43

Use Table 9.4 to calculate an approximate enthalpy or heat of reaction for the combustion of one mole of methane gas (CH4) to form gaseous H2O and CO2. What Volume (in L) of methane is needed to produce 3098 kJ of heat for methane gas at 31 oC and 7.89 atm pressure? Hint given in feedback
Correct

Explanation

2) Balanced chemiacl reaction is CH4 + 2O2 CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 35 CO2 + 2H2O CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 36H0 = CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 37 CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 36H0 (Reactant bonds) – CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 37CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 36H0 (product bonds) CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 36H0 =. [4 CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 36H0(C – H) + 2CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 36H0(O = O) ] – [2 CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 36H0(C = O) + 4CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 36H0(O – H) ] Substitute the bond enthalpy in above equation [4(414 kJ/mol) + 2(498 KJ/mpl)] – [2(799 kJ/mol) + 4( 463 kJ/mol) ] (2652 kJ/mol) – (3450) = -798 kJ/mol CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 36H0 for your given reaction is = -798 kJ/mol 1 mole methane produce -798 kJ heat then to produce 3098 kJ heat required methane = 3098 / 798 = 3.88 mole of methane volume fo methane in liter = We know that PV = nRT V = nRT/P n = 3.88 mole, T = 190C = 31+273.15 = 304 K, P= 7.89 atm, R = 0.08206 L atm mol-1 K-1 ( R = gas constant) V = ? Substitute these value in above equation. V = 3.88 X 0.08205 X 304/7.89 = 12.27 L 12.27 L methane required

Question 44

What molecular compound is expected to be form from nitrogen and chlorine based on Lewis structures.. For NxCly, what is x+y?
Correct

Explanation

CHEM125 - CHAPTER 09 - REGULAR HOMEWORK 47
NCl3: 1 + 3 = 4