CHEM 133 Week 16 Final Exam

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CHEM 133 Week 16 Final Exam | American Public University System

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CHEM 133 Week 16 Final Exam | American Public University System

CHEM 133 Final Exam

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Comment:Hi Suzy, Your final exam is graded. Please review and message me if you have questions. -Gloria

Part 1 of 11 - 0.0/ 0.0 Points

No Questions

Part 2 of 11 - 133 MC Chapter 07 10.0/ 10.0 Points

Question 1 of 35

2.0/ 2.0 Points

Calculate the Lattice Energy of KCl(s) given the following data using the Born-Haber cycle:

ΔHsublimation K = 79.2 kJ/mol

IE1 K = 418.7 kJ/mol

Bond EnergyCl–Cl = 242.8 kJ/mol

EACl = 348 kJ/mol

ΔHKCl(s) = –435.7 kJ/mol

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288 kJ/mol

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629 kJ/mol

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828 kJ/mol

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-165 kJ/mol

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707 kJ/mol

Feedback:

See Chapter 07

Question 2 of 35

2.0/ 2.0 Points

A chlorine atom in Cl2 should have a

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partial charge δ-

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partial charge δ+

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charge of 0

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charge of 1-

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Question 3 of 35

2.0/ 2.0 Points

The electron pair in a C-F bond could be considered

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closer to C because carbon has a larger radius and thus exerts greater control over the shared electron pair

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closer to F because fluorine has a higher electronegativity than carbon

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an inadequate model since the bond is ionic

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closer to C because carbon has a lower electronegativity than fluorine

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centrally located directly between the C and F

Feedback:

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Question 4 of 35

2.0/ 2.0 Points

Estimate the heat of combustion for one mole of acetylene:

C2H2(g) + O2(g) → 2CO2(g) + H2O(g)

Bond Bond Energy (kJ/mol)

C≡C 839

C–H 413

O=O 495

C=O 799

O–H 467

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1228 kJ

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-447 kJ

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-1228 kJ

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447 kJ

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365 kJ

Feedback:

See Chapter 07

Question 5 of 35

2.0/ 2.0 Points

Which of the following groups contains no ionic compounds?

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HCN, NO2, Ca(NO3)2

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CH2O, H2S, NH3

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NaH, CaF2, NaNH2

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KOH, CCl4, SF4

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PCl5, LiBr, Zn(OH)2

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Part 3 of 11 - 133 MC Chapter 08 OpenStax 8.0/ 10.0 Points

Question 6 of 35

2.0/ 2.0 Points

Which one of the following molecules has an atom with an incomplete octet?

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AsCl3

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BF3

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H2O

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NF3

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GeH4

Feedback:

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Question 7 of 35

2.0/ 2.0 Points

When comparing Be2 and H2:

I. Be2 is more stable because it contains both bonding and antibonding valence electrons.

II. H2 has a higher bond order than Be2.

III. H2 is more stable because it only contains 1s electrons.

IV. H2 is more stable because it is diamagnetic, whereas Be2 is paramagnetic.

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II, III

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II, III, IV

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III, IV

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III only

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I, II

Feedback:

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Question 8 of 35

0.0/ 2.0 Points

What is the bond order of He2+?

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0.0

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1.0

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1.5

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2.0

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0.5

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Question 9 of 35

2.0/ 2.0 Points

Which of the elements listed below is most likely to exhibit an expanded octet in its compounds?

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Question 10 of 35

2.0/ 2.0 Points

How many lone pairs are on the Br atom in BrF2-?

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Part 4 of 11 - 133 MC Chapter 10 OpenStax 8.0/ 10.0 Points

Question 11 of 35

2.0/ 2.0 Points

Which of the following compounds is expected to have the strongest ionic bonds?

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A. LiF

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B. NaBr

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C. RbI

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D. CsF

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E. KCl

Question 12 of 35

2.0/ 2.0 Points

Which is classified as an amorphous solid?

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platinum

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phosphorus tetrachloride

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palladium(II) chloride

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plastic

Question 13 of 35

0.0/ 2.0 Points

The experimentally measured vapor pressure of SO2 at -25oC is 49.5 kPa and at 0oC is 155 kPa. Determine the enthalpy of vaporization, ΔHvap, at 100 oC.

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8549 torr

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4150 torr

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3230 torr

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9821 torr

Feedback:

See Chapter 10

Question 14 of 35

2.0/ 2.0 Points

Ethyl chloride, C2H5Cl, is used as a local anesthetic. It works by cooling tissue as it vaporizes. The heat of vaporization is 26.4 kJ/mol. How much heat could be removed by 10.0 g of ethyl chloride?

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264 kJ

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4.09 kJ

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170. kJ

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1.70 x 103 kJ

Feedback:

See Chapter 10

Question 15 of 35

2.0/ 2.0 Points

The freezing point of helium is –270°C. The freezing point of xenon is –112°C. Both of these are in the noble gas family. Which of the following statements is supported by these data?

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The London dispersion forces between the helium molecules are greater than the London dispersion between the xenon molecules.

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Helium and xenon form highly polar molecules.

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As the molecular weight of the noble gas increases, the freezing point decreases.

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The London dispersion forces between the helium molecules are less than the London dispersion forces between the xenon molecules.

Part 5 of 11 - 133 MC Chapter 05 OpenStax 10.0/ 10.0 Points

Question 16 of 35

2.0/ 2.0 Points

A 0.1326 g sample of magnesium was burned in an oxygen bomb calorimeter. The total heat capacity of the calorimeter plus water was 5,760 J/°C. If the temperature rise of the calorimeter with water was 0.570°C, calculate the enthalpy of combustion of magnesium.

Mg(s) + 1/2O2(g) → MgO(s)

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–3280 kJ/mol

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106 kJ/mol

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–24.8 kJ/mol

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–602 kJ/mol

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E. 435 kJ/mol

Feedback:See Chapter 05

Question 17 of 35

2.0/ 2.0 Points

How much heat is required to raise the temperature of 1.5 x 103 g of water from 45°F to 130.°F? The specific heat of water is 4.184 J/g•°C.

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8.2 x 102 kJ

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3.0 x 102 kJ

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3.0 x 101 kJ

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3.4 x 102 kJ

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5.3 x 102 kJ

Feedback:

See Chapter 05

Question 18 of 35

2.0/ 2.0 Points

Which of the following processes is exothermic?

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CO2(s) → CO2(g)

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CO2(g) + 2 H2O(l) → CH4(g) + 2 O2(g)

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6 H2O(g) + 4 CO2(g) → 2 C2H6(g) + 7 O2(g)

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CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

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H2O(l) → H2O(g)

Feedback:

See Chapter 05

Question 19 of 35

2.0/ 2.0 Points

How much heat is absorbed/released when 20.00 g of NH3(g) reacts in the presence of excess O2(g) to produce NO(g) and H2O(l) according to the following chemical equation?

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l) ΔH° = +1168 kJ

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1372 kJ of heat are released.

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342.9 kJ of heat are released.

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342.9 kJ of heat are absorbed.

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1372 kJ of heat are absorbed.

Feedback:

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Question 20 of 35

2.0/ 2.0 Points

Aluminum metal has a specific heat of 0.900 J/g•°C. Calculate the amount of heat required to raise the temperature of 10.5 moles of Al from 30.5 °C to 225°C.

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A. 57.3 kJ

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49.6 kJ

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2.41 kJ

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65.1 kJ

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1.84 kJ

Feedback:See Chapter 05

Part 6 of 11 - 133 MC Chapter 09 OpenStax 8.0/ 10.0 Points

Question 21 of 35

0.0/ 2.0 Points

Pressure is

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A. measured in grams

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measured in Newtons

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defined as the number of moles of substance divided by the mass of the substance

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defined as the force per unit area

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defined as the mass that an object exerts when at rest

Feedback:See Chapter 09

Question 22 of 35

2.0/ 2.0 Points

A small bubble rises from the bottom of a lake, where the temperature and pressure are 4°C and 3.0 atm, to the water's surface, where the temperature is 25°C and the pressure is 0.95 atm. Calculate the final volume of the bubble if its initial volume was 2.1 mL.

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0.72 mL

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B. 41.4 mL

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6.2 mL

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7.1 mL

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22.4 mL

Feedback:See Chapter 09

Question 23 of 35

2.0/ 2.0 Points

A sample of pure nitrogen has a temperature of 15 °C. What is the temperature of the nitrogen in units of Kelvin?

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290 K

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300 K

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288.15 K

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288 K

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288.2 K

Question 24 of 35

It is found that 250. mL of a gas at STP has a mass of 0.677 g. What is the molar mass?

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2.71 g/mol

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16.5 g/mol

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11.2 g/mol

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60.7 g/mol

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132 g/mol

Question 25 of 35

Consider a sample of gas in a container on a comfortable spring day. The Celsius temperature suddenly doubles, and you transfer the gas to a container with twice the volume of the first container. If the original pressure was 12 atm, what is a good estimate for the new pressure?

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5.5 atm

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3 atm

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6.4 atm

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12 atm

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15 atm

See Chapter 09

Question 26 of 35

For the electronic transition from n = 2 to n = 4 in the hydrogen atom.

a) Calculate the energy.

b) Calculate the wavelength (in nm).

a) Calculate the energy.

RH = Rydberg constant

Energy = h * c / wavelength = h * c * RH * Z^2 [1 / nf^2 - 1 / ni^2]

Energy = 6.626 * 10^-34 * 3.0 * 10^8 * 10 973 731.6 * (1)^2 * [1 / 2^2 - 1 / 4^2]

Energy = 4.1*10^0-19J

b) Calculate the wavelength (in nm).

1 / wavelength = RH * Z^2 [1 / nf^2 - 1 / ni^2] = 10 973 731.6 * (1)^2 * [1 / 2^2 - 1 / 4^2]

wavelength = 486nm

Question 27 of 35

For the electronic transition from n = 3 to n = 5 in the hydrogen atom.

a) Calculate the energy.

b) Calculate the wavelength (in nm).

a) Calculate the energy.

RH = Rydberg constant

Energy = h * c / wavelength = h * c * RH * Z^2 [1 / nf^2 - 1 / ni^2]

Energy = 6.626 * 10^-34 * 3.0 * 10^8 * 10 973 731.6 * (1)^2 * [1 / 3^2 - 1 / 5^2]

Energy = 2.18 * 10^-18 * 0.071

Energy = 1.55 * 10^-19 J

b) Calculate the wavelength (in nm).

1 / wavelength = RH * Z^2 [1 / nf^2 - 1 / ni^2] = 10 973 731.6 * (1)^2 * [1 / 3^2 - 1 / 5^2]

wavelength = 1.281 * 10^-6 m = 1281 nm

Question 28 of 35

5.0/ 5.0 Points

Given N22- using molecular orbital and valence bond theory,

a) Write molecular orbital configuration ______________________

b) Determine bond order ______________________

c) Stability ______________________

d) Paramagnetic or Diamagnetic? ______________________

a) σ21sσ*21sσ22sσ*22sπ22pπ22pσ22pπ*12pπ*1 p

b) Bond order=(10-6)/2 =2

c) it is less stable as compared to N2

d) It is Paramagnetic as there are two unpaired electrons

Feedback:See Chapter 08

Question 29 of 35

Consider the following molecules, H2, He2, and O2.

a) Which one has a bond order of 2? ___________

b) Which one has a single bond? ___________

c) Which is the most unstable? ___________

d) Write them in order of increasing stability.

a)0-2 has order of 2

b) H2 has a single bond

c) He2

d He2,H2 , O2.

Question 30 of 35

Substances with high lattice energies tend to be less soluble than substances with low lattice energies. On that basis predict the relative aqueous solubility at 20 °C, from highest to lowest, of the following ionic compounds: Ce2(SO4)3, K2SO4, KBr, NaCl.

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Ce2(SO4)3 > K2SO4 >KBr>NaCl

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NaCl>KBr> K2SO4 > Ce2(SO4)3

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Ce2(SO4)3 > K2SO4 >NaCl>KBr

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KBr>NaCl> K2SO4 > Ce2(SO4)3

Question 31 of 35

A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. Calculate the number of atoms in the unit cell and diameter of the metal atom.

for face centered cubic:

sqrt(2) *a = 4r {a is edge length}

sqrt(2)*408 = 4*r

r=144.625x2

diameter=288.5pm

Ans: 288.5pm

number of atoms =6+8+14 atoms

Feedback:See Chapter 10

Comment:wrong # atoms

Part 10 of 11 - 133 MC Chapter 05 OpenStax

Question 32 of 35

5.0/ 5.0 Points

At 25°C, the following heats of reaction are known:

2ClF(g) + O2(g) → Cl2O(g) + F2O(g) ΔH°rxn = 167.4 kJ/mol

2ClF3(g) + 2O2(g) → Cl2O(g) + 3F2O(g) ΔH°rxn = 341.4 kJ/mol

2F2(g) + O2(g) → 2F2O(g) ΔH°rxn = –43.4 kJ/mol

At the same temperature, use Hess's law to calculate ΔH°rxn for the reaction:

ClF(g) + F2(g) → ClF3(g)

ClF(g) + F2(g) ------------------> ClF3(g)

half the first reaction.

ClF(g) + 1/2O2(g) ------------> 1/2ClO(g) + 1/2F2O(g) ∆H°rxn = 83.7 KJ/mol.

revers and half the first reaction.

1.5 F2O(g) + 1/2Cl2O(g) -----------> ClF3(g) + O2(g) ∆H°rxn = -170.7 KJ/mol

half the third reaction

F2(g) + 1/2O2(g) -----------> F2O(g) ∆H°rxn = -21.7 KJ/mol

add all 3 equation.

ClF(g) + 1/2O2(g) ------------> 1/2ClO(g) + 1/2F2O(g) ∆H°rxn = 83.7 KJ/mol.

1.5 F2O(g) + 1/2Cl2O(g) -----------> ClF3(g) + O2(g) ∆H°rxn = -170.7 KJ/mol

F2(g) + 1/2O2(g) -----------> F2O(g) ∆H°rxn = -21.7 KJ/mol

ClF(g) + F2(g) ------------------> ClF3(g) ∆H°rxn = - 108.7 KJ/mol

Feedback:See Chapter 05

Question 33 of 35

5.0/ 5.0 Points

Equation I shows phosphorous pentachloride preparation from PCl3 and Cl2:

I) PCl3 (l) + Cl2(g) → PCl5(s)

Use equation II and III to calculate ∆Hrxs of equation I:

II) P4 (s) + 6 Cl2 (g) → 4 PCl3 (l) ∆H = -1280 KJ

III) P4 (s) + 10 Cl2 (g) → 4 PCl5 (s) ∆H = -1774 KJ

4 (s) + 6 Cl2 (g) ------> 4 PCl3 ΔH2 = -1280 kJ ------- eqn 2

P4 (s) + 10 Cl2 (g) ------> 4 PCl5 ΔH3 = -1774 kJ -------- eqn 3

Doing operation eqn 3 - eqn 2

4 PCl3 + 4 PCl2 ------> 4 PCl5

ΔH = ΔH3 - ΔH2 = - 494 kJ

=> PCl3 + PCl2 ------> PCl5

ΔH1 = ΔH/4 = -123.5 kJ

Feedback:See Chapter 05

Part 11 of 11 - 133 MC Chapter 09 OpenStax 5.0/ 10.0 Points

Question 34 of 35

5.0/ 5.0 Points

What volume of O2(g), measured at 27 °C and 743 torr, is consumed in the combustion of 12.50 L of C2H6(g)?

2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l)

2 C2H6(g) + 7 O2(g) ==> 4 CO2(g) + 6 H2O(l)

at STP, 2 moles of C2H6(g) needs 7 moles of O2

1 mole of any gas at STP = 22.4 liters

Then at STP, (2x 22.4 =) 44.8 liters of C2H6(g) needs (7x22.4) = 156.8 liters of O2

So at STP, 1 liter of C2H6(g) needs (156.8/44.8=) 3.5 liters of O2

so at STP, 12.50 liter of C2H6(g) needs (3.5x12,5=) 43.75 liters of O2

At stp we got volume = 43.75 liters, At STP, temperature (T) = 273.15 K and Pressure(p) = 760 torr

we need to find the volume of O2(g), at measured at 27 °C= 27 + 273 = 300K and and pressure P = 743 torr,

Let us assume volume (V1) = 43.75 liters, At STP, temperature (T1) = 273.15 K and Pressure(P1) = 760 torr

and volume of O2(g) = ?, temperature (T2)= 27 °C= 27 + 273 = 300K and and pressure (P2) = 743 torr

According to gas law [P1V1/T1] = [P2V2/T2]

P1V1T2 = P2V2T1

(760)(43.75)(300) = (743)(V2)(273.15)

V2 = 49.1499 =49.15 L

Feedback:See Chapter 09

Question 35 of 35

0.0/ 5.0 Points

A mixture of 14.0 grams of H2, 84.0 grams of N2, and 64.0 grams of O2 are placed in a flask. The partial pressure of the O2 is 78.00 torr.

1. What is the total pressure in the flask?

2. What are the partial pressures of the remaining gases, H2 & N2

H2 N2 O2

w (g) 14.0 84.0 64.0

MW (g/mole) 2 28 32

mole=w/MW 7 3 2 Total = 12

mole fraction 7/12= 0.58 3/12= 0.25 2/12= 0.17

partial pressure (torr) 266 115 78

(Total pressure = partial pressure/mole fraction = 78/0.17 =458.82 torr)

Feedback:See Chapter 09

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