chemistry, the central science , 10th edition

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Donna Narsavage Heald Siena College Loudonville, NY 2006, Prentice-Hall Midterm Review Nuclear Chemistry Gas Laws Intermolecular Forces Colligative Properties Chemical Equilibrium Chemistry, The Central Science, 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten

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0. Midterm Review Nuclear Chemistry Gas Laws Intermolecular Forces Colligative Properties Chemical Equilibrium. Chemistry, The Central Science , 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten. Donna Narsavage Heald Siena College Loudonville, NY - PowerPoint PPT Presentation

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Page 1: Chemistry, The Central Science , 10th edition

Donna Narsavage HealdSiena CollegeLoudonville, NY 2006, Prentice-Hall

Midterm ReviewNuclear Chemistry

Gas LawsIntermolecular ForcesColligative PropertiesChemical Equilibrium

Chemistry, The Central Science, 10th edition

Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten

Page 2: Chemistry, The Central Science , 10th edition

What change in the mass number of a nucleus occurs when the nucleus emits an alpha particle?

1. Mass number decreases by 1.

2. Mass number decreases by 2.

3. Mass number decreases by 4.

4. Mass number does not change.

Page 3: Chemistry, The Central Science , 10th edition

What change in the mass number of a nucleus occurs when the nucleus emits an alpha particle?

1. Mass number decreases by 1.

2. Mass number decreases by 2.

3. Mass number decreases by 4.

4. Mass number does not change.

Page 4: Chemistry, The Central Science , 10th edition

1. Yes

2. No

Page 5: Chemistry, The Central Science , 10th edition

1. Yes

2. No

Page 6: Chemistry, The Central Science , 10th edition

Bonus:

1. Alpha rays most efficiently pass through the window.

2. Beta rays most efficiently pass through the window.

3. Gamma rays most efficiently pass through the window.

Page 7: Chemistry, The Central Science , 10th edition

Bonus:

1. Alpha rays most efficiently pass through the window.

2. Beta rays most efficiently pass through the window.

3. Gamma rays most efficiently pass through the window.

Page 8: Chemistry, The Central Science , 10th edition

Indicate the number of protons (p) and neutrons (n) in the radioactive nucleus represented by: 60

27 Co1. 60 p, 27 n2. 33 p, 27 n3. 27 p, 33 n4. 27 p, 60 n5. None of the above

Page 9: Chemistry, The Central Science , 10th edition

Correct Answer:

1. 60 p, 27 n2. 33 p, 27 n3. 27 p, 33 n4. 27 p, 60 n5. None of the above

The difference between the mass number and the atomic number is the number of neutrons (60 27) = 33.

Page 10: Chemistry, The Central Science , 10th edition

If the nucleus:

undergoes positron emission it will become the isotope:

1. Potassium392. Argon383. Argon394. Calcium385. Calcium39

3819 K

Page 11: Chemistry, The Central Science , 10th edition

Correct Answer:

The nuclear equation describing this process is given below:

3819 K

01 e

3818 Ar +

1. Potassium392. Argon383. Argon394. Calcium385. Calcium39

Page 12: Chemistry, The Central Science , 10th edition

Which of the following nuclei would you expect not to be radioactive? A band of stability is provided for reference.

1. Carbon-142. Phosphorus-313. Iron-594. Radium-2265. Uranium-235

Page 13: Chemistry, The Central Science , 10th edition

Correct Answer:

Only phosphorus-31 is not outside the band of stability; its position is indicated by the red arrow.

1. Carbon-142. Phosphorus-313. Iron-594. Radium-2265. Uranium-235

Page 14: Chemistry, The Central Science , 10th edition

How many (X) neutrons are required to balance the nuclear transmutation equation shown below?

1. 12. 23. 34. 45. 6

nXKrBaUn 10

9136

14256

23592

10

Page 15: Chemistry, The Central Science , 10th edition

Correct Answer:

1. 12. 23. 34. 45. 6

n3KrBaUn1

0

91

36

142

56

235

92

1

0

The balanced nuclear equation is given below:

Page 16: Chemistry, The Central Science , 10th edition

The isotope carbon-14 has a half-life of 5,715 years. How much of a 1.00 g sample of carbon-14 will exist after 34,290 years?

1. 0.0156 g2. 0.0225 g3. 0.143 g

4. 0.166 g5. None (all would have

decayed).

Page 17: Chemistry, The Central Science , 10th edition

Correct Answer:

1. 0.0156 g2. 0.0225 g3. 0.143 g4. 0.166 g5. None (all would

have decayed).

34,290 yr/5,715 yr = 6

Thus, six half-lives occur over this time.

1.00 g ( ) 6 = 1.00 g (1/64) = 0.0156 g1 2

Page 18: Chemistry, The Central Science , 10th edition

1. Loss of a neutron.2. Gain of a proton.3. Loss of electrons.4. Loss of an alpha particle.5. Gain of a positron.

The radioactive decay of Ra to Rn occurs with:

226Ra 222Rn

Page 19: Chemistry, The Central Science , 10th edition

1. Loss of a neutron.2. Gain of a proton.3. Loss of electrons.4. Loss of an alpha particle.5. Gain of a positron.

The radioactive decay of Ra to Rn occurs with:

226Ra 222Rn

Page 20: Chemistry, The Central Science , 10th edition

Oxygen-15 is used as a radioactive tracer in medical imaging and has a half-life of 2.0 min. If the initial

concentration of 15O is 16 mM, how long will it take for the concentration to be reduced to 1 mM?

1. 4 min2. 6 min3. 8 min4. 10 min5. 12 min

Page 21: Chemistry, The Central Science , 10th edition

Oxygen-15 is used as a radioactive tracer in medical imaging and has a half-life of 2.0 min. If the initial

concentration of 15O is 16 mM, how long will it take for the concentration to be reduced to 1 mM?

1. 4 min2. 6 min3. 8 min4. 10 min5. 12 min

Page 22: Chemistry, The Central Science , 10th edition

If 250 mL of NO is placed in a flask with O2, what volume of O2 is needed for complete reaction?

1. 100 mL2. 125 mL3. 200 mL4. 250 mL5. Cannot be determined from the given information.

2 NO (g) + O2 (g) 2 NO2 (g)

Page 23: Chemistry, The Central Science , 10th edition

If 250 mL of NO is placed in a flask with O2, what volume of O2 is needed for complete reaction?

1. 100 mL2. 125 mL3. 200 mL4. 250 mL5. Cannot be determined from the given information.

2 NO (g) + O2 (g) 2 NO2 (g)

Page 24: Chemistry, The Central Science , 10th edition

If equal masses of CH4, C2H6, and C3H8 are placedin a flask, which of the following is true?

1. PCH4 = PC2H6

= PC3H8

2. PCH4 ~ PC2H6

~ PC3H8

3. PCH4 > PC2H6

> PC3H8

4. PCH4 < PC2H6

< PC3H8

5. None of the above

Page 25: Chemistry, The Central Science , 10th edition

If equal masses of CH4, C2H6, and C3H8 are placedin a flask, which of the following is true?

1. PCH4 = PC2H6

= PC3H8

2. PCH4 ~ PC2H6

~ PC3H8

3. PCH4 > PC2H6

> PC3H8

4. PCH4 < PC2H6

< PC3H8

5. None of the above

Page 26: Chemistry, The Central Science , 10th edition

1. Gases are generally of very low molecular mass.

2. Gases have high kinetic energies.

3. In gases, the distance between molecules is relatively large.

4. In gases, the attractions between molecules are large.

Page 27: Chemistry, The Central Science , 10th edition

1. Gases are generally of very low molecular mass.

2. Gases have high kinetic energies.

3. In gases, the distance between molecules is relatively large.

4. In gases, the attractions between molecules are large.

Page 28: Chemistry, The Central Science , 10th edition

1. Densities cannot be compared without additional amount and volume information.

2. Water vapor has the same density as N2 under

the same conditions of temperature and pressure.

3. Water vapor is less dense than N2.

4. Water vapor is more dense than N2.

Page 29: Chemistry, The Central Science , 10th edition

1. Densities cannot be compared without additional amount and volume information.

2. Water vapor has the same density as N2 under

the same conditions of temperature and pressure.

3. Water vapor is less dense than N2.

4. Water vapor is more dense than N2.

Page 30: Chemistry, The Central Science , 10th edition

1. Additional pressure information is needed to compare average kinetic energies.

2. The heaviest gas (HCl) has the highest average kinetic energy and the other average kinetic energies decrease with molar mass.

3. HCl (298 K) and H2 (298 K) have the same

average kinetic energy while O2 (350 K) has a

lower average kinetic energy.

4. HCl (298 K) and H2 (298 K) have the same

average kinetic energy while O2 (350 K) has a

higher average kinetic energy.

Page 31: Chemistry, The Central Science , 10th edition

1. Additional pressure information is needed to compare average kinetic energies.

2. The heaviest gas (HCl) has the highest average kinetic energy and the other average kinetic energies decrease with molar mass.

3. HCl (298 K) and H2 (298 K) have the same

average kinetic energy while O2 (350 K) has a

lower average kinetic energy.

4. HCl (298 K) and H2 (298 K) have the same

average kinetic energy while O2 (350 K) has a

higher average kinetic energy.

Page 32: Chemistry, The Central Science , 10th edition

1. Gases deviate from ideal behavior if their molar mass exceeds that of H2O.

2. Gases deviate from ideal behavior if the temperature is lower than room 298 K.

3. Gases deviate from ideal behavior because the molecules have finite sizes and there are some attractions between the molecules.

4. Gases deviate from ideal behavior if their normal boiling points are below room temperature and greater than the freezing point of H2O.

Page 33: Chemistry, The Central Science , 10th edition

1. Gases deviate from ideal behavior if their molar mass exceeds that of H2O.

2. Gases deviate from ideal behavior if the temperature is lower than room 298 K.

3. Gases deviate from ideal behavior because the molecules have finite sizes and there are some attractions between the molecules.

4. Gases deviate from ideal behavior if their normal boiling points are below room temperature and greater than the freezing point of H2O.

Page 34: Chemistry, The Central Science , 10th edition

A gas initially at 2.0 atm is in an adjustable volume container of 10. L in volume. If the pressure is decreased to 0.50 atm, what is the new volume?

1. 40. L2. 20. L3. 10. L4. 5.0 L

Page 35: Chemistry, The Central Science , 10th edition

Correct Answer:

PV

1constant

constantPV

Thus,

2.00 atm(10. L) = 0.50 atm (Vfinal)Vfinal = 2.00 atm(10. L)/0.50 atm = 40. L

1. 40. L2. 20. L3. 10. L4. 5.0 L

Page 36: Chemistry, The Central Science , 10th edition

Assuming pressure is held constant, to what volume will a balloon initially at 1.0 L change if its temperature is decreased from 300 K to 75 K?

1. 1.0 L2. 2.0 L3. 0.25 L4. 4.0 L

Page 37: Chemistry, The Central Science , 10th edition

Correct Answer:

TV constant

constantT

V

Thus,

1.0 L/300 K = (Vfinal)/75 KVfinal = 75 K/(1.0 L)300 K = 0.25 L

1. 1.0 L2. 2.0 L3. 0.25 L4. 4.0 L

Page 38: Chemistry, The Central Science , 10th edition

At standard temperature and pressure, how many moles of gas are present in a box with a volume of 112 L?

1. 1.00 moles2. 2.00 moles3. 5.00 moles4. 0.200 moles

Page 39: Chemistry, The Central Science , 10th edition

Correct Answer:

L 22.41

atm 1.000

K 273.15KL·atm/mol· 0.08206mol 1

P

nRTV

nRTPV

Thus, at STP

22.41 L = 112 L1.00 mol n

n = 5.00 moles

1. 1.00 moles2. 2.00 moles3. 5.00 moles4. 0.200 moles

Page 40: Chemistry, The Central Science , 10th edition

A container holds a mixture of oxygen, neon, and helium gases whose partial pressures are 150 torr, 300 torr, and 450 torr, respectively. The mole fraction of neon is

1. 0.172. 0.333. 0.504. 0.67

totali PP i

Page 41: Chemistry, The Central Science , 10th edition

Correct Answer:

Xi = Pi/Ptotal

Xi = (300 torr)/(150 + 300 + 450) torr

Xi = 300 torr/900 torr = 0.33

totali PP i1. 0.172. 0.333. 0.504. 0.67

Page 42: Chemistry, The Central Science , 10th edition

An unknown gas effuses at half the rate of helium. This gas is likely to be which of the following?

1. H2

2. CH4

3. Ne4. O2

5. Ar

Page 43: Chemistry, The Central Science , 10th edition

1

2

M

M

2

1

r

r

Correct Answer:

(r1/2)2 =M2/M1

M2= (r1/r2)2M1

M2= (2/1)2(4.0 g/mol) = 16.0 g/mol

Therefore it could be CH4

1. H2

2. CH4

3. Ne4. O2

5. Ar

Page 44: Chemistry, The Central Science , 10th edition

Real gases deviate from ideal behavior at __________ and _________.

1. High temperature; low pressure

2. Low temperature; high pressure

3. High temperature; high pressure

4. Low temperature; low pressure

Page 45: Chemistry, The Central Science , 10th edition

Correct Answer:

At low temperature and high pressure, intermolecular forces increase as the molecules get closer together.

1. High temperature; low pressure

2. Low temperature; high pressure

3. High temperature; high pressure

4. Low temperature; low pressure

Page 46: Chemistry, The Central Science , 10th edition

Which of the following molecules can hydrogen bond with itself?

1. 1, 22. 2, 33. 3, 44. 1, 2, 35. 1, 2, 3, 4

CH2F2 NH3 CH3-O-H CH3C CH3

O

1 2 3 4

Page 47: Chemistry, The Central Science , 10th edition

Which of the following molecules can hydrogen bond with itself?

1. 1, 22. 2, 33. 3, 44. 1, 2, 35. 1, 2, 3, 4

CH2F2 NH3 CH3-O-H CH3C CH3

O

1 2 3 4

Page 48: Chemistry, The Central Science , 10th edition

Arrange the following according to increasing vapor pressure.

1. NH3 < I2 < Br2 < CH4

2. I2 < Br2 < NH3 < CH4

3. NH3 < CH4 < I2 < Br2

4. NH3 < I2 < CH4 < Br2

5. CH4 < NH3 < Br2 < I2

NH3 I2 Br2 CH4

Page 49: Chemistry, The Central Science , 10th edition

Arrange the following according to increasing vapor pressure.

1. NH3 < I2 < Br2 < CH4

2. I2 < Br2 < NH3 < CH4

3. NH3 < CH4 < I2 < Br2

4. NH3 < I2 < CH4 < Br2

5. CH4 < NH3 < Br2 < I2

NH3 I2 Br2 CH4

Page 50: Chemistry, The Central Science , 10th edition

Which statement is true?

1. Boiling point ~120°C2. Boiling point ~95°C 3. Boiling point ~75°C4. Melting point ~95°C5. Melting point ~75°C

0

200

400

600

800

Vap

or

Pre

ssu

re (

mm

Hg

)

25 50 75 100Temperature (° C)

0

Page 51: Chemistry, The Central Science , 10th edition

Which statement is true?

1. Boiling point ~120°C2. Boiling point ~95°C 3. Boiling point ~75°C4. Melting point ~95°C5. Melting point ~75°C

0

200

400

600

800

Vap

or

Pre

ssu

re (

mm

Hg

)

25 50 75 100Temperature (° C)

0

Page 52: Chemistry, The Central Science , 10th edition

Of the following substances, predict which has the highest boiling point based on intermolecular forces.

1. Propane, C3H8

2. Dimethyl ether, CH3OCH3

3. Methyl chloride, CH3Cl4. Acetaldehyde, CH3CHO5. Acetonitrile, CH3CN

Page 53: Chemistry, The Central Science , 10th edition

Correct Answer:

Each of these molecules has almost the same molecular weight; however, acetonitrile has the largest dipole moment (3.9 D) and hence the largest dipole-dipole forces. Thus it has the highest boiling point.

1. Propane, C3H8

2. Dimethyl ether, CH3OCH3

3. Methyl chloride, CH3Cl4. Acetaldehyde, CH3CHO5. Acetonitrile, CH3CN

Page 54: Chemistry, The Central Science , 10th edition

Of the following substances, predict which has the lowest boiling point based on London dispersion forces.

1. He2. Ne3. Ar4. Kr5. Xe

Page 55: Chemistry, The Central Science , 10th edition

Correct Answer:

More massive species have more polarizability and stronger London dispersion forces; consequently, amongst the noble gases He has the lowest boiling point.

1. He2. Ne3. Ar4. Kr5. Xe

Page 56: Chemistry, The Central Science , 10th edition

Which one of the following phase changes is an exothermic process?

1. Sublimation2. Vaporization3. Condensation4. Melting

Page 57: Chemistry, The Central Science , 10th edition

Correct Answer:

All the other phase-change processes listed are endothermic.

1. Sublimation2. Vaporization3. Condensation4. Melting

Page 58: Chemistry, The Central Science , 10th edition

How much energy is required to raise the temperature of 1800. g ice at 0°C to 10°C? Hfus = 6.01 kJ/mol, heat capacity of water is 75.2 J/mol-K.

1. 60.1 kJ2. 75.2 kJ3. 135 kJ4. 601 kJ5. 676 kJ

Page 59: Chemistry, The Central Science , 10th edition

Correct Answer:

There are 100 moles of ice, so the enthalpy of fusion is:H =(6.01 kJ/mol)(100 mol) = 601 kJ.To raise the water temperature 10°C requires q = (75.2 J/mol-K)(100 mol)(10°C ) = 75.2 kJ. Total energy = 601 kJ +75 kJ = 676 kJ

1. 60.1 kJ2. 75.2 kJ3. 135 kJ4. 601 kJ5. 676 kJ

Page 60: Chemistry, The Central Science , 10th edition

Temperature

Pre

ssu

reAA

DD

CC

BB

In the unlabeled phase diagram below, the line segment from A to B separates which two phases?

1. Gas-liquid2. Liquid-solid3. Solid-gas

Page 61: Chemistry, The Central Science , 10th edition

Correct Answer:

1. Gas-liquid2. Liquid-solid3. Solid-gas

Temperature

Pre

ssu

reAA

DD

CC

BB

Page 62: Chemistry, The Central Science , 10th edition

Diamond and graphite are examples of which type of crystalline solids?

1. Molecular2. Covalent network3. Ionic4. Metallic

Page 63: Chemistry, The Central Science , 10th edition

Correct Answer:

Diamond and graphite are both forms of carbon, and consist of a network of covalent bonds (hence covalent network).

1. Molecular2. Covalent network3. Ionic4. Metallic

Page 64: Chemistry, The Central Science , 10th edition

a. gas

1. No comparison can be made between kinetic and attraction energy without temperature information.

2. The average kinetic energy is greater than the energy of attraction.

3. The average kinetic energy is equal to the energy of attraction.

4. The average kinetic energy is less than the energy of attraction.

Page 65: Chemistry, The Central Science , 10th edition

a. gas

1. No comparison can be made between kinetic and attraction energy without temperature information.

2. The average kinetic energy is greater than the energy of attraction.

3. The average kinetic energy is equal to the energy of attraction.

4. The average kinetic energy is less than the energy of attraction.

Page 66: Chemistry, The Central Science , 10th edition

b. solid

1. No comparison can be made between kinetic and attraction energy without temperature information.

2. The average kinetic energy is equal to the energy of attraction.

3. The average kinetic energy is greater than the energy of attraction.

4. The average kinetic energy is less than the energy of attraction.

Page 67: Chemistry, The Central Science , 10th edition

b. solid

1. No comparison can be made between kinetic and attraction energy without temperature information.

2. The average kinetic energy is equal to the energy of attraction.

3. The average kinetic energy is greater than the energy of attraction.

4. The average kinetic energy is less than the energy of attraction.

Page 68: Chemistry, The Central Science , 10th edition

1. Ion-dipole forces are encountered in both solutions.

2. Ion-dipole forces are encountered in neither solution.

3. Ion-dipole forces only are encountered in Ca(NO3)2 in water.

4. Ion-dipole forces only are encountered in CH3OH in water.

Page 69: Chemistry, The Central Science , 10th edition

1. Ion-dipole forces are encountered in both solutions.

2. Ion-dipole forces are encountered in neither solution.

3. Ion-dipole forces only are encountered in Ca(NO3)2 in water.

4. Ion-dipole forces only are encountered in CH3OH in water.

Page 70: Chemistry, The Central Science , 10th edition

1. Crystalline solids always melt at a higher temperatures than amorphous solids.

2. Crystalline solids melt at a specific temperature, whereas amorphous ones tend to melt over a temperature range.

3. Amorphous solids always melt at a higher temperatures than crystalline solids.

4. Amorphous solids melt at a specific temperature, whereas crystalline ones tend to melt over a temperature range.

Page 71: Chemistry, The Central Science , 10th edition

1. Crystalline solids always melt at a higher temperatures than amorphous solids.

2. Crystalline solids melt at a specific temperature, whereas amorphous ones tend to melt over a temperature range.

3. Amorphous solids always melt at a higher temperatures than crystalline solids.

4. Amorphous solids melt at a specific temperature, whereas crystalline ones tend to melt over a temperature range.

Page 72: Chemistry, The Central Science , 10th edition

The curve describes the solubility properties of

1. A liquid

2. An ionic solid

3. A covalent solid

4. A gas

5. A gas or liquid0

1.0

2.0

So

lub

ilit

y m

M

15 30 45 60Temperature (° C)

0

Page 73: Chemistry, The Central Science , 10th edition

The curve describes the solubility properties of

1. A liquid

2. An ionic solid

3. A covalent solid

4. A gas

5. A gas or liquid0

1.0

2.0

So

lub

ilit

y m

M

15 30 45 60Temperature (° C)

0

Page 74: Chemistry, The Central Science , 10th edition

Predict which aqueous solution will have the lowest freezing point.

1. 0.25 m C2H5OH

2. 0.15 m CaCl2

3. 0.20 m NaCl

4. 0.15 m NH4NO3

5. 0.15 m Na3PO4

Page 75: Chemistry, The Central Science , 10th edition

Predict which aqueous solution will have the lowest freezing point.

1. 0.25 m C2H5OH

2. 0.15 m CaCl2

3. 0.20 m NaCl

4. 0.15 m NH4NO3

5. 0.15 m Na3PO4

Page 76: Chemistry, The Central Science , 10th edition

Arrange the aqueous solutions according to increasing boiling point.

1. AlCl3 < KNO3 < Na2SO4

2. Na2SO4 < AlCl3 < KNO3

3. Na2SO4 < KNO3 < AlCl3

4. KNO3 < AlCl3 < Na2SO4

5. KNO3 < Na2SO4 < AlCl3

0.10 m Na2SO4

0.15 m AlCl3

0.20 m KNO3

Page 77: Chemistry, The Central Science , 10th edition

Arrange the aqueous solutions according to increasing boiling point.

1. AlCl3 < KNO3 < Na2SO4

2. Na2SO4 < AlCl3 < KNO3

3. Na2SO4 < KNO3 < AlCl3

4. KNO3 < AlCl3 < Na2SO4

5. KNO3 < Na2SO4 < AlCl3

0.10 m Na2SO4

0.15 m AlCl3

0.20 m KNO3

Page 78: Chemistry, The Central Science , 10th edition

1. 23 ppm

2. 2.3 ppm

3. 230 ppm

4. 2300 ppm

Page 79: Chemistry, The Central Science , 10th edition

1. 23 ppm

2. 2.3 ppm

3. 230 ppm

4. 2300 ppm

Page 80: Chemistry, The Central Science , 10th edition

Determine the mass percentage of hexane in a solution containing 11 g of butane in 110 g of hexane.

1. 9.0 %2. 10. %3. 90.%4. 91 %

Page 81: Chemistry, The Central Science , 10th edition

Correct Answer:

Thus,

110 g(110 g + 11 g)

100solution of mass total

solution incomponent of masscomponent of % mass

100 = 91%

1. 9.0 %2. 10. %3. 90.%4. 91 %

Page 82: Chemistry, The Central Science , 10th edition

What is the molality of 6.4 g of methanol (CH3OH) dissolved in 50. moles of water?

1. 0.040 m2. 0.22 m3. 0.064 m4. 0.11 m

Page 83: Chemistry, The Central Science , 10th edition

Correct Answer:

solvent of kg

solute moles Molality, m

kg/1000g) g/mol)(1 .0 water)(18mol (50

g/mol) (32.0methanol)/ g (6.4m

mm 0.22kg) (0.90

mol) (0.20

1. 0.040 m2. 0.22 m3. 0.064 m4. 0.11 m

Page 84: Chemistry, The Central Science , 10th edition

How many moles of solute are there in 240 g of a solution that is 5.0% glucose (C6H12O6) by mass?

1. 0.033 moles2. 0.067 moles3. 0.10 moles4. 0.12 moles5. 0.20 moles

Page 85: Chemistry, The Central Science , 10th edition

Correct Answer:

5.0 % glucose means 5.0 g glucose/100 g solution

(5.0 g glucose/100 g solution)(240 g solution) = 12 g

(12 g glucose) (1 mol glucose/180 g glucose) = 0.067

moles glucose

1. 0.033 moles2. 0.067 moles3. 0.10 moles4. 0.12 moles5. 0.20 moles

Page 86: Chemistry, The Central Science , 10th edition

Ethanol normally boils at 78.4°C. The boiling point elevation constant for ethanol is 1.22°C/m. What is the boiling point of a 1.0 m solution of CaCl2 in ethanol?

1. 77.2°C2. 79.6°C3. 80.8°C

4. 82.1°C5. 83.3°C

Page 87: Chemistry, The Central Science , 10th edition

Correct Answer:

The increase in boiling point is determined by the molality of total particles in the solution. Thus, a 1.0 m solution of CaCl2 contains 1.0 m Ca2+ and

2.0 m Cl for a total of 3.0 m. Thus, the boiling point is elevated 3.7°C, so it is 78.4°C + 3.7°C = 82.1°C.

mKT bb

1. 77.2°C2. 79.6°C3. 80.8°C4. 82.1°C5. 83.3°C

Page 88: Chemistry, The Central Science , 10th edition

Which of the following is not an example of a colloid?

1. Fog2. Smoke3. Paint4. Milk5. Carbonated water

Page 89: Chemistry, The Central Science , 10th edition

Correct Answer:

Carbonated water is a solution; all the other substances in the list are excellent examples of colloids.

1. Fog2. Smoke3. Paint4. Milk5. Carbonated water

Page 90: Chemistry, The Central Science , 10th edition

What is Kp in terms of Kc for the following reaction ?

1. Kp = KcRT

2. Kp = Kc(RT)

3. Kp = KcR/T

4. Kp = Kc

5. Kp = Kc/(RT)2

2NO (g) + O2 (g) 2NO2 (g)

Page 91: Chemistry, The Central Science , 10th edition

What is Kp in terms of Kc for the following reaction ?

1. Kp = KcRT

2. Kp = Kc(RT)

3. Kp = KcR/T

4. Kp = Kc

5. Kp = Kc/(RT)2

2NO (g) + O2 (g) 2NO2 (g)

Page 92: Chemistry, The Central Science , 10th edition

What is the correct equilibrium constant expression for the reaction:

P4 (s) + 6Cl2 (g) 4PCl3 (l)

Page 93: Chemistry, The Central Science , 10th edition

What is the correct equilibrium constant expression for the reaction:

P4 (s) + 6Cl2 (g) 4PCl3 (l)

Page 94: Chemistry, The Central Science , 10th edition

Which accurately reflects the changes in concentration that will occur if O2 is added to disturb the equilibrium?

[NO] [O2] [NO2]

1. Increase Increase Increase

2. Increase Increase Decrease

3. Decrease Decrease Decrease

4. Decrease Decrease Increase

5. Decrease Increase Increase

2NO (g) + O2 (g) 2NO2 (g)

NO NO2

Page 95: Chemistry, The Central Science , 10th edition

Which accurately reflects the changes in concentration that will occur if O2 is added to disturb the equilibrium?

[NO] [O2] [NO2]

1. Increase Increase Increase

2. Increase Increase Decrease

3. Decrease Decrease Decrease

4. Decrease Decrease Increase

5. Decrease Increase Increase

2NO (g) + O2 (g) 2NO2 (g)

NO NO2

Page 96: Chemistry, The Central Science , 10th edition

Which of the following will result in an equilibrium shift to the right?

1. Increase temperature/increase volume

2. Increase temperature/decrease volume

3. Decrease temperature/increase volume

4. Decrease temperature/decrease volume

5. None of the above

PCl3 (g) + Cl2 (g) PCl5 (g) H° = -87.9 kJ/mol

PCl5

Page 97: Chemistry, The Central Science , 10th edition

Which of the following will result in an equilibrium shift to the right?

1. Increase temperature/increase volume

2. Increase temperature/decrease volume

3. Decrease temperature/increase volume

4. Decrease temperature/decrease volume

5. None of the above

PCl3 (g) + Cl2 (g) PCl5 (g) H° = -87.9 kJ/mol

PCl5

Page 98: Chemistry, The Central Science , 10th edition

What is the value of Kc for the reaction?

1. Kc = 1/(5.0 x 1018)1/2

2. Kc = 1/(2.5 x 1018)

3. Kc = -(5.0 x 1018)/2

4. Kc = -(5.0 x 1018)1/2

5. Kc = 2/(5.0 x 1018)1/2

2CO(g) + O2(g) 2CO2(g) Kc = 5.0 x 1018 at 25 °C

CO2(g) CO(g) + 1/2 O2(g) Kc = ?? at 25 °C

Page 99: Chemistry, The Central Science , 10th edition

What is the value of Kc for the reaction?

1. Kc = 1/(5.0 x 1018)1/2

2. Kc = 1/(2.5 x 1018)

3. Kc = -(5.0 x 1018)/2

4. Kc = -(5.0 x 1018)1/2

5. Kc = 2/(5.0 x 1018)1/2

2CO(g) + O2(g) 2CO2(g) Kc = 5.0 x 1018 at 25 °C

CO2(g) CO(g) + 1/2 O2(g) Kc = ?? at 25 °C

Page 100: Chemistry, The Central Science , 10th edition

For the gas-phase reaction A Bthe forward reaction rate is 3.0 104 s1 and the reverse reaction rate is 1.5 102 s1. What is the value of the equilibrium constant, Keq?

1. 0.022. 503. 0.0004

4. 2500

Page 101: Chemistry, The Central Science , 10th edition

Correct Answer:

r

fc k

kK

0.021.5x10

3x102-

-4

cK

1. 0.022. 503. 0.0004

4. 2500

Page 102: Chemistry, The Central Science , 10th edition

1. no bubbles are observed in the reaction mix.

2. the temperature of the reaction mix cools.

3. the concentrations of reactants and products no longer change.

4. the color of the reaction mix remains constant.

Page 103: Chemistry, The Central Science , 10th edition

1. no bubbles are observed in the reaction mix.

2. the temperature of the reaction mix cools.

3. the concentrations of reactants and products no longer change.

4. the color of the reaction mix remains constant.

Page 104: Chemistry, The Central Science , 10th edition

1. Kc is 2nd order with respect to NO2 and 1st order with

respect to N2O4.

2. Kc is independent of the starting concentrations of

reactants and products.

3. Kc is 2nd order with respect to NO2 and inversely

related to N2O4.

4. Kc is directly related to NO2 and inversely related to

N2O4.

Page 105: Chemistry, The Central Science , 10th edition

1. Kc is 2nd order with respect to NO2 and 1st order with

respect to N2O4.

2. Kc is independent of the starting concentrations of

reactants and products.

3. Kc is 2nd order with respect to NO2 and inversely

related to N2O4.

4. Kc is directly related to NO2 and inversely related to

N2O4.

Page 106: Chemistry, The Central Science , 10th edition

1. Kp = 1/[H2Og]

2. Kp = PH2O

3. Kp = [H2Og]

4. Kp = 1/PH2O

Page 107: Chemistry, The Central Science , 10th edition

1. Kp = 1/[H2Og]

2. Kp = PH2O

3. Kp = [H2Og]

4. Kp = 1/PH2O

Page 108: Chemistry, The Central Science , 10th edition

1. cannot determine without information about energy of reaction.

2. cannot determine without information about reaction rate constant.

3. no.

4. yes.

Page 109: Chemistry, The Central Science , 10th edition

1. cannot determine without information about energy of reaction.

2. cannot determine without information about reaction rate constant.

3. no.

4. yes.