a) CH 3 (CH 2 ) 29 CH 3 , a component found in paraffin wax is a solid at room temperature while octane is a liquid. Interpretation: The observation that, CH 3 (CH 2 ) 29 CH 3 , a component found in paraffin wax is a solid at room temperature while octane is a liquid is to be explained. Concept introduction: Even in non polar molecules, the attractive dispersion forces are possible, caused by temporary dipoles arising due to the costant change in electon distribution wihin the molecule. These temporary dipoles, though have a fleeting existence are constantly changing. But their cumulative effect may be strong enough to hold the molecule close together so that a substance can be a solid or liquid. To explain: The observation that, CH 3 (CH 2 ) 29 CH 3 , a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Question

Want to see the full answer?

Answer 1

Question

Chapter 2.SE, Problem 65AP

Interpretation Introduction

a) CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Interpretation:

The observation that, CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid is to be explained.

Concept introduction:

Even in non polar molecules, the attractive dispersion forces are possible, caused by temporary dipoles arising due to the costant change in electon distribution wihin the molecule. These temporary dipoles, though have a fleeting existence are constantly changing. But their cumulative effect may be strong enough to hold the molecule close together so that a substance can be a solid or liquid.

To explain:

The observation that, CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Interpretation Introduction

b) CH₃CH₂CH₂OH has a higher boiling point than CH₄

Interpretation:

The observation CH₃CH₂CH₂OH has a higher boiling point than CH₄.

Concept introduction:

An attractive interaction between a hydrogen bonded to an electronegative oxygen or nitrogen atom and an unshared electron pair on another oxygen or nitrogrn atom is termed as hydrogen bonding. It is a very strong dipole-dipole interaction that leads to molecular association. These molecules associated through hydrogen bonding will have a high boiling point.

To explain:

The observation CH₃CH₂CH₂OH has a higher boiling point than CH₄.

Interpretation Introduction

c) CH₃CO₂H, which is found in vinegar, will dissolve in water but not in oil- for simplicity you may assume oil is CH₃(CH₂)₄CH₃

Interpretation:

The observation that CH₃CO₂H, which is found in vinegar, will dissolve in water but not in oil with molecular formula CH₃(CH₂)₄CH₃.

Concept introduction:

Like dissolves like. A polar molecule will dissolve in another polar solvent. The solubility can also be explained using the formation of hydrogen bonding between the two substances.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Which has the greater polarizability? Explain.(a) Br⁻ or I⁻ (b) CH₂=CH₂ or CH₃—CH₃(c) H₂O or H₂Se

Which type of intermolecular force accounts for each of these differences? (a) CH3OH boils at 65 °C; CH3SH boils at 6 °C. (b) Xe is a liquid at atmospheric pressure and 120 K, whereas Ar is a gas under the same conditions. (c) Kr, atomic weight 84 amu, boils at 120.9 K, whereas Cl2, molecular weight about 71 amu, boils at 238 K. (d) Acetone boils at 56 °C, whereas 2-methylpropane boils at -12 °C. CH3 CH3 O C CH3 CH3 CH3 CH Acetone 2-Methylpropane

In degrees C, what is the approximate boiling point of (CH3CH2)2O?

Answer 2

Question

Chapter 2.SE, Problem 65AP

Interpretation Introduction

a) CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Interpretation:

The observation that, CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid is to be explained.

Concept introduction:

Even in non polar molecules, the attractive dispersion forces are possible, caused by temporary dipoles arising due to the costant change in electon distribution wihin the molecule. These temporary dipoles, though have a fleeting existence are constantly changing. But their cumulative effect may be strong enough to hold the molecule close together so that a substance can be a solid or liquid.

To explain:

The observation that, CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Interpretation Introduction

b) CH₃CH₂CH₂OH has a higher boiling point than CH₄

Interpretation:

The observation CH₃CH₂CH₂OH has a higher boiling point than CH₄.

Concept introduction:

An attractive interaction between a hydrogen bonded to an electronegative oxygen or nitrogen atom and an unshared electron pair on another oxygen or nitrogrn atom is termed as hydrogen bonding. It is a very strong dipole-dipole interaction that leads to molecular association. These molecules associated through hydrogen bonding will have a high boiling point.

To explain:

The observation CH₃CH₂CH₂OH has a higher boiling point than CH₄.

Interpretation Introduction

c) CH₃CO₂H, which is found in vinegar, will dissolve in water but not in oil- for simplicity you may assume oil is CH₃(CH₂)₄CH₃

Interpretation:

The observation that CH₃CO₂H, which is found in vinegar, will dissolve in water but not in oil with molecular formula CH₃(CH₂)₄CH₃.

Concept introduction:

Like dissolves like. A polar molecule will dissolve in another polar solvent. The solubility can also be explained using the formation of hydrogen bonding between the two substances.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 2.SE, Problem 65AP

Interpretation Introduction

a) CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Interpretation:

The observation that, CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid is to be explained.

Concept introduction:

Even in non polar molecules, the attractive dispersion forces are possible, caused by temporary dipoles arising due to the costant change in electon distribution wihin the molecule. These temporary dipoles, though have a fleeting existence are constantly changing. But their cumulative effect may be strong enough to hold the molecule close together so that a substance can be a solid or liquid.

To explain:

The observation that, CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Interpretation Introduction

b) CH₃CH₂CH₂OH has a higher boiling point than CH₄

Interpretation:

The observation CH₃CH₂CH₂OH has a higher boiling point than CH₄.

Concept introduction:

An attractive interaction between a hydrogen bonded to an electronegative oxygen or nitrogen atom and an unshared electron pair on another oxygen or nitrogrn atom is termed as hydrogen bonding. It is a very strong dipole-dipole interaction that leads to molecular association. These molecules associated through hydrogen bonding will have a high boiling point.

To explain:

The observation CH₃CH₂CH₂OH has a higher boiling point than CH₄.

Interpretation Introduction

c) CH₃CO₂H, which is found in vinegar, will dissolve in water but not in oil- for simplicity you may assume oil is CH₃(CH₂)₄CH₃

Interpretation:

The observation that CH₃CO₂H, which is found in vinegar, will dissolve in water but not in oil with molecular formula CH₃(CH₂)₄CH₃.

Concept introduction:

Like dissolves like. A polar molecule will dissolve in another polar solvent. The solubility can also be explained using the formation of hydrogen bonding between the two substances.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 2.SE, Problem 65AP

Interpretation Introduction

a) CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Interpretation:

The observation that, CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid is to be explained.

Concept introduction:

Even in non polar molecules, the attractive dispersion forces are possible, caused by temporary dipoles arising due to the costant change in electon distribution wihin the molecule. These temporary dipoles, though have a fleeting existence are constantly changing. But their cumulative effect may be strong enough to hold the molecule close together so that a substance can be a solid or liquid.

To explain:

The observation that, CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Interpretation Introduction

b) CH₃CH₂CH₂OH has a higher boiling point than CH₄

Interpretation:

The observation CH₃CH₂CH₂OH has a higher boiling point than CH₄.

Concept introduction:

An attractive interaction between a hydrogen bonded to an electronegative oxygen or nitrogen atom and an unshared electron pair on another oxygen or nitrogrn atom is termed as hydrogen bonding. It is a very strong dipole-dipole interaction that leads to molecular association. These molecules associated through hydrogen bonding will have a high boiling point.

To explain:

The observation CH₃CH₂CH₂OH has a higher boiling point than CH₄.

Interpretation Introduction

c) CH₃CO₂H, which is found in vinegar, will dissolve in water but not in oil- for simplicity you may assume oil is CH₃(CH₂)₄CH₃

Interpretation:

The observation that CH₃CO₂H, which is found in vinegar, will dissolve in water but not in oil with molecular formula CH₃(CH₂)₄CH₃.

Concept introduction:

Like dissolves like. A polar molecule will dissolve in another polar solvent. The solubility can also be explained using the formation of hydrogen bonding between the two substances.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 2.SE, Problem 65AP

Interpretation Introduction

a) CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Interpretation:

The observation that, CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid is to be explained.

Concept introduction:

Even in non polar molecules, the attractive dispersion forces are possible, caused by temporary dipoles arising due to the costant change in electon distribution wihin the molecule. These temporary dipoles, though have a fleeting existence are constantly changing. But their cumulative effect may be strong enough to hold the molecule close together so that a substance can be a solid or liquid.

To explain:

The observation that, CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Interpretation Introduction

b) CH₃CH₂CH₂OH has a higher boiling point than CH₄

Interpretation:

The observation CH₃CH₂CH₂OH has a higher boiling point than CH₄.

Concept introduction:

An attractive interaction between a hydrogen bonded to an electronegative oxygen or nitrogen atom and an unshared electron pair on another oxygen or nitrogrn atom is termed as hydrogen bonding. It is a very strong dipole-dipole interaction that leads to molecular association. These molecules associated through hydrogen bonding will have a high boiling point.

To explain:

The observation CH₃CH₂CH₂OH has a higher boiling point than CH₄.

Interpretation Introduction

c) CH₃CO₂H, which is found in vinegar, will dissolve in water but not in oil- for simplicity you may assume oil is CH₃(CH₂)₄CH₃

Interpretation:

The observation that CH₃CO₂H, which is found in vinegar, will dissolve in water but not in oil with molecular formula CH₃(CH₂)₄CH₃.

Concept introduction:

Like dissolves like. A polar molecule will dissolve in another polar solvent. The solubility can also be explained using the formation of hydrogen bonding between the two substances.

Answer 6

Chapter 2.SE, Problem 65AP

Interpretation Introduction

a) CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Interpretation:

The observation that, CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid is to be explained.

Concept introduction:

Even in non polar molecules, the attractive dispersion forces are possible, caused by temporary dipoles arising due to the costant change in electon distribution wihin the molecule. These temporary dipoles, though have a fleeting existence are constantly changing. But their cumulative effect may be strong enough to hold the molecule close together so that a substance can be a solid or liquid.

To explain:

The observation that, CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Answer 7

Chapter 2.SE, Problem 65AP

Interpretation Introduction

a) CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Interpretation:

The observation that, CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid is to be explained.

Concept introduction:

Even in non polar molecules, the attractive dispersion forces are possible, caused by temporary dipoles arising due to the costant change in electon distribution wihin the molecule. These temporary dipoles, though have a fleeting existence are constantly changing. But their cumulative effect may be strong enough to hold the molecule close together so that a substance can be a solid or liquid.

To explain:

The observation that, CH₃(CH₂)₂₉CH₃, a component found in paraffin wax is a solid at room temperature while octane is a liquid.

Answer 8

Interpretation Introduction

b) CH₃CH₂CH₂OH has a higher boiling point than CH₄

Interpretation:

The observation CH₃CH₂CH₂OH has a higher boiling point than CH₄.

Concept introduction:

An attractive interaction between a hydrogen bonded to an electronegative oxygen or nitrogen atom and an unshared electron pair on another oxygen or nitrogrn atom is termed as hydrogen bonding. It is a very strong dipole-dipole interaction that leads to molecular association. These molecules associated through hydrogen bonding will have a high boiling point.

To explain:

The observation CH₃CH₂CH₂OH has a higher boiling point than CH₄.

Answer 9

Interpretation Introduction

b) CH₃CH₂CH₂OH has a higher boiling point than CH₄

Interpretation:

The observation CH₃CH₂CH₂OH has a higher boiling point than CH₄.

Concept introduction:

An attractive interaction between a hydrogen bonded to an electronegative oxygen or nitrogen atom and an unshared electron pair on another oxygen or nitrogrn atom is termed as hydrogen bonding. It is a very strong dipole-dipole interaction that leads to molecular association. These molecules associated through hydrogen bonding will have a high boiling point.

To explain:

The observation CH₃CH₂CH₂OH has a higher boiling point than CH₄.

Answer 10

Interpretation Introduction

c) CH₃CO₂H, which is found in vinegar, will dissolve in water but not in oil- for simplicity you may assume oil is CH₃(CH₂)₄CH₃

Interpretation:

The observation that CH₃CO₂H, which is found in vinegar, will dissolve in water but not in oil with molecular formula CH₃(CH₂)₄CH₃.

Concept introduction:

Like dissolves like. A polar molecule will dissolve in another polar solvent. The solubility can also be explained using the formation of hydrogen bonding between the two substances.

Answer 11

Interpretation Introduction

c) CH₃CO₂H, which is found in vinegar, will dissolve in water but not in oil- for simplicity you may assume oil is CH₃(CH₂)₄CH₃

Interpretation:

The observation that CH₃CO₂H, which is found in vinegar, will dissolve in water but not in oil with molecular formula CH₃(CH₂)₄CH₃.

Concept introduction:

Like dissolves like. A polar molecule will dissolve in another polar solvent. The solubility can also be explained using the formation of hydrogen bonding between the two substances.

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 15

Ch. 2.1 - Prob. 1P Ch. 2.1 - Prob. 2P Ch. 2.1 - Use the electronegativity values shown in Figure...Ch. 2.1 - Look at the following electrostatic potential map...Ch. 2.2 - Ethylene glycol, HOCH2CH2OH, may look nonpolar...Ch. 2.2 - Make three-dimensional drawings of the following...Ch. 2.3 - Calculate formal charges for the nonhydrogen atoms...Ch. 2.3 - Organic phosphate groups occur commonly in...Ch. 2.6 - Which of the following pairs of structures...Ch. 2.6 - Draw the indicated number of resonance forms for...

Want to see the full answer?

Chapter 2 Solutions