Chemistry 2012 Student Edition (hard Cover) Grade 11

12th Edition

ISBN: 9780132525763

Author: Prentice Hall

Publisher: Prentice Hall

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Question

Chapter 13, Problem 61A

(a)

Interpretation Introduction

Interpretation: The physical state of phenol is to be determined.

Concept introduction:

The boiling point of a liquid is the temperature at which the pressure exerted by its surroundings equals the vapor pressure present at the liquid's surface.

The temperature at which a substance transforms from a solid to a liquid is known as its melting point.

(a)

Expert Solution

Answer to Problem 61A

Liquid

Explanation of Solution

Phenol is a white crystalline solid at normal temperature. As given in the table, its melting point is 40.9°C and boiling point is 181.9°C . The given temperature of phenol is 99°C which is higher than its melting point but lower than its boiling point.

Thus, phenol will exist as a liquid at 99°C .

(b)

Interpretation Introduction

Interpretation: The physical state of ammonia is to be determined.

Concept introduction:

The boiling point of a liquid is the temperature at which the pressure exerted by its surroundings equals the vapor pressure present at the liquid's surface.

The temperature at which a substance transforms from a solid to a liquid is known as its melting point.

(b)

Expert Solution

Answer to Problem 61A

Liquid

Explanation of Solution

Ammonia is a colorless gas at normal temperatures. As given in the table, its melting point is −77.7°C and boiling point is −33.4°C . The given temperature of phenol is −25°C which is higher than its melting point, so it condenses at this temperature.

Thus, ammonia will exist as a liquid at −25°C .

(c)

Interpretation Introduction

Interpretation: The physical state of methanol is to be determined.

Concept introduction:

The boiling point of a liquid is the temperature at which the pressure exerted by its surroundings equals the vapor pressure present at the liquid's surface.

The temperature at which a substance transforms from a solid to a liquid is known as its melting point.

(c)

Expert Solution

Answer to Problem 61A

Liquid

Explanation of Solution

Methanol is a colorless liquid at normal temperatures. As given in the table, its melting point is −97.7°C and boiling point is 64.7°C . The given temperature of methanol is 0°C as it is in ab ice water bath which is higher than the melting point and lower than the boiling point.

Thus, methanol in an ice water bath will exist as a liquid.

(d)

Interpretation Introduction

Interpretation: The physical state of methanol is to be determined.

Concept introduction:

The boiling point of a liquid is the temperature at which the pressure exerted by its surroundings equals the vapor pressure present at the liquid's surface.

The temperature at which a substance transforms from a solid to a liquid is known as its melting point.

(d)

Expert Solution

Answer to Problem 61A

Gaseous state

Explanation of Solution

Methanol is a colorless liquid at normal temperatures. As given in the table, its melting point is −97.7°C and boiling point is 64.7°C . The given temperature of methanol is 100°C as it is in ab boiling water bath which is higher than the boiling point.

Thus, methanol in a hot water bath will exist as a gas.

(e)

Interpretation Introduction

Interpretation: The physical state of ammonia is to be determined.

Concept introduction:

The boiling point of a liquid is the temperature at which the pressure exerted by its surroundings equals the vapor pressure present at the liquid's surface.

The temperature at which a substance transforms from a solid to a liquid is known as its melting point.

(e)

Expert Solution

Answer to Problem 61A

Solid

Explanation of Solution

Ammonia is a colorless gas at normal temperatures. As given in the table, its melting point is −77.7°C and boiling point is −33.4°C . The given temperature of phenol is −100°C which is lower than its melting point, so it will exist as a solid.

Thus, ammonia will exist as a solid at −100°C .

(d)

Interpretation Introduction

Interpretation: The physical state of phenol is to be determined.

Concept introduction:

The boiling point of a liquid is the temperature at which the pressure exerted by its surroundings equals the vapor pressure present at the liquid's surface.

The temperature at which a substance transforms from a solid to a liquid is known as its melting point.

(d)

Expert Solution

Answer to Problem 61A

solid

Explanation of Solution

Phenol is a white crystalline solid at normal temperature. As given in the table, its melting point is 40.9°C and boiling point is 181.9°C . The given temperature of phenol is 25°C which is lower than its melting point.

Thus, phenol will exist as a solid at 25°C .

Chapter 13, Problem 60A

Chapter 13, Problem 62A

Chapter 13 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 13.1 - Prob. 1SP Ch. 13.1 - Prob. 2SP Ch. 13.1 - Prob. 3LC Ch. 13.1 - Prob. 4LC Ch. 13.1 - Prob. 5LC Ch. 13.1 - Prob. 6LC Ch. 13.1 - Prob. 7LC Ch. 13.1 - Prob. 8LC Ch. 13.1 - Prob. 9LC Ch. 13.2 - Prob. 10LC

Ch. 13.2 - Prob. 11LC Ch. 13.2 - Prob. 12LC Ch. 13.2 - Prob. 13LC Ch. 13.2 - Prob. 15LC Ch. 13.2 - Prob. 16LC Ch. 13.2 - Prob. 17LC Ch. 13.3 - Prob. 18LC Ch. 13.3 - Prob. 19LC Ch. 13.3 - Prob. 20LC Ch. 13.3 - Prob. 21LC Ch. 13.3 - Prob. 22LC Ch. 13.3 - Prob. 23LC Ch. 13.3 - Prob. 24LC Ch. 13.4 - Prob. 25LC Ch. 13.4 - Prob. 26LC Ch. 13.4 - Prob. 27LC Ch. 13.4 - Prob. 29LC Ch. 13.4 - Prob. 30LC Ch. 13 - Prob. 31A Ch. 13 - Prob. 32A Ch. 13 - Prob. 33A Ch. 13 - Prob. 34A Ch. 13 - Prob. 35A Ch. 13 - Prob. 36A Ch. 13 - Prob. 37A Ch. 13 - Prob. 38A Ch. 13 - Prob. 39A Ch. 13 - Prob. 40A Ch. 13 - Prob. 41A Ch. 13 - Prob. 42A Ch. 13 - Prob. 43A Ch. 13 - Prob. 44A Ch. 13 - Prob. 45A Ch. 13 - Prob. 46A Ch. 13 - Prob. 47A Ch. 13 - Prob. 49A Ch. 13 - Prob. 50A Ch. 13 - Prob. 51A Ch. 13 - Prob. 52A Ch. 13 - Prob. 53A Ch. 13 - Prob. 54A Ch. 13 - Prob. 55A Ch. 13 - Prob. 57A Ch. 13 - Prob. 58A Ch. 13 - Prob. 60A Ch. 13 - Prob. 61A Ch. 13 - Prob. 62A Ch. 13 - Prob. 63A Ch. 13 - Prob. 64A Ch. 13 - Prob. 67A Ch. 13 - Prob. 68A Ch. 13 - Prob. 69A Ch. 13 - Prob. 70A Ch. 13 - Prob. 71A Ch. 13 - Prob. 72A Ch. 13 - Prob. 73A Ch. 13 - Prob. 74A Ch. 13 - Prob. 75A Ch. 13 - Prob. 76A Ch. 13 - Prob. 77A Ch. 13 - Prob. 78A Ch. 13 - Prob. 79A Ch. 13 - Prob. 80A Ch. 13 - Prob. 81A Ch. 13 - Prob. 82A Ch. 13 - Prob. 83A Ch. 13 - Prob. 84A Ch. 13 - Prob. 85A Ch. 13 - Prob. 86A Ch. 13 - Prob. 87A Ch. 13 - Prob. 88A Ch. 13 - Prob. 89A Ch. 13 - Prob. 90A Ch. 13 - Prob. 91A Ch. 13 - Prob. 92A Ch. 13 - Prob. 93A Ch. 13 - Prob. 94A Ch. 13 - Prob. 95A Ch. 13 - Prob. 96A Ch. 13 - Prob. 97A Ch. 13 - Prob. 98A Ch. 13 - Prob. 99A Ch. 13 - Prob. 100A Ch. 13 - Prob. 101A Ch. 13 - Prob. 102A Ch. 13 - Prob. 1STP Ch. 13 - Prob. 2STP Ch. 13 - Prob. 3STP Ch. 13 - Prob. 4STP Ch. 13 - Prob. 5STP Ch. 13 - Prob. 6STP Ch. 13 - Prob. 7STP Ch. 13 - Prob. 8STP Ch. 13 - Prob. 9STP Ch. 13 - Prob. 10STP

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