Living By Chemistry: First Edition Textbook
Living By Chemistry: First Edition Textbook
1st Edition
ISBN: 9781559539418
Author: Angelica Stacy
Publisher: MAC HIGHER
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Chapter U3.5, Problem 9E

(a)

Interpretation Introduction

Interpretation: The given temperature in Fahrenheit scale that is 95 oF needs to be converted into Kelvin scale.

Concept Introduction:

The Kelvin and degree Fahrenheit are scales of measuring temperature. The relation between two scales is as follows:

  (32°F  32) × 5/9 + 273.15 = 273.15K

Here, K is Kelvin and oF is degree Fahrenheit.

(a)

Expert Solution
Check Mark

Answer to Problem 9E

  308.15 K

Explanation of Solution

The relation between two scales is as follows:

  (32°F  32) × 5/9 + 273.15 = 273.15K

The given temperature is 95 oF. It can be converted into Kelvin as follows:

  (95°F  32) × 5/9 + 273.15 = x

On solving,

  x=308.15 K

Thus, the temperature 95 oF is equal to 308.15 K.

(b)

Interpretation Introduction

Interpretation: The given temperature in Fahrenheit scale that is 350 oF needs to be converted into Kelvin scale.

Concept Introduction:

The Kelvin and degree Fahrenheit are scales of measuring temperature. The relation between two scales is as follows:

  (32°F  32) × 5/9 + 273.15 = 273.15K

Here, K is Kelvin and oF is degree Fahrenheit.

(b)

Expert Solution
Check Mark

Answer to Problem 9E

  449.82 K

Explanation of Solution

The relation between two scales is as follows:

  (32°F  32) × 5/9 + 273.15 = 273.15K

The given temperature is 350 oF. It can be converted into Kelvin as follows:

  (350°F  32) × 5/9 + 273.15 = x

On solving,

  x=449.82 K

Thus, the temperature 350 oF is equal to 449.82 K.

(c)

Interpretation Introduction

Interpretation: The given temperature in Fahrenheit scale that is 5 oF needs to be converted into Kelvin scale.

Concept Introduction:

The Kelvin and degree Fahrenheit are scales of measuring temperature. The relation between two scales is as follows:

  (32°F  32) × 5/9 + 273.15 = 273.15K

Here, K is Kelvin and oF is degree Fahrenheit.

(c)

Expert Solution
Check Mark

Answer to Problem 9E

  258.15 K

Explanation of Solution

The relation between two scales is as follows:

  (32°F  32) × 5/9 + 273.15 = 273.15K

The given temperature is 5 oF. It can be converted into Kelvin as follows:

  (5°F  32) × 5/9 + 273.15 = x

On solving

  x=258.15 K

Thus, the temperature 5 oF is equal to 258.15 K.

Chapter U3 Solutions

Living By Chemistry: First Edition Textbook

Ch. U3.2 - Prob. 4ECh. U3.2 - Prob. 5ECh. U3.2 - Prob. 6ECh. U3.2 - Prob. 7ECh. U3.3 - Prob. 1TAICh. U3.3 - Prob. 1ECh. U3.3 - Prob. 2ECh. U3.3 - Prob. 3ECh. U3.3 - Prob. 4ECh. U3.3 - Prob. 5ECh. U3.3 - Prob. 6ECh. U3.3 - Prob. 7ECh. U3.3 - Prob. 8ECh. U3.3 - Prob. 9ECh. U3.4 - Prob. 1TAICh. U3.4 - Prob. 1ECh. U3.4 - Prob. 2ECh. U3.4 - Prob. 3ECh. U3.4 - Prob. 4ECh. U3.4 - Prob. 5ECh. U3.4 - Prob. 6ECh. U3.4 - Prob. 7ECh. U3.4 - Prob. 8ECh. U3.4 - Prob. 9ECh. U3.5 - Prob. 1TAICh. U3.5 - Prob. 1ECh. U3.5 - Prob. 2ECh. U3.5 - Prob. 3ECh. U3.5 - Prob. 4ECh. U3.5 - Prob. 5ECh. U3.5 - Prob. 6ECh. U3.5 - Prob. 7ECh. U3.5 - Prob. 8ECh. U3.5 - Prob. 9ECh. U3.5 - Prob. 10ECh. U3.5 - Prob. 11ECh. U3.5 - Prob. 12ECh. U3.6 - Prob. 1TAICh. U3.6 - Prob. 1ECh. U3.6 - Prob. 2ECh. U3.6 - Prob. 3ECh. U3.6 - Prob. 4ECh. U3.6 - Prob. 5ECh. U3.6 - Prob. 6ECh. U3.6 - Prob. 7ECh. U3.6 - Prob. 8ECh. U3.7 - Prob. 1TAICh. U3.7 - Prob. 1ECh. U3.7 - Prob. 2ECh. U3.7 - Prob. 3ECh. U3.7 - Prob. 4ECh. U3.7 - Prob. 5ECh. U3.7 - Prob. 6ECh. U3.8 - Prob. 1TAICh. U3.8 - Prob. 1ECh. U3.8 - Prob. 2ECh. U3.8 - Prob. 3ECh. U3.8 - Prob. 4ECh. U3.8 - Prob. 5ECh. U3.8 - Prob. 6ECh. U3.8 - Prob. 7ECh. U3.8 - Prob. 8ECh. U3.8 - Prob. 9ECh. U3.8 - Prob. 10ECh. U3.9 - Prob. 1TAICh. U3.9 - Prob. 1ECh. U3.9 - Prob. 2ECh. U3.9 - Prob. 3ECh. U3.9 - Prob. 4ECh. U3.9 - Prob. 5ECh. U3.9 - Prob. 6ECh. U3.9 - Prob. 7ECh. U3.9 - Prob. 8ECh. U3.9 - Prob. 10ECh. U3.10 - Prob. 1TAICh. U3.10 - Prob. 1ECh. U3.10 - Prob. 2ECh. U3.10 - Prob. 4ECh. U3.10 - Prob. 5ECh. U3.10 - Prob. 6ECh. U3.10 - Prob. 7ECh. U3.11 - Prob. 1TAICh. U3.11 - Prob. 1ECh. U3.11 - Prob. 2ECh. U3.11 - Prob. 3ECh. U3.11 - Prob. 4ECh. U3.11 - Prob. 5ECh. U3.12 - Prob. 1TAICh. U3.12 - Prob. 1ECh. U3.12 - Prob. 2ECh. U3.12 - Prob. 3ECh. U3.12 - Prob. 4ECh. U3.12 - Prob. 5ECh. U3.12 - Prob. 6ECh. U3.12 - Prob. 7ECh. U3.12 - Prob. 8ECh. U3.13 - Prob. 1TAICh. U3.13 - Prob. 1ECh. U3.13 - Prob. 2ECh. U3.13 - Prob. 3ECh. U3.13 - Prob. 4ECh. U3.13 - Prob. 5ECh. U3.13 - Prob. 6ECh. U3.13 - Prob. 7ECh. U3.14 - Prob. 1TAICh. U3.14 - Prob. 1ECh. U3.14 - Prob. 2ECh. U3.15 - Prob. 1TAICh. U3.15 - Prob. 1ECh. U3.15 - Prob. 2ECh. U3.15 - Prob. 3ECh. U3.15 - Prob. 4ECh. U3.15 - Prob. 5ECh. U3.15 - Prob. 6ECh. U3.15 - Prob. 7ECh. U3.15 - Prob. 8ECh. U3.16 - Prob. 1TAICh. U3.16 - Prob. 1ECh. U3.16 - Prob. 2ECh. U3.16 - Prob. 3ECh. U3.16 - Prob. 4ECh. U3.16 - Prob. 5ECh. U3.16 - Prob. 6ECh. U3.16 - Prob. 7ECh. U3.17 - Prob. 1TAICh. U3.17 - Prob. 1ECh. U3.17 - Prob. 2ECh. U3.17 - Prob. 3ECh. U3.17 - Prob. 4ECh. U3.17 - Prob. 5ECh. U3.17 - Prob. 6ECh. U3.17 - Prob. 7ECh. U3.17 - Prob. 8ECh. U3.18 - Prob. 1TAICh. U3.18 - Prob. 1ECh. U3.18 - Prob. 2ECh. U3.18 - Prob. 3ECh. U3.18 - Prob. 4ECh. U3.18 - Prob. 5ECh. U3.18 - Prob. 6ECh. U3.18 - Prob. 7ECh. U3.18 - Prob. 8ECh. U3.18 - Prob. 9ECh. U3.18 - Prob. 10ECh. U3.18 - Prob. 11ECh. U3.18 - Prob. 12ECh. U3.19 - Prob. 1TAICh. U3.19 - Prob. 1ECh. U3.19 - Prob. 2ECh. U3.19 - Prob. 4ECh. U3 - Prob. SI1RECh. U3 - Prob. SI2RECh. U3 - Prob. SI3RECh. U3 - Prob. SI4RECh. U3 - Prob. SI5RECh. U3 - Prob. SII1RECh. U3 - Prob. SII2RECh. U3 - Prob. SII3RECh. U3 - Prob. SII4RECh. U3 - Prob. SII5RECh. U3 - Prob. SIII1RECh. U3 - Prob. SIII2RECh. U3 - Prob. SIII3RECh. U3 - Prob. SIII4RECh. U3 - Prob. SIII5RECh. U3 - Prob. 1RECh. U3 - Prob. 2RECh. U3 - Prob. 3RECh. U3 - Prob. 4RECh. U3 - Prob. 5RECh. U3 - Prob. 6RECh. U3 - Prob. 7RECh. U3 - Prob. 8RECh. U3 - Prob. 9RECh. U3 - Prob. 10RECh. U3 - Prob. 11RECh. U3 - Prob. 12RECh. U3 - Prob. 13RE
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