FOUND.OF COLLEGE CHEMISTRY
FOUND.OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781119234555
Author: Hein
Publisher: WILEY
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 2, Problem 54PE

(a)

Interpretation Introduction

Interpretation:

The temperature in Celsius has to be converted into degree Fahrenheit including an equation.

Concept Introduction:

Temperature:  The measurement of hotness or coldness of an object is temperature.  The three temperature scales as,

  1. 1) Celsius (oC)
  2. 2) Kelvin (K)
  3. 3) Fahrenheit (oF)

Celsius to Fahrenheit conversion:

Celsius can be converted to Fahrenheit using the formula,

  oF=1.8(oC)+32

(a)

Expert Solution
Check Mark

Answer to Problem 54PE

The temperature in degree Fahrenheit is 90.oF.

Explanation of Solution

Given,

The temperature is 32oC.

The temperature in Fahrenheit is calculated as,

  oF=1.8(oC)+32oF=1.8(32oC)+32oF=90.oF

The temperature in degree Fahrenheit is 90.oF.

(b)

Interpretation Introduction

Interpretation:

The temperature in Fahrenheit has to be converted into Celsius including an equation.

Concept Introduction:

Fahrenheit to Celsius conversion:

Fahrenheit can be converted to Celsius using the formula,

  oC=oF-321.8

(b)

Expert Solution
Check Mark

Answer to Problem 54PE

The temperature in Celsius is -22.6oC.

Explanation of Solution

Given,

The temperature is -8.6oF.

The temperature in Celsius is calculated as,

  oC=oF-321.8oC=-8.6oF-321.8oC=-22.6oC

The temperature in Celsius is -22.6oC.

(c)

Interpretation Introduction

Interpretation:

The temperature in degree Celsius has to be converted into Kelvin including an equation.

Concept Introduction:

Celsius to Kelvin conversion:

Celsius can be converted to Kelvin using the formula,

  K=oC+273.15

(c)

Expert Solution
Check Mark

Answer to Problem 54PE

The temperature in Kelvin is 546K.

Explanation of Solution

Given,

The temperature is 273oC.

The temperature in Kelvin is calculated as,

  K=oC+273.15K=273oC+273.15K=546.15K

The temperature in Kelvin is 546K.

(d)

Interpretation Introduction

Interpretation:

The temperature in Kelvin has to be converted into degree Fahrenheit including an equation.

Concept Introduction:

Kelvin to Celsius conversion:

Kelvin can be converted to Celsius using the formula,

  oC=K-273.15

Celsius to Fahrenheit conversion:

Celsius can be converted to Fahrenheit using the formula,

  oF=1.8(oC)+32

(d)

Expert Solution
Check Mark

Answer to Problem 54PE

The temperature in degree Fahrenheit is -300oF.

Explanation of Solution

Given,

The temperature is 100K.

The temperature in degree Fahrenheit is calculated as,

  oC=K-273.15oC=100-273.15oC=-173oC

From the calculated degree celsius, the Fahrenheit is calculated as,

  oF=1.8(oC)+32oF=1.8(-173oC)+32oF=-300oF

The temperature in degree Fahrenheit is -300oF.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
4. Make the following conversions and include an equation for each one:(a) 32 ℃ to ℉ (c) 273 ℃ to K(b) -8.6 ℉ to ℃ (d) 100 K to ℉
Which of the following include exact numbers?(a) The speed of light in a vacuum is a physical constant; to six significant figures, it is 2.99792×108 m/s.(b) The density of mercury at 25°C is 13.53 g/mL.(c) There are 3600 s in 1 h.(d) In 2010, the United States had 50 states.
Write the following numbers in standard notation. Use a ter-minal decimal point when needed:(a) 6.500*10^3(b) 3.46*10^5(c) 7.5*10^2(d) 1.8856*10^2

Chapter 2 Solutions

FOUND.OF COLLEGE CHEMISTRY

Ch. 2.6 - Prob. 2.11PCh. 2.6 - Prob. 2.12PCh. 2.6 - Prob. 2.13PCh. 2.6 - Prob. 2.14PCh. 2.6 - Prob. 2.15PCh. 2.7 - Prob. 2.16PCh. 2.7 - Prob. 2.17PCh. 2.7 - Prob. 2.18PCh. 2.7 - Prob. 2.19PCh. 2.8 - Prob. 2.20PCh. 2.8 - Prob. 2.21PCh. 2.9 - Prob. 2.22PCh. 2.9 - Prob. 2.23PCh. 2 - Prob. 1RQCh. 2 - Prob. 2RQCh. 2 - Prob. 3RQCh. 2 - Prob. 4RQCh. 2 - Prob. 5RQCh. 2 - Prob. 6RQCh. 2 - Prob. 7RQCh. 2 - Prob. 8RQCh. 2 - Prob. 9RQCh. 2 - Prob. 10RQCh. 2 - Prob. 11RQCh. 2 - Prob. 12RQCh. 2 - Prob. 13RQCh. 2 - Prob. 14RQCh. 2 - Prob. 15RQCh. 2 - Prob. 16RQCh. 2 - Prob. 17RQCh. 2 - Prob. 18RQCh. 2 - Prob. 19RQCh. 2 - Prob. 20RQCh. 2 - Prob. 21RQCh. 2 - Prob. 1PECh. 2 - Prob. 2PECh. 2 - Prob. 3PECh. 2 - Prob. 4PECh. 2 - Prob. 5PECh. 2 - Prob. 6PECh. 2 - Prob. 7PECh. 2 - Prob. 8PECh. 2 - Prob. 9PECh. 2 - Prob. 10PECh. 2 - Prob. 11PECh. 2 - Prob. 12PECh. 2 - Prob. 13PECh. 2 - Prob. 14PECh. 2 - Prob. 15PECh. 2 - Prob. 16PECh. 2 - Prob. 17PECh. 2 - Prob. 18PECh. 2 - Prob. 19PECh. 2 - Prob. 20PECh. 2 - Prob. 21PECh. 2 - Prob. 22PECh. 2 - Prob. 23PECh. 2 - Prob. 24PECh. 2 - Prob. 25PECh. 2 - Prob. 26PECh. 2 - Prob. 27PECh. 2 - Prob. 28PECh. 2 - Prob. 29PECh. 2 - Prob. 30PECh. 2 - Prob. 31PECh. 2 - Prob. 32PECh. 2 - Prob. 33PECh. 2 - Prob. 34PECh. 2 - Prob. 35PECh. 2 - Prob. 36PECh. 2 - Prob. 37PECh. 2 - Prob. 38PECh. 2 - Prob. 39PECh. 2 - Prob. 40PECh. 2 - Prob. 41PECh. 2 - Prob. 42PECh. 2 - Prob. 43PECh. 2 - Prob. 44PECh. 2 - Prob. 45PECh. 2 - Prob. 46PECh. 2 - Prob. 47PECh. 2 - Prob. 48PECh. 2 - Prob. 49PECh. 2 - Prob. 50PECh. 2 - Prob. 51PECh. 2 - Prob. 52PECh. 2 - Prob. 53PECh. 2 - Prob. 54PECh. 2 - Prob. 55PECh. 2 - Prob. 56PECh. 2 - Prob. 57PECh. 2 - Prob. 58PECh. 2 - Prob. 59PECh. 2 - Prob. 60PECh. 2 - Prob. 61PECh. 2 - Prob. 62PECh. 2 - Prob. 63PECh. 2 - Prob. 64PECh. 2 - Prob. 65PECh. 2 - Prob. 66PECh. 2 - Prob. 67PECh. 2 - Prob. 68PECh. 2 - Prob. 69PECh. 2 - Prob. 70PECh. 2 - Prob. 71AECh. 2 - Prob. 72AECh. 2 - Prob. 73AECh. 2 - Prob. 74AECh. 2 - Prob. 75AECh. 2 - Prob. 76AECh. 2 - Prob. 77AECh. 2 - Prob. 78AECh. 2 - Prob. 79AECh. 2 - Prob. 80AECh. 2 - Prob. 81AECh. 2 - Prob. 82AECh. 2 - Prob. 83AECh. 2 - Prob. 84AECh. 2 - Prob. 85AECh. 2 - Prob. 86AECh. 2 - Prob. 87AECh. 2 - Prob. 88AECh. 2 - Prob. 89AECh. 2 - Prob. 90AECh. 2 - Prob. 91AECh. 2 - Prob. 92AECh. 2 - Prob. 93AECh. 2 - Prob. 94AECh. 2 - Prob. 95AECh. 2 - Prob. 96AECh. 2 - Prob. 97AECh. 2 - Prob. 98AECh. 2 - Prob. 99AECh. 2 - Prob. 100AECh. 2 - Prob. 101AECh. 2 - Prob. 102AECh. 2 - Prob. 103AECh. 2 - Prob. 104AECh. 2 - Prob. 105AECh. 2 - Prob. 106CECh. 2 - Prob. 108CECh. 2 - Prob. 109CECh. 2 - Prob. 110CECh. 2 - Prob. 111CECh. 2 - Prob. 112CE
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Types of Matter: Elements, Compounds and Mixtures; Author: Professor Dave Explains;https://www.youtube.com/watch?v=dggHWvFJ8Xs;License: Standard YouTube License, CC-BY