Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 2, Problem 20E
Interpretation Introduction

(a)

Interpretation:

The given metric-English equivalent is to be stated.

Concept introduction:

The metric system unit is the unit system which is accepted as the standard system of unit internationally. The nonmetric units are the common unit system which is not standard for the parameter to be calculated. The unit factors are calculated by taking the ratio of two equivalent measurement systems and these are used in unit conversions. The English unit system is common in United States and therefore it is used in the conversion factors.

Expert Solution
Check Mark

Answer to Problem 20E

The given metric-English equivalent is 1m=1.09yd.

Explanation of Solution

The conversion given is 1m=?yd. “ yd” is the unit denoting yard in the English system for the measurement of length. The English system of units can be converted to the metric system of units by using the equivalent factors. The relation between yard and meter is given below as,

1yd=0.914m

Thus, the equivalent for the conversion of m into yd can be calculated using the above relation as given below.

1yd=0.914m1m=1yd0.914m×1m=1.09yd

Conclusion

The given metric-English equivalent is 1m=1.09yd.

Interpretation Introduction

(b)

Interpretation:

The given metric-English equivalent is to be stated.

Concept introduction:

The metric system unit is the unit system which is accepted as the standard system of unit internationally. The nonmetric units are the common unit system which is not standard for the parameter to be calculated. The unit factors are calculated by taking the ratio of two equivalent measurement systems and these are used in unit conversions. The English unit system is common in United States and therefore it is used in the conversion factors.

Expert Solution
Check Mark

Answer to Problem 20E

The given metric-English equivalent is 1kg=2.20lb.

Explanation of Solution

The conversion given is 1kg=?lb. “ lb” is the unit denoting pound in the English system for the measurement of mass. The English system of units can be converted to the metric system of units by using the equivalent factors. The relation between g and lb is given below as,

1lb=454g

The relation between g and kg is given below as,

1kg=1000g

Thus, the equivalent for the conversion of kg into lb can be calculated using the above relations as given below.

1kg=1000g×1lb454g1kg=2.20lb

Conclusion

The given metric-English equivalent is 1kg=2.20lb.

Interpretation Introduction

(c)

Interpretation:

The given metric-English equivalent is to be stated.

Concept introduction:

The metric system unit is the unit system which is accepted as the standard system of unit internationally. The nonmetric units are the common unit system which is not standard for the parameter to be calculated. The unit factors are calculated by taking the ratio of two equivalent measurement systems and these are used in unit conversions. The English unit system is common in United States and therefore it is used in the conversion factors.

Expert Solution
Check Mark

Answer to Problem 20E

The given metric-English equivalent is 1L=1.06qt.

Explanation of Solution

The conversion given is 1L=?qt. “ qt” is the unit denoting quart in the English system for the measurement of volume. The English system of units can be converted to the metric system of units by using the equivalent factors. The relation between mL and qt is given below as,

1qt=946mL

The relation between mL and L is given below as,

1L=1000mL

Thus, the equivalent for the conversion of L into qt can be calculated using the above relations as given below.

1L=1000mL×1qt946mL1L=1.06qt

Conclusion

The given metric-English equivalent is 1L=1.06qt.

Interpretation Introduction

(d)

Interpretation:

The given metric-English equivalent is to be stated.

Concept introduction:

The metric system unit is the unit system which is accepted as the standard system of unit internationally. The nonmetric units are the common unit system which is not standard for the parameter to be calculated. The unit factors are calculated by taking the ratio of two equivalent measurement systems and these are used in unit conversions. The English unit system is common in United States and therefore it is used in the conversion factors.

Expert Solution
Check Mark

Answer to Problem 20E

The given metric-English equivalent is 1s=1sec.

Explanation of Solution

The conversion given is 1s=?sec. “ sec” is the unit denoting seconds in the English system for the measurement of time. The English system of units can be converted to the metric system of units by using the equivalent factors. The seconds in the metric system of units are denoted by s. Thus, the equivalent given for the conversion of sec in s is given below as,

1s=1sec

Conclusion

The given metric-English equivalent is 1s=1sec.

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
Complete the following volume equivalents. (a) 1 mL = ? cm³ (b) 1 in.³= ? cm³
Carry out the following conversions. Report your answers to the correct number of significant figures. (a) 37 mm to m (b) 357 kg to g  (c) 77.00 mL to L  (d) 0.0084650 m to cm  (f) 36.7 in to cm (g) 162 lb to g (h) 911 qt to L (i) 44.2 cm to in  (j) 236.4 g to lb  (k) 4.0 L to qt
Complete the following volume equivalents. (a) 1 L = ? cm³ (b)1 in.³ = ? mL

Chapter 2 Solutions

Introductory Chemistry: Concepts and Critical Thinking (8th Edition)

Ch. 2 - Prob. 2.11CECh. 2 - Prob. 2.12CECh. 2 - Prob. 2.13CECh. 2 - Prob. 2.14CECh. 2 - Prob. 2.15CECh. 2 - Prob. 2.16CECh. 2 - Prob. 2.17CECh. 2 - Prob. 2.18CECh. 2 - Prob. 1KTCh. 2 - Prob. 2KTCh. 2 - Prob. 3KTCh. 2 - Prob. 4KTCh. 2 - Prob. 5KTCh. 2 - Prob. 6KTCh. 2 - Prob. 7KTCh. 2 - Prob. 8KTCh. 2 - Prob. 9KTCh. 2 - Prob. 10KTCh. 2 - Prob. 11KTCh. 2 - Prob. 12KTCh. 2 - Prob. 13KTCh. 2 - Prob. 14KTCh. 2 - Prob. 15KTCh. 2 - Prob. 16KTCh. 2 - Prob. 17KTCh. 2 - Prob. 18KTCh. 2 - Prob. 19KTCh. 2 - Prob. 20KTCh. 2 - Prob. 21KTCh. 2 - Prob. 22KTCh. 2 - Prob. 23KTCh. 2 - Prob. 24KTCh. 2 - Prob. 25KTCh. 2 - Prob. 1ECh. 2 - Prob. 2ECh. 2 - Prob. 3ECh. 2 - Prob. 4ECh. 2 - Prob. 5ECh. 2 - Prob. 6ECh. 2 - Prob. 7ECh. 2 - Prob. 8ECh. 2 - Prob. 9ECh. 2 - Prob. 10ECh. 2 - Prob. 11ECh. 2 - Prob. 12ECh. 2 - Prob. 13ECh. 2 - Prob. 14ECh. 2 - Prob. 15ECh. 2 - Prob. 16ECh. 2 - Prob. 17ECh. 2 - Prob. 18ECh. 2 - Prob. 19ECh. 2 - Prob. 20ECh. 2 - Prob. 21ECh. 2 - Prob. 22ECh. 2 - Prob. 23ECh. 2 - Prob. 24ECh. 2 - Prob. 25ECh. 2 - Prob. 26ECh. 2 - Prob. 27ECh. 2 - Prob. 28ECh. 2 - Prob. 29ECh. 2 - Prob. 30ECh. 2 - Prob. 31ECh. 2 - Prob. 32ECh. 2 - Prob. 33ECh. 2 - Prob. 34ECh. 2 - Prob. 35ECh. 2 - Prob. 36ECh. 2 - Prob. 37ECh. 2 - Prob. 38ECh. 2 - Prob. 39ECh. 2 - Prob. 40ECh. 2 - Prob. 41ECh. 2 - Prob. 42ECh. 2 - Prob. 43ECh. 2 - Prob. 44ECh. 2 - Prob. 45ECh. 2 - Prob. 46ECh. 2 - Prob. 47ECh. 2 - Prob. 48ECh. 2 - Prob. 49ECh. 2 - Prob. 50ECh. 2 - Prob. 51ECh. 2 - Prob. 52ECh. 2 - Prob. 53ECh. 2 - Prob. 54ECh. 2 - Prob. 55ECh. 2 - Prob. 56ECh. 2 - Prob. 57ECh. 2 - Prob. 58ECh. 2 - Prob. 59ECh. 2 - Prob. 60ECh. 2 - Prob. 61ECh. 2 - Prob. 62ECh. 2 - Prob. 63ECh. 2 - Prob. 64ECh. 2 - Prob. 65ECh. 2 - Prob. 66ECh. 2 - Prob. 67ECh. 2 - Prob. 68ECh. 2 - Prob. 69ECh. 2 - Prob. 70ECh. 2 - Prob. 71ECh. 2 - Prob. 72ECh. 2 - Prob. 73ECh. 2 - Prob. 74ECh. 2 - Prob. 75ECh. 2 - Prob. 76ECh. 2 - Prob. 77ECh. 2 - Prob. 78ECh. 2 - Prob. 79ECh. 2 - Prob. 80ECh. 2 - Prob. 81ECh. 2 - Prob. 82ECh. 2 - Prob. 83ECh. 2 - Prob. 84ECh. 2 - Prob. 85ECh. 2 - Prob. 86ECh. 2 - Prob. 87ECh. 2 - Prob. 88ECh. 2 - Prob. 89ECh. 2 - Prob. 90ECh. 2 - Prob. 91ECh. 2 - Prob. 92ECh. 2 - Prob. 93ECh. 2 - Prob. 94ECh. 2 - Prob. 95ECh. 2 - Prob. 96ECh. 2 - Prob. 1STCh. 2 - Prob. 2STCh. 2 - Prob. 3STCh. 2 - Prob. 4STCh. 2 - Prob. 5STCh. 2 - Prob. 6STCh. 2 - Prob. 7STCh. 2 - Prob. 8STCh. 2 - Prob. 9STCh. 2 - Prob. 10STCh. 2 - Prob. 11STCh. 2 - Prob. 12STCh. 2 - Prob. 13STCh. 2 - Prob. 14ST
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
Measurement and Significant Figures; Author: Professor Dave Explains;https://www.youtube.com/watch?v=Gn97hpEkTiM;License: Standard YouTube License, CC-BY
Trigonometry: Radians & Degrees (Section 3.2); Author: Math TV with Professor V;https://www.youtube.com/watch?v=U5a9e1J_V1Y;License: Standard YouTube License, CC-BY