Chapter 1, Problem 1.73QE

(a)

Interpretation Introduction

Interpretation:

The given conversions have to be performed to fill the blanks.  The operations have to be done without calculator or spreadsheet.

  $\text{6}\text{.39}\text{cm}\text{=}\underset{\_}{}\text{m}\text{=}\underset{\_}{}\text{mm}\text{=}\underset{\_}{}\text{nm}$

Explanation of Solution

Given,

  $\text{6}\text{.39}\text{cm}\text{=}\underset{\_}{}\text{m}\text{=}\underset{\_}{}\text{mm}\text{=}\underset{\_}{}\text{nm}$

The conversion factor used to convert centimeters to meters is

  $\begin{array}{l}\text{1}\text{cm}\text{=}{\text{10}}^{\text{-2}}\text{m}\\ \\ \text{then,}\\ \text{6}\text{.39}\text{cm}\text{=}\text{6}\text{.39}\text{×}{\text{10}}^{\text{-2}}\text{m}\\ \text{=}\text{0}\text{.0639}\text{m}\end{array}$

The conversion factor used to convert centimeters to millimeters is

  $\begin{array}{l}\text{1}\text{cm}\text{=}{\text{10}}^{\text{-2}}\text{m}\\ \text{1}\text{mm}\text{=}{\text{10}}^{\text{-3}}\text{m}\\ \\ \text{then,}\\ \text{6}\text{.39}\text{cm}\text{=}\text{0}\text{.0639}\text{m}\text{=}\text{0}\text{.0639}\text{×}{\text{10}}^{\text{3}}\text{mm}\\ \text{=}\text{63}\text{.9}\text{mm}\end{array}$

The conversion factor used to convert centimeters to nanometers is

  $\begin{array}{l}\text{1}\text{cm}\text{=}{\text{10}}^{\text{-2}}\text{m}\\ \text{1}\text{nm}\text{=}{\text{10}}^{\text{-9}}\text{m}\\ \\ \text{then,}\\ \text{6}\text{.39}\text{cm}\text{=}\text{0}\text{.0639}\text{m}\text{=}\text{0}\text{.0639}\text{×}{\text{10}}^{\text{9}}\text{nm}\\ \text{=}\text{63900000}\text{nm}\\ \text{=}\text{6}\text{.39}\text{×}{\text{10}}^{\text{7}}\text{nm}\end{array}$

The given blanks can be filled as follows:

  $\text{6}\text{.39}\text{cm}\text{=}\underset{\_}{\text{0}\text{.0639}}\text{m}\text{=}\underset{\_}{\text{63}\text{.9}}\text{mm}\text{=}\underset{\_}{\text{6}\text{.39}\text{×}{\text{10}}^{\text{7}}}\text{nm}$

(b)

Interpretation Introduction

Interpretation:

The given conversions have to be performed to fill the blanks.  The operations have to be done without calculator or spreadsheet.

  $\text{55}\text{.0}{\text{cm}}^{\text{3}}\text{=}\underset{\_}{}{\text{dm}}^{\text{3}}\text{=}\underset{\_}{}\text{mL}\text{=}\underset{\_}{}\text{L}\text{=}\underset{\_}{}{\text{m}}^{\text{3}}$

Explanation of Solution

Given,

  $\text{55}\text{.0}{\text{cm}}^{\text{3}}\text{=}\underset{\_}{}{\text{dm}}^{\text{3}}\text{=}\underset{\_}{}\text{mL}\text{=}\underset{\_}{}\text{L}\text{=}\underset{\_}{}{\text{m}}^{\text{3}}$

The conversion factor used to convert cubic centimeters to cubic decimeters is

  $\begin{array}{l}\text{1}{\text{cm}}^{\text{3}}\text{=}{\text{10}}^{\text{-3}}{\text{dm}}^{\text{3}}\\ \\ \text{then,}\\ \text{55}\text{.0}{\text{cm}}^{\text{3}}\text{=}\text{55}\text{.0}\text{×}{\text{10}}^{\text{-3}}{\text{dm}}^{\text{3}}\\ \text{=}\text{0}\text{.0550}{\text{dm}}^{\text{3}}\end{array}$

The conversion factor used to convert cubic centimeters to milliliters is

  $\begin{array}{l}\text{1}{\text{cm}}^{\text{3}}\text{=}\text{1}\text{mL}\\ \\ \text{then,}\\ \text{55}\text{.0}{\text{cm}}^{\text{3}}\text{=}\text{55}\text{.0}\text{mL}\end{array}$

The conversion factor used to convert cubic centimeters to liters is

  $\begin{array}{l}\text{1}{\text{cm}}^{\text{3}}\text{=}\text{1}\text{mL}\\ \text{1}\text{mL}\text{=}{\text{10}}^{\text{-3}}\text{L}\\ \\ \text{then,}\\ \text{55}\text{.0}{\text{cm}}^{\text{3}}\text{=}\text{55}\text{.0}\text{mL}\text{=}\text{55}\text{.0}\text{×}{\text{10}}^{\text{-3}}\text{L}\\ \text{=}\text{0}\text{.055}\text{L}\end{array}$

The conversion factor used to convert cubic centimeters to cubic meters is

  $\begin{array}{l}\text{1}{\text{cm}}^{\text{3}}\text{=}{\text{10}}^{\text{-6}}{\text{m}}^{\text{3}}\\ \\ \text{then,}\\ \text{55}\text{.0}{\text{cm}}^{\text{3}}\text{=}\text{55}\text{.0}\text{×}{\text{10}}^{\text{-6}}{\text{m}}^{\text{3}}\\ \text{=}\text{5}\text{.5}\text{×}{\text{10}}^{\text{-5}}{\text{m}}^{\text{3}}\end{array}$

The given blanks can be filled as follows:

  $\text{55}\text{.0}{\text{cm}}^{\text{3}}\text{=}\underset{\_}{\text{0}\text{.0550}}{\text{dm}}^{\text{3}}\text{=}\underset{\_}{\text{55}\text{.0}}\text{mL}\text{=}\underset{\_}{\text{0}\text{.055}}\text{L}\text{=}\underset{\_}{\text{5}\text{.5}\text{×}{\text{10}}^{\text{-5}}}{\text{m}}^{\text{3}}$

(c)

Interpretation Introduction

Interpretation:

The given conversions have to be performed to fill the blanks.  The operations have to be done without calculator or spreadsheet.

  $\text{23}\text{.1}\text{g}\text{=}\underset{\_}{}\text{mg}\text{=}\underset{\_}{}\text{kg}$

Explanation of Solution

Given,

  $\text{23}\text{.1}\text{g}\text{=}\underset{\_}{}\text{mg}\text{=}\underset{\_}{}\text{kg}$

The conversion factor used to convert grams to milligrams is

  $\begin{array}{l}\text{1}\text{g}\text{=}{\text{10}}^{\text{3}}\text{mg}\\ \\ \text{then,}\\ \text{23}\text{.1}\text{g}\text{=}\text{23}\text{.1}\text{×}{\text{10}}^{\text{3}}\text{mg}\\ \text{=}\text{2}\text{.31}\text{×}{\text{10}}^{\text{4}}\text{mg}\end{array}$

The conversion factor used to convert grams to kilograms is

  $\begin{array}{l}\text{1}\text{g}\text{=}{\text{10}}^{\text{-3}}\text{kg}\\ \\ \text{then,}\\ \text{23}\text{.1}\text{g}\text{=}\text{23}\text{.1}\text{×}{\text{10}}^{\text{-3}}\text{kg}\\ \text{=}\text{0}\text{.0231}\text{kg}\end{array}$

The given blanks can be filled as follows:

  $\text{23}\text{.1}\text{g}\text{=}\underset{\_}{\text{2}\text{.31}\text{×}{\text{10}}^{\text{4}}}\text{mg}\text{=}\underset{\_}{\text{0}\text{.0231}}\text{kg}$

(d)

Interpretation Introduction

Interpretation:

The given conversions have to be performed to fill the blanks.  The operations have to be done without calculator or spreadsheet.

  $\text{98}{\text{.6}}^{\text{°}}\text{F}\text{=}\underset{\_}{}{}^{\text{°}}\text{C}\text{=}\underset{\_}{}\text{K}$

Explanation of Solution

Given,

  $\text{98}{\text{.6}}^{\text{°}}\text{F}\text{=}\underset{\_}{}{}^{\text{°}}\text{C}\text{=}\underset{\_}{}\text{K}$

The conversion factor used to convert degree Fahrenheit to degree Celsius is

  $\begin{array}{l}{\text{T}}_{\text{C}}\text{=}\left({\text{T}}_{\text{F}}\text{-}{\text{32}}^{\text{°}}\text{F}\right)\text{×}\left(\frac{\text{1}{\text{.0}}^{\text{°}}\text{C}}{\text{1}{\text{.8}}^{\text{°}}\text{F}}\right)\\ \\ \text{then,}\\ \text{98}{\text{.6}}^{\text{°}}\text{F}\text{=}{\text{37}}^{\text{°}}\text{C}\end{array}$

The conversion factor used to convert degree Fahrenheit to Kelvin is

  $\begin{array}{l}{\text{T}}_{\text{C}}\text{=}\left({\text{T}}_{\text{F}}\text{-}{\text{32}}^{\text{°}}\text{F}\right)\text{×}\left(\frac{\text{1}{\text{.0}}^{\text{°}}\text{C}}{\text{1}{\text{.8}}^{\text{°}}\text{F}}\right)\\ {\text{T}}_{\text{K}}\text{=}{\text{T}}_{\text{C}}\text{+}\text{273}\text{.15}\\ \\ \text{then,}\\ \text{98}{\text{.6}}^{\text{°}}\text{F}\text{=}{\text{37}}^{\text{°}}\text{C}\text{=}\text{310}\text{.15}\text{K}\end{array}$

The given blanks can be filled as follows:

  $\text{98}{\text{.6}}^{\text{°}}\text{F}\text{=}\underset{\_}{\text{37}}{}^{\text{°}}\text{C}\text{=}\underset{\_}{\text{310}\text{.15}}\text{K}$