Chapter 1, Problem 1.72QE

(a)

Interpretation Introduction

Interpretation:

The given conversion has to be performed.

  $\text{25}\text{.5}\text{m}\text{to}\text{km}$

Explanation of Solution

Given, $\text{25}\text{.5}\text{m}\text{to}\text{km}$.

The conversion factor used to convert meters to kilometers is

  $\begin{array}{l}\text{1}\text{km}\text{=}\text{1000}\text{m}\\ \\ \text{conversion}\text{factor}\text{=}\left(\frac{\text{1}\text{km}}{\text{1000}\text{m}}\right)\end{array}$

The conversion can be carried out as follows:

  $\begin{array}{l}\text{=}\text{25}\text{.5}\text{m}\text{×}\left(\frac{\text{1}\text{km}}{\text{1000}\text{m}}\right)\\ \text{=}\text{0}\text{.0255}\text{km}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The given conversion has to be performed.

  $\text{36}\text{.3}\text{km}\text{to}\text{m}$

Explanation of Solution

Given, $\text{36}\text{.3}\text{km}\text{to}\text{m}$.

The conversion factor used to convert kilometers to meters is

  $\begin{array}{l}\text{1}\text{km}\text{=}\text{1000}\text{m}\\ \\ \text{conversion}\text{factor}\text{=}\left(\frac{\text{1000}\text{m}}{\text{1}\text{km}}\right)\end{array}$

The conversion can be carried out as follows:

  $\begin{array}{l}\text{=}\text{36}\text{.3}\text{km}\text{×}\left(\frac{\text{1000}\text{m}}{\text{1}\text{km}}\right)\\ \text{=}\text{36300}\text{m}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The given conversion has to be performed.

  $\text{487}\text{kg}\text{to}\text{g}$

Explanation of Solution

Given, $\text{487}\text{kg}\text{to}\text{g}$.

The conversion factor used to convert kilograms to grams is

  $\begin{array}{l}\text{1}\text{kg}\text{=}\text{1000}\text{g}\\ \\ \text{conversion}\text{factor}\text{=}\left(\frac{\text{1000}\text{g}}{\text{1}\text{kg}}\right)\end{array}$

The conversion can be carried out as follows:

  $\begin{array}{l}\text{=}\text{487}\text{kg}\text{×}\left(\frac{\text{1000}\text{g}}{\text{1}\text{kg}}\right)\\ \text{=}\text{487000}\text{g}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The given conversion has to be performed.

  $\text{1}\text{.32}\text{L}\text{to}\text{mL}$

Explanation of Solution

Given, $\text{1}\text{.32}\text{L}\text{to}\text{mL}$.

The conversion factor used to convert Liters to milliliters is

  $\begin{array}{l}\text{1}\text{L}\text{=}\text{1000}\text{mL}\\ \\ \text{conversion}\text{factor}\text{=}\left(\frac{\text{1000}\text{mL}}{\text{1}\text{L}}\right)\end{array}$

The conversion can be carried out as follows:

  $\begin{array}{l}\text{=}\text{1}\text{.32}\text{L}\text{×}\left(\frac{\text{1000}\text{mL}}{\text{1}\text{L}}\right)\\ \text{=}\text{1320}\text{mL}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The given conversion has to be performed.

  $\text{55}\text{.9}\text{dL}\text{to}\text{L}$

Explanation of Solution

Given, $\text{55}\text{.9}\text{dL}\text{to}\text{L}$.

The conversion factor used to convert deciliters to Liters is

  $\begin{array}{l}\text{1}\text{dL}\text{=}{\text{10}}^{\text{-1}}\text{L}\\ \\ \text{conversion}\text{factor}\text{=}\left(\frac{{\text{10}}^{\text{-1}}\text{L}}{\text{1}\text{dL}}\right)\end{array}$

The conversion can be carried out as follows:

  $\begin{array}{l}\text{=}\text{55}\text{.9}\text{dL}\text{×}\left(\frac{{\text{10}}^{\text{-1}}\text{L}}{1\text{dL}}\right)\\ \text{=}\text{5}\text{.59}\text{L}\end{array}$

(f)

Interpretation Introduction

Interpretation:

The given conversion has to be performed.

  $\text{6251}\text{L}\text{to}{\text{cm}}^{\text{3}}$

Explanation of Solution

Given, $\text{6251}\text{L}\text{to}{\text{cm}}^{\text{3}}$.

The conversion factor used to convert Liters to cubic centimeters is

  $\begin{array}{l}\text{1}\text{mL}\text{=}\text{0}\text{.001}\text{L}\\ \text{1}{\text{cm}}^{\text{3}}\text{=}\text{1}\text{mL}\\ \\ \text{conversion}\text{factor}\text{=}\left(\frac{\text{1}\text{mL}}{\text{0}\text{.001}\text{L}}\right)\text{×}\left(\frac{\text{1}{\text{cm}}^{\text{3}}}{\text{1}\text{mL}}\right)\end{array}$

The conversion can be carried out as follows:

  $\begin{array}{l}\text{=}\text{6251}\text{L}\text{×}\left(\frac{\text{1}\text{mL}}{\text{0}\text{.001}\text{L}}\right)\text{×}\left(\frac{\text{1}{\text{cm}}^{\text{3}}}{\text{1}\text{mL}}\right)\\ \text{=}\text{6251000}{\text{cm}}^{\text{3}}\end{array}$