Chapter 2, Problem 20PE

(a)

Interpretation Introduction

Interpretation:

The quantity $\text{4}\text{.5}\text{cm}$ has to be converted into $\stackrel{\text{o}}{\text{A}}\text{.}$

Answer to Problem 20PE

The quantity $\text{4}\text{.5}\text{cm}$ in $\stackrel{\text{o}}{\text{A}}$ is $\text{4}{\text{.5×10}}^{\text{8}}\stackrel{\text{o}}{\text{A}}\text{.}$

Explanation of Solution

The given quantity is $\text{4}\text{.5}\text{cm}\text{.}$

Centimeters is converted into $\stackrel{\text{o}}{\text{A}}$ using the conversion factor,

  $\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\left(\frac{\text{1}\stackrel{\text{o}}{\text{A}}}{{\text{10}}^{\text{-10}}\text{m}}\right)$

The quantity $\text{4}\text{.5}\text{cm}$ is converted as,

  $\begin{array}{l}\stackrel{\text{o}}{\text{A}}\text{=4}\text{.5}\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\left(\frac{\text{1}\stackrel{\text{o}}{\text{A}}}{{\text{10}}^{\text{-10}}\text{m}}\right)\\ \stackrel{\text{o}}{\text{A}}\text{=4}{\text{.5×10}}^{\text{8}}\stackrel{\text{o}}{\text{A}}\end{array}$

The quantity $\text{4}\text{.5}\text{cm}$ in $\stackrel{\text{o}}{\text{A}}$ is $\text{4}{\text{.5×10}}^{\text{8}}\stackrel{\text{o}}{\text{A}}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The quantity $\text{12}\text{nm}$ has to be converted into centimeters.

Answer to Problem 20PE

The quantity $\text{12}\text{nm}$ in centimeters is $\text{1}{\text{.2×10}}^{\text{-6}}\text{cm}\text{.}$

Explanation of Solution

The given quantity is $\text{12}\text{nm}\text{.}$

Nanometers is converted into centimeters using the conversion factor,

  $\left(\frac{{\text{10}}^{\text{-9}}\text{m}}{\text{1}\text{nm}}\right)\left(\frac{\text{100}\text{cm}}{\text{1}\text{m}}\right)$

The quantity $\text{12}\text{nm}$ is converted as,

  $\begin{array}{l}\text{cm=12}\text{nm}\left(\frac{{\text{10}}^{\text{-9}}\text{m}}{\text{1}\text{nm}}\right)\left(\frac{\text{100}\text{cm}}{\text{1}\text{m}}\right)\\ \text{cm=1}{\text{.2×10}}^{\text{-6}}\text{cm}\end{array}$

The quantity $\text{12}\text{nm}$ in centimeters is $\text{1}{\text{.2×10}}^{\text{-6}}\text{cm}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The quantity $\text{8}\text{.0}\text{km}$ has to be converted into millimeters.

Answer to Problem 20PE

The quantity $\text{8}\text{.0}\text{km}$ in millimeters is $\text{8}{\text{.0×10}}^{\text{6}}\text{mm}\text{.}$

Explanation of Solution

The given quantity is $\text{8}\text{.0}\text{km}\text{.}$

Kilometers is converted into millimeters using the conversion factor,

  $\left(\frac{\text{1000}\text{m}}{\text{1km}}\right)\left(\frac{\text{1000}\text{mm}}{\text{1}\text{m}}\right)$

The quantity $\text{8}\text{.0}\text{km}$ is converted as,

  $\begin{array}{l}\text{mm=8}\text{.0km}\left(\frac{\text{1000}\text{m}}{\text{1km}}\right)\left(\frac{\text{1000}\text{mm}}{\text{1}\text{m}}\right)\\ \text{mm=8}{\text{.0×10}}^{\text{6}}\text{mm}\end{array}$

The quantity $\text{8}\text{.0}\text{km}$ in millimeters is $\text{8}{\text{.0×10}}^{\text{6}}\text{mm}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The quantity $\text{164}\text{mg}$ has to be converted into grams.

Answer to Problem 20PE

The quantity $\text{164}\text{mg}$ in grams is $\text{0}\text{.164 g}\text{.}$

Explanation of Solution

The given quantity is $\text{164}\text{mg}\text{.}$

Milligrams is converted into grams using the conversion factor,

  $\left(\frac{\text{1}\text{g}}{\text{1000}\text{mg}}\right)$

The quantity $\text{164}\text{mg}$ is converted as,

  $\begin{array}{l}\text{g=164}\text{mg}\left(\frac{\text{1}\text{g}}{\text{1000}\text{mg}}\right)\\ \text{g=0}\text{.164}\text{g}\end{array}$

The quantity $\text{164}\text{mg}$ in grams is $\text{0}\text{.164 g}\text{.}$

(e)

Interpretation Introduction

Interpretation:

The quantity $\text{0}\text{.65}\text{kg}$ has to be converted into milligrams.

Answer to Problem 20PE

The quantity $\text{0}\text{.65}\text{kg}$ in milligrams is $\text{6}{\text{.5×10}}^{\text{5}}\text{mg}\text{.}$

Explanation of Solution

The given quantity is $\text{0}\text{.65}\text{kg}\text{.}$

Kilograms is converted into milligrams using the conversion factor,

  $\left(\frac{\text{1000}\text{g}}{\text{1kg}}\right)\left(\frac{\text{1000}\text{mg}}{\text{1}\text{g}}\right)$

The quantity $\text{0}\text{.65}\text{kg}$ is converted as,

  $\begin{array}{l}\text{mg=0}\text{.65 kg}\left(\frac{\text{1000}\text{g}}{\text{1kg}}\right)\left(\frac{\text{1000}\text{mg}}{\text{1}\text{g}}\right)\\ \text{mg=6}{\text{.5×10}}^{\text{5}}\text{mg}\end{array}$

The quantity $\text{0}\text{.65}\text{kg}$ in milligrams is $\text{6}{\text{.5×10}}^{\text{5}}\text{mg}\text{.}$

(f)

Interpretation Introduction

Interpretation:

The quantity $\text{5}\text{.5}\text{kg}$ has to be converted into grams.

Answer to Problem 20PE

The quantity $\text{5}\text{.5}\text{kg}$ in grams is $\text{5}{\text{.5×10}}^{\text{3}}\text{g}\text{.}$

Explanation of Solution

The given quantity is $\text{5}\text{.5}\text{kg}\text{.}$

Kilograms is converted into grams using the conversion factor,

  $\left(\frac{\text{1000 g}}{\text{1}\text{kg}}\right)$

The quantity $\text{5}\text{.5}\text{kg}$ is converted as,

  $\begin{array}{l}\text{g=5}\text{.5 kg}\left(\frac{\text{1000 g}}{\text{1}\text{kg}}\right)\\ \text{g=5}{\text{.5×10}}^{\text{3}}\text{g}\end{array}$

The quantity $\text{5}\text{.5}\text{kg}$ in grams is $\text{5}{\text{.5×10}}^{\text{3}}\text{g}\text{.}$

(g)

Interpretation Introduction

Interpretation:

The quantity $\text{0}\text{.468 L}$ has to be converted into milliliters.

Answer to Problem 20PE

The quantity $\text{0}\text{.468 L}$ in milliliters is $\text{468 mL}\text{.}$

Explanation of Solution

The given quantity is $\text{0}\text{.468 L}\text{.}$

Liters is converted into milliliters using the conversion factor,

  $\left(\frac{\text{1000 mL}}{\text{1L}}\right)$

The quantity $\text{0}\text{.468 L}$ is converted as,

  $\begin{array}{l}\text{mL=0}\text{.468 L}\left(\frac{\text{1000 mL}}{\text{1L}}\right)\\ \text{mL=468}\text{mL}\end{array}$

The quantity $\text{0}\text{.468 L}$ in milliliters is $\text{468 mL}\text{.}$

(h)

Interpretation Introduction

Interpretation:

The quantity $\text{9}\text{.0}\text{μL}$ has to be converted into milliliters.

Answer to Problem 20PE

The quantity $\text{9}\text{.0}\text{μL}$ in milliliters is $\text{9}{\text{.0×10}}^{\text{-3}}\text{mL}\text{.}$

Explanation of Solution

The given quantity is $\text{9}\text{.0}\text{μL}\text{.}$

Microliters is converted into milliliters using the conversion factor,

  $\left(\frac{\text{1L}}{{\text{10}}^{\text{6}}\text{μL}}\right)\left(\frac{\text{1000}\text{mL}}{\text{1}\text{L}}\right)$

The quantity $\text{9}\text{.0}\text{μL}$ is converted as,

  $\begin{array}{l}\text{mL=9}\text{.0}\text{μL}\left(\frac{\text{1L}}{{\text{10}}^{\text{6}}\text{μL}}\right)\left(\frac{\text{1000}\text{mL}}{\text{1}\text{L}}\right)\\ \text{mL=9}{\text{.0×10}}^{\text{-3}}\text{mL}\end{array}$

The quantity $\text{9}\text{.0}\text{μL}$ in milliliters is $\text{9}{\text{.0×10}}^{\text{-3}}\text{mL}\text{.}$