Chapter 2, Problem 19PE

(a)

Interpretation Introduction

Interpretation:

The quantity $\text{28}\text{.0}\text{cm}$ has to be converted into meters.

Answer to Problem 19PE

The quantity $\text{28}\text{.0}\text{cm}$ in meters is $\text{0}\text{.280}\text{m}\text{.}$

Explanation of Solution

The given quantity is $\text{28}\text{.0}\text{cm}\text{.}$

Centimeters is converted into meters using the conversion factor,

  $\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)$

The quantity $\text{28}\text{.0}\text{cm}$ is converted as,

  $\begin{array}{l}\text{m=28}\text{.0}\text{cm×}\frac{\text{1}\text{m}}{\text{100}\text{cm}}\\ \text{m=0}\text{.280}\text{m}\end{array}$

The quantity $\text{28}\text{.0}\text{cm}$ in meters is $\text{0}\text{.280}\text{m}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The quantity $\text{1000}\text{.}\text{m}$ has to be converted into kilometers.

Answer to Problem 19PE

The quantity $\text{1000}\text{.}\text{m}$ in kilometers is $\text{1}\text{.000}\text{km}\text{.}$

Explanation of Solution

The given quantity is $\text{1000}\text{.}\text{m}\text{.}$

Meters is converted into kilometers using the conversion factor,

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

The quantity $\text{1000}\text{.0}\text{m}$ is converted as,

  $\begin{array}{l}\text{km=1000}\text{.}\text{m×}\frac{\text{1}\text{km}}{\text{1000 m}}\\ \text{km=1}\text{.000}\text{km}\end{array}$

The quantity $\text{1000}\text{.}\text{m}$ in kilometers is $\text{1}\text{.000}\text{km}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The quantity $\text{9}\text{.28}\text{cm}$ has to be converted into millimeters.

Answer to Problem 19PE

The quantity $\text{9}\text{.28}\text{cm}$ in millimeters is $\text{92}\text{.8}\text{mm}\text{.}$

Explanation of Solution

The given quantity is $\text{9}\text{.28}\text{cm}\text{.}$

Centimeters is converted into millimeters using the conversion factor,

  $\left(\frac{\text{10}\text{mm}}{\text{1cm}}\right)$

The quantity $\text{9}\text{.28}\text{cm}$ is converted as,

  $\begin{array}{l}\text{mm=9}\text{.28}\text{cm×}\frac{\text{10}\text{mm}}{\text{1 cm}}\\ \text{mm=92}\text{.8}\text{mm}\end{array}$

The quantity $\text{9}\text{.28}\text{cm}$ in millimeters is $\text{92}\text{.8}\text{mm}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The quantity $\text{10}\text{.68}\text{g}$ has to be converted into milligrams.

Answer to Problem 19PE

The quantity $\text{10}\text{.68}\text{g}$ in milligrams is $\text{1}{\text{.068×10}}^{\text{4}}\text{mg}\text{.}$

Explanation of Solution

The given quantity is $\text{10}\text{.68}\text{g}\text{.}$

Grams is converted into milligrams using the conversion factor,

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

The quantity $\text{10}\text{.68}\text{g}$ is converted as,

  $\begin{array}{l}\text{mg=10}\text{.68}\text{g×}\frac{\text{1000}\text{mg}}{\text{1}\text{g}}\\ \text{mg=1}{\text{.068×10}}^{\text{4}}\text{mg}\end{array}$

The quantity $\text{10}\text{.68}\text{g}$ in milligrams is $\text{1}{\text{.068×10}}^{\text{4}}\text{mg}\text{.}$

(e)

Interpretation Introduction

Interpretation:

The quantity $\text{6}{\text{.8×10}}^{\text{4}}\text{mg}$ has to be converted into kilograms.

Answer to Problem 19PE

The quantity $\text{6}{\text{.8×10}}^{\text{4}}\text{mg}$ in kilograms is $\text{6}{\text{.8×10}}^{\text{-2}}\text{kg}\text{.}$

Explanation of Solution

The given quantity is $\text{6}{\text{.8×10}}^{\text{4}}\text{mg}\text{.}$

Milligrams is converted into kilograms using the conversion factor,

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

The quantity $\text{6}{\text{.8×10}}^{\text{4}}\text{mg}$ is converted as,

  $\begin{array}{l}\text{kg=6}{\text{.8×10}}^{\text{4}}\text{mg×}\frac{\text{1}\text{g}}{\text{1000}\text{mg}}\text{×}\frac{\text{1}\text{kg}}{\text{1000}\text{g}}\\ \text{kg=6}{\text{.8×10}}^{\text{-2}}\text{kg}\end{array}$

The quantity $\text{6}{\text{.8×10}}^{\text{4}}\text{mg}$ in kilograms is $\text{6}{\text{.8×10}}^{\text{-2}}\text{kg}\text{.}$

(f)

Interpretation Introduction

Interpretation:

The quantity $\text{8}\text{.54}\text{g}$ has to be converted into kilograms.

Answer to Problem 19PE

The quantity $\text{8}\text{.54}\text{g}$ in kilograms is $\text{0}\text{.00854}\text{kg}\text{.}$

Explanation of Solution

The given quantity is $\text{8}\text{.54}\text{g}\text{.}$

Grams is converted into kilograms using the conversion factor,

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

The quantity $\text{8}\text{.54}\text{g}$ is converted as,

  $\begin{array}{l}\text{kg=8}\text{.54 g×}\frac{\text{1}\text{kg}}{\text{1000}\text{g}}\\ \text{kg=0}\text{.00854}\text{kg}\end{array}$

The quantity $\text{8}\text{.54}\text{g}$ in kilograms is $\text{0}\text{.00854}\text{kg}\text{.}$

(g)

Interpretation Introduction

Interpretation:

The quantity $\text{25}\text{.0}\text{mL}$ has to be converted into liters.

Answer to Problem 19PE

The quantity $\text{25}\text{.0}\text{mL}$ in liters is $\text{2}{\text{.50×10}}^{\text{-2}}\text{L}\text{.}$

Explanation of Solution

The given quantity is $\text{25}\text{.0}\text{mL}\text{.}$

Milliliters is converted into liters using the conversion factor,

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

The quantity $\text{25}\text{.0}\text{mL}$ is converted as,

  $\begin{array}{l}\text{L=25}\text{.0}\text{mL×}\frac{\text{1L}}{\text{1000}\text{mL}}\\ \text{L=2}{\text{.50×10}}^{\text{-2}}\text{L}\end{array}$

The quantity $\text{25}\text{.0}\text{mL}$ in liters is $\text{2}{\text{.50×10}}^{\text{-2}}\text{L}\text{.}$

(h)

Interpretation Introduction

Interpretation:

The quantity $\text{22}\text{.4}\text{L}$ has to be converted into microliters.

Answer to Problem 19PE

The quantity $\text{22}\text{.4}\text{L}$ in microliters is $\text{2}{\text{.24×10}}^{\text{7}}\text{μL}\text{.}$

Explanation of Solution

The given quantity is $\text{22}\text{.4}\text{L}\text{.}$

Liters is converted into microliters using the conversion factor,

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

The quantity $\text{22}\text{.4}\text{L}$ is converted as,

  $\begin{array}{l}\text{μL=22}\text{.4}\text{L×}\frac{{\text{10}}^{\text{6}}\text{μL}}{\text{1 L}}\\ \text{μL=2}{\text{.24×10}}^{\text{7}}\text{μL}\end{array}$

The quantity $\text{22}\text{.4}\text{L}$ in microliters is $\text{2}{\text{.24×10}}^{\text{7}}\text{μL}\text{.}$