Chapter F, Problem A.30E

(a)

Interpretation Introduction

Interpretation:

The measurement of $\text{36}\text{L}$ has to be converted into ${\text{m}}^{\text{3}}$.

Concept Introduction:

Measurement is the measure of quantity, capacity, dimensions or extent of something.  Mass, length, pressure, temperature, volume, time and concentration are some common type of measurements. English system and metric system are two types of measurement systems that are in use in the United States.  Measurement wise both English system and metric system are precise.  There are instance where measurements from one system has to be changed to their equivalent in other system.  This is done by dimensional analysis, which involves conversion factors.  Conversion factor is a ratio that indicates how one unit of measurement is related to another unit of measurement.  This can be expressed as shown below;

    $\text{Conversion}\text{factor}=\frac{\text{Units}\text{required}}{\text{units}\text{given}}$

Explanation of Solution

Given measurement is $\text{36}\text{L}$.  Some of the conversion units given are shown below;

    $\begin{array}{c}\text{1L}={10}^3{\text{cm}}^{\text{3}}\\ \text{1cm}=1{\text{0}}^{-2}\text{m}\end{array}$

Therefore, the conversion factor that has to be used to convert the given measurement into ${\text{m}}^{\text{3}}$ is;

    $\begin{array}{c}\text{Conversion}\text{factors:}\\ \frac{\text{1L}}{{10}^3{\text{cm}}^{\text{3}}}=\frac{{10}^3{\text{cm}}^{\text{3}}}{\text{1}\text{L}}\\ \frac{\text{1}\text{cm}}{{\text{10}}^{-2}\text{m}}=\frac{{\text{10}}^{-2}\text{m}}{\text{1}\text{cm}}\end{array}$

The given measurement is converted into ${\text{m}}^{\text{3}}$ as shown below;

    $\begin{array}{c}\text{36}\text{L}=\text{36}\text{L}\times \frac{{10}^3{\text{cm}}^{\text{3}}}{\text{1}\text{L}}\times {\left(\frac{{\text{10}}^{-2}\text{m}}{\text{1}\text{cm}}\right)}^3\\ \text{=36}\text{L}\times \frac{{10}^3{\text{cm}}^{\text{3}}}{\text{1}\text{L}}\times \frac{{\text{10}}^{-6}{\text{m}}^{\text{3}}}{\text{1}{\text{cm}}^{\text{3}}}\\ =3.6\times {10}^{-3}{\text{m}}^{\text{3}}\end{array}$

Therefore, the converted measurement is $3.6\times {10}^{-3}{\text{m}}^{\text{3}}$.

(b)

Interpretation Introduction

Interpretation:

The measurement of $\text{45}\text{g}\cdot {\text{L}}^{\text{-1}}$ has to be converted into $\text{mg}\cdot {\text{mL}}^{\text{-1}}$.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given measurement is $\text{45}\text{g}\cdot {\text{L}}^{\text{-1}}$.  Some of the conversion units given are shown below;

    $\begin{array}{c}\text{1mg}={10}^{-3}\text{g}\\ \text{1L}={10}^3\text{mL}\end{array}$

Therefore, the conversion factor that has to be used to convert the given measurement into $\text{mg}\cdot {\text{mL}}^{\text{-1}}$ is;

    $\begin{array}{c}\text{Conversion}\text{factors:}\\ \frac{\text{1mg}}{{10}^{-3}\text{g}}=\frac{{10}^{-3}\text{g}}{\text{1}\text{mg}}\\ \frac{\text{1}\text{L}}{{10}^3\text{mL}}=\frac{{10}^3\text{mL}}{\text{1}\text{L}}\end{array}$

The given measurement is converted into $\text{mg}\cdot {\text{mL}}^{\text{-1}}$ as shown below;

    $\begin{array}{c}\text{45}\text{g}\cdot {\text{L}}^{\text{-1}}=\text{45}\text{g}\cdot {\text{L}}^{\text{-1}}\times \frac{\text{1mg}}{{10}^{-3}\text{g}}\times \frac{\text{1}\text{L}}{{10}^3\text{mL}}\\ =45\text{mg}\cdot {\text{mL}}^{\text{-1}}\end{array}$

Therefore, the converted measurement is $45\text{mg}\cdot {\text{mL}}^{\text{-1}}$.

(c)

Interpretation Introduction

Interpretation:

The measurement of $\text{1}\text{.54}\text{mm}\cdot {\text{s}}^{\text{-1}}$ has to be converted into $\text{nm}\cdot \mu {\text{s}}^{-1}$.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given measurement is $\text{1}\text{.54}\text{mm}\cdot {\text{s}}^{\text{-1}}$.  Some of the conversion units given are shown below;

    $\begin{array}{c}\text{1mm}={10}^{-3}\text{m}\\ \text{1nm}={10}^{-9}\text{m}\\ \text{1}\mu \text{s}={10}^{-6}\text{s}\end{array}$

Therefore, the conversion factor that has to be used to convert the given measurement into $\text{nm}\cdot \mu {\text{s}}^{-1}$ is;

    $\begin{array}{c}\text{Conversion}\text{factors:}\\ \frac{\text{1mm}}{{10}^{-3}\text{m}}=\frac{{10}^{-3}\text{m}}{\text{1}\text{mm}}\\ \frac{\text{1nm}}{{10}^{-9}\text{m}}=\frac{{10}^{-9}\text{m}}{\text{1}\text{nm}}\\ \frac{\text{1}\mu \text{s}}{{10}^{-6}\text{s}}=\frac{{10}^{-6}\text{s}}{\text{1}\mu \text{s}}\end{array}$

The given measurement is converted into $\text{nm}\cdot \mu {\text{s}}^{-1}$ as shown below;

    $\begin{array}{c}\text{1}\text{.54}\text{mm}\cdot {\text{s}}^{\text{-1}}=\text{1}\text{.54}\text{mm}\cdot {\text{s}}^{\text{-1}}\times \frac{{10}^{-3}\text{m}}{\text{1}\text{mm}}\times \frac{\text{1nm}}{{10}^{-9}\text{m}}\times \frac{{10}^{-6}\text{s}}{\text{1}\mu \text{s}}\\ =1.54\text{nm}\cdot \mu {\text{s}}^{-1}\end{array}$

Therefore, the converted measurement is $1.54\text{nm}\cdot \mu {\text{s}}^{-1}$.

(d)

Interpretation Introduction

Interpretation:

The measurement of $\text{7}\text{.01}\text{cm}\cdot {\text{s}}^{\text{-1}}$ has to be converted into $\text{km}\cdot {\text{h}}^{\text{-1}}$.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given measurement is $\text{7}\text{.01}\text{cm}\cdot {\text{s}}^{\text{-1}}$.  Some of the conversion units given are shown below;

    $\begin{array}{c}\text{1h}=\text{60}\text{min}\\ \text{1min}=\text{60}\text{s}\\ \text{1m}={10}^2\text{cm}\\ \text{1km}={10}^2\text{m}\end{array}$

Therefore, the conversion factor that has to be used to convert the given measurement into $\text{km}\cdot {\text{h}}^{\text{-1}}$ is;

    $\begin{array}{c}\text{Conversion}\text{factors:}\\ \frac{\text{1h}}{60\min }=\frac{60\min }{\text{1}\text{h}}\\ \frac{\text{1min}}{60\text{s}}=\frac{60\text{s}}{\text{1}\text{min}}\\ \frac{\text{1}\text{m}}{{10}^2\text{cm}}=\frac{{10}^2\text{cm}}{\text{1}\text{m}}\\ \frac{\text{1}\text{km}}{{10}^3\text{m}}=\frac{{10}^3\text{m}}{\text{1}\text{km}}\end{array}$

The given measurement is converted into $\text{km}\cdot {\text{h}}^{\text{-1}}$ as shown below;

    $\begin{array}{c}\text{7}\text{.01}\text{cm}\cdot {\text{s}}^{\text{-1}}=\text{7}\text{.01}\text{cm}\cdot {\text{s}}^{\text{-1}}\times \frac{\text{1}\text{m}}{{10}^2\text{cm}}\times \frac{\text{1}\text{km}}{{10}^2\text{m}}\times \frac{60\text{s}}{\text{1}\text{min}}\times \frac{60\min }{\text{1}\text{h}}\\ =\text{9576}\times {10}^{-4}\text{km}\cdot {\text{h}}^{\text{-1}}\\ =0.\text{9576}\text{km}\cdot {\text{h}}^{\text{-1}}\end{array}$

Therefore, the converted measurement is $0.\text{9576}\text{km}\cdot {\text{h}}^{\text{-1}}$.

(e)

Interpretation Introduction

Interpretation:

The measurement of $\text{\$3}\text{.50/gallon}$ has to be converted into $\text{peso/liter}$.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given measurement is $\text{\$3}\text{.50/gallon}$.  Some of the conversion units given are shown below;

    $\begin{array}{c}\text{1dollar}\approx 13.1\text{peso}\\ \text{1}\text{gallon}=\text{3}\text{.78541}\text{L}\end{array}$

Therefore, the conversion factor that has to be used to convert the given measurement into $\text{peso/liter}$ is;

    $\begin{array}{c}\text{Conversion}\text{factors:}\\ \frac{\text{1dollar}}{13.1\text{peso}}=\frac{13.1\text{peso}}{\text{1}\text{dollar}}\\ \frac{\text{1}\text{gallon}}{\text{3}\text{.78541}\text{L}}=\frac{\text{3}\text{.78541}\text{L}}{\text{1}\text{gallon}}\end{array}$

The given measurement is converted into $\text{peso/liter}$ as shown below;

    $\begin{array}{c}\text{\$3}\text{.50/gallon}=\text{\$3}\text{.50/gallon}\times \frac{13.1\text{peso}}{\text{1}\text{dollar}}\times \frac{\text{1}\text{gallon}}{\text{3}\text{.78541}\text{L}}\\ =\text{12}\text{.1}\text{peso}\cdot {\text{L}}^{\text{-1}}\end{array}$

Therefore, the converted measurement is $\text{12}\text{.1}\text{peso}\cdot {\text{L}}^{\text{-1}}$.