Chapter 1, Problem 21QP

Carry out the following conversions. Report your answers to the correct number of signi�cant �gures.

(a) 947 m to mi

(b) 6.74 kg to lb

(c) 250.4 mL to gal

(d) 2.30 dL to mL

(e) 0.000450 cm to nm

(f) 37.5 in to m

(g) 689 lb to kg

(h) 0.5 qt to mL

(i) 125 cm to ft

(j) 542 mg to lb

(k) 25 nL to gal

Interpretation Introduction

Interpretation:

The conversion of $\text{947 m}$ into $\text{mi}$ .

Explanation of Solution

The relation between $\text{km}$ and $\text{m}$ is shown below.

$1\text{ km}=1000\text{ m}$

The conversion factors used to convert $\text{m}$ into $\text{km}$ are shown below.

$\frac{1\text{ km}}{1000\text{ m}}\text{or}\frac{1000\text{ m}}{1\text{ km}}$

The relation between $\text{mi}$ and $\text{km}$ is shown below.

$1\text{ mi}=\text{1}\text{.609 km}$

The conversion factors used to convert $\text{km}$ into $\text{mi}$ are shown below.

$\frac{1\text{ mi}}{1.609\text{ km}}\text{or}\frac{1.609\text{ km}}{1\text{ mi}}$

$\text{947 m}$ is converted into $\text{mi}$ using the conversion factor shown below.

$\begin{array}{c}\text{947 m}=\text{947 m}\times \frac{1\text{ km}}{1000\text{ m}}\times \frac{1\text{ mi}}{1.609\text{ km}}\\ =0.5885\text{ mi}\\ =0.588\text{ mi}\end{array}$

The given number, $\text{947 m}$ , has three significant figures. Therefore, the answer will be reported up to three significant figures. The number is rounded off by dropping the last digit, which is equal to $5$ , and the digit present on the left is even. Therefore, the last significant figure will remain unchanged. As a result, the final answer, after rounding off up to three significant figures, is $0.589\text{ mi}$ .

Interpretation Introduction

Interpretation:

The conversion of $\text{6}\text{.74 kg}$ into $\text{lb}$ .

Explanation of Solution

The relation between kilogram $\left(\text{kg}\right)$ into gram $\left(\text{g}\right)$ is shown below.

$1\text{ kg}=\text{1000 g}$

The conversion factors used to convert $\text{kg}$ into $\text{g}$ are shown below.

$\frac{1\text{ kg}}{1000\text{ g}}\text{or}\frac{1000\text{ g}}{1\text{ kg}}$

The relation between $\text{lb}$ and $\text{g}$ is shown below.

$1\text{ lb}=\text{453}\text{.6 g}$

The conversion factors used to convert $\text{g}$ into $\text{lb}$ are shown below.

$\frac{1\text{ lb}}{\text{453}\text{.6 g}}\text{or}\frac{\text{453}\text{.6 g}}{1\text{ lb}}$

$\text{6}\text{.74 kg}$ is converted into $\text{lb}$ using the conversion factor shown below.

$\begin{array}{c}\text{6}\text{.74 kg}=\text{6}\text{.74 kg}\times \frac{1000\text{ g}}{1\text{ kg}}\times \frac{1\text{ lb}}{\text{453}\text{.6 g}}\\ =14.85\text{ lb}\\ =14.8\text{ lb}\end{array}$

The given number, $\text{6}\text{.74 kg}$ , has three significant figures. Therefore, the answer will be reported up to three significant figures. The number is rounded off by dropping the last digit, which is equal to $5$ , and the digit present on the left is even. Therefore, the last significant figure will remain unchanged. As a result, the final answer, after rounding off up to three significant figures, is $14.9\text{ lb}$ .

Interpretation Introduction

Interpretation:

The conversion of $\text{250}\text{.4 mL}$ into $\text{gal}$ .

Explanation of Solution

The relation between milliliter $\left(\text{mL}\right)$ and liter $\left(\text{L}\right)$ is shown below.

$1\text{ mL}={10}^{-3}\text{L}$

The conversion factors used to convert $\text{mL}$ into $\text{L}$ are shown below.

$\frac{1\text{ mL}}{{10}^{-3}\text{L}}\text{or}\frac{{10}^{-3}\text{L}}{1\text{ mL}}$

The relation between liter $\left(\text{L}\right)$ and galloon $\left(\text{gal}\right)$ is shown below.

$1\text{ gal}=3.785\text{ L}$

The conversion factors used to convert $\text{L}$ into $\text{gal}$ are shown below.

$\frac{1\text{ gal}}{\text{3}\text{.785 L}}\text{or}\frac{\text{3}\text{.785 L}}{1\text{ gal}}$

$\text{250}\text{.4 mL}$ is converted into $\text{gal}$ using the conversion factor shown below.

$\begin{array}{c}\text{250}\text{.4 mL}=\text{250}\text{.4 mL}\times \frac{{10}^{-3}\text{L}}{1\text{ mL}}\times \frac{1\text{ gal}}{\text{3}\text{.785 L}}\\ =0.066155\text{ gal}\\ =0.06616\text{ gal}\end{array}$

The given number, $\text{250}\text{.4 mL}$ , has four significant figures. Therefore, the answer will be reported up to four significant figures. The number is rounded off by dropping the last digit, which is equal to $5$ , and the digit present on the left is odd. Therefore, the last significant figure will be increased by one unit. As a result, the final answer, after rounding off up to four significant figures, is $0.06616\text{ gal}$ .

Interpretation Introduction

Interpretation:

The conversion of $\text{2}\text{.30 dL}$ into $\text{mL}$ .

Explanation of Solution

The relation between deciliter $\left(\text{dL}\right)$ and liter $\left(\text{L}\right)$ is shown below.

$1\text{ L}=10\text{ dL}$

The conversion factors used to convert $\text{dL}$ into $\text{L}$ are shown below.

$\frac{1\text{ L}}{10\text{ dL}}\text{or}\frac{10\text{ dL}}{1\text{ L}}$

The relation between milliliter $\left(\text{mL}\right)$ and liter $\left(\text{L}\right)$ is shown below.

$1\text{ mL}={10}^{-3}\text{L}$

The conversion factors used to convert $\text{L}$ into $\text{mL}$ are shown below.

$\frac{1\text{ mL}}{{10}^{-3}\text{L}}\text{or}\frac{{10}^{-3}\text{L}}{1\text{ mL}}$

$\text{2}\text{.30 dL}$ is converted into $\text{mL}$ using the conversion factor shown below.

$\begin{array}{c}\text{2}\text{.30 dL}=\text{2}\text{.30 dL}\times \frac{1\text{ L}}{10\text{ dL}}\times \frac{1\text{ mL}}{{10}^{-3}\text{L}}\\ =230\text{ mL}\\ =2.30\times {10}^2\text{mL}\end{array}$

The given number, $\text{2}\text{.30 dL}$ , has three significant figures. Therefore, the answer will be reported up to three significant figures. The answer, $230\text{ mL}$ , is larger than $1$ , which means that the exponent will be a positive integer. The decimal shifts towards the left up to $2$ places to obtain the simplest and shortest value.

Therefore, the final answer, up to three significant figures, is $2.30\times {10}^2\text{mL}$ .

Interpretation Introduction

Interpretation:

The conversion of $0.000450\text{ cm}$ into $\text{nm}$ .

Explanation of Solution

The relation between meter $\left(\text{m}\right)$ and centimeter $\left(\text{cm}\right)$ is shown below.

$1\text{ m}=100\text{ cm}$

The conversion factors used to convert $\text{cm}$ into $\text{m}$ are shown below.

$\frac{1\text{ m}}{100\text{ cm}}\text{or}\frac{100\text{ cm}}{1\text{ m}}$

The relation between nanometer $\left(\text{nm}\right)$ and meter $\left(\text{m}\right)$ is shown below.

$1\text{ mm}={10}^{-9}\text{ m}$

The conversion factors used to convert $\text{m}$ into $\text{nm}$ are shown below.

$\frac{1\text{ nm}}{{\text{10}}^{-9}\text{ m}}\text{or}\frac{{\text{10}}^{-9}\text{ m}}{1\text{ nm}}$

$0.000450\text{ cm}$ is converted into $\text{nm}$ using the conversion factor shown below.

$\begin{array}{c}0.000450\text{ cm}=0.000450\text{ cm}\times \frac{1\text{ m}}{100\text{ cm}}\times \frac{1\text{ nm}}{{10}^{-9}\text{m}}\\ =4.50\times {10}^9\times {10}^{-6}\text{nm}\\ =4.50\times {10}^{\left(9+\left(-6\right)\right)}\text{ nm}\\ =4.50\times {10}^3\text{ nm}\end{array}$

The given number, $0.000450\text{ cm}$ , has three significant figures. Therefore, the answer will be reported up to three significant figures. $0.000450\text{ cm}$ is smaller than $1$ , which means that the exponent will be a negative integer. The decimal shifts towards the right up to $6$ places to obtain the simpler and shortest value.

In the multiplication method, the addition of exponents takes place in order to get the final answer. Therefore, the final answer, up to three significant figures, is $4.50\times {10}^3\text{ nm}$ .

Interpretation Introduction

Interpretation:

The conversion of $37.5\text{ in}$ into $\text{m}$ .

Explanation of Solution

The relation between $\text{in}$ and $\text{cm}$ is shown below.

$1\text{ in}=2.54\text{ cm}$

The conversion factors used to convert $\text{in}$ into $\text{cm}$ are shown below.

$\frac{1\text{ in}}{\text{2}\text{.54 cm}}\text{or}\frac{\text{2}\text{.54 cm}}{1\text{ in}}$

The relation between meter $\left(\text{m}\right)$ and centimeter $\left(\text{cm}\right)$ is shown below.

$1\text{ m}=100\text{ cm}$

The conversion factors used to convert $\text{cm}$ into $\text{m}$ are shown below.

$\frac{1\text{ m}}{100\text{ cm}}\text{or}\frac{100\text{ cm}}{1\text{ m}}$

$37.5\text{ in}$ is converted into $\text{m}$ using the conversion factor shown below.

$\begin{array}{c}37.5\text{ in}=37.5\text{ in}\times \frac{\text{2}\text{.54 cm}}{1\text{ in}}\times \frac{1\text{ m}}{100\text{ cm}}\\ =0.9525\text{ m}\\ =0.952\text{ m}\end{array}$

The given number, $37.5\text{ in}$ , has three significant figures. Therefore, the answer will be reported up to three significant figures. The number is rounded off by dropping the last digit, which is equal to $5$ , and the digit present on the left is even. Therefore, the last significant figure will remain unchanged. As a result, the final answer, after rounding off up to three significant figures, is $0.952\text{ m}$ .

.

Interpretation Introduction

Interpretation:

The conversion of $689\text{ lb}$ into $\text{kg}$ .

Explanation of Solution

The relation between $\text{lb}$ and $\text{g}$ is shown below.

$1\text{ lb}=\text{453}\text{.6 g}$

The conversion factors used to convert $\text{lb}$ into $\text{g}$ are shown below.

$\frac{1\text{ lb}}{\text{453}\text{.6 g}}\text{or}\frac{\text{453}\text{.6 g}}{1\text{ lb}}$

The relation between kilogram $\left(\text{kg}\right)$ into gram $\left(\text{g}\right)$ is shown below.

$1\text{ kg}=\text{1000 g}$

The conversion factors used to convert $\text{g}$ into $\text{kg}$ are shown below.

$\frac{1\text{ kg}}{1000\text{ g}}\text{or}\frac{1000\text{ g}}{1\text{ kg}}$

$689\text{ lb}$ is converted into $\text{kg}$ using the conversion factor shown below.

$\begin{array}{c}689\text{ lb}=689\text{ lb}\times \frac{\text{453}\text{.6 g}}{1\text{ lb}}\times \frac{1\text{ kg}}{1000\text{ g}}\\ =312.5\text{ kg}\\ =312\text{ kg}\end{array}$

The given number, $689\text{ lb}$ , has three significant figures. Therefore, the answer will be reported up to three significant figures. The number is rounded off by dropping the last digit, which is equal to $5$ , and the digit present on the left is even. Therefore, the last significant figure will remain unchanged. As a result, the final answer, after rounding off up to three significant figures, is $312\text{ kg}$ .

Interpretation Introduction

Interpretation:

The conversion of $\text{0}\text{.5 qt}$ into $\text{mL}$ .

Explanation of Solution

The relation between Quart $\left(\text{qt}\right)$ and Liter $\left(\text{L}\right)$ is shown below.

$1\text{ qt}=0.9464\text{ L}$

The conversion factors used to convert $\text{qt}$ into $\text{L}$ are shown below.

$\frac{1\text{ qt}}{0.9464\text{ L}}\text{or}\frac{0.9464\text{ L}}{1\text{ qt}}$

The relation between milliliter $\left(\text{mL}\right)$ and liter $\left(\text{L}\right)$ is shown below.

$1\text{ mL}={10}^{-3}\text{L}$

The conversion factors used to convert $\text{L}$ into $\text{mL}$ are shown below.

$\frac{1\text{ mL}}{{10}^{-3}\text{L}}\text{or}\frac{{10}^{-3}\text{L}}{1\text{ mL}}$

$\text{0}\text{.5 qt}$ is converted into $\text{mL}$ using the conversion factor shown below.

$\begin{array}{c}\text{0}\text{.5 qt}=\text{0}\text{.5 qt}\times \frac{0.9464\text{ L}}{1\text{ qt}}\times \frac{1\text{ mL}}{{10}^{-3}\text{L}}\\ =473.2\text{ mL}\\ =4.7\times {10}^2\text{mL}\\ =5\times {10}^2\text{mL}\end{array}$

The given number, $\text{0}\text{.5 qt}$ , has one significant figure. Therefore, the answer will be reported up to one significant figure. The answer, $473.2\text{ mL}$ , is larger than $1$ , which means that the exponent will be a positive integer. The decimal shifts towards the left up to $2$ places to obtain the simpler and shortest value.

Therefore, the answer becomes $4.7\times {10}^2\text{mL}$ .

After that, the number is rounded off by dropping the last digit, which is greater than $5$ . Therefore, the last significant figure will be increased by one unit. As a result, the final answer, after rounding off up to one significant figure, is $5\times {10}^2\text{ mL}$ .

Interpretation Introduction

Interpretation:

The conversion of $125\text{ cm}$ into $\text{ft}$ .

Explanation of Solution

The relation between $\text{in}$ and $\text{cm}$ is shown below.

$1\text{ in}=2.54\text{ cm}$

The conversion factors used to convert $\text{cm}$ into $\text{in}$ are shown below.

$\frac{1\text{ in}}{\text{2}\text{.54 cm}}\text{or}\frac{\text{2}\text{.54 cm}}{1\text{ in}}$

The relation between Foot $\left(\text{ft}\right)$ and inch $\left(\text{in}\right)$ is shown below.

$1\text{ ft}=12\text{ in}$

The conversion factors used to convert $\text{in}$ into $\text{ft}$ are shown below.

$\frac{1\text{ ft}}{12\text{ in}}\text{or}\frac{12\text{ in}}{1\text{ ft}}$

$125\text{ cm}$ is converted into $\text{ft}$ using the conversion factor shown below.

$\begin{array}{c}125\text{ cm}=125\text{ cm}\times \frac{1\text{ in}}{\text{2}\text{.54 cm}}\times \frac{1\text{ ft}}{12\text{ in}}\\ =4.10\text{ ft}\end{array}$

The given number, $125\text{ cm}$ , has three significant figures, which means that the answer will be reported up to three significant figures. Therefore, the final answer up to three significant figures is $4.10\text{ ft}$ .

Interpretation Introduction

Interpretation:

The conversion of $542\text{ mg}$ into $\text{lb}$ .

Explanation of Solution

The relation between gram $\left(\text{g}\right)$ into milligram $\left(\text{mg}\right)$ is shown below.

$1\text{ g}=\text{1000 mg}$

The conversion factors used to convert $\text{mg}$ into $\text{g}$ are shown below.

$\frac{1\text{ g}}{1000\text{ mg}}\text{or}\frac{1000\text{ mg}}{1\text{ g}}$

The relation between $\text{lb}$ and $\text{g}$ is shown below.

$1\text{ lb}=\text{453}\text{.6 g}$

The conversion factors used to convert $\text{g}$ into $\text{lb}$ are shown below.

$\frac{1\text{ lb}}{\text{453}\text{.6 g}}\text{or}\frac{\text{453}\text{.6 g}}{1\text{ lb}}$

$542\text{ mg}$ is converted into $\text{lb}$ using the conversion factor shown below.

$\begin{array}{c}542\text{ mg}=542\text{ mg}\times \frac{1\text{ g}}{1000\text{ mg}}\times \frac{1\text{ lb}}{\text{453}\text{.6 g}}\\ =0.00119\text{ lb}\\ =1.19\times {10}^{-3}\text{ lb}\end{array}$

The given number, $542\text{ mg}$ , has three significant figures. Therefore, the answer will be reported up to three significant figures. The answer, $0.00119\text{ lb}$ , is smaller than $1$ , which means that the exponent will be a negative integer. The decimal shifts towards the right up to $3$ places to obtain the simpler and shortest value.

Therefore, the final answer is $1.19\times {10}^{-3}\text{ lb}$ .

Interpretation Introduction

Interpretation:

The conversion of $\text{25 nL}$ into $\text{gal}$ .

Explanation of Solution

The relation between nano liter $\left(\text{nL}\right)$ and liter $\left(\text{L}\right)$ is shown below.

$1\text{ nL}={10}^{-9}\text{ L}$

The conversion factors used to convert $\text{nL}$ into $\text{L}$ are shown below.

$\frac{1\text{ nL}}{{\text{10}}^{-9}\text{ L}}\text{or}\frac{{\text{10}}^{-9}\text{ L}}{1\text{ nL}}$

The relation between liter $\left(\text{L}\right)$ and galloon $\left(\text{gal}\right)$ is shown below.

$1\text{ gal}=3.785\text{ L}$

The conversion factors used to convert $\text{L}$ into $\text{gal}$ are shown below.

$\frac{1\text{ gal}}{\text{3}\text{.785 L}}\text{or}\frac{\text{3}\text{.785 L}}{1\text{ gal}}$

The $\text{25 nL}$ is converted into $\text{gal}$ using conversion factor is shown below.

$\begin{array}{c}\text{25 nL}=\text{25 nL}\times \frac{{\text{10}}^{-9}\text{ L}}{1\text{ nL}}\times \frac{1\text{ gal}}{\text{3}\text{.785 L}}\\ =6.6\times {10}^{-9}\text{gal}\end{array}$

The given number, $\text{25 nL}$ , has two significant figures. Therefore, the answer will be reported up to two significant figures. Therefore, the final answer is $6.6\times {10}^{-9}\text{gal}$ .