Chapter 1, Problem 20QP

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

(a) 175.5km to m

(b) 25.7 g to mg

(c) 10.516 L to dL

(d) 5.2 cm to m

(e) 0.000000450 m to nm

(f) 12 into cm

(g) 25.6 lb to g

(h) 4.005 qt to L

(i) 934 cm to in

(j) 155 g to 1b

(k) 22.4 L to qt

Interpretation Introduction

Interpretation:

The conversion of $\text{75}\text{.5 km}$ into $\text{m}$ .

Explanation of Solution

The relation between kilometer $\left(\text{km}\right)$ and meter $\left(\text{m}\right)$ is shown below:

$1\text{ km}={10}^3\text{ m}$

The conversion factors used to convert $\text{mm}$ into $\text{m}$ is shown below:

$\frac{1\text{ km}}{{\text{10}}^3\text{ m}}\text{or}\frac{{\text{10}}^3\text{ m}}{1\text{ km}}$

$\text{75}\text{.5 km}$ is converted into $\text{m}$ using the conversion factor as follows:

$\begin{array}{c}\text{75}\text{.5 km}=\text{75}\text{.5 km}\times \frac{{10}^3\text{ m}}{1\text{ km}}\\ =75500\text{ m}\\ =7.55\times {10}^4\text{ m}\end{array}$

The given number $\text{75}\text{.5 km}$ has three significant figures. Therefore, the answer will be reported up to three significant figures. The answer $75500\text{ m}$ is larger than $1$ that means the exponent will be a positive integer. The decimal shifts to left up to $4$ places to obtain simpler and the shortest value. Therefore, the final answer is $7.55\times {10}^4\text{ m}$ .

Interpretation Introduction

Interpretation:

The conversion of $\text{25}\text{.7 g}$ into $\text{mg}$ .

Explanation of Solution

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

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

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

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

$\text{25}\text{.7 g}$ is converted into $\text{mg}$ using conversion factor that is shown below.

$\begin{array}{c}\text{25}\text{.7 g}=\text{25}\text{.7 g}\times \frac{1000\text{ mg}}{1\text{ g}}\\ =25700\text{ mg}\\ =2.57\times {10}^4\text{mg}\end{array}$

The given number $\text{25}\text{.7 g}$ has three significant figures. Therefore, the answer will be reported up to three significant figures. The answer $25700\text{ mg}$ is larger than $1$ that means the exponent will be a positive integer. The decimal shifts to left up to $4$ places to obtain simpler and the shortest value.

Therefore, the final answer up to three significant figures is $2.57\times {10}^4\text{mg}$ .

Interpretation Introduction

Interpretation:

The conversion of $\text{0}\text{.516 L}$ into $\text{dL}$ .

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 $\text{L}$ into $\text{dL}$ is shown below:

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

$\text{0}\text{.516 L}$ is converted into $\text{dL}$ using conversion factor that is shown below.

$\begin{array}{c}\text{0}\text{.516 L}=\text{0}\text{.516 L}\times \frac{10\text{ dL}}{1\text{ L}}\\ =5.16\text{ dL}\end{array}$

The given number $\text{0}\text{.516 L}$ has three significant figures that means the answer will be reported up to three significant figures. Therefore, the final answer up to three significant figures is $\text{5}\text{.16 dL}$ .

Interpretation Introduction

Interpretation:

The conversion of $\text{5}\text{.2 cm}$ into $\text{m}$ .

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 $\text{m}$ is shown below:

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

$\text{5}\text{.2 cm}$ is converted into $\text{m}$ using the conversion factor that is shown below.

$\begin{array}{c}\text{5}\text{.2 cm}=\text{5}\text{.2 cm}\times \frac{\text{1m}}{100\text{ cm}}\\ =0.052\text{ m}\end{array}$

The given number $\text{5}\text{.2 cm}$ has two significant figures which means that the answer will be reported up to two significant figures. Therefore, the final answer up to two significant figures is $0.052\text{ m}$ .

Interpretation Introduction

Interpretation:

The conversion of $\text{0}\text{.000000450 m}$ into $\text{nm}$ .

Explanation of Solution

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 $\text{nm}$ is shown below:

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

$\text{0}\text{.000000450 m}$ is converted into $\text{m}$ using the conversion factor as follows:

$\begin{array}{c}\text{0}\text{.000000450 m}=\text{0}\text{.000000450 m}\times \frac{1\text{ nm}}{{10}^{-9}\text{m}}\\ =\text{4}\text{.50}\times {10}^9\times {10}^{-7}\text{ m}\\ =4.50\times {10}^{\left(9+\left(-7\right)\right)}\text{ m}\\ =4.50\times {10}^2\text{ m}\end{array}$

The given number $\text{0}\text{.000000450 m}$ has three significant figures. Therefore, the answer will be reported up to three significant figures. $\text{0}\text{.000000450 m}$ is smaller than $1$ which means that the exponent will be a negative integer. The decimal shifts to right up to $7$ places to obtain simpler and the shortest value. In multiplication method, addition of exponents takes place.

In multiplication method, addition of exponents takes place in order to get the final answer. Therefore, the final answer up to three significant figures is $4.50\times {10}^2\text{ m}$ .

Interpretation Introduction

Interpretation:

The conversion of $\text{12 in}$ into $\text{cm}$ .

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 $\text{cm}$ is shown below:

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

$\text{12 in}$ is converted into $\text{cm}$ using the conversion factor as follows:

$\begin{array}{c}\text{12 in}=\text{12 in}\times \frac{\text{2}\text{.54 cm}}{1\text{ in}}\\ =30.48\text{ cm}\\ =3.0\times {10}^1\text{ cm}\end{array}$

The given number $\text{12 in}$ has two significant figures. The answer will be reported up to two significant figures. The answer $30.48\text{ cm}$ is larger than $1$ which means that the exponent will be a positive integer. The decimal shifts to left up to $1$ place to obtain simpler and the shortest value.

Therefore, the final answer is $\text{3}\text{.0}\times {\text{10}}^1\text{ cm}$ .

Interpretation Introduction

Interpretation:

The conversion of $\text{25}\text{.6 lb}$ into $\text{g}$ .

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 $\text{g}$ is shown below:

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

$\text{25}\text{.6 lb}$ is converted into $\text{g}$ using the conversion factor as follows:

$\begin{array}{c}\text{25}\text{.6 lb}=\text{25}\text{.6 lb}\times \frac{\text{453}\text{.6 g}}{1\text{ lb}}\\ =11612.16\text{ g}\\ =1.16\times {10}^4\text{ g}\end{array}$

The given number $\text{25}\text{.6 lb}$ has three significant figures. Therefore, the answer will be reported up to three significant figures. The answer $11612.16$ is larger than $1$ which means that the exponent will be a positive integer. The decimal shifts to left up to $4$ places to obtain simpler and the shortest value.

Therefore, the final answer up to three significant figures is $1.16\times {10}^4\text{ g}$ .

Interpretation Introduction

Interpretation:

The conversion of $\text{4}\text{.005 qt}$ into $\text{L}$ .

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 $\text{L}$ is shown below:

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

$\text{4}\text{.005 qt}$ is converted into $\text{L}$ using the conversion factor as follows:

$\begin{array}{c}\text{4}\text{.005 qt}=\text{4}\text{.005 qt}\times \frac{0.9464\text{ L}}{1\text{ qt}}\\ =3.790\text{ L}\end{array}$

The given number $\text{4}\text{.005 qt}$ has four significant figures which means that the answer will be reported up to four significant figures. Therefore, the final answer is $\text{3}\text{.790 L}$ .

Interpretation Introduction

Interpretation:

The conversion of $\text{934 cm}$ into $\text{in}$ .

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 $\text{in}$ is shown below:

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

$\text{934 cm}$ is converted into $\text{cm}$ using the conversion factor as follows:

$\begin{array}{c}\text{934 cm}=\text{934 cm}\times \frac{1\text{ in}}{\text{2}\text{.54 cm}}\\ =367.7\text{ in}\\ =368\text{ in}\end{array}$

The given number $\text{934 cm}$ has three significant figures which means that the answer will be reported up to three significant figures. The number is rounded off by dropping the last number which is more than $5$ . Therefore, the last significant figure will be increased by one unit. As a result, the final answer after rounding off up to three significant figures is $\text{368 in}$ .

Interpretation Introduction

Interpretation:

The conversion of $\text{155 g}$ into $\text{lb}$ .

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 $\text{g}$ into $\text{lb}$ is shown below:

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

$\text{155 g}$ is converted into $\text{g}$ using the conversion factor as follows:

$\begin{array}{c}\text{155 g}=\text{155 g}\times \frac{1\text{ lb}}{\text{453}\text{.6 g}}\\ =0.34171\text{ lb}\\ =0.3417\text{ lb}\\ =0.342\text{ lb}\end{array}$

The given number $\text{155 g}$ has three significant figures which means that the answer will be reported up to three significant figures due to which the last digit is removed. After that, the number is rounded off by dropping the last number which is more than $5$ . Therefore, the last significant figure will be increased by one unit. As a result, the final answer after rounding off up to three significant figures is $0.342\text{ lb}$ .

Interpretation Introduction

Interpretation:

The conversion of $22.4\text{ L}$ into $\text{qt}$ .

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 $\text{L}$ into $\text{qt}$ is shown below:

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

$22.4\text{ L}$ is converted into $\text{qt}$ using the conversion factor as follows:

$\begin{array}{c}22.4\text{ L}=22.4\text{ L}\times \frac{1\text{ qt}}{0.9464\text{ L}}\\ =23.66\text{ qt}\\ =23.7\text{ qt}\end{array}$

The given number $22.4\text{ L}$ has three significant figures which means that the answer will be reported up to three significant figures. The number is rounded off by dropping the last number which is more than $5$ . Therefore, the last significant figure will be increased by one unit. As a result, the final answer after rounding off up to three significant figures is $23.7\text{ qt}$ .