Chapter 1, Problem 1.136QP

Convert:

1. a 7.19 μg of cyanocobalamin (vitamin B₁₂) to milligrams
2. b 104 pm, the radius of a sulfur atom, to angstroms
3. c 0.010 mm, the diameter of a typical blood capillary, to centimeters
4. d 0.0605 kPa (the pascal, Pa, is a unit of pressure) to centipascals

Interpretation Introduction

Interpretation:

The mass in microgram has to be converted to milligrams.

Concept Introduction:

Conversion-factor method is the one which can be used to convert one metric unit into another.  With this, any unit can be converted to another unit by means of ratio.  The ratio that is used to convert unit is known as conversion-factor.  For example, one kilogram can be converted into gram by multiplying with 1000.

Answer to Problem 1.136QP

The given mass in microgram is converted to milligrams as $\text{7}\text{.19 }\text{×}{\text{10}}^{\text{-3}}\text{mg}$.

Explanation of Solution

Given mass in problem statement is $\text{7}\text{.19 μg}$.

Since, $\text{1}\text{μg}\text{=}{\text{10}}^{\text{-6}}\text{g}$ and $\text{1}\text{mg}\text{=}{\text{10}}^{\text{-3}}\text{g}$, we can use this to convert the above unit in micrograms,

$\text{7}\text{.19 μg × }\frac{{\text{10}}^{\text{-6}}\text{ g}}{\text{1 μg}}\text{ × }\frac{\text{1 mg}}{{\text{10}}^{\text{-3}}\text{ g}}\text{=}\text{7}\text{.19 }\text{×}{\text{10}}^{\text{-3}}\text{mg}$

The given mass was converted into milligrams as shown above.

Interpretation Introduction

Interpretation:

The distance in picometer has to be converted to angstrom.

Concept Introduction:

Conversion-factor method is the one which can be used to convert one metric unit into another.  With this, any unit can be converted to another unit by means of ratio.  The ratio that is used to convert unit is known as conversion-factor.  For example, one kilogram can be converted into gram by multiplying with 1000.

Answer to Problem 1.136QP

The given distance in picometer is converted to angstrom as $\text{1}\text{.04}\text{Å}$.

Explanation of Solution

Given distance in problem statement is $\text{104 pm}$.

Since, $\text{1}\text{pm}\text{=}{\text{10}}^{\text{-12}}\text{m}$ and $\text{1}\text{Å}\text{=}{\text{10}}^{\text{-10}}\text{m}$, we can use this to convert the above unit in angstrom,

$\text{104 pm × }\frac{{\text{10}}^{\text{-12}}\text{ m}}{\text{1 pm}}\text{ × }\frac{\text{1 Å}}{{\text{10}}^{\text{-10}}\text{ m}}\text{=}\text{1}\text{.04Å}$

The given distance was converted into angstrom as shown above.

Interpretation Introduction

Interpretation:

The length in millimeter has to be converted to centimeter.

Concept Introduction:

Conversion-factor method is the one which can be used to convert one metric unit into another.  With this, any unit can be converted to another unit by means of ratio.  The ratio that is used to convert unit is known as conversion-factor.  For example, one kilogram can be converted into gram by multiplying with 1000.

Answer to Problem 1.136QP

The given millimeter is converted to centimeter as $\text{1}\text{.0 }\times {\text{ 10}}^{\text{-3}}\text{ cm}$.

Explanation of Solution

Given length in problem statement is $\text{0}\text{.010 mm}$.

Since, $\text{1}\text{mm}\text{=}{\text{10}}^{\text{-3}}\text{m}$ and $\text{1 cm}\text{=}{\text{10}}^{-2}\text{m}$, we can use this to convert the above unit in centimeter,

$\text{0}\text{.010 mm × }\frac{{\text{10}}^{\text{-3}}\text{ m}}{\text{1 mm}}\text{ × }\frac{\text{1 cm}}{{\text{10}}^{\text{-2}}\text{ m}}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-3}}\text{cm}$

The given length was converted into centimeter as shown above.

Interpretation Introduction

Interpretation:

The pressure in kilopascal has to be converted to centipascal.

Concept Introduction:

Conversion-factor method is the one which can be used to convert one metric unit into another.  With this, any unit can be converted to another unit by means of ratio.  The ratio that is used to convert unit is known as conversion-factor.  For example, one kilogram can be converted into gram by multiplying with 1000.

Answer to Problem 1.136QP

The given kilopascal is converted to centipascal as $\text{6}{\text{.05 × 10}}^3\text{ cPa}$.

Explanation of Solution

Given pressure in problem statement is $\text{0}\text{.0605 kPa}$.

Since, $\text{1}\text{kPa}\text{=}{\text{10}}^{\text{3}}\text{Pa}$ and $\text{1}\text{cPa}\text{=}{\text{10}}^{\text{-2}}\text{Pa}$, we can use this to convert the above unit in kilojoules,

$\text{0}\text{.0605 kPa × }\frac{{\text{10}}^{\text{3}}\text{ Pa}}{\text{1 kPa}}\text{ × }\frac{\text{1 cPa}}{{\text{10}}^{\text{-2}}\text{ Pa}}\text{=}6.05{\text{ × 10}}^3\text{ cPa}$

The given pressure was converted into centipascals as shown above.