Question

Convert each quantity to the indicated units. \(\begin{array}{ll}{\text { a. } 3.01 \mathrm{g} \rightarrow \mathrm{cg}} & {\text { d. } 0.2 \mathrm{L} \rightarrow \mathrm{dm}^{3}} \\ {\text { b. } 6200 \mathrm{m} \rightarrow \mathrm{km}} & {\text { e. } 0.13 \mathrm{cal} / \mathrm{g} \rightarrow \mathrm{kcal} / \mathrm{g}} \\ {\text { c. } 6.24 \times 10^{-7} \mathrm{g} \rightarrow \mu \mathrm{g}} & {\text { f. } 3.21 \mathrm{mL} \rightarrow \mathrm{L}}\end{array}\)

Short answer

a. \(301\,\text{cg}\) b. \(6.2\,\text{km}\) c. \(624\,\mu\text{g}\) d. \(0.2\,\text{dm}^3\) e. \(1.3\, \times\, 10^{-4}\,\frac{\text{kcal}}{\text{g}}\) f. \(0.00321\,\text{L}\)

Step-by-step solution

a. 3.01 g to cg

Multiply by the conversion factor: \(3.01\,\text{g}\, \times\, 100\,\frac{\text{cg}}{\text{g}} = 301\,\text{cg}\).

b. 6200 m to km

Multiply by the conversion factor: \(6200\,\text{m}\, \times\, 0.001\,\frac{\text{km}}{\text{m}} = 6.2\,\text{km}\).

c. \(6.24\, \times\, 10^{-7}\) g to µg

Multiply by the conversion factor: \(6.24\, \times\, 10^{-7}\,\text{g}\, \times\, 10^9\,\frac{\mu\text{g}}{\text{g}} = 624\,\mu\text{g}\).

d. 0.2 L to dm³

Since 1 L = 1 dm³, we don't need to apply any conversion factor: \(0.2\,\text{L} = 0.2\,\text{dm}^3\).

e. 0.13 cal/g to kcal/g

Multiply by the conversion factor: \(0.13\,\frac{\text{cal}}{\text{g}}\, \times\, 0.001\,\frac{\text{kcal}}{\text{cal}} = 1.3\, \times\, 10^{-4}\,\frac{\text{kcal}}{\text{g}}\).

f. 3.21 mL to L

Multiply by the conversion factor: \(3.21\,\text{mL}\, \times\, 0.001\,\frac{\text{L}}{\text{mL}} = 0.00321\,\text{L}\).

Key concepts

Metric System

When studying chemistry, the metric system acts as the foundation of measurement. It is an internationally agreed decimal system of measurement, used almost universally for scientific purposes. The metric system uses units such as the meter for length, the gram for mass, and the liter for volume.

For students grappling with the metric system, the key is to understand the base units and the prefixes that modify them. Prefixes like milli- (m), centi- (c), and kilo- (k) represent fractions or multiples of the base units. For instance, a kilometer (km) is 1,000 meters (m), and a milligram (mg) is one-thousandth of a gram (g).

Let's apply these concepts to an exercise from the textbook: To convert 6200 meters (m) to kilometers (km), we use the prefix indicating 'kilo' means thousand, so we multiply 6200 meters by the conversion factor 0.001 (or 1/1000) to get 6.2 km, simplifying the process of conversion immensely.

Scientific Notation

Scientific notation is a way of expressing numbers that are too large or too small to be conveniently written in decimal form. It's particularly useful in chemistry, which often involves working with extremely small or large numbers. A number in scientific notation is expressed as the product of a number between 1 and 10 and a power of 10. For example, the number 6200 can be written as 6.2 x 10³.

Understanding this notation eases the handling of cumbersome measurements, like the conversion of 6.24 x 10^-7 grams to micrograms (µg). Here, remember that 'micro-' signifies one-millionth. Therefore, by multiplication with 10⁹ (the inverse of one-millionth when converting to micrograms), we get 624 µg. Importantly, scientific notation maintains accuracy in expressing and converting between these vast differences in magnitude.

Measurement in Chemistry

Measurement in chemistry is vital for precision and accuracy in experiments. It often involves using the metric system and may require conversions between different units. Precision in measurement is what allows chemists to replicate experiments and validate results.

In our textbook exercise, careful unit conversion is exemplified when 0.13 calories per gram (cal/g) is converted to kilocalories per gram (kcal/g). As calories and kilocalories are units of energy, understanding that 1 kilocalorie equals 1,000 calories allows us to use the conversion factor of 0.001 to convert 0.13 cal/g to 0.00013 kcal/g or 1.3 x 10^-4 kcal/g in scientific notation. This practice emphasizes the importance of both accurate measurement and unit conversion, which are foundational to quantitative chemistry.