Problem 46 - Chemistry: The Central Science

Question

Carry out the following conversions: (a) \(0.105\) in. to \(\mathrm{mm}\), (b) \(0.650\) qt to \(\mathrm{mL}\), (c) \(8.75 \mu \mathrm{m} / \mathrm{s}\) to \(\mathrm{km} / \mathrm{hr}\), (d) \(1.955 \mathrm{~m}^{3}\) to \(\mathrm{yd}^{3}\), (e) \(\$ 3.99 / \mathrm{lb}\) to dollars per \(\mathrm{kg}\), (f) \(8.75 \mathrm{lb} / \mathrm{ft}^{3}\) to \(\mathrm{g} / \mathrm{mL}\).

Step-by-Step Solution

Answer

(a) \(2.667~\mathrm{mm}\), (b) \(615.329~\mathrm{mL}\), (c) \(0.0000247~\frac{\mathrm{km}}{\mathrm{hr}}\), (d) \(2.555~\mathrm{yd}^{3}\), (e) \(\$~8.80~\frac{1}{\mathrm{kg}}\), (f) \(0.0014~\frac{\mathrm{g}}{\mathrm{mL}}\)

Step 1: (a) Conversion from inches to millimeters

To convert inches to millimeters, we need the conversion factor that relates the two units: \( 1~\text{in.} = 25.4~\text{mm}\). We can use this conversion factor to convert \(0.105~\text{in.}\) to millimeters. \[0.105~\mathrm{in.} \times \frac{25.4~\mathrm{mm}}{1~\mathrm{in.}} = 2.667~\mathrm{mm}\]

Step 2: (b) Conversion from quarts to milliliters

To convert quarts to milliliters, we need the conversion factor: \(1~\text{qt} = 946.353~\text{mL}\). We can use this conversion factor to convert \(0.650~\text{qt}\) to milliliters. \[0.650~\mathrm{qt} \times \frac{946.353~\mathrm{mL}}{1~\mathrm{qt}} \approx 615.329~\mathrm{mL}\]

Step 3: (c) Conversion from micrometers per second to kilometers per hour

To convert micrometers per second to kilometers per hour, we need two conversion factors: \(1~\text{µm} = 10^{-6}~\text{km}\) and \(1~\text{s} = \frac{1}{3600}~\text{hr}\). We can use these conversion factors to convert \(8.75~\text{µm/s}\) to kilometers per hour. \[8.75~\frac{\mu \mathrm{m}}{\mathrm{s}} \times \frac{10^{-6}~\mathrm{km}}{1~\mu \mathrm{m}} \times \frac{1~\mathrm{hr}}{3600~\mathrm{s}} = 0.0000247~\frac{\mathrm{km}}{\mathrm{hr}}\]

Step 4: (d) Conversion from cubic meters to cubic yards

To convert cubic meters to cubic yards, we need the conversion factor: \(1~\text{m}^3 = 1.30795~\text{yd}^3\). We can use this conversion factor to convert \(1.955~\text{m}^3\) to cubic yards. \[1.955~\mathrm{m}^{3} \times \frac{1.30795~\mathrm{yd}^{3}}{1~\mathrm{m}^{3}} \approx 2.555~\mathrm{yd}^{3}\]

Step 5: (e) Conversion from dollars per pound to dollars per kilogram

To convert dollars per pound to dollars per kilogram, we need the conversion factor: \(1~\text{lb} = 0.453592~\text{kg}\). We can use this conversion factor to convert \(\$3.99~\text{lb}\) to dollars per kilogram. \[\$~3.99~\frac{1}{\mathrm{lb}} \times \frac{1~\mathrm{kg}}{0.453592~\mathrm{lb}} \approx \$~8.80~\frac{1}{\mathrm{kg}}\]

Step 6: (f) Conversion from pounds per cubic feet to grams per milliliter

To convert pounds per cubic feet to grams per milliliter, we need two conversion factors: \(1~\text{lb} = 453.592~\text{g}\) and \(1~\text{ft}^3 = 28316.8466~\text{mL}\). We can use these conversion factors to convert \(8.75~\text{lb/ft}^3\) to grams per milliliter. \[8.75~\frac{\mathrm{lb}}{\mathrm{ft}^{3}} \times \frac{453.592~\mathrm{g}}{1~\mathrm{lb}} \times \frac{1~\mathrm{ft}^{3}}{28316.8466~\mathrm{mL}} \approx 0.0014~\frac{\mathrm{g}}{\mathrm{mL}}\]

Key Concepts

Conversion Factors

Conversion factors are essential tools for converting units from one system to another. They serve as multipliers that you use to transform a measurement in one unit into its equivalent in another unit. Generally, a conversion factor is a simple ratio or fraction that equals one. This feature helps in retaining the physical quantity unchanged even after conversion. For example, to convert inches to millimeters, the factor is \(1~\text{in} = 25.4~\text{mm}\). This ratio allows you to multiply any measurement in inches by \(25.4\) to get its value in millimeters. This process means that if you had 0.105 inches, multiplying by the conversion factor yields \(0.105 \times 25.4 = 2.667\) mm.

Using conversion factors correctly requires a good understanding of the relationship between the two units, which is often based on a defined or measured standard. The choice of the correct conversion factor is critical for ensuring the accuracy of the conversion.

Measurement Systems

Measurement systems provide a standardized way to express and quantify physical quantities. Around the world, the main systems in use are the Imperial system and the Metric system. Each system has its unique set of units. The Imperial system uses units like inches, feet, and pounds, while the Metric system uses units such as meters, liters, and grams.

In everyday life, you might encounter both systems, depending on where you live or work. For example, the United States predominantly uses the Imperial system, while most other countries use the Metric system. Understanding these systems is crucial in fields like science, engineering, and cooking, as you often need to convert measurements from one system to the other to ensure precision and consistency.

• Imperial System: Used mainly in the US and a few other countries. Example units include inches (in), feet (ft), and pounds (lb).
• Metric System: Used internationally. It's uniform and decimal-based, using units such as meters (m), liters (L), and grams (g).

Metric System

The Metric system is widely used globally and is the official system of measurement for most countries. It is a decimal-based system of measurement that uses a set of base units, such as the meter for length, the liter for volume, and the gram for mass. Because of its decimal nature, it is easy to scale units up or down by powers of ten. For instance, 1 kilometer (km) is 1000 meters (m), and 1 millimeter (mm) is 0.001 meters.

The simplicity and uniformity of the Metric system make it particularly useful in scientific and technical fields. This is because calculations are often straightforward, given that they can leverage base 10 multiplication and division, which aligns with the way the Metric system is structured.

• Length: meter (m) is the base unit. Multiples include kilometer (km) and millimeter (mm).
• Volume: liter (L) is the base unit. It easily converts to milliliters (mL).
• Mass: gram (g) is the base unit, allowing for conversions to kilograms (kg) or milligrams (mg).

Dimensional Analysis

Dimensional analysis is a powerful method used to convert units or to solve problems involving physical quantities. It is especially useful because it involves the use of conversion factors and ensures that your final answer has the correct dimensions or units.

By treating dimensions as algebraic quantities, dimensional analysis helps in checking the consistency of units within equations. For example, when converting micrometers per second to kilometers per hour, you use dimensional analysis to systematically multiply by the appropriate conversion factors: \(\frac{10^{-6}~\text{km}}{1~\mu\text{m}}\) and \(\frac{3600~\text{hr}}{1~\text{s}}\). This guarantees that units properly cancel out where necessary, leaving you with the desired units of kilometers per hour in the final result.

• Identify the units you have and the units you need.
• Set up conversion factors as fractions that equal one.
• Multiply through, canceling units, to arrive at the desired unit.

Dimensional analysis is not only useful for conversions but also helps in confirming the validity of derived equations in scientific contexts.