Problem 48 - Chemistry

Question

Which quantity in each of the following pairs is larger? (a) \(5.63 \times 10^{6} \mathrm{~cm}\) or \(6.02 \times 10^{1} \mathrm{~km}\) (b) \(46 \mu\) s or \(3.2 \times 10^{-2} \mathrm{~ms}\) (c) \(200,098 \mathrm{~g}\) or \(17 \times 10^{1} \mathrm{~kg}\)

Step-by-Step Solution

Answer

For (a), 6.02 x 10^1 km; for (b), 3.2 x 10^-2 ms; for (c), 200,098 g.

Step 1: Convert kilometers to centimeters

Convert the given quantity in kilometers to centimeters. Use the conversion factor: \(1 \text{ km} = 10^5 \text{ cm}\).For \(6.02 \times 10^1 \text{ km} = 6.02 \times 10^6 \text{ cm}\).

Step 2: Compare quantities in centimeters

Now compare the two quantities: \(5.63 \times 10^6 \text{ cm}\) and \(6.02 \times 10^6 \text{ cm}\).

Step 3: Conclusion for (a)

Since \(6.02 \times 10^6 \text{ cm} > 5.63 \times 10^6 \text{ cm}\), the larger quantity is \(6.02 \times 10^1 \text{ km}\).

Step 3: Convert microseconds to milliseconds

Convert the microseconds into milliseconds. Use the conversion factor: \(1 \mu\text{s} = 10^{-3} \text{ ms}\).For \(46 \underline{\phantom{xxx}} \mu\text{s} = 0.046 \text{ ms}\).

Step 4: Compare quantities in milliseconds

Now compare the two quantities: \(0.046 \text{ ms}\) and \(3.2 \times 10^{-2} \text{ ms}\).

Step 6: Conclusion for (b)

Since \(0.032 \text{ ms} = 3.2 \times 10^{-2} \text{ ms}\), the larger quantity is \(3.2 \times 10^{-2} \text{ ms}\).

Step 5: Convert kilograms to grams

Convert the kilograms into grams. Use the conversion factor: \(1 \text{ kg} = 10^3 \text{ g}\).For \(17 \times 10^{1} \text{ kg} = 1,700 \text{ g}\).

Step 6: Compare quantities in grams

Now compare the two quantities: \(200,098 \text{ g}\) and \(1,700 \text{ g}\).

Step 9: Conclusion for (c)

Since \(200,098 \text{ g} > 1,700 \text{ g}\), the larger quantity is \(200,098 \text{ g}\).

Key Concepts

Centimeters to Kilometers Conversion

Understanding how to convert centimeters to kilometers is a handy skill when tackling different scientific and mathematical problems. Essentially, you're changing a small unit of length (centimeters) into a much larger one (kilometers).
To convert centimeters into kilometers, remember the conversion factor:

• 1 kilometer = 100,000 centimeters

This means to change centimeters to kilometers, you divide the number of centimeters by 100,000. For instance, if you have 5,000,000 centimeters, it becomes:\[5,000,000 \text{ cm} \div 100,000 = 50 \text{ km}\]This way, you'll precisely find how many kilometers those centimeters add up to. Converting accurately helps in comparing measurements effectively, especially when dealing with large quantities as shown in the solution above.

Microseconds to Milliseconds Conversion

When you're working with time measurements, converting between units like microseconds and milliseconds can be essential for precision.
Let's begin with understanding the size of these units:

• 1 microsecond is a millionth of a second
• 1 millisecond is a thousandth of a second

As a result, to convert microseconds to milliseconds, one needs to divide the value by 1,000. The conversion factor here is:

• 1 microsecond = 0.001 milliseconds

For instance, if you have 46 microseconds, converting it to milliseconds makes it:\[46 \underline{\phantom{xxx}} \mu\text{s} = 0.046 \text{ ms}\]This step is crucial for ensuring comparisons between such time measurements are accurate, particularly in time-sensitive fields like computing or physics.

Kilograms to Grams Conversion

Weight measurements often require conversions between kilograms and grams, which is straightforward but important for precision. Grams are the unit used for smaller weights, while kilograms are suitable for larger weights.
The relationship between them is defined by the conversion factor:

• 1 kilogram = 1,000 grams

Therefore, to convert from kilograms to grams, you multiply the number of kilograms by 1,000. For example, to convert 17 kilograms to grams:\[17 \times 1,000 = 17,000 \text{ g}\]Understanding this conversion not only helps in calculations but also ensures comparison of weights are done correctly, as demonstrated in the solution above. This practical knowledge is valuable in many scientific and day-to-day applications.