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Carry out these conversions: (a) 242 lb to milligrams, (b) \(68.3 \mathrm{~cm}^{3}\) to cubic meters.

Short Answer

Expert verified
(a) 242 lb is equivalent to 109669093.34 mg. (b) 68.3 cm^3 is equivalent to 0.0000683 m^3.

Step by step solution

01

Convert pounds (lb) to milligrams (mg)

Using the conversion factor of 1 pound (lb) being equivalent to 453592.37 milligrams (mg), you will multiply 242 lb by 453592.37 mg/lb. Thus: \(242 \text{ lb} \times 453592.37 \text{ mg/lb} = 109669093.34 \text{ mg}\)
02

Convert cubic centimeters (cm^3) to cubic meters (m^3)

Using the conversion factor of 1 cubic meter (m^3) being equivalent to \(1,000,000 \text{ cubic centimeters (cm^3)}\), you will divide 68.3 cm^3 by 1,000,000 cm^3/m^3. Thus: \(68.3 \text{ cm}^3 ÷ 1,000,000 \text{ cm}^3/\text{m}^3 = 0.0000683 \text{ m}^3\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pound to Milligram Conversion
Converting units from pounds to milligrams involves understanding the relationship between these two measures of mass. In the metric system, conversion factors are essential building blocks that help bridge different units.

When converting from pounds to milligrams, you need to recognize the large difference in size and scale between these units. A pound (lb) is a unit of weight commonly used in the United States and is part of the Imperial system. A milligram (mg) is a much smaller unit of weight and is a part of the metric system.
  • 1 pound is equal to 453,592.37 milligrams.
  • To convert from pounds to milligrams, you multiply the number of pounds by 453,592.37.
For instance, if you have 242 pounds and want to convert it to milligrams, the calculation is as follows: \( 242 \text{ lb} \times 453,592.37 \text{ mg/lb} = 109,669,093.34 \text{ mg} \)By multiplying, you transition from a larger to a smaller unit, effectively scaling your measurements to milligrams.
Cubic Centimeters to Cubic Meters Conversion
When converting from cubic centimeters to cubic meters, you're working with a conversion that involves volume rather than mass. It can be puzzling but, thankfully, the metric system uses base units that simplify conversions.

A cubic centimeter (cm³) is 1,000,000 times smaller than a cubic meter (m³), as the liter-based metric system is devised around powers of ten. This power of ten relationship makes metric conversions straightforward.
  • 1 cubic meter equals 1,000,000 cubic centimeters.
  • To convert cubic centimeters to cubic meters, divide the volume in cubic centimeters by 1,000,000.
Take for example 68.3 cubic centimeters that you wish to convert into cubic meters:\( 68.3 \text{ cm}^3 \div 1,000,000 = 0.0000683 \text{ m}^3 \)This division simplifies your volume from a smaller to a larger unit, allowing you to express it more suitably in cubic meters.
Metric System Conversions
The metric system is a decimal-based system of measurement used worldwide as the standard for scientific and technical purposes. It provides a unified method for conversions based on powers of ten. This simplicity makes it highly user-friendly for students and professionals alike.

Remember these core aspects when handling metric conversions:
  • Units are interrelated by factors of ten.
  • Prefixes indicate size: milli- (1/1,000), centi- (1/100), and kilo- (1,000).
  • Simple arithmetic operations like multiplication or division allow fluid conversions between units.
Whether you are converting weight, length, or volume, the fundamental principle remains the same: utilize the respective conversion factor and apply either multiplication or division. This method not only ensures accuracy but also facilitates an understanding of the scale and scope of measurements in the real world. Embracing metric system conversions can significantly enhance your ability to work across different scientific and practical settings with ease.

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Most popular questions from this chapter

Magnesium (Mg) is a valuable metal used in alloys, in batteries, and in chemical synthesis. It is obtained mostly from seawater, which contains about \(1.3 \mathrm{~g}\) of Mg for every kilogram of seawater. Calculate the volume of seawater (in liters) needed to extract \(8.0 \times 10^{4}\) tons of \(\mathrm{Mg}\), which is roughly the annual production in the United States. (Density of seawater = \(1.03 \mathrm{~g} / \mathrm{mL} .)\)

Give an example of an intensive property and an example of an extensive property.

A silver \((\mathrm{Ag})\) object weighing \(194.3 \mathrm{~g}\) is placed in a graduated cylinder containing \(242.0 \mathrm{~mL}\) of water. The volume of water now reads \(260.5 \mathrm{~mL}\). From these data calculate the density of silver.

Carry out these operations as if they were calculations of experimental results, and express each answer in the correct units and with the correct number of significant figures: (a) \(7.310 \mathrm{~km} \div 5.70 \mathrm{~km}\) (b) \(\left(3.26 \times 10^{-3} \mathrm{mg}\right)-\left(7.88 \times 10^{-5} \mathrm{mg}\right)\) (c) \(\left(4.02 \times 10^{6} \mathrm{dm}\right)+\left(7.74 \times 10^{7} \mathrm{dm}\right)\) (d) \((7.8 \mathrm{~m}-0.34 \mathrm{~m}) /(1.15 \mathrm{~s}+0.82 \mathrm{~s})\)

(a) Convert the following temperatures to kelvin: (i) \(113^{\circ} \mathrm{C},\) the melting point of sulfur, (ii) \(37^{\circ} \mathrm{C}\), the normal body temperature, (iii) \(357^{\circ} \mathrm{C},\) the boiling point of mercury. (b) Convert the following temperatures to degrees Celsius: (i) \(77 \mathrm{~K}\), the boiling point of liquid nitrogen, (ii) \(4.2 \mathrm{~K},\) the boiling point of liquid helium, (iii) \(601 \mathrm{~K}\), the melting point of lead.

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