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Ch.1 - Matter, Measurement & Problem Solving

Tro - Chemistry: A Molecular Approach 5th Edition

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Tro 5th Edition

Ch.1 - Matter, Measurement & Problem Solving

Problema 127

Capitolo 1, Problema 127

The diameter of a hydrogen atom is 212 pm. Find the length in kilometers of a row of 6.02×10²³ hydrogen atoms. The diameter of a ping pong ball is 4.0 cm. Find the length in kilometers of a row of 6.02×10²³ ping pong balls.

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1

Convert the diameter of a hydrogen atom from picometers (pm) to meters (m) using the conversion factor: 1 pm = 1 x 10^{-12} m.

Calculate the total length in meters of a row of 6.02 x 10^{23} hydrogen atoms by multiplying the diameter of one hydrogen atom (in meters) by Avogadro's number (6.02 x 10^{23}).

Convert the total length from meters to kilometers by using the conversion factor: 1 km = 1000 m.

Convert the diameter of a ping pong ball from centimeters (cm) to meters (m) using the conversion factor: 1 cm = 0.01 m.

Calculate the total length in meters of a row of 6.02 x 10^{23} ping pong balls by multiplying the diameter of one ping pong ball (in meters) by Avogadro's number (6.02 x 10^{23}), and then convert this length to kilometers.

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Avogadro's Number

Avogadro's number, approximately 6.02 x 10^23, is the number of atoms, ions, or molecules in one mole of a substance. This constant is fundamental in chemistry for converting between the number of particles and the amount of substance in moles, allowing for calculations involving large quantities of atoms or molecules.

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Avogadro's Law

Unit Conversion

Unit conversion is the process of converting a quantity expressed in one set of units to another set of units. In this problem, it involves converting diameters measured in picometers (pm) and centimeters (cm) to kilometers (km) to find the total length of a row of atoms or balls. Mastery of unit conversion is essential for solving problems in chemistry and physics.

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Conversion Factors

Volume and Density Relationships

Understanding the relationship between volume, density, and dimensions is crucial in this context. The volume of an object can be calculated from its dimensions, and knowing the density allows for conversions between mass and volume. In this case, while not directly needed, recognizing how these concepts relate can help in visualizing the scale of the objects being compared.

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Relationship of Volume and Moles Example

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