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Ch.2 - Atoms, Molecules, and Ions
Chapter 2, Problem 89b

A cube of gold that is 1.00 cm on a side has a mass of 19.3 g. A single gold atom has a mass of 197.0 u. (b) From the information given, estimate the diameter in Å of a single gold atom.

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Step 1: First, we need to calculate the volume of the gold cube. The volume of a cube is given by the formula $V = a^3$, where $a$ is the length of one side of the cube. In this case, $a = 1.00$ cm.
Step 2: Next, we need to calculate the number of gold atoms in the cube. We know that the mass of the cube is 19.3 g and the mass of a single gold atom is 197.0 u. We can convert the mass of the cube to atomic mass units (u) by using the conversion factor 1 g = 6.022 x 10^23 u (Avogadro's number). Then, we divide the total mass of the cube by the mass of a single atom to find the number of atoms.
Step 3: Now, we can calculate the volume occupied by a single gold atom. We divide the total volume of the cube by the number of atoms to get the volume of a single atom.
Step 4: We know that atoms are roughly spherical, so we can use the formula for the volume of a sphere $V = \frac{4}{3}πr^3$ to find the radius of a gold atom. We can solve this equation for $r$ to get $r = \left(\frac{3V}{4π}\right)^{\frac{1}{3}}$.
Step 5: Finally, we can find the diameter of a gold atom by multiplying the radius by 2. Remember to convert the diameter from cm to Å by using the conversion factor 1 cm = 10^8 Å.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Density

Density is defined as mass per unit volume and is a crucial property of materials. In this case, the density of gold can be calculated using the mass of the cube and its volume. Understanding density allows us to relate the macroscopic properties of a substance to its atomic structure, which is essential for estimating atomic dimensions.
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Avogadro's Number

Avogadro's number, approximately 6.022 x 10²³, is the number of atoms or molecules in one mole of a substance. This concept is vital for converting between the macroscopic scale (grams) and the atomic scale (individual atoms). By using Avogadro's number, we can determine how many gold atoms are present in the given mass, which aids in estimating the size of a single atom.
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Atomic Radius

The atomic radius is a measure of the size of an atom, typically defined as the distance from the nucleus to the outermost electron shell. In this context, estimating the diameter of a gold atom involves understanding its atomic radius and how it relates to the number of atoms in a given volume. This concept is fundamental for translating the mass and volume of gold into the dimensions of individual atoms.
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Related Practice
Textbook Question

The natural abundance of 3He is 0.000137%. (b) Based on the sum of the masses of their subatomic particles, which is expected to be more massive, an atom of 3He or an atom of 3H (which is also called tritium)?

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Textbook Question

The natural abundance of 3He is 0.000137%. (c) Based on your answer to part (b), what would need to be the precision of a mass spectrometer that is able to differentiate between peaks that are due to 3He+ and 3H+?

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Textbook Question

A cube of gold that is 1.00 cm on a side has a mass of 19.3 g. A single gold atom has a mass of 197.0 u. (a) How many gold atoms are in the cube?

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Textbook Question

The diameter of a rubidium atom is 495 pm We will consider two different ways of placing the atoms on a surface. In arrangement A, all the atoms are lined up with one another to form a square grid. Arrangement B is called a close-packed arrangement because the atoms sit in the 'depressions' formed by the previous row of atoms: (a) Using arrangement A, how many Rb atoms could be placed on a square surface that is 1.0 cm on a side?

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Textbook Question

(b) How many molecules of C13H18O2 are in this tablet?

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Textbook Question

"The diameter of a rubidium atom is 495 pm We will consider two different ways of placing the atoms on a surface. In arrangement A, all the atoms are lined up with one another to form a square grid. Arrangement B is called a close-packed arrangement because the atoms sit in the 'depressions' formed by the previous row of atoms:

(c) By what factor has the number of atoms on the surface increased in going to arrangement B from arrangement A?

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