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Atoms are spherical and so when silver atoms pack together to form silver metal, they cannot fill all the available space. In a sample of silver metal, approximately \(26.0 \%\) of the sample is empty space. Given that the density of silver metal is \(10.5 \mathrm{g} / \mathrm{cm}^{3}\), what is the radius of a silver atom? Express your answer in picometers.

Short Answer

Expert verified
The radius of a silver atom is approximately \(144.4 \mathrm{pm}\).

Step by step solution

01

Identify the used constants

The atomic weight of silver is \(107.87 \mathrm{~g/mole}\) and the Avogadro number is \(6.022 \times 10^{23} \mathrm{atoms/mole}\). The fraction of silver sphere occupied is \(74 \%\) which is \(0.74\) as a decimal.
02

Calculate the number of atoms

The number of atoms in \(1 \mathrm{cm}^{3}\) of silver can be calculated based on the fraction of space they occupy. This is done by the formula \(\text{{Number of atoms}} = \frac{{\text{{Density}} \times \text{{Avogadro's number}}}}{{\text{{Atomic weight}}}} = \frac{{10.5 \mathrm{g/cm^{3}} \times 6.022 \times 10^{23} \mathrm{atoms/mole}}}{{107.87 \mathrm{g/mole}}} \approx 5.855 \times 10^{22} \mathrm{atoms/cm^{3}}\).
03

Calculate the volume occupied by a single atom

The volume a single atom occupies is simply the total volume divided by the number of atoms. Since we know that the total volume for the silver atoms is \(74 \%\) of \(1 \mathrm{~cm^{3}}\), the volume of a single atom is \(\frac{{0.74 \mathrm{cm^{3}}}}{{5.855 \times 10^{22} \mathrm{atoms}}} \approx 1.263 \times 10^{-23} \mathrm{cm^{3/atom}}\).
04

Calculate the radius of the atom

Since the silver atom is spherical, the volume can be represented as \(\frac{4}{3}\pi r^{3}\). Solving this equation for \(r\) and substituting the previously calculated volume into this equation, we have \(r = \left(\frac{3V}{4\pi}\right)^{1/3} = \left(\frac{3 \times 1.263 \times 10^{-23} \mathrm{cm^{3/atom}}}{4\pi}\right)^{1/3} \approx 1.444 \times 10^{-8} \mathrm{cm}\). To convert centimeters to picometers, 1 centimeter is equal to \(1\times 10^{10}\) picometers. Hence, the radius of the silver atom is \(1.444 \times 10^{-8} \mathrm{cm} \times 1\times 10^{10} \mathrm{pm/cm} \approx144.4 \mathrm{pm}\).

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Key Concepts

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

Silver Atom
Silver atoms are tiny particles that make up the element silver, known as a metal. A silver atom is spherical, meaning it has a round shape similar to a ball. This geometric property is crucial in determining how atoms pack together in solid forms of silver. When packed, only 74% of the space is occupied by atoms, and 26% is empty space. This is because spheres cannot fill space completely without leaving gaps.
A detailed understanding of the volume that a silver atom occupies and its spherical nature allows scientists to estimate its radius. By knowing the volume of a single silver atom, calculated through its density and atomic properties, it is possible to back-calculate the radius using mathematical formulas for spheres.
Understanding the size and geometry of silver atoms provides insights into their physical and chemical characteristics, such as electrical and thermal conductivity, which are important in manufacturing and technology application.
Density of Silver
The density of silver is a measure of how much mass is contained in a given volume, typically expressed in grams per cubic centimeter (g/cm³). For silver, this value is 10.5 g/cm³. This relatively high density is due to the closely packed arrangement of silver atoms in its metallic form.
Density is an important physical property that can tell us a lot about a material. In the case of silver, it helps in understanding how tightly its atoms are packed in a given space and aids in calculations involving the volume and weight of silver in various applications.
  • The formula for density is: \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \).
  • This relationship helps in determining the number of atoms per cubic centimeter in a sample.
By knowing the density and relating it to the atomic weight and Avogadro's number, scientists can calculate the number of atoms present in a cubic centimeter of silver.
Volume of a Sphere
The volume of a sphere is the three-dimensional space occupied by the sphere. For a sphere, the volume \( V \) can be calculated using the formula \( V = \frac{4}{3}\pi r^3 \), where \( r \) is the radius of the sphere. This formula is essential when calculating the size of spherical objects like atoms, including silver atoms.
In problems involving atomic structures, knowing the volume of a sphere helps determine the size of an individual atom when packing into different structures, such as a crystal lattice. By dividing the total occupied volume by the number of atoms, we can find the volume of a single atom, and thus calculate its radius.
  • Volume calculations are crucial for determining the physical characteristics of materials at the atomic level.
  • Knowledge of volume helps to infer other properties like density and packing efficiency.
These insights are valuable for scientific research and industrial applications where precise material specifications are required.
Avogadro's Number
Avogadro's number, \(6.022 \times 10^{23}\), is a fundamental constant useful in chemistry for quantifying the number of particles (like atoms or molecules) in one mole of a substance. Named after the scientist Amedeo Avogadro, it links the macroscopic scale of materials that we can manipulate physically with the atomic scale that describes individual atoms or molecules.
This number is pivotal for calculating the number of atoms in a given sample of silver, allowing us to translate between the mass of a sample and the number of atoms it contains. By multiplying Avogadro's number with the inverse of atomic weight and density, we determine how many atoms fit in a defined volume.
  • It provides a bridge between quantities measured in the laboratory and those that describe individual atomic levels.
  • It is crucial for mole-based calculations in chemistry.
Understanding Avogadro's number allows chemists to compute precise measurements and conversions necessary for chemical reactions and material characterizations.

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Most popular questions from this chapter

Without doing detailed calculations, determine which of the following samples occupies the largest volume: (a) 25.5 mol of sodium metal \(\left(d=0.971 \mathrm{g} / \mathrm{cm}^{3}\right)\) (b) 0.725 L of liquid bromine \((d=3.12 \mathrm{g} / \mathrm{mL})\) (c) \(1.25 \times 10^{25}\) atoms of chromium metal \(\left(d=9.4 \mathrm{g} / \mathrm{cm}^{3}\right)\) (d) \(2.15 \mathrm{kg}\) of plumber's solder \(\left(d=9.4 \mathrm{g} / \mathrm{cm}^{3}\right), \mathrm{a}\) lead-tin alloy with a 2: 1 atom ratio of lead to tin

Within the limits of experimental error, show that the law of conservation of mass was obeyed in the following experiment: \(10.00 \mathrm{g}\) calcium carbonate (found in limestone) was dissolved in 100.0 mL hydrochloric acid \((d=1.148 \mathrm{g} / \mathrm{mL}) .\) The products were \(120.40 \mathrm{g}\) solution (a mixture of hydrochloric acid and calcium chloride) and 2.22 L carbon dioxide gas \((d=1.9769 \mathrm{g} / \mathrm{L})\)

Without doing detailed calculations, determine which of the following samples has the greatest number of atoms: (a) a cube of iron with a length of \(10.0 \mathrm{cm}\) \(\left(d=7.86 \mathrm{g} / \mathrm{cm}^{3}\right)\) (b) \(1.00 \mathrm{kg}\) of hydrogen contained in a \(10,000 \mathrm{L}\) balloon (c) a mound of sulfur weighing \(20.0 \mathrm{kg}\) (d) a 76 lb sample of liquid mercury \((d=13.5 \mathrm{g} / \mathrm{mL})\)

Refer to the periodic table inside the front cover and identify (a) the element that is in group 14 and the fourth period (b) one element similar to and one unlike sulfur (c) the alkali metal in the fifth period (d) the halogen element in the sixth period

When a solid mixture consisting of \(10.500 \mathrm{g}\) calcium hydroxide and \(11.125 \mathrm{g}\) ammonium chloride is strongly heated, gaseous products are evolved and \(14.336 \mathrm{g}\) of a solid residue remains. The gases are passed into \(62.316 \mathrm{g}\) water, and the mass of the resulting solution is \(69.605 \mathrm{g}\). Within the limits of experimental error, show that these data conform to the law of conservation of mass.

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