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A sample of a certain metal has a volume of \(4.0 \times 10^{-5} \mathrm{~m}^{3}\). The metal has a density of \(9.0 \mathrm{~g} / \mathrm{cm}^{3}\) and a molar mass of \(60 \mathrm{~g} / \mathrm{mol} .\) The atoms are bivalent. How many conduction electrons (or valence electrons) are in the sample?

Short Answer

Expert verified
There are approximately \(7.2264 \times 10^{24}\) conduction electrons in the sample.

Step by step solution

01

Convert Volume to cm鲁

First, convert the volume from m鲁 to cm鲁. Since 1 m鲁 = 10鈦 cm鲁, the volume in cm鲁 is \(4.0 \times 10^{-5} \, \text{m}^{3} \times 10^{6} \text{ cm}^{3} = 40 \text{ cm}^{3}\).
02

Calculate the Mass of the Sample

Using the density \(9.0 \, \text{g/cm}^{3}\) and volume \(40 \, \text{cm}^{3}\), calculate the mass: \(\text{Mass} = \text{Density} \times \text{Volume} = 9.0 \, \text{g/cm}^{3} \times 40 \, \text{cm}^{3} = 360 \, \text{g}\).
03

Calculate the Number of Moles

With the mass \(360 \, \text{g}\) and molar mass \(60 \, \text{g/mol}\), calculate the number of moles: \(\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{360 \, \text{g}}{60 \, \text{g/mol}} = 6 \, \text{mol}\).
04

Calculate Total Number of Atoms

Use Avogadro's number \(6.022 \times 10^{23} \, \text{atoms/mol}\) to calculate the total number of atoms: \(\text{Total Atoms} = 6 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 3.6132 \times 10^{24} \, \text{atoms}\).
05

Calculate the Number of Conduction Electrons

Given that the metal atoms are bivalent, each atom provides 2 conduction electrons. Therefore, calculate the total number of conduction electrons: \(\text{Conduction Electrons} = 2 \times 3.6132 \times 10^{24} = 7.2264 \times 10^{24} \, \text{electrons}\).

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Key Concepts

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

Understanding Density
Density is a fundamental concept in both physics and chemistry, representing mass per unit volume. For example, in this exercise, the metal's density is given as 9.0 g/cm鲁.
To use density in calculations, remember this formula:
  • Density = Mass / Volume
Essentially, density tells us how compact the matter is in a substance. A higher density indicates tighter packing of atoms, and vice versa.
In the context of the exercise, understanding how to manipulate and convert density units is crucial. You should be comfortable converting volume units, as volume may initially be given in cubic meters instead of cubic centimeters. Knowing these conversions helps in applying the density formula correctly, allowing the determination of mass from a given volume.
Exploring Molar Mass
Molar mass is a chemical property that indicates the mass of one mole of a substance, measured in grams/mol. For the given metal, the molar mass is 60 g/mol.
  • Molar Mass = Mass / Number of Moles
This concept is vital when converting the mass of a substance to the number of moles, which are crucial for further calculations involving atoms or molecules.
Understanding molar mass ensures you can link mass to the atomic scale, working through quantities typically used in chemical reactions and stoichiometry. To use molar mass effectively, you typically have to know the atomic species involved and use this value to calculate the corresponding moles from mass, as demonstrated in the exercise.
The Significance of Avogadro's Number
Avogadro's number is a constant that defines the number of atoms, ions, or molecules in one mole of a substance, approximately \( 6.022 \times 10^{23} \).
  • Atoms in a Sample = Moles 脳 Avogadro's Number
It is central for converting between atomic-scale measurements and macroscopic quantities. By using Avogadro's number, we bridge the gap between grams (a macroscopic measure) and atoms (a microscopic measure).
In exercises like this, we use Avogadro's number to answer important questions about the quantity of atoms present in a given mass. This facilitates further analysis, like determining the total conduction electrons present in the metallic sample.
Understanding Bivalent Atoms
Bivalent atoms are atoms that have a valency of two, meaning they can form two bonds with other atoms. In the context of conduction electrons, this denotes that each atom of the metal contributes two electrons for electrical conduction.
Knowing whether atoms are mono-, bi-, tri-, or tetravalent helps in understanding the electron contribution to conduction bands in metals.
  • Bivalent: each atom contributes 2 electrons
This concept plays a crucial role in electronics and materials science, as it influences the electrical properties of materials.
In the exercise, identifying the atoms as bivalent is key to calculating the total number of conduction electrons. By recognizing the contribution from each metal atom, we multiply the total number of atoms by two to find the total conduction electrons available.

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Most popular questions from this chapter

Pure silicon at room temperature has an electron number density in the conduction band of about \(5 \times 10^{15} \mathrm{~m}^{-3}\) and an equal density of holes in the valence band. Suppose that one of every \(10^{7}\) silicon atoms is replaced by a phosphorus atom. (a) Which type will the doped semiconductor be, \(n\) or \(p ?\) (b) What charge carrier number density will the phosphorus add? (c) What is the ratio of the charge carrier number density (electrons in the conduction band and holes in the valence band) in the doped silicon to that in pure silicon?

In a particular crystal, the highest occupied band is full. The crystal is transparent to light of wavelengths longer than \(295 \mathrm{nm}\) but opaque at shorter wavelengths. Calculate, in electron-volts, the gap between the highest occupied band and the next higher (empty) band for this material.

Doping changes the Fermi energy of a semiconductor. Consider silicon, with a gap of \(1.11 \mathrm{eV}\) between the top of the valence band and the bottom of the conduction band. At \(300 \mathrm{~K}\) the Fermi level of the pure material is nearly at the midpoint of the gap. Suppose that silicon is doped with donor atoms, each of which has a state \(0.15 \mathrm{eV}\) below the bottom of the silicon conduction band, and suppose further that doping raises the Fermi level to \(0.11 \mathrm{eV}\) below the bottom of that band (Fig. \(41-22\) ). For (a) pure and (b) doped silicon, calculate the probability that a state at the bottom of the silicon conduction band is occupied. (c) Calculate the probability that a state in the doped material (at the donor level) is occupied.

(a) What maximum light wavelength will excite an electron in the valence band of diamond to the conduction band? The energy gap is \(5.50 \mathrm{eV}\). (b) In what part of the electromagnetic spectrum does this wavelength lie?

A potassium chloride crystal has an energy band gap of \(7.6 \mathrm{eV}\) above the topmost occupied band, which is full. Is this crystal opaque or transparent to light of wavelength \(140 \mathrm{nm} ?\)

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