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Chapter 2: Problem 65

Calculate the average density of a single Al-27 atom by assuming that it is a sphere with a radius of \(0.143 \mathrm{~nm}\). The masses of a proton, electron, and neutron are \(1.6726 \times 10^{-24} \mathrm{~g}, 9.1094 \times 10^{-28} \mathrm{~g}\), and \(1.6749 \times 10^{-24} \mathrm{~g}\), respectively. The volume of a sphere is \(4 \pi r^{3} / 3\), where \(r\) is its radius. Express the answer in grams per cubic centimeter. The density of aluminum is found experimentally to be \(2.70 \mathrm{~g} / \mathrm{cm}^{3} .\) What does that suggest about the packing of aluminum atoms in the metal?

Short Answer

Expert verified
Answer: Based on the comparison, the calculated average density of a single Al-27 atom is greater than the experimental density of aluminum metal. This implies that the atoms in aluminum metal are not tightly packed and there is some empty space between the atoms.

Step by step solution

01

Calculate the number of protons, neutrons, and electrons in Al-27

Since Al-27 has an atomic number of 13, it means there are 13 protons and 13 electrons in it. Also, given that its mass number is 27, there would be (27-13) = 14 neutrons.
02

Calculate the mass of a single Al-27 atom

To find the mass of a single Al-27 atom, we need to add up the masses of its protons, neutrons, and electrons. Using the given masses: Mass of protons = 13 protons * 1.6726e-24 g/proton = \$2.1744 \times 10^{-23}\mathrm{~g}\$ Mass of neutrons = 14 neutrons * 1.6749e-24 g/neutron = \$2.3449 \times 10^{-23}\mathrm{~g}\$ Mass of electrons = 13 electrons * 9.1094e-28 g/electron = \$1.1842 \times 10^{-26}\mathrm{~g}\$ Total mass = Mass of protons + Mass of neutrons + Mass of electrons = \$2.1744 \times 10^{-23}\mathrm{~g}\$ + \$2.3449 \times 10^{-23}\mathrm{~g}\$ + \$1.1842 \times 10^{-26}\mathrm{~g}\$ = \$4.5194 \times 10^{-23} \mathrm{~g}\$
03

Calculate the volume of Al-27 atom sphere

Since we are given the sphere radius as \$0.143\mathrm{~nm}\$, we can find the volume using the formula for the volume of a sphere. Note that we need to convert the radius from nanometers to centimeters: Radius in cm = \$0.143\mathrm{~nm}\$ * (1e-9 m/nm) * (1e2 cm/m) = \$0.143 \times 10^{-7}\mathrm{~cm}\$ Volume = \$(4/3) \pi r^{3}\$ = \$(4/3) * \pi * (0.143 \times 10^{-7}\mathrm{~cm})^{3}\$ = \$1.220 \times 10^{-23}\mathrm{~cm}^{3}\$
04

Calculate the average density of Al-27 atom

Density is calculated as mass divided by volume. Using the mass and volume found in previous steps, we get: Density = Mass/Volume = \$(4.5194 \times 10^{-23} \mathrm{~g}) / (1.220 \times 10^{-23}\mathrm{~cm}^{3})\$ = \$3.70\mathrm{~g}/\mathrm{cm}^{3}\$
05

Compare calculated density with experimental density

The calculated average density of Al-27 atom is \$3.70\mathrm{~g}/\mathrm{cm}^{3}\$, while the experimental density of aluminum is given as \$2.70\mathrm{~g}/\mathrm{cm}^{3}\$. Since the calculated density is greater than the experimental density, it suggests that the atoms in the aluminum metal are not tightly packed and there is some empty space between the atoms.

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Most popular questions from this chapter

Write the names of the following ionic compounds. (a) \(\mathrm{HCl}(a q)\) (b) \(\mathrm{HClO}_{3}(a q)\) (c) \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{3}\right)_{3}\) (d) \(\mathrm{Ba}\left(\mathrm{NO}_{2}\right)_{2}\) (e) \(\mathrm{NaClO}\)

Ethane and ethylene are two gases containing only hydrogen and carbon atoms. In a certain sample of ethane, \(4.53 \mathrm{~g}\) of hydrogen is combined with \(18.0 \mathrm{~g}\) of carbon. In a sample of ethylene, \(7.25 \mathrm{~g}\) of hydrogen is combined with \(43.20 \mathrm{~g}\) of carbon. (a) Show how the data illustrate the law of multiple proportions. (b) Suggest reasonable formulas for the two compounds.

Selenium-75 is used in the diagnosis of disorders related to the pan- creas. How many (a) protons are in its nucleus? (b) neutrons are in its nucleus? (c) electrons are in a selenium atom?

Nuclei with the same mass number but different atomic numbers are called isobars. Consider Ca-40, Ca-41, K-41 and Ar-41. (a) Which of these are isobars? Which are isotopes? (b) What do \(\mathrm{Ca}-40\) and \(\mathrm{Ca}-41\) have in common? (c) Correct the statement (if it is incorrect): Atoms of \(\mathrm{Ca}-41, \mathrm{~K}-41\), and Ar-41 have the same number of neutrons.

Give the formulas of all the compounds containing no ions other than \(\mathrm{K}^{+}, \mathrm{Ca}^{2+}, \mathrm{Cl}^{-}\), and \(\mathrm{S}^{2-} .\)
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