/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 SOLUTION Analyze We must write... [FREE SOLUTION] | 91Ó°ÊÓ

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SOLUTION Analyze We must write balanced nuclear equations in which the masses and charges of reactants and products are equal. Plan We can begin by writing the complete chemical symbols for the nuclei and decay particles that are given in the problem. Solve (a) The information given in the question can be summarized as $$ { }_{80}^{201} \mathrm{Hg}+{ }_{-1}^{0} \mathrm{e} \longrightarrow{ }_{Z}^{A} \mathrm{X} $$ The mass numbers must have the same sum on both sides of the equation: $$ 201+0=A $$ Thus, the product nucleus must have a mass number of 201. Similarly, balancing the atomic numbers gives $$ 80-1=Z $$ Thus, the atomic number of the product nucleus must be 79 , which identifies it as gold (Au): $$ { }_{80}^{201} \mathrm{Hg}+{ }_{-1}^{0} \mathrm{e} \longrightarrow{ }_{79}^{201} \mathrm{Au} $$ (b) In this case we must determine what type of particle is emitted in the course of the radioactive decay: $$ { }_{90}^{231} \mathrm{Th} \longrightarrow{ }_{91}^{231} \mathrm{~Pa}+{ }_{\mathrm{Z}}^{\mathrm{X}} \mathrm{X} $$ From \(231=231+A\) and \(90=91+Z\), we deduce \(A=0\) and \(Z=-1\). According to Table \(21.2\), the particle with these characteristics is the beta particle (electron). We therefore write $$ { }_{90}^{231} \mathrm{Th} \longrightarrow{ }_{91}^{231} \mathrm{~Pa}+{ }_{-1}^{0} \mathrm{e} \text { or } \quad{ }_{90}^{231} \mathrm{Th} \longrightarrow{ }_{91}^{231} \mathrm{~Pa}+\beta^{-} $$

Short Answer

Expert verified
(a) In the electron capture decay of \(_{80}^{201}\textrm{Hg}\), the balanced nuclear equation is: \[ _{80}^{201}\textrm{Hg} + _{-1}^{0}\textrm{e} \longrightarrow _{79}^{201}\textrm{Au} \] (b) In the radioactive decay of \(_{90}^{231}\textrm{Th}\) to \(_{91}^{231}\textrm{Pa}\), the balanced nuclear equation is: \[ _{90}^{231}\textrm{Th} \longrightarrow _{91}^{231}\textrm{Pa} + _{-1}^{0}\textrm{e} \] or \[ _{90}^{231}\textrm{Th} \longrightarrow _{91}^{231}\textrm{Pa} + \beta^{-} \]

Step by step solution

01

(a) Writing the initial given equation of decay

Given information: \(_{80}^{201}\textrm{Hg}\) decays by electron capture (\(_{-1}^{0}\textrm{e}\)). We can represent this as: \[ _{80}^{201}\textrm{Hg}+_{-1}^{0}\textrm{e} \longrightarrow _{Z}^{A}\textrm{X} \]
02

(a) Balancing equations and finding the product nucleus

The mass numbers and atomic numbers should be equal on both sides of the equation. For mass number (A): \[201+0=A\] Which means A=201. For atomic number (Z): \[80-1=Z\] Which means Z=79. The atomic number 79 corresponds to gold (Au). The balanced equation is: \[ _{80}^{201}\textrm{Hg} + _{-1}^{0}\textrm{e} \longrightarrow _{79}^{201}\textrm{Au} \]
03

(b) Writing the initial given equation of decay

Given information: \(_{90}^{231}\textrm{Th}\) decays to \(_{91}^{231}\textrm{Pa}\). We need to find the emitted particle (X). The equation can be written as: \[ _{90}^{231}\textrm{Th} \longrightarrow _{91}^{231}\textrm{Pa} + _{Z}^{A}\textrm{X} \]
04

(b) Balancing equations and finding the emitted particle

Balancing mass and atomic numbers: For mass number (A): \[231=231+A\] Which means A=0. For atomic number (Z): \[90=91+Z\] Which means Z=-1. A particle with a mass number of 0 and an atomic number of -1 is a beta particle (electron, e). Therefore, the balanced equation is: \[ _{90}^{231}\textrm{Th} \longrightarrow _{91}^{231}\textrm{Pa} + _{-1}^{0}\textrm{e} \] or \[ _{90}^{231}\textrm{Th} \longrightarrow _{91}^{231}\textrm{Pa} + \beta^{-} \]

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Key Concepts

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

Nuclear Chemistry
Nuclear chemistry is an area of chemistry that involves the study of changes that occur within atomic nuclei. This field explores various types of nuclear reactions, including radioactive decay, nuclear fission, and nuclear fusion. Understanding nuclear chemistry is pivotal in many applications such as nuclear power generation, medical diagnostics and treatments, as well as in radiocarbon dating.

Nuclear reactions can lead to changes in an atom's nucleus, resulting in the production of a different element or a different isotope of the same element. These transformations are governed by the laws of conservation of mass and energy, implying that the total mass and total charge before and after a nuclear reaction must be equal. Writing balanced nuclear equations is essential in nuclear chemistry to depict these reactions accurately.
Radioactive Decay
Radioactive decay is a spontaneous process by which unstable atomic nuclei release energy by emitting radiation in the form of particles or electromagnetic waves. There are several types of radioactive decay, including alpha decay, beta decay, gamma decay, and electron capture. Each type of decay results in the transmutation of elements and can be represented by balanced nuclear equations.

In radioactive decay, atoms of one element can change into atoms of another element altogether, or into a different isotope of the same element. The rate of decay is characterized by half-lives, which indicate the time required for half of the radioactive atoms in a sample to decay. Radioactive decay is a natural and random process, and cannot be influenced by changes in temperature, pressure, or chemical state.
Electron Capture
Electron capture is a type of beta decay, which is a process where an unstable atom becomes more stable. During electron capture, a proton within the atomic nucleus captures an orbiting electron and combines with it to form a neutron. As a result, the atomic number decreases by one while the mass number remains the same, because the proton and neutron have approximately equal mass.

The captured electron typically comes from the innermost electron shell (called the K shell), and the process can be represented in a nuclear equation by showing an electron being added to the nucleus of the parent atom. Electron capture is an important process in the balance of nuclear equations since it helps to illustrate the transformation of an element into another with a lower atomic number.
Beta Particle Emission
Beta particle emission is another form of radioactive decay. In beta-minus (β−) decay, a neutron in the nucleus is transformed into a proton, an electron (called a beta particle), and an antineutrino. The beta particle is ejected from the nucleus, resulting in an increase in the atomic number by one, while the mass number remains unchanged.

Beta particles are high-energy, high-speed electrons or positrons emitted by certain types of radioactive nuclei such as potassium-40. The nuclear equation representing beta particle emission shows the nucleus of the parent atom on one side and the nucleus of the new atom and the emitted beta particle on the other. Beta particle emission is central to many applications, including medical therapies and radiometric dating. Balancing nuclear equations for beta particle emission helps in visualizing the identity of the newly formed atom after this type of decay.

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Most popular questions from this chapter

Account for the following observations: (a) Phosphorus forms a pentachloride, but nitrogen does not. (b) \(\mathrm{H}_{3} \mathrm{PO}_{2}\) is a monoprotic acid. (c) Phosphonium salts, such as \(\mathrm{PH}_{4} \mathrm{Cl}\), can be formed under anhydrous conditions, but they cannot be made in aqueous solution. (d) White phosphorus is more reactive than red phosphorus.

Write the Lewis structure for each of the following species, and indicate the structure of each: (a) \(\mathrm{SeO}_{3}{ }^{2-} ;\) (b) \(\mathrm{S}_{2} \mathrm{Cl}_{2}\); (c) chlorosulfonic acid, \(\mathrm{HSO}_{3} \mathrm{Cl}\) (chlorine is bonded to sulfur).

Write a balanced equation for each of the following reactions. (You may have to guess at one or more of the reaction products, but you should be able to make a reasonable guess, based on your study of this chapter.) (a) Hydrogen selenide can be prepared by reaction of an aqueous acid solution on aluminum selenide. (b) Sodium thiosulfate is used to remove excess \(\mathrm{Cl}_{2}\) from chlorine-bleached fabrics. The thiosulfate ion forms \(\mathrm{SO}_{4}{ }^{2-}\) and elemental sulfur, while \(\mathrm{Cl}_{2}\) is reduced to \(\mathrm{Cl}^{-}\). Nitrogen and the Other Group 5A Elements (Sections 22.7 and 22.8)

(a) Determine the number of calcium ions in the chemical formula of the mineral hardystonite, \(\mathrm{Ca}_{x} \mathrm{Zn}\left(\mathrm{Si}_{2} \mathrm{O}_{7}\right)\). (b) Determine the number of hydroxide ions in the chemical formula of the mineral pyrophyllite, \(\mathrm{Al}_{2}\left(\mathrm{Si}_{2} \mathrm{O}_{5}\right)_{2}(\mathrm{OH})_{x}\).

Write the chemical formula for each of the following compounds, and indicate the oxidation state of nitrogen in each: (a) nitric oxide, (b) hydrazine, (c) potassium cyanide, (d) sodium nitrite, (e) ammonium chloride, (f) lithium nitride.

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