/*! 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 80 Seaborgium (Sg, element 106 ) is... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Seaborgium (Sg, element 106 ) is prepared by the bombardment of curium-248 with neon-22, which produces two isotopes, \(^{265} \mathrm{Sg}\) and \(^{266} \mathrm{Sg}\). Write balanced nuclear reactions for the formation of both isotopes. Are these reactions better described as fusion or fission processes?

Short Answer

Expert verified
Answer: The reactions for the formation of isotopes \(^{265}\mathrm{Sg}\) and \(^{266}\mathrm{Sg}\) when curium-248 is bombarded with neon-22 are fusion processes.

Step by step solution

01

Determine the Mass and Charge of Reactants

The mass and charge of the two reactants, Curium-248 and Neon-22, should be identified. The symbol for Curium is \(\mathrm{Cm}\) and its atomic number (charge) is 96. For Neon, the symbol is \(\mathrm{Ne}\), and its atomic number (charge) is 10. So we have: 1. Curium-248 : \(_{96}^{248}\mathrm{Cm}\) - this isotope has mass number 248 and atomic number (charge) 96. 2. Neon-22 : \(_{10}^{22}\mathrm{Ne}\) - this isotope has mass number 22 and atomic number (charge) 10.
02

Determine Mass and Charge of Products for First Reaction

We are asked to find a balanced reaction that results in the formation of \(^{265}\mathrm{Sg}\). The atomic number (charge) of \(\mathrm{Sg}\) is 106. In order to conserve mass, \(248 + 22 = 265 + A\) Where A is the mass number of the other product formed in the reaction. Solving for A, we get: \(A = 248 + 22 - 265 = 5\) To conserve charge (atomic number), \(96 + 10 = 106 + Z\) Where Z is the atomic number of the other product formed in the reaction. Solving for Z, we get: \(Z = 96 + 10 - 106 = 0\) So the balanced reaction will involve a product with mass number 5 and atomic number 0.
03

Write First Balanced Nuclear Reaction (for \(^{265} \mathrm{Sg}\))

The product with mass number 5 and atomic number 0 is a neutron represented by \(_{0}^{5}\mathrm{n}\). The balanced nuclear reaction for the formation of \(^{265}\mathrm{Sg}\) is: \(_{96}^{248}\mathrm{Cm} + _{10}^{22}\mathrm{Ne} \rightarrow _{106}^{265}\mathrm{Sg} + _{0}^{5}\mathrm{n}\)
04

Determine Mass and Charge of Products for Second Reaction

Similarly, we need to find a balanced reaction that results in the formation of \(_{106}^{266}\mathrm{Sg}\). To conserve mass and charge, we can repeat Steps 2 and 3: To conserve mass: \(248 + 22 = 266 + B\) Where B is the mass number of the other product formed in the reaction. Solving for B, we get: \(B = 248 + 22 - 266 = 4\) To conserve charge (atomic number), \(96 + 10 = 106 + Z\) Where Z is the atomic number of the other product formed in the reaction. Solving for Z, we get: \(Z = 96 + 10 - 106 = 0\)
05

Write Second Balanced Nuclear Reaction (for \(^{266} \mathrm{Sg}\))

The product with mass number 4 and atomic number 0 is \(_{0}^{4}\mathrm{n}\). So the balanced nuclear reaction for the formation of \(^{266}\mathrm{Sg}\) is: \(_{96}^{248}\mathrm{Cm} + _{10}^{22}\mathrm{Ne} \rightarrow _{106}^{266}\mathrm{Sg} + _{0}^{4}\mathrm{n}\)
06

Determine if the Reactions are Fusion or Fission Processes

In fusion reactions, light nuclei combine to form a heavier nucleus, whereas, in fission reactions, a heavy nucleus breaks apart to form lighter nuclei. Since in both the reactions given, a heavy nucleus (Curium-248) is bombarded with a lighter nucleus (Neon-22) to form a product of heavier nucleus (Seaborgium), these reactions can be classified as fusion processes.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Seaborgium
Seaborgium, with the chemical symbol Sg and atomic number 106, is a synthetic element named after the famous scientist Glenn T. Seaborg. It is part of the transactinide series and is renowned for its challenging synthesis and study due to its high radioactivity and short half-life. This element does not occur in nature and must be created in a laboratory setting. The creation of Seaborgium involves the bombardment of other heavy elements, typically using a particle accelerator. This makes the study and understanding of Seaborgium crucial in nuclear chemistry as it helps scientists explore the boundaries of the periodic table and the properties of superheavy elements. The research into Seaborgium and its isotopes provides unique insights into the structure of atomic nuclei and helps to refine models of atomic theory.
Isotopes
Isotopes are different forms of the same element that have the same number of protons but different numbers of neutrons. This change in the neutron count alters the mass number, thus forming different isotopes of an element. For Seaborgium, the isotopes of interest in nuclear reactions are \(^{265}\mathrm{Sg}\) and \(^{266}\mathrm{Sg}\).

When curium-248 and neon-22 are used to synthesize Seaborgium, the reactions produce these two isotopes. Each isotope has unique properties and stability characteristics. Isotopes like \(^{265}\mathrm{Sg}\) and \(^{266}\mathrm{Sg}\) are instrumental in research because their properties can reveal information about nuclear forces and the stability of atomic nuclei. Additionally, studying isotopes helps to understand radioactive decay processes, which have implications in fields such as medical imaging and nuclear energy.
Fusion Processes
Fusion processes involve the combining of light atomic nuclei to form a heavier nucleus. This is in contrast to fission, where a heavy nucleus splits into lighter nuclei. In the creation of Seaborgium isotopes, the fusion process occurs when neon-22 nuclei are bombarded into curium-248 nuclei.

This process requires significant energy to overcome the repulsive forces between the positively charged nuclei. Once the nuclei fuse, they form a new element, in this case, Seaborgium, along with a neutron or other byproducts.

The fusion process is a natural phenomenon that powers the stars, including the sun, and is a key area of research for sustainable energy. In the laboratory, creating new elements through fusion provides critical insights into chemical behavior and nuclear stability, helping scientists to refine models of atomic interaction and further extend the periodic table.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Complete the following nuclear equations describing the preparation of isotopes for nuclear medicine. a. \(^{197} \mathrm{Au}+? \rightarrow^{199} \mathrm{Hg}+\beta\) b. \(^{64} \mathrm{Ni}+^{1} \mathrm{H} \rightarrow^{64} \mathrm{Cu}+?\) c. \(^{63} \mathrm{Cu}+? \rightarrow^{66} \mathrm{Ga}+^{1} \mathrm{n}\) d. \(^{67} \mathrm{Zn}+^{1} \mathrm{n} \rightarrow^{67} \mathrm{Cu}+?\)

In Section 21.6 we state that "no energy would be released if two \(^{4} \mathrm{He}\) nuclei were to fuse together to form \(^{8} \mathrm{Be}\). Similarly, \(^{8} \mathrm{Be}\) nuclei require no energy to spontaneously decompose into \(^{4}\) He nuclei, so they would immediately do so." Verify this statement by calculating the binding energy of \(^{8} \mathrm{Be}\) and comparing it to that of \(^{4} \mathrm{He}\).

Rhodium-105 is an isotope currently under investigation in diagnostic applications. The half-life of \(^{105} \mathrm{Rh}\) is \(35.4 \mathrm{h}\) which is sufficiently long for transport from the supplier a hospital. A supplier ships \(250 \mathrm{mg}\) of \(^{105} \mathrm{RhCl}_{3}\) overnight (12 hours). a. What percentage of the \(^{105}\) Rh remains upon arrival? b. How long will it take for \(95 \%\) of the \(^{105} \mathrm{Rh}\) to decay?

Dating Cave Paintings Cave paintings in Gua Saleh Cave in Borneo have been dated by measuring the amount of \(^{14} \mathrm{C}\) in calcium carbonate deposits that formed over the pigments used in the paint. The source of the carbonate ion was atmospheric \(\mathrm{CO}_{2}\).a. What is the ratio of the \(^{14} \mathrm{C}\) radioactivity in calcium carbonate that formed 9900 years ago to that in calcium carbonate formed today? b. The archaeologists also used a second method, uranium-thorium dating, to confirm the age of the paintings by measuring trace quantities of these elements present as contaminants in the calcium carbonate. Shown below are two candidates for the U-Th dating method. Which isotope of uranium do you suppose was chosen? Explain your answer. $$\begin{aligned} &t_{1 / 2}=^{233} \mathrm{U} \quad 7.04 \times 10^{8} \mathrm{yr} \quad^{231} \mathrm{Th} \quad \rightarrow \quad^{231} \mathrm{p}_{2} \quad 3 \quad \rightarrow \quad\\\ &t_{1 / 2}=^{234} \mathrm{U} \quad \begin{array}{rl}\rightarrow & ^{230} \mathrm{Th} \\ 2.44 \times 10^{5} \mathrm{yr} & 7.7 \times 10^{4} \mathrm{hr}\end{array} \quad \begin{array}{rl} 226 \mathrm{p}_{2} & \rightarrow \\\1600 & \mathrm{yr}\end{array}\end{aligned}$$

All of the following fusion reactions produce \(^{32}\) S. Calculate the energy released in each reaction from the masses of the isotopes: \(^{4} \mathrm{He}(4.00260 \mathrm{amu}),^{6} \mathrm{Li}(6.01512 \mathrm{amu}),^{12} \mathrm{C}\),\((12.000 \mathrm{amu}),^{14} \mathrm{N}(14.00307 \mathrm{amu}),^{16} \mathrm{O}(15.99491 \mathrm{amu})\),\(^{24} \mathrm{Mg}(23.98504 \mathrm{amu}),^{28} \mathrm{Si}(27.97693 \mathrm{amu}),^{32} \mathrm{S}\),\((31.97207 \mathrm{amu})\). a. \(16+^{16} \mathrm{O} \rightarrow^{32} \mathrm{S}\) b. \(^{23} \mathrm{Si}+^{4} \mathrm{He} \rightarrow^{32} \mathrm{S}\) \(c_{1}^{14} N+^{12} C+^{6} L i \rightarrow^{32} S\) d. \(^{24} \mathrm{Mg}+2^{4} \mathrm{He} \rightarrow^{32} \mathrm{S}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.