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Chapter 31: Problem 3

In each of the following cases, what element does the symbol \(\mathrm{X}\) represent and how many neutrons are in the nucleus: (a) 195 78 \(\mathrm{X}\), (b) \(\frac{32}{16} \mathrm{X}\), (c) \({ }_{29}^{63} \mathrm{X}\), (d) \({ }_{5}^{11} \mathrm{X}\), and (e) \(\frac{239}{94} X\) ? Use the periodic table on the in side of the back cover as needed.

Short Answer

Expert verified
(a) Pt, 117 neutrons; (b) S, 16 neutrons; (c) Cu, 34 neutrons; (d) B, 6 neutrons; (e) Pu, 145 neutrons.

Step by step solution

01

Identify the Element Using Atomic Number

The symbol for each element includes two numbers: the atomic number (bottom number) which indicates the number of protons, and the mass number (top number) which is the sum of protons and neutrons. To identify the element, match the atomic number to the periodic table entry.
02

Calculate the Number of Neutrons

The number of neutrons in an atom is determined by subtracting the atomic number (number of protons) from the mass number. Use the formula: \( \text{Neutrons} = \text{Mass number} - \text{Atomic number} \).
03

Solve for Case (a)

For 195 \(_{78}^{ }\mathrm{X}\), the atomic number is 78, which corresponds to Platinum (Pt) on the periodic table. The mass number is 195. Thus, the number of neutrons is \(195 - 78 = 117\).
04

Solve for Case (b)

For \(\frac{32}{16} \mathrm{X}\), the atomic number is 16, which represents Sulfur (S). The mass number is 32. Hence, the number of neutrons is \(32 - 16 = 16\).
05

Solve for Case (c)

For \(_{29}^{63}\mathrm{X}\), the atomic number is 29, which is Copper (Cu) on the periodic table. The mass number is 63. Therefore, the number of neutrons is \(63 - 29 = 34\).
06

Solve for Case (d)

For \(_{5}^{11}\mathrm{X}\), the atomic number is 5, corresponding to Boron (B). The mass number is 11. The number of neutrons is \(11 - 5 = 6\).
07

Solve for Case (e)

For \(\frac{239}{94} X\), the atomic number is 94, which is Plutonium (Pu). The mass number is 239. Thus, the number of neutrons is \(239 - 94 = 145\).

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Key Concepts

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

Elements and Atomic Numbers
In chemistry, elements are the simplest substances in nature, consisting of only one type of atom. Each element is uniquely identified by its atomic number. The atomic number is a crucial piece of information that represents the number of protons in an element's atom. Protons are positively charged particles located in the atom's nucleus.
The atomic number is often denoted as the subscript (bottom number) in an element symbol. For example, in \( _{5}^{11}\mathrm{X} \,\), the atomic number is 5, indicating that the element is Boron, which consists of atoms with five protons each.
  • Protons define the identity of an element.
  • {$title=Positive charge of each proton neutralizes one electron's negative charge.
  • The atomic number is consistent on the periodic table.
Understanding atomic numbers helps in identifying elements on the periodic table and plays a vital role in determining the behavior of atoms in chemical reactions.
Mass Number
The mass number of an element is another vital figure alongside the atomic number. It represents the total number of protons and neutrons in an atom's nucleus. Unlike the atomic number, the mass number may vary amongst atoms of the same element due to different numbers of neutrons. These variations are known as isotopes.
The mass number is represented as a superscript (top number) in the elemental symbol. For instance, in \( _{29}^{63}\mathrm{X} \,\), the mass number is 63, indicating that the copper nucleus contains a combined total of 63 protons and neutrons. To find the amount of neutrons, subtract the atomic number from the mass number.
  • Neutrons equal mass number minus atomic number.
  • Isotopes affect the mass number but not the chemical properties.
  • Mass number may differ in radioactive elements because of decay processes.
Recognizing mass numbers is crucial for calculating molecular masses and understanding isotope differences.
Periodic Table Usage
The periodic table is an essential tool in chemistry, providing a wealth of information about elements and their properties. Each element is placed based on its atomic number, which helps quickly identify the element by matching it with the atomic number in a problem.
For example, when dealing with \(^{195}_{78}\mathrm{X} \,\), you can locate the element Platinum (Pt) on the table by finding the atomic number 78. Using the periodic table allows for easy recognition of elements, saving from potential errors in elemental identification.
  • Useful for finding atomic number and symbol of elements.
  • Check atomic mass for isotopic abundance insights.
  • Provides information on element groups and periods, leading to chemical property insights.
Mastery in using the periodic table is a fundamental skill for anyone aiming to excel in chemistry and related fields.

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Most popular questions from this chapter

(a) Energy is required to separate a nucleus into its constituent nucleons, as Figure \(31-3\) indicates; this energy is the total binding energy of the nucleus. In a similar way one can speak of the energy that binds a single nucleon to the remainder of the nucleus. For example, separating nitrogen \({ }_{7}^{14} N\) into nitrogen \({ }_{7}^{13} \mathrm{~N}\) and a neutron takes energy equal to the binding energy of the neutron, as shown below: $$ \frac{14}{7} N+\text { Energy } \rightarrow \frac{13}{7} N+\frac{1}{0} n $$ Find the energy (in MeV) that binds the neutron to the \(\frac{14}{7} N\) nucleus by considering the mass of \(13 \mathrm{~N}\) (atomic mass \(=13.005738 \mathrm{u}\) ) and the mass of \(\frac{1}{0^{\mathrm{n}}}\) (atomic mass \(=1.008\) \(665 \mathrm{u}\) ), as compared to the mass of \({ }_{7}^{14} \mathrm{~N}\) ( atomic mass \(=14.003074 \mathrm{u}\) ). (b) Similarly, one can speak of the energy that binds a single proton to the\(\frac{14}{7} \mathrm{~N}\) nucleus: $$ { }_{7}^{14} \mathrm{~N}+\text { Energy } \rightarrow{ }_{6}^{13} \mathrm{C}+{ }_{1}^{1} \mathrm{H} $$ Following the procedure outlined in part (a), determine the energy (in MeV) that binds the proton (atomic mass \(=1.007825 \mathrm{u}\) ) to the \(\frac{14}{7} \mathrm{~N}\) nucleus. The atomic mass of carbon \(\frac{13}{6} \mathrm{C}\) is \(13.003335 \mathrm{u}\) (c) Which nucleon is more tightly bound, the neutron or the proton?

Osmium \(\frac{191}{76} \mathrm{Os}\) (atomic mass \(\left.=190.960920 \mathrm{u}\right)\) is converted into Iridium \(\frac{191}{77} \mathrm{Ir}\) (atomic mass \(=190.960584 \mathrm{u}\) ) via \(\beta^{-}\) decay. What is the energy (in \(\mathrm{MeV}\) ) released in this process?

A device used in radiation therapy for cancer contains \(0.50 \mathrm{~g}\) of cobalt \(\frac{60}{27} \mathrm{Co}\) \((59.933819 \mathrm{u}) .\) The half-life of \(\underset{27}{60} \mathrm{C} \circ\) is \(5.27 \mathrm{yr}\). Determine the activity of the radioactive material.

Determine the mass defect (in atomic mass units) for (a) helium \({ }_{2}^{3} \mathrm{He},\) which has an atomic mass of \(3.016030 \mathrm{u},\) and \((\mathrm{b})\) the isotope of hydrogen known as tritium \(\frac{3}{1} \mathrm{~T}\) which has an atomic mass of \(3.016050 \mathrm{u}\). (c) On the basis of your answers to parts (a) and (b), state which nucleus requires more energy to disassemble it into its separate and stationary constituent nucleons. Give your reasoning.

An unknown nucleus contains 70 neutrons and has twice the volume of the nickel \(60 \mathrm{Ni}\) nucleus. Identify the unknown nucleus in the form \(A_{X}\). Use the periodic table on the inside of the back cover as needed.
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