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Chapter 2: Problem 74

Determine the only possible isotope (E) for which the following conditions are met: \(\bullet\)The mass number of \(\mathrm{E}\) is 2.50 times its atomic number. \(\bullet\)The atomic number of \(\mathrm{E}\) is equal to the mass number of another isotope (Y). In turn, isotope Y has a neutron number that is 1.33 times the atomic number of \(Y\) and equal to the neutron number of selenium- 82 .

Short Answer

Expert verified
The isotope E that fits the given conditions is silicon-36 (Si-36).

Step by step solution

01

Define the known and unknown variables

Let's denote the atomic number of isotope E as Z1 and its mass number as A1. The atomic number 'Z' and the mass number 'A' for Y are Z2 and A2 respectively. From the given conditions, the following relationships are obtained: A1 = 2.5 * Z1, Z2 = A1, and Number of neutrons in Y = 1.33 * Z2 = 48.
02

Solve the equations

By plugging Z2 = A1 into the third equation gives 1.33 * A1 = 48. Solving for A1 gives A1 = 48 / 1.33, which results in A1 = 36 (rounded to the nearest whole number).
03

Determine the atomic number of E

Knowing the value of A1, we can use the equation A1 = 2.5 * Z1 to calculate Z1. Substituting A1 = 36 into the equation gives Z1 = A1 / 2.5, which results in Z1 = 36 / 2.5 = 14.4, roughly rounded to the nearest whole number gives Z1 = 14.
04

Identify isotope E

From the periodic table, the element with the atomic number 14 is silicon (Si). So, the isotope E is silicon-36 (Si-36).

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Key Concepts

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

Mass number
The mass number is a fundamental concept in understanding isotopes. It denotes the total number of protons and neutrons present in the nucleus of an atom. Protons and neutrons are collectively known as nucleons, and they reside in the nucleus at the center of an atom.
  • Mass number is represented by the symbol \( A \).
  • For any isotope, the mass number is the sum of the number of protons (atomic number) and the number of neutrons.
  • For example, if an atom has 6 protons and 6 neutrons, its mass number is \( A = 6 + 6 = 12 \).
It's crucial in differentiating isotopes because isotopes of the same element have the same number of protons but different numbers of neutrons. The mass number is specific for each isotope, contributing to its identity and properties. It is not typically found on the periodic table, as it varies between isotopes.
Atomic number
The atomic number is an essential element of chemistry and one of the primary identifiers of an element. It represents the number of protons present in the nucleus of an atom.
  • The atomic number is symbolized by \( Z \).
  • It directly influences the chemical properties of an element.
  • All atoms of a given element have the same atomic number, ensuring that they share similar chemical characteristics.
For instance, in our example of isotope E, if the atomic number is 14, this indicates that the element is silicon, no matter the number of neutrons. The periodic table is organized by atomic numbers, providing a systematic way to categorize and locate elements based on this property.
Neutron number
The neutron number is the count of neutrons within an atomic nucleus. This number varies among isotopes of a particular element, resulting in differences in their mass numbers.
  • The neutron number is given by the mass number minus the atomic number: \(\text{Number of neutrons} = A - Z\).
  • Knowing the neutron number helps in identifying isotopes and understanding their properties.
  • Unlike protons, which carry a charge, neutrons are neutral, yet they play a crucial role in the stability of an atom's nucleus.
For example, in the isotope Y mentioned in the problem, the neutron number is determined by multiplying its atomic number by 1.33. Different neutron numbers result in isotopes that can have slightly varied physical properties, even though they may behave identically chemically.

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Most popular questions from this chapter

Without doing detailed calculations, determine which of the following samples occupies the largest volume: (a) 25.5 mol of sodium metal \(\left(d=0.971 \mathrm{g} / \mathrm{cm}^{3}\right)\) (b) 0.725 L of liquid bromine \((d=3.12 \mathrm{g} / \mathrm{mL})\) (c) \(1.25 \times 10^{25}\) atoms of chromium metal \(\left(d=9.4 \mathrm{g} / \mathrm{cm}^{3}\right)\) (d) \(2.15 \mathrm{kg}\) of plumber's solder \(\left(d=9.4 \mathrm{g} / \mathrm{cm}^{3}\right), \mathrm{a}\) lead-tin alloy with a 2: 1 atom ratio of lead to tin

Appendix E describes a useful study aid known as concept mapping. Using the method presented in Appendix \(\mathrm{E}\), construct a concept map illustrating the different concepts in Sections \(2-7\) and \(2-8\).

A 0.406 g sample of magnesium reacts with oxygen, producing \(0.674 \mathrm{g}\) of magnesium oxide as the only product. What mass of oxygen was consumed in the reaction?

Within the limits of experimental error, show that the law of conservation of mass was obeyed in the following experiment: \(10.00 \mathrm{g}\) calcium carbonate (found in limestone) was dissolved in 100.0 mL hydrochloric acid \((d=1.148 \mathrm{g} / \mathrm{mL}) .\) The products were \(120.40 \mathrm{g}\) solution (a mixture of hydrochloric acid and calcium chloride) and 2.22 L carbon dioxide gas \((d=1.9769 \mathrm{g} / \mathrm{L})\)

When \(3.06 \mathrm{g}\) hydrogen was allowed to react with an excess of oxygen, \(27.35 \mathrm{g}\) water was obtained. In a second experiment, a sample of water was decomposed by electrolysis, resulting in \(1.45 \mathrm{g}\) hydrogen and 11.51 g oxygen. Are these results consistent with the law of constant composition? Demonstrate why or why not.
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