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Chapter 2: Problem 11

Why do we call \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) barium nitrate, but we call \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}\) iron(II) nitrate?

Short Answer

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We call \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) barium nitrate because barium, an alkaline earth metal, has only one possible charge (\(\mathrm{Ba}^{2+}\)). However, we call \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}\) iron(II) nitrate because iron, a transition metal, can have different charges. In this case, it has a \(\mathrm{Fe}^{2+}\) charge, so we specify the oxidation state using a Roman numeral in parentheses – iron(II).

Step by step solution

01

Analyze the chemical formula for barium nitrate

Since barium is a member of the alkaline earth metals group, it only has one possible charge, \(\mathrm{Ba}^{2+}\). So in the compound \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\), there will be no confusion regarding the charge of barium.
02

Naming barium nitrate

As there is no ambiguity in the oxidation state of barium, we can simply name the compound by combining the names of cation and anion: barium nitrate.
03

Analyze the chemical formula for iron(II) nitrate

In the compound \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}\), the cation is iron, which is a transition metal that can have different charges. In this case, it has a charge of \(\mathrm{Fe}^{2+}\). To indicate the specific oxidation state of iron, we will use a Roman numeral in parentheses after the cation's name.
04

Naming iron(II) nitrate

Given that the iron cation has a \(2+\) charge, we write the name as iron(II), followed by the name of the anion, nitrate. So the compound is named iron(II) nitrate.

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Most popular questions from this chapter

A chemist in a galaxy far, far away performed the Millikan oil drop experiment and got the following results for the charges on various drops. Use these data to calculate the charge of the electron in zirkombs. \(2.56 \times 10^{-12}\) zirkombs \(\quad 7.68 \times 10^{-12}\) zirkombs \(3.84 \times 10^{-12}\) zirkombs \(\quad 6.40 \times 10^{-13}\) zirkombs

A sample of chloroform is found to contain \(12.0 \mathrm{~g}\) of carbon, \(106.4 \mathrm{~g}\) of chlorine, and \(1.01 \mathrm{~g}\) of hydrogen. If a second sample of chloroform is found to contain \(30.0 \mathrm{~g}\) of carbon, what is the total mass of chloroform in the second sample?

Four \(\mathrm{Fe}^{2+}\) ions are key components of hemoglobin, the protein that transports oxygen in the blood. Assuming that these ions are \({ }^{53} \mathrm{Fe}^{2+}\), how many protons and neutrons are present in each nucleus, and how many electrons are present in each ion?

Each of the following compounds is incorrectly named. What is wrong with each name, and what is the correct name for each compound? a. \(\mathrm{FeCl}_{3}\), iron chloride b. \(\mathrm{NO}_{2}\), nitrogen(IV) oxide c. \(\mathrm{CaO}\), calcium(II) monoxide d. \(\mathrm{Al}_{2} \mathrm{~S}_{3}\), dialuminum trisulfide

You take three compounds, each consisting of two elements \((\mathrm{X}, \mathrm{Y}\), and \(/ \mathrm{or} \mathrm{Z})\), and decompose them to their respective elements. To determine the relative masses of \(\mathrm{X}, \mathrm{Y}\), and \(\mathrm{Z}\), you collect and weigh the elements, obtaining the following data: $$ \begin{array}{|ll|} \hline \text { Elements in Compound } & \text { Masses of Elements } \\ \hline \text { 1. } \mathrm{X} \text { and } \mathrm{Y} & \mathrm{X}=0.4 \mathrm{~g}, \mathrm{Y}=4.2 \mathrm{~g} \\ \text { 2. Y and } \mathrm{Z} & \mathrm{Y}=1.4 \mathrm{~g}, \mathrm{Z}=1.0 \mathrm{~g} \\ \text { 3. } \mathrm{X} \text { and } \mathrm{Y} & \mathrm{X}=2.0 \mathrm{~g}, \mathrm{Y}=7.0 \mathrm{~g} \\ \hline \end{array} $$ a. What are the assumptions needed to solve this problem? b. What are the relative masses of \(\mathrm{X}, \mathrm{Y}\), and \(\mathrm{Z} ?\) c. What are the chemical formulas of the three compounds? d. If you decompose \(21 \mathrm{~g}\) of compound \(\mathrm{XY}\), how much of each element is present?
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