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Chapter 2: Problem 56

Predict the chemical formulas of the compounds formed by (b) \(\mathrm{Fe}^{3+}\) and the following pairs of ions: (a) \(\mathrm{Cr}^{3+}\) and \(\mathrm{Br}^{-}\), \(\mathrm{O}^{2-},(\mathrm{c}) \mathrm{Hg}_{2}^{2+}\) and \(\mathrm{CO}_{3}^{2-},(\mathrm{d}) \mathrm{Ca}^{2+}\) and \(\mathrm{ClO}_{3}^{-},(\mathrm{e}) \mathrm{NH}_{4}^{+}\) and \(\mathrm{PO}_{4}{ }^{3-}\).

Short Answer

Expert verified
The chemical formulas of the compounds formed by the given pairs of ions are: (a) \(\mathrm{CrBr_3}\) (b) \(\mathrm{Fe_2O_3}\) (c) \(\mathrm{Hg_2CO_3}\) (d) \(\mathrm{Ca(ClO_3)_2}\) (e) \(\mathrm{(NH_4)_3PO_4}\)

Step by step solution

01

Pair (a): \(\mathrm{Cr}^{3+}\) and \(\mathrm{Br}^{-}\)

Since both Cr and Br have equal but opposite charges, they will combine in a 1:1 ratio. The chemical formula for the compound formed by \(\mathrm{Cr}^{3+}\) and \(\mathrm{Br}^{-}\) is \(\mathrm{CrBr_3}\).
02

Pair (b): \(\mathrm{Fe}^{3+}\) and \(\mathrm{O}^{2-}\)

In this pair, since the charges on Fe and O ions are not equal, we will need to find the least common multiple (LCM) of their absolute charge values, which is 6. Thus, the ratio of \(\mathrm{Fe}^{3+}\) to \(\mathrm{O}^{2-}\) should be 2:3 to achieve charge balance. The chemical formula for the compound formed is \(\mathrm{Fe_2O_3}\).
03

Pair (c): \(\mathrm{Hg}_{2}^{2+}\) and \(\mathrm{CO}_{3}^{2-}\)

In this pair, the charges on \(\mathrm{Hg}_{2}^{2+}\) and \(\mathrm{CO}_3^{2-}\) are equal but opposite. Therefore, they will combine in a 1:1 ratio. The chemical formula for the compound formed is \(\mathrm{Hg_2CO_3}\).
04

Pair (d): \(\mathrm{Ca}^{2+}\) and\(\mathrm{ClO}_{3}^{-}\)

Here, the positive charge on \(\mathrm{Ca}^{2+}\) is twice the negative charge on \(\mathrm{ClO}_3^{-}\). Therefore, we need two \(\mathrm{ClO}_3^{-}\) ions to balance the positive charge of one \(\mathrm{Ca}^{2+}\) ion. The chemical formula for the compound formed will be \(\mathrm{Ca(ClO_3)_2}\).
05

Pair (e): \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{PO}_{4}^{3-}\)

In this case, the positive charge on \(\mathrm{NH}_4^+\) is three times smaller than the negative charge on \(\mathrm{PO}_4^{3-}\). Hence, we require three \(\mathrm{NH}_4^+\) ions to balance the negative charge of one \(\mathrm{PO}_4^{3-}\) ion. The chemical formula for the compound formed is \(\mathrm{(NH_4)_3PO_4}\).

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Most popular questions from this chapter

Name the following ionic compounds: (a) \(\mathrm{KCN},\) (b) \(\mathrm{NaBrO}_{2}\), (c) \(\mathrm{Sr}(\mathrm{OH})_{2}\), (d) CoS, (e) \(\mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3}\) (f) \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}, \quad(\mathrm{~g})\) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{3},(\mathrm{~h}) \mathrm{NaH}_{2} \mathrm{PO}_{4},\) (i) \(\mathrm{KMnO}_{4},(\mathrm{j}) \mathrm{Ag}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\)

Give the name or chemical formula, as appropriate, for each of the following acids: (a) \(\mathrm{HBrO}_{3},\) (b) \(\mathrm{HBr}\), (c) \(\mathrm{H}_{3} \mathrm{PO}_{4}\), (d) hypochlorous acid, \((e)\) iodic acid, \((f)\) sulfurous acid.

Answer the following questions without referring to Table 2.1: (a) What are the main subatomic particles that make up the atom? (b) What is the relative charge (in multiples of the electronic charge) of each of the particles? (c) Which of the particles is the most massive? (d) Which is the least massive?

Draw the structural formulas for three isomers of pentane, \(\mathrm{C}_{5} \mathrm{H}_{12}\) $$ \begin{array}{cc} \hline \text { Droplet } & \text { Calculated Charge (wa) } \\ \hline \text { A } & 3.84 \times 10^{-8} \\ \text {B } & 4.80 \times 10^{-8} \\ \text {C } & 2.88 \times 10^{-8} \\ \text {D } & 8.64 \times 10^{-8} \\ \hline \end{array} $$ (c) Based on your answer to part (b), how many electrons are there on each of the droplets? (d) What is the conversion factor between warmombs and coulombs?

Determine the molecular and empirical formulas of the following: (a) the organic solvent benzene, which has six carbon atoms and six hydrogen atoms; (b) the compound silicon tetrachloride, which has a silicon atom and four chlorine atoms and is used in the manufacture of computer chips; (c) the reactive substance diborane, which has two boron atoms and six hydrogen atoms; (d) the sugar called glucose, which has six carbon atoms, twelve hydrogen atoms, and six oxygen atoms.
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