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Chapter 2: Problem 76

Name the compounds in parts a–d and write the formulas for the compounds in parts e–h. a. \(\mathrm{Hg}_{2} \mathrm{O}\) b. \(\mathrm{FeBr}_{3}\) c. \(\mathrm{CoS}\) d. \(\mathrm{TiCl}_{4}\) e. tin(II) nitride f. cobalt(III) iodide g. mercury(II) oxide h. chromium(VI) sulfide

Short Answer

Expert verified
a. mercury(I) oxide b. iron(III) bromide c. cobalt(II) sulfide d. titanium(IV) chloride e. \(\mathrm{Sn}_{3} \mathrm{N}_{2}\) f. \(\mathrm{CoI}_{3}\) g. \(\mathrm{HgO}\) h. \(\mathrm{CrS}_{3}\)

Step by step solution

01

Part a: Naming \(\mathrm{Hg}_{2} \mathrm{O}\)

The compound \(\mathrm{Hg}_{2} \mathrm{O}\) consists of mercury (Hg) and oxygen (O). Oxygen has a charge of -2. Since there are two mercury atoms, each mercury atom in this compound has a charge of +1. Therefore, the compound is named mercury(I) oxide.
02

Part b: Naming \(\mathrm{FeBr}_{3}\)

The compound \(\mathrm{FeBr}_{3}\) consists of iron (Fe) and bromine (Br). Bromine has a charge of -1, and since there are 3 bromine atoms, the overall charge on iron in this compound should be +3. Therefore, the compound is named iron(III) bromide.
03

Part c: Naming \(\mathrm{CoS}\)

The compound \(\mathrm{CoS}\) consists of cobalt (Co) and sulfur (S). Sulfur has a charge of -2, and the cobalt atom has a charge of +2. Therefore, the compound is named cobalt(II) sulfide.
04

Part d: Naming \(\mathrm{TiCl}_{4}\)

The compound \(\mathrm{TiCl}_{4}\) consists of titanium (Ti) and chlorine (Cl). Chlorine has a charge of -1, and since there are 4 chlorine atoms in the compound, the overall charge on titanium should be +4. Therefore, the compound is named titanium(IV) chloride.
05

Part e: Writing the formula for tin(II) nitride

Tin(II) nitride contains tin (Sn) with a charge of +2 and nitrogen (N) with a charge of -3. To balance the charges, we need three tin atoms with a total charge of +6 and two nitrogen atoms with a total charge of -6. Therefore, the formula for tin(II) nitride is \(\mathrm{Sn}_{3} \mathrm{N}_{2}\).
06

Part f: Writing the formula for cobalt(III) iodide

Cobalt(III) iodide contains cobalt (Co) with a charge of +3 and iodine (I) with a charge of -1. To balance the charges, we need one cobalt atom with a charge of +3 and three iodine atoms with a total charge of -3. Therefore, the formula for cobalt(III) iodide is \(\mathrm{CoI}_{3}\).
07

Part g: Writing the formula for mercury(II) oxide

Mercury(II) oxide contains mercury (Hg) with a charge of +2 and oxygen (O) with a charge of -2. The charges are already balanced, so the formula for mercury(II) oxide is \(\mathrm{HgO}\).
08

Part h: Writing the formula for chromium(VI) sulfide

Chromium(VI) sulfide contains chromium (Cr) with a charge of +6 and sulfur (S) with a charge of -2. To balance the charges, we need three sulfur atoms with a total charge of -6 and one chromium atom with a charge of +6. Therefore, the formula for chromium(VI) sulfide is \(\mathrm{CrS}_{3}\).

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Most popular questions from this chapter

In a combustion reaction, 46.0 g of ethanol reacts with 96.0 g of oxygen to produce water and carbon dioxide. If 54.0 g of water is produced, what mass of carbon dioxide is produced?

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When mixtures of gaseous \(\mathrm{H}_{2}\) and gaseous \(\mathrm{Cl}_{2}\) react, a product forms that has the same properties regardless of the relative amounts of \(\mathrm{H}_{2}\) and \(\mathrm{Cl}_{2}\) used. a. How is this result interpreted in terms of the law of definite proportion? b. When a volume of \(\mathrm{H}_{2}\) reacts with an equal volume of \(\mathrm{Cl}_{2}\) at the same temperature and pressure, what volume of product having the formula HCl is formed?
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