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

Short Answer

Expert verified
The chemical formulas for the given pairs of ions are: (a) \(\mathrm{CuBr_2}\), (b) \(\mathrm{Fe_2O_3}\), (c) \(\mathrm{Hg_2CO_3}\), (d) \(\mathrm{Ca_3(AsO_4)_2}\), and (e) \(\mathrm{(NH_4)_2CO_3}\).

Step by step solution

01

(a) 颁耻虏鈦 and 叠谤鈦

First, we will find the least common multiple (LCM) of the charges of both ions to balance them. For 颁耻虏鈦 (Copper) and 叠谤鈦 (Bromine), the LCM of 2 and 1 is 2. Multiply each ion by the required factor to reach the LCM: Cu x 1 and Br x 2, so we have one 颁耻虏鈦 ion combined with two 叠谤鈦 ions. The chemical formula is: \(\mathrm{CuBr_2}\)
02

(b) Fe鲁鈦 and O虏鈦

For Fe鲁鈦 (Iron) and O虏鈦 (Oxygen), the LCM of 3 and 2 is 6. Multiply each ion by the necessary factor: Fe x 2 and O x 3. Combining two Fe鲁鈦 ions with three O虏鈦 ions, we get the chemical formula: \(\mathrm{Fe_2O_3}\)
03

(c) Hg鈧偮测伜 and CO鈧兟测伝

For Hg鈧偮测伜 (Mercury) and CO鈧兟测伝 (Carbonate), the charges are already balanced (虏鈦 and 虏鈦). Thus, we can directly combine them without any extra factor, resulting in the chemical formula: \(\mathrm{Hg_2CO_3}\)
04

(d) Ca虏鈦 and AsO鈧劼斥伝

For Ca虏鈦 (Calcium) and AsO鈧劼斥伝 (Arsenate), the LCM of 2 and 3 is 6. Multiply each ion by the necessary factor: Ca x 3 and AsO鈧 x 2. Combining three Ca虏鈦 ions with two AsO鈧劼斥伝 ions, the chemical formula is: \(\mathrm{Ca_3(AsO_4)_2}\)
05

(e) NH鈧勨伜 and CO鈧兟测伝

For NH鈧勨伜 (Ammonium) and CO鈧兟测伝 (Carbonate), the LCM of 1 and 2 is 2. Multiply each ion by the necessary factor: NH鈧 x 2 and CO鈧 x 1. Combining two NH鈧勨伜 ions with one CO鈧兟测伝 ion, we arrive at the chemical formula: \(\mathrm{(NH_4)_2CO_3}\)

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Key Concepts

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

Ion Combination
Creating a chemical formula involves understanding how ions with opposite charges combine. Ions are atoms or molecules with a net electric charge due to the loss or gain of one or more electrons. In a compound, cations (positively charged ions) and anions (negatively charged ions) are attracted to each other to form a neutral compound.
For instance, in the step involving 颁耻虏鈦 and 叠谤鈦, the ions combine to balance their charges, resulting in a neutral compound,
  • 颁耻虏鈦 (Copper ion with a +2 charge)
  • 叠谤鈦 (Bromine ion with a -1 charge)
The compound is two bromine ions being needed to balance the two positive charges of copper, leading to the formation of CuBr鈧. This process of combining oppositely charged ions is essential for understanding how chemical compounds come together.
Balancing Charges
Balancing charges is a fundamental step in forming neutral chemical compounds from ions. The principle is that the total positive charge from the cations must equal the total negative charge from the anions.
When predicting the formula for a compound from ions, it's crucial to adjust the number of each type of ion to reach overall charge neutrality. In step (b) with Fe鲁鈦 and O虏鈦, we multiply the Fe ion by 2 and the O ion by 3 to get
  • 2 Fe鲁鈦 ions (total charge of +6)
  • 3 O虏鈦 ions (total charge of -6)
Thus, the charges balance to form Fe鈧侽鈧. Balancing charges is central to deriving proper chemical formulas.
Least Common Multiple (LCM)
In chemistry, the Least Common Multiple (LCM) is used to determine the smallest number to balance opposing ionic charges. This ensures the compound's neutrality.
In step (d), involving Ca虏鈦 and AsO鈧劼斥伝, the LCM of 2 and 3 is calculated to be 6.
This tells us how to scale the ions:
  • Ca虏鈦 must be multiplied by 3, giving +6 charge in total
  • AsO鈧劼斥伝 must be multiplied by 2, giving -6 charge in total
The result is a balanced compound: Ca鈧(AsO鈧)鈧. Recognizing the LCM allows us to accurately and efficiently figure out the ratios of ions needed for a neutral compound.
Inorganic Chemistry
Inorganic chemistry deals with compounds that are not based on a carbon-hydrogen structure. This field involves a wide range of compounds, including ionic compounds formed through the transfer of electrons between atoms.
The formation of ionic compounds like Fe鈧侽鈧 and Ca鈧(AsO鈧)鈧 in the exercises illustrates key principles of inorganic chemistry:
  • Understanding how ions interact and form compounds
  • Balancing charges to ensure neutral compounds
Inorganic compounds are vital, making up minerals, ores, and materials essential for technology and everyday life. Grasping the basics of ion combination and balancing charges is pivotal to mastering inorganic chemistry.
Compound Formation
Compound formation is the process by which elements come together to form chemical compounds. This involves the interaction of atoms or ions with opposite charges combining to create stable, neutral compounds.
In the exercise, compound formation is exemplified by NH鈧勨伜 and CO鈧兟测伝 to form (NH鈧)鈧侰O鈧, showing a typical example of ionic compound formation. The process includes:
  • Identifying the ions involved 鈥 Knowing the type of ions is key to combination.
  • Balancing the charges 鈥 Neutrality is achieved by ensuring the total charge is zero.
  • Writing the chemical formula 鈥 Final step involves writing down the correct chemical formula based on the combination of ions.
Mastering compound formation helps understand the properties and reactions of diverse materials.

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Most popular questions from this chapter

Fill in the gaps in the following table: $$ \begin{array}{|l|l|c|c|c|} \hline \text { Symbol } & { }^{59} \mathrm{Co}^{3+} & & & \\ \hline \text { Protons } & & 34 & 76 & 80 \\ \hline \text { Neutrons } & & 46 & 116 & 120 \\ \hline \text { Electrons } & & 36 & & 78 \\ \hline \text { Net charge } & & & 2+ & \\ \hline \end{array} $$

How did Rutherford interpret the following observations made during his \(\alpha\) -particle scattering experiments? (a) Most \(\alpha\) particles were not appreciably deflected as they passed through the gold foil. (b) A few \(\alpha\) particles were deflected at very large angles. (c) What differences would you expect if beryllium foil were used instead of gold foil in the \(\alpha\) -particle scattering experiment?

The compound cyclohexane is an alkane in which six carbon atoms form a ring. The partial structural formula of the compound is as follows: (a) Complete the structural formula for cyclohexane. (b) Is the molecular formula for cyclohexane the same as that for \(n\) -hexane, in which the carbon atoms are in a straight line? If possible, comment on the source of any differences. (c) Propose a structural formula for cyclohexanol, the alcohol derived from cycIohexane.

Give the chemical formula for each of the following ionic compounds: (a) sodium phosphate, (b) zinc nitrate, (c) barium bromate, (d) iron(II) perchlorate, (e) cobalt(II) hydrogen carbonate, (f) chromium(III) acetate, \((\mathrm{g})\) potassium dichromate.

Assume that you encounter the following sentences in your reading. What is the chemical formula for each substance mentioned? (a) Sodium hydrogen carbonate is used as a deodorant. (b) Calcium hypochlorite is used in some bleaching solutions. (c) Hydrogen cyanide is a very poisonous gas. (d) Magnesium hydroxide is used as a cathartic. (e) Tin(II) fluoride has been used as a fluoride additive in toothpastes. (f) When cadmium sulfide is treated with sulfuric acid, fumes of hydrogen sulfide are given off.

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