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Give the formula and the number of each ion that makes up each of the following compounds: (a) \(\mathrm{K}_{2} \mathrm{S}\) (b) \(\cos \mathrm{O}_{4}\) (c) \(\mathrm{KMnO}_{4}\) (d) \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4}\) (e) \(\mathrm{Ca}(\mathrm{ClO})_{2}\) (f) \(\mathrm{NaCH}_{3} \mathrm{CO}_{2}\)

Short Answer

Expert verified
(a) 2 K鈦, 1 S虏鈦; (b) Co and O ions; (c) 1 K鈦, 1 MnO鈧勨伝; (d) 3 NH鈧勨伜, 1 PO鈧劼斥伝; (e) 1 Ca虏鈦, 2 ClO鈦; (f) 1 Na鈦, 1 CH鈧僀O鈧傗伝.

Step by step solution

01

Understanding Compound Ionization

To solve this problem, you need to identify the ions present in each of the chemical compounds listed. This involves breaking down each compound into its constituent ions and understanding their ratios within the compound.
02

Identifying Ions in Potassium Sulfide

For the compound \( \mathrm{K}_2\mathrm{S} \): Potassium (K) is a metal ion with a charge of +1, represented as \( \mathrm{K}^+ \). Sulfide (S) is a non-metal ion with a charge of -2, represented as \( \mathrm{S}^{2-} \). The formula \( \mathrm{K}_2\mathrm{S} \) indicates there are two potassium ions for every sulfide ion.
03

Identifying Ions in Cobalt Tetraoxide

For the compound \( \cos \mathrm{O}_4 \): Cobalt (Co) is typically found in +2 or +3 ionic forms. Since oxygen is usually found in the -2 form, the compound could suggest that cobalt is acting in such a way for charge balance, but \( \cos \mathrm{O}_4 \) isn't a conventional nomenclature for known cobalt oxides; it's more likely miswritten. For illustrative purposes, assuming \( \mathrm{Co}_3\mathrm{O}_4 \) or similar would mean each cobalt ion balances with oxygen in terms of total charge.
04

Identifying Ions in Potassium Permanganate

For the compound \( \mathrm{KMnO}_4 \): Potassium (K) as \( \mathrm{K}^+ \) and permanganate as \( \mathrm{MnO}_4^- \). The formula indicates one potassium ion per permanganate ion.
05

Identifying Ions in Ammonium Phosphate

For the compound \( \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \): Ammonium is \( \mathrm{NH}_4^+ \), and phosphate is \( \mathrm{PO}_4^{3-} \). The formula \( \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \) indicates three ammonium ions for every phosphate ion.
06

Identifying Ions in Calcium Hypochlorite

For the compound \( \mathrm{Ca} (\mathrm{ClO})_2 \): Calcium is \( \mathrm{Ca}^{2+} \), and hypochlorite is \( \mathrm{ClO}^- \). The formula \( \mathrm{Ca} (\mathrm{ClO})_2 \) indicates one calcium ion for every two hypochlorite ions.
07

Identifying Ions in Sodium Acetate

For the compound \( \mathrm{NaCH}_{3}\mathrm{CO}_{2} \): Sodium is \( \mathrm{Na}^+ \), and acetate is \( \mathrm{CH}_3\mathrm{CO}_2^- \). The formula \( \mathrm{NaCH}_3\mathrm{CO}_2 \) suggests one sodium ion per acetate ion.

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Key Concepts

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

Potassium Sulfide
Potassium sulfide is a chemical compound with the formula \( \mathrm{K}_{2} \mathrm{S} \). It is composed of two potassium ions and one sulfide ion. Potassium is an alkali metal and tends to lose one electron to form a \( \mathrm{K}^{+} \) ion. This results from its position in Group 1 of the periodic table, which makes it highly reactive. On the other hand, the sulfide ion \( \mathrm{S}^{2-} \) is formed when sulfur gains two electrons, due to its inclination to fill its outer shell. Thus, in \( \mathrm{K}_{2} \mathrm{S} \):
  • There are two \( \mathrm{K}^{+} \) ions.
  • There is one \( \mathrm{S}^{2-} \) ion.
Together, they create a stable ionic compound by balancing the charges鈥2 positive from potassium and 2 negative from sulfide, resulting in electrical neutrality.
Cobalt Oxides
Cobalt oxides are a group of compounds between cobalt and oxygen. The formula presented, \( \cos \mathrm{O}_4 \), requires clarification as it seems to be a misnomer for a common cobalt oxide like \( \mathrm{Co}_3 \mathrm{O}_4 \). Cobalt can exist in multiple oxidation states, often +2 and +3.
  • In \( \mathrm{Co}_3 \mathrm{O}_4 \), the compound is made of a mix of \( \mathrm{Co}^{2+} \) and \( \mathrm{Co}^{3+} \) ions.
  • Oxygen remains as \( \mathrm{O}^{2-} \) ions.
This mixed oxide, \( \mathrm{Co}_3 \mathrm{O}_4 \), is a common form, reflecting both cobalt鈥檚 ability to participate in different charge states and the need for charge neutrality in oxides.
Potassium Permanganate
Potassium permanganate is a vibrant purple compound with the formula \( \mathrm{KMnO}_{4} \). It consists of potassium ions and permanganate ions. Potassium, as always, appears as \( \mathrm{K}^{+} \) because it's keen to lose its single valence electron.
  • The permanganate ion, \( \mathrm{MnO}_{4}^{-} \), is a polyatomic ion where manganese adopts a high oxidation state of +7, which is balanced by four oxygen atoms each bearing a \( \mathrm{O}^{2-} \) charge.
  • The compound thus contains one \( \mathrm{K}^{+} \) ion and one \( \mathrm{MnO}_{4}^{-} \) ion.
This matchup creates a robust oxidizing agent used widely in chemical reactions as well as in water treatment processes.
Ammonium Phosphate
Ammonium phosphate is represented by the formula \( \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \). This compound combines ammonium and phosphate ions:
  • \( \mathrm{NH}_{4}^{+} \) is the ammonium ion, and it's a common ion formed by the combination of nitrogen and hydrogen.
  • \( \mathrm{PO}_{4}^{3-} \) is the phosphate ion, a tetrahedral structure where phosphorus is surrounded by four oxygen atoms, maintaining a 3- charge.
In the compound's formula, the three ammonium ions \( (3 \times +1) \) balance out the charge of a single phosphate ion \( (-3) \), ensuring the overall charge is zero, crucial for maintaining the stability of the compound.
Calcium Hypochlorite
Calcium hypochlorite appears in the formula \( \mathrm{Ca} (\mathrm{ClO})_{2} \). It is widely used as a bleaching agent and disinfectant, due to its capacity to release chlorine when dissolved in water.
  • Calcium forms a \( \mathrm{Ca}^{2+} \) ion, losing two electrons.
  • The hypochlorite ion, \( \mathrm{ClO}^{-} \), involves chlorine and oxygen, where the single negative charge is crucial for its bleaching ability.
In this compound, one \( \mathrm{Ca}^{2+} \) ion pairs with two \( \mathrm{ClO}^{-} \) ions to achieve charge balance and form a stable compound, which decomposes to release useful free chlorine.
Sodium Acetate
Sodium acetate is an ionic compound marked by the formula \( \mathrm{NaCH}_{3} \mathrm{CO}_{2} \), and it's frequently encountered as a buffering agent in laboratories.
  • Sodium enters the scene as \( \mathrm{Na}^{+} \), surrendering one electron.
  • The acetate ion, \( \mathrm{CH}_{3} \mathrm{CO}_{2}^{-} \), is a derivative of acetic acid, where the carboxylic group \( (COO) \) holds the negative charge.
The pairing of \( \mathrm{Na}^{+} \) and \( \mathrm{CH}_{3} \mathrm{CO}_{2}^{-} \) results in a neutral compound and highlights its role in maintaining pH balance through its weak acid-base nature.

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Most popular questions from this chapter

Estimating the radius of a lead atom. (a) You are given a cube of lead that is \(1.000 \mathrm{cm}\) on each side. The density of lead is \(11.35 \mathrm{g} /\) \(\mathrm{cm}^{3} .\) How many atoms of lead are in the sample? (b) Atoms are spherical; therefore, the lead atoms in this sample cannot fill all the available space. As an approximation, assume that \(60 \%\) of the space of the cube is filled with spherical lead atoms. Calculate the volume of one lead atom from this information. From the calculated volume (V) and the formula \((4 / 3) \pi r^{3}\) for the volume of a sphere, estimate the radius \((r)\) of a lead atom.

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