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Write a balanced equation for the reaction that occurs in each of the following cases: (a) Ozone decomposes to dioxygen. (b) Xenon reacts with fluorine. (Write three different equations.) (c) Sulfur reacts with hydrogen gas. (d) Fluorine reacts with water.

Short Answer

Expert verified
The short answers for each chemical reaction are: (a) Ozone decomposes to dioxygen: \[ 2O_3 \rightarrow 3O_2 \] (b) Xenon reacts with fluorine: 1. Xenon hexafluoride (XeF₆): \[ Xe + 3F_2 \rightarrow XeF_6 \] 2. Xenon tetrafluoride (XeF₄): \[ Xe + 2F_2 \rightarrow XeF_4 \] 3. Xenon difluoride (XeF₂): \[ Xe + F_2 \rightarrow XeF_2 \] (c) Sulfur reacts with hydrogen gas: \[ S_8 + 8H_2 \rightarrow 8H_2S \] (d) Fluorine reacts with water: \[ 2F_2 + 2H_2O \rightarrow 4HF + O_2 \]

Step by step solution

01

a) Ozone Decomposition

To write a balanced chemical equation for the decomposition of ozone (O₃) into dioxygen (O₂), follow these steps: 1. Identify the correct chemical formulas for the reactants and products: Ozone (O₃) decomposes into dioxygen (O₂). 2. Write the unbalanced equation using the correct chemical formulas: O₃ → O₂ 3. Balance the equation by adding coefficients as needed to ensure the same number of atoms for each element on both sides: 2O₃ → 3O₂ 4. Check the number of atoms in the reactants and products, and verify that the equation is balanced. The balanced equation for the decomposition of ozone is: \[ 2O_3 \rightarrow 3O_2 \]
02

b) Xenon Reacting with Fluorine

To write three different balanced chemical equations for the reaction of xenon (Xe) with fluorine (F₂), we need to consider the formation of different xenon fluorides: 1. Xenon hexafluoride (XeF₆) 2. Xenon tetrafluoride (XeF₄) 3. Xenon difluoride (XeF₂) Balanced equations for the formation of these compounds from Xenon and Fluorine are: 1. Xenon hexafluoride (XeF₆): \[ Xe + 3F_2 \rightarrow XeF_6 \] 2. Xenon tetrafluoride (XeF₄): \[ Xe + 2F_2 \rightarrow XeF_4 \] 3. Xenon difluoride (XeF₂): \[ Xe + F_2 \rightarrow XeF_2 \]
03

c) Sulfur Reacting with Hydrogen Gas

To write a balanced chemical equation for the reaction between sulfur (S) and hydrogen gas (H₂), follow these steps: 1. Identify the correct chemical formulas for the reactants and products: Sulfur (S₈) reacts with hydrogen gas (H₂) to form hydrogen sulfide (H₂S). 2. Write the unbalanced equation using the correct chemical formulas: S₈ + H₂ → H₂S 3. Balance the equation by adding coefficients as needed to ensure the same number of atoms for each element on both sides: S₈ + 8H₂ → 8H₂S 4. Check the number of atoms in the reactants and products, and verify that the equation is balanced. The balanced equation for the reaction between sulfur and hydrogen gas is: \[ S_8 + 8H_2 \rightarrow 8H_2S \]
04

d) Fluorine Reacting with Water

To write a balanced chemical equation for the reaction between fluorine (F₂) and water (H₂O), follow these steps: 1. Identify the correct chemical formulas for the reactants and products: Fluorine (F₂) reacts with water (H₂O) to form hydrogen fluoride (HF) and oxygen (O₂). 2. Write the unbalanced equation using the correct chemical formulas: F₂ + H₂O → HF + O₂ 3. Balance the equation by adding coefficients as needed to ensure the same number of atoms for each element on both sides: 2F₂ + 2H₂O → 4HF + O₂ 4. Check the number of atoms in the reactants and products, and verify that the equation is balanced. The balanced equation for the reaction between fluorine and water is: \[ 2F_2 + 2H_2O \rightarrow 4HF + O_2 \]

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Key Concepts

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

Understanding Chemical Reactions
Chemical reactions are fundamental processes where reactants are transformed into products through the breaking and forming of chemical bonds. For example, in the decomposition of ozone (O₃) into dioxygen (O₂), the ozone molecules break apart to form dioxygen molecules.

Each chemical reaction can be represented by a chemical equation, which uses chemical formulas to denote the substances involved. The equation must be balanced to comply with the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction.

This balance means having an equal number of atoms for each element on both sides of the equation. Like solving a puzzle, coefficients are added before the chemical formulas to achieve this balance.
The Role of Stoichiometry in Chemical Equations
Stoichiometry is akin to a recipe for a chemical reaction. It provides the exact proportions of reactants and products involved in a chemical process. Using the decomposition of ozone as an example, stoichiometry tells us that two ozone molecules will produce three molecules of dioxygen.

Understanding stoichiometry is crucial because it allows chemists to predict the amounts of substances consumed and produced in a reaction. This knowledge is invaluable in various applications, such as manufacturing, where precise quantities of materials are required to ensure a successful and economical reaction.

In the classroom, practicing stoichiometry helps students develop problem-solving skills by combining concepts of chemistry with mathematical calculations.
Interpreting Chemical Formulas
Chemical formulas are a shorthand notation to describe the composition of molecules and compounds. They provide information on the types and numbers of atoms that make up a substance. For instance, the formula for water, H₂O, indicates that it consists of two hydrogen atoms bonded to a single oxygen atom.

Recognizing and writing correct chemical formulas is essential for balancing chemical equations. Incorrect formulas can lead to unbalanced equations and misunderstandings of the reaction chemistry. The formulas are also the foundation for identifying the types of bonds, visualizing molecular structures, and understanding the physical and chemical properties of substances.

For students working on homework exercises like the reactions of xenon with fluorine, practice with chemical formulas is not only about getting the right answer but also about seeing the bigger picture in chemical compound formation.

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Most popular questions from this chapter

The As - As bond length in elemental arsenic is \(2.48 \AA\). The \(\mathrm{Cl}-\mathrm{Cl}\) bond length in \(\mathrm{Cl}_{2}\) is \(1.99 \AA\). (a) Based on these data, what is the predicted As - Cl bond length in arsenic trichloride, \(\mathrm{AsCl}_{3}\), in which each of the three \(\mathrm{Cl}\) atoms is bonded to the As atom? (b) What bond length is predicted for \(\mathrm{AsCl}_{3}\), using the atomic radii in Figure 7.7?

The electron affinities, in \(\mathrm{kJ} / \mathrm{mol}\), for the group \(1 \mathrm{~B}\) and group 2B metals are as follows: (a) Why are the electron affinities of the group \(2 B\) elements greater than zero? (b) Why do the electron affinities of the group \(1 \mathrm{~B}\) elements become more negative as we move down the group? [Hint: Examine the trends in the electron affinities of other groups as we proceed down the periodic table.]

(a) Why does xenon react with fluorine, whereas neon does not? (b) Using appropriate reference sources, look up the bond lengths of Xe-F bonds in several molecules. How do these numbers compare to the bond lengths calculated from the atomic radii of the elements?

Which of the following chemical equations is connected to the definitions of (a) the first ionization energy of oxygen, (b) the second ionization energy of oxygen, and (c) the electron affinity of oxygen? (i) \(\mathrm{O}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{O}^{-}(g)\) (ii) \(\mathrm{O}(g) \longrightarrow \mathrm{O}^{+}(g)+\mathrm{e}^{-}\) (iii) \(\mathrm{O}(g)+2 \mathrm{e}^{-} \longrightarrow \mathrm{O}^{2-}(g)\) (iv) \(\mathrm{O}(g) \longrightarrow \mathrm{O}^{2+}(g)+2 \mathrm{e}^{-}\) (v) \(\mathrm{O}^{+}(g) \longrightarrow \mathrm{O}^{2+}(g)+\mathrm{e}^{-}\)

Note from the following table that there is a significant increase in atomic radius upon moving from \(\mathrm{Y}\) to La whereas the radii of \(\mathrm{Zr}\) to \(\mathrm{Hf}\) are the same. Suggest an explanation for this effect. $$ \begin{array}{llll} \hline & {\text { Atomic Radii (Å) }} \\ \hline \mathrm{Sc} & 1.70 & \mathrm{Ti} & 1.60 \\ \mathrm{Y} & 1.90 & \mathrm{Zr} & 1.75 \\ \mathrm{La} & 2.07 & \text { Hf } & 1.75 \\ \hline \end{array} $$

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