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In most compounds, the solid phase is denser than the liquid phase. Why isn't this true for water?

Short Answer

Expert verified
In water, the solid phase (ice) has a lower density than the liquid phase due to its molecular structure and hydrogen bonding properties. When water freezes, its molecules form a stable crystalline structure known as hexagonal ice, where hydrogen bonds arrange water molecules in hexagonal patterns. This lattice occupies more space and creates larger empty spaces between the molecules, resulting in a lower density for the solid phase (ice) than for the liquid phase.

Step by step solution

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1. Introduction to density and its relation to water's phases

Density is a measure of how much mass is in a unit volume of a substance, usually represented by the Greek letter ρ (rho), with the formula: \[ρ = \frac{m}{V}\] where m is mass, and V is volume. In general, most compounds become denser as they transition from liquid to solid. This is because their molecules arrange themselves in a more compact structure upon solidifying, thus occupying lesser volume for the same amount of mass. However, water is an exception to this trend, as its solid phase (ice) has a lower density than its liquid phase.
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2. The structure of water molecules

Water is composed of two hydrogen atoms covalently bonded to an oxygen atom, thus forming a molecule with the formula H2O. Due to the difference in electronegativity between hydrogen and oxygen, the oxygen atom attracts the shared electrons more, resulting in a net negative charge around the oxygen atom and a net positive charge around the two hydrogen atoms. This imbalance creates a dipole moment, making water a polar molecule.
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3. Hydrogen bonding

The polarity of water molecules leads to hydrogen bonding, an attractive force between the positively charged hydrogen atoms of one water molecule and the negatively charged oxygen atoms of another water molecule. These relatively strong intermolecular forces create an organized structure in water where each molecule is attracted to its neighbors, affecting the physical properties of water, such as its density.
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4. Density differences in the phases of water

In the liquid phase, water molecules are constantly in motion and the hydrogen bonds are continuously forming and breaking. This allows water molecules to be close together, leading to a relatively high density. However, when water freezes and becomes a solid, a more stable crystalline structure forms, known as hexagonal ice. The hydrogen bonds cause the water molecules to arrange themselves in a hexagonal shape, which occupies more space and thus creates larger empty spaces between the molecules. This results in a lower density for ice compared to liquid water as mass is distributed over a larger volume.
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5. Conclusion

Water is unique due to its molecular structure and hydrogen bonding properties. Unlike most compounds, water's solid phase (ice) is less dense than its liquid phase because of the hexagonal lattice formed by the hydrogen bonds when it freezes. This ordered structure occupies more space and creates larger empty spaces between the molecules, resulting in a lower density for ice than for liquid water.

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Most popular questions from this chapter

The U.S. Public Health Service recommends the fluoridation of water as a means for preventing tooth decay. The recommended concentration is \(1 \mathrm{mg} \mathrm{F}^{-}\) per liter. The presence of calcium ions in hard water can precipitate the added fluoride. What is the maximum molarity of calcium ions in hard water if the fluoride concentration is at the USPHS recommended level? ( \(K_{\text {sp }}\) for \(\left.\mathrm{CaF}_{2}=4.0 \times 10^{-11} .\right)\)

Fluorine reacts with sulfur to form several different covalent compounds. Three of these compounds are \(\mathrm{SF}_{2}, \mathrm{SF}_{4}\), and \(\mathrm{SF}_{6}\). Draw the Lewis structures for these compounds, and predict the molecular structures (including bond angles). Would you expect \(\mathrm{OF}_{4}\) to be a stable compound?

One pathway for the destruction of ozone in the upper atmosphere is $$\begin{array}{l}\mathrm{O}_{3}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad \text { Slow } \\\\\mathrm{NO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g) \quad \text { Fast } \\ \text { Overall reaction: } \mathrm{O}_{3}(g)+\mathrm{O}(g) \rightarrow 2 \mathrm{O}_{2}(g)\end{array}$$ a. Which species is a catalyst? b. Which species is an intermediate? c. The activation energy \(E_{\mathrm{a}}\) for the uncatalyzed reaction $$\mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g)$$ is \(14.0 \mathrm{~kJ} . E_{\mathrm{a}}\) for the same reaction when catalyzed by the presence of \(\mathrm{NO}\) is \(11.9 \mathrm{~kJ} .\) What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at \(25^{\circ} \mathrm{C}\) ? Assume that the frequency factor \(A\) is the same for each reaction. d. One of the concerns about the use of Freons is that they will migrate to the upper atmosphere, where chlorine atoms can be generated by the reaction $$\mathrm{CCl}_{2} \mathrm{~F}_{2} \stackrel{\mathrm{hr}}{\longrightarrow} \mathrm{CF}_{2} \mathrm{Cl}+\mathrm{Cl}$$ Freon- 12 Chlorine atoms also can act as a catalyst for the destruction of ozone. The first step of a proposed mechanism for chlorinecatalyzed ozone destruction is $$\mathrm{Cl}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g)$$ Slow Assuming a two-step mechanism, propose the second step in the mechanism and give the overall balanced equation. e. The activation energy for Cl-catalyzed destruction of ozone is \(2.1 \mathrm{~kJ} / \mathrm{mol}\). Estimate the efficiency with which \(\mathrm{Cl}\) atoms destroy ozone as compared with NO molecules at \(25^{\circ} \mathrm{C}\). Assume that the frequency factor \(A\) is the same for each catalyzed reaction and assume similar rate laws for each catalyzed reaction.

Trisodium phosphate (TSP) is an effective grease remover. Like many cleaners, TSP acts as a base in water. Write a balanced equation to account for this basic behavior.

Use bond energies to estimate the maximum wavelength of light that will cause the reaction $$\mathrm{O}_{3} \stackrel{\mathrm{hr}}{\longrightarrow} \mathrm{O}_{2}+\mathrm{O}$$

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