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Chapter 4: Problem 30

Commercial cold packs and hot packs are available for treating athletic injuries. Both types contain a pouch of water and a dry chemical. When the pack is struck, the pouch of water breaks, dissolving the chemical, and the solution becomes either hot or cold. Many hot packs use magnesium sulfate, and many cold packs use ammonium nitrate. Write reactions to show how these strong electrolytes break apart when they dissolve in water

Short Answer

Expert verified
Magnesium sulfate (\(MgSO_4\)) and ammonium nitrate (\(NH_4NO_3\)) are strong electrolytes that dissociate in water. The dissociation reactions are: 1. \(MgSO_4(s) \xrightarrow{H_2O} Mg^{2+}(aq) + SO_4^{2-}(aq)\) 2. \(NH_4NO_3(s) \xrightarrow{H_2O} NH_4^{+}(aq) + NO_3^{-}(aq)\)

Step by step solution

01

1. Magnesium sulfate dissolving in water

1. Write the chemical formula of magnesium sulfate: Magnesium sulfate has the chemical formula \(MgSO_4\). 2. Identify the ions that magnesium sulfate breaks into: When dissolved in water, magnesium sulfate breaks into magnesium ions and sulfate ions. The ions are: - Magnesium ion(\(Mg^{2+}\)) - Sulfate ion(\(SO_4^{2-}\)) 3. Write the balanced dissociation reaction: When magnesium sulfate dissolves in water, it dissociates into its ions as follows: \[MgSO_4(s) \xrightarrow{H_2O} Mg^{2+}(aq) + SO_4^{2-}(aq)\]
02

2. Ammonium nitrate dissolving in water

1. Write the chemical formula of ammonium nitrate: Ammonium nitrate has the chemical formula \(NH_4NO_3\). 2. Identify the ions that ammonium nitrate breaks into: When dissolved in water, ammonium nitrate breaks into ammonium ions and nitrate ions. The ions are: - Ammonium ion(\(NH_4^+\)) - Nitrate ion(\(NO_3^-\)) 3. Write the balanced dissociation reaction: When ammonium nitrate dissolves in water, it dissociates into its ions as follows: \[NH_4NO_3(s) \xrightarrow{H_2O} NH_4^{+}(aq) + NO_3^{-}(aq)\]

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Most popular questions from this chapter

Chromium has been investigated as a coating for steel cans. The thickness of the chromium film is determined by dissolving a sample of a can in acid and oxidizing the resulting \(\mathrm{Cr}^{3+}\) to $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ with the peroxydisulfate ion: $$\begin{aligned} \mathrm{S}_{2} \mathrm{O}_{\mathrm{x}}^{2-}(a q)+\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \\ &+\mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q)(\text { Unbalanced }) \end{aligned} $$ After removal of unreacted \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-},\) an excess of ferrous ammonium sulfate $\left[\mathrm{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\right]$ is added, reacting with \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) produced from the first reaction. The unreacted \(\mathrm{Fe}^{2+}\) from the excess ferrous ammonium sulfate is titrated with a separate \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution. The reaction is: $$\begin{array}{c}{\mathrm{H}^{+}(a q)+\mathrm{Fe}^{2+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)} \\\ {+\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(I) \text { (Unbalanced) }}\end{array}$$ a. Write balanced chemical equations for the two reactions. b. In one analysis, a \(40.0-\mathrm{cm}^{2}\) sample of a chromium-plated can was treated according to this procedure. After dissolution and removal of excess \(\mathrm{S}_{2} \mathrm{O}_{8}^{2}-, 3.000 \mathrm{g}\) of $\operatorname{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\( was added. It took 8.58 \)\mathrm{mL}$ of 0.0520 \(\mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution to completely react with the excess \(\mathrm{Fe}^{2+}\) . Calculate the thickness of the chromium film on the can. (The density of chromium is 7.19 $\mathrm{g} / \mathrm{cm}^{3} .$ )

Separate samples of a solution of an unknown soluble ionic compound are treated with \(\mathrm{KCl}, \mathrm{Na}_{2} \mathrm{SO}_{4},\) and \(\mathrm{NaOH}\) . A precipitate forms only when $\mathrm{Na}_{2} \mathrm{SO}_{4}$ is added. Which cations could be present in the unknown soluble ionic compound?

A 0.500 -L sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution was analyzed by taking a 100.0 -mL aliquot and adding 50.0 \(\mathrm{mL}\) of 0.213 \(\mathrm{M}\) NaOH. After the reaction occurred, an excess of \(\mathrm{OH}^{-}\) ions remained in the solution. The excess base required 13.21 \(\mathrm{mL}\) of 0.103 \(\mathrm{M}\) \(\mathrm{HCl}\) for neutralization. Calculate the molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) . Sulfuric acid has two acidic hydrogens.

Carminic acid, a naturally occurring red pigment extracted from the cochineal insect, contains only carbon, hydrogen, and oxygen. It was commonly used as a dye in the first half of the nineteenth century. It is 53.66% C and 4.09% H by mass. A titration required 18.02 mL of 0.0406 M NaOH to neutralize 0.3602 g carminic acid. Assuming that there is only one acidic hydrogen per molecule, what is the molecular formula of carminic acid?

The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$\mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }}$$ In the case of very dilute aqueous solutions, a concentration of 1.0 \(\mathrm{ppm}\) is equal to 1.0\(\mu \mathrm{g}\) of solute per 1.0 \(\mathrm{mL}\) , which equals 1.0 \(\mathrm{g}\) solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. 5.0 \(\mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. 1.0 \(\mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. 10.0 \(\mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. 0.10 $\mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{C} 1_{5}\right)\( in \)\mathrm{H}_{2} \mathrm{O}$
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