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Chapter 15: Problem 71

Eggshells are composed mostly of calcium carbonate \(\left(\mathrm{CaCO}_{3}\right)\) formed by the reaction $$ \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftharpoons \mathrm{CaCO}_{3}(s) $$ The carbonate ions are supplied by carbon dioxide produced as a result of metabolism. Explain why eggshells are thinner in the summer, when the rate of chicken panting is greater. Suggest a remedy for this situation.

Short Answer

Expert verified
Eggshells are thinner in the summer because increased chicken panting excretes more carbon dioxide, reducing the availability of carbonate ions for the calcium carbonate \(\mathrm{CaCO}_{3}\) eggshell formation. To remedy this, chickens can be kept cool and given a calcium-rich diet.

Step by step solution

01

Understand the chemical equation

The equation describes how calcium ions \(\mathrm{Ca}^{2+}\) react with carbonate ions \(\mathrm{CO}_{3}^{2-}\) to form calcium carbonate \(\mathrm{CaCO}_{3}\), the main substance in eggshells. This reaction is reversible, meaning it can proceed in both directions.
02

Connect panting to eggshell formation

When chickens pant, they exhale more carbon dioxide (CO2), which is the source of the carbonate ions in the eggshell formation reaction. Thus, increased panting can deplete carbonate ions and reduce the production of calcium carbonate \(\mathrm{CaCO}_{3}\), resulting in thinner eggshells.
03

Suggest a remedy

One possible remedy is to ensure that chickens are kept cool during the summer, with methods such as providing shade or using fans. Furthermore, providing a diet rich in calcium can also help with eggshell formation, as it could help make up for the reduced carbonate ion concentration.

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Most popular questions from this chapter

Consider the dissociation of iodine: $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ A \(1.00-\mathrm{g}\) sample of \(\mathrm{I}_{2}\) is heated at \(1200^{\circ} \mathrm{C}\) in a 500 -mL flask. At equilibrium the total pressure is 1.51 atm. Calculate \(K_{P}\) for the reaction. [Hint: Use the result in problem \(15.65(\mathrm{a}) .\) The degree of dissociation \(\alpha\) can be obtained by first calculating the ratio of observed pressure over calculated pressure, assuming no dissociation.

The "boat" form and "chair" form of cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) interconverts as shown here: In this representation, the \(\mathrm{H}\) atoms are omitted and a \(\mathrm{C}\) atom is assumed to be at each intersection of two lines (bonds). The conversion is first order in each direction. The activation energy for the chair \(\longrightarrow\) boat conversion is \(41 \mathrm{~kJ} / \mathrm{mol}\). If the frequency factor is \(1.0 \times 10^{12} \mathrm{~s}^{-1}\), what is \(k_{1}\) at \(298 \mathrm{~K}\) ? The equilibrium constant \(K_{c}\) for the reaction is \(9.83 \times 10^{3}\) at \(298 \mathrm{~K}\).

Consider this reaction at equilibrium in a closed container: $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ What would happen if (a) the volume is increased, (b) some \(\mathrm{CaO}\) is added to the mixture, (c) some \(\mathrm{CaCO}_{3}\) is removed, (d) some \(\mathrm{CO}_{2}\) is added to the mixture, (e) a few drops of an \(\mathrm{NaOH}\) solution are added to the mixture, (f) a few drops of an \(\mathrm{HCl}\) solution are added to the mixture (ignore the reaction between \(\mathrm{CO}_{2}\) and water \(),(\mathrm{g})\) the temperature is increased?

A quantity of \(6.75 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) was placed in a 2.00 -L flask. At \(648 \mathrm{~K}\), there is 0.0345 mole of \(\mathrm{SO}_{2}\) present. Calculate \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$

Industrially, sodium metal is obtained by electrolyzing molten sodium chloride. The reaction at the cathode is \(\mathrm{Na}^{+}+e^{-} \longrightarrow \mathrm{Na}\). We might expect that potassium metal would also be prepared by electrolyzing molten potassium chloride. However, potassium metal is soluble in molten potassium chloride and therefore is hard to recover. Furthermore, potassium vaporizes readily at the operating temperature, creating hazardous conditions. Instead, potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor at \(892^{\circ} \mathrm{C}\) : $$ \mathrm{Na}(g)+\mathrm{KCl}(l) \rightleftharpoons \mathrm{NaCl}(l)+\mathrm{K}(g) $$ In view of the fact that potassium is a stronger reducing agent than sodium, explain why this approach works. (The boiling points of sodium and potassium are \(892^{\circ} \mathrm{C}\) and \(770^{\circ} \mathrm{C}\), respectively.)
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