When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates
\(\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \operatorname{AgCl}(s) \quad \Delta H=-65.5 \mathrm{~kJ}\)
(a) Calculate \(\Delta H\) for the production of 0.450 mol of AgCl by this reaction.
(b) Calculate \(\Delta H\) for the production of 9.00 g of AgCl.
(c) Calculate \(\Delta H\) when \(9.25 \times 10^{-4}\) mol of AgCl dissolves in water.
Text Transcription:
Ag+(aq) + Cl -(aq) \longrightarrow AgCl(s) \Delta H = -65.5 kJ
\Delta H
9.25 X 10-4
Step 1 of 5) Every element exhibits a large increase in ionization energy when the first of its inner-shell electrons is removed. This observation supports the idea that only the outermost electrons are involved in the sharing and transfer of electrons that give rise to chemical bonding and reactions. As we will see when we talk about chemical bonds in Chapters 8 and 9, the inner electrons are too tightly bound to the nucleus to be lost from the atom or even shared with another atom.Analyze and Plan The locations of the elements in the periodic table allow us to predict the electron configurations. The greatest ionization energies involve removal of core electrons. Thus, we should look first for an element with only one electron in the outermost occupied shell. The red box represents Na, which has one valence electron. The second ionization energy of this element is associated, therefore, with the removal of a core electron. The other elements indicated, S (green) and Ca (blue), have two or more valence electrons. Thus, Na should have the largest second ionization energy.
