Chapter 21: Problem 21
What causes high-altitude sickness, and what is high-altitude acclimatization?
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Chapter 21: Problem 21
What causes high-altitude sickness, and what is high-altitude acclimatization?
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Use standard reduction potentials to calculate \(\mathscr{C}^{\circ}, \Delta G^{\circ},\) and \(K\) (at 298 K) for the reaction that is used in production of gold: $$2 \mathrm{Au}(\mathrm{CN})_{2}-(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}(a q)$$ The relevant half-reactions are $$\begin{aligned} \operatorname{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-0.60 \mathrm{V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{C}^{\circ}=-1.26 \mathrm{V} \end{aligned}$$
Will 0.10 mol of AgBr completely dissolve in 1.0 \(\mathrm{L}\) of 3.0 \(\mathrm{M} \mathrm{NH}_{3} ?\) The \(K_{\mathrm{sp}}\) value for \(\mathrm{AgBr}(s)\) is \(5.0 \times 10^{-13},\) and the overall formation constant for the complex ion \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+\) is \(1.7 \times 10^{7}\) , that is, $$\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}$$
For the process $$\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}^{+}(a q)+\mathrm{NH}_{3}(a q)$$ what would be the expected ratio of cis to trans isomers in the product?
When aqueous KI is added gradually to mercury(II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. \(\left(\text {Hint} : \mathrm{Hg}^{2+} \text { reacts with } \mathrm{I}^{-} \text { to form } \mathrm{HgI}_{4}^{2-} .\right)\) Would you expect \(\mathrm{Hg} \mathrm{I}_{4}^{2-}\) to form colored solutions? Explain.
A compound related to acetylacetone is 1,1,1-trifluoroacetylacetone (abbreviated Htfa): Htfa forms complexes in a manner similar to acetylacetone. (See Exercise 49.) Both Be \(^{2+}\) and \(\mathrm{Cu}^{2+}\) form complexes with tfa - having the formula \(\mathrm{M}(\mathrm{tfa})_{2}\) . Two isomers are formed for each metal complex. a. The Be \(^{2+}\) complexes are tetrahedral. Draw the two isomers of \(\mathrm{Be}(\mathrm{tfa})_{2} .\) What type of isomerism is exhibited by \(\mathrm{Be}(\mathrm{tfa})_{2} ?\) b. The \(\mathrm{Cu}^{2+}\) complexes are square planar. Draw the two isomers of \(\mathrm{Cu}(\mathrm{tfa})_{2} .\) What type of isomerism is exhibited by \(\mathrm{Cu}(\mathrm{tfa})_{2} ?\)
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