Chapter 19: Problem 7
In most compounds, the solid phase is denser than the liquid phase. Why isn't this true for water?
Chapter 19: Problem 7
In most compounds, the solid phase is denser than the liquid phase. Why isn't this true for water?
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Get started for freeThere is evidence that radon reacts with fluorine to form compounds similar to those formed by xenon and fluorine. Predict the formulas of these \(\operatorname{RnF}_{x}\) compounds. Why is the chemistry of radon difficult to study?
The compound \(\mathrm{Pb}_{3} \mathrm{O}_{4}\) (red lead) contains a mixture of lead(II) and lead(IV) oxidation states. What is the mole ratio of lead(II) to lead(IV) in \(\mathrm{Pb}_{3} \mathrm{O}_{4} ?\)
Many structures of phosphorus-containing compounds are drawn with some \(P=O\) bonds. These bonds are not the typical \(\pi\) bonds we've considered, which involve the overlap of two \(p\) orbitals. Instead, they result from the overlap of a \(d\) orbital on the phosphorus atom with a \(p\) orbital on oxygen. This type of \(\pi\) bonding is sometimes used as an explanation for why \(\mathrm{H}_{3} \mathrm{PO}_{3}\) has the first structure below rather than the second:Draw a picture showing how a \(d\) orbital and a \(p\) orbital overlap to form a \(\pi\) bond.
The atmosphere contains \(9.0 \times 10^{-6 \%}\) Xe by volume at \(1.0 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) a. Calculate the mass of Xe in a room \(7.26 \mathrm{~m}\) by \(8.80 \mathrm{~m}\) by \(5.67 \mathrm{~m}\) b. A typical person takes in about \(2 \mathrm{~L}\) of air during a breath. How many Xe atoms are inhaled in each breath?
While selenic acid has the formula \(\mathrm{H}_{2} \mathrm{SeO}_{4}\) and thus is directly related to sulfuric acid, telluric acid is best visualized as \(\mathrm{H}_{6} \mathrm{TeO}_{6}\) or \(\mathrm{Te}(\mathrm{OH})_{6}\) a. What is the oxidation state of tellurium in \(\operatorname{Te}(\mathrm{OH})_{6} ?\) b. Despite its structural differences with sulfuric and selenic acid, telluric acid is a diprotic acid with \(\mathrm{p} K_{\mathrm{a}_{1}}=7.68\) and \(\mathrm{p} K_{\mathrm{a}_{2}}=11.29 .\) Telluric acid can be prepared by hydrolysis of tellurium hexafluoride according to the equation $$\operatorname{TeF}_{6}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\operatorname{Te}(\mathrm{OH})_{6}(a q)+6 \mathrm{HF}(a q)$$ Tellurium hexafluoride can be prepared by the reaction of elemental tellurium with fluorine gas:$$\operatorname{Te}(s)+3 \mathrm{F}_{2}(g) \longrightarrow \operatorname{Te} \mathrm{F}_{6}(g)$$.If a cubic block of tellurium (density \(=6.240 \mathrm{g} / \mathrm{cm}^{3}\) ) measuring \(0.545 \mathrm{cm}\) on edge is allowed to react with 2.34 L fluorine gas at 1.06 atm and \(25^{\circ} \mathrm{C}\), what is the \(\mathrm{pH}\) of a solution of \(\mathrm{Te}(\mathrm{OH})_{6}\) formed by dissolving the isolated \(\operatorname{Te} \mathrm{F}_{6}(g)\) in \(115 \mathrm{mL}\) solution? Assume \(100 \%\) yield in all reactions.
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