Chapter 20: Problem 46
Tin forms compounds in the \(+2\) and \(+4\) oxidation states. Therefore, when tin reacts with fluorine, two products are possible. Write balanced equations for the production of the two tin halide compounds and name them.
Chapter 20: Problem 46
Tin forms compounds in the \(+2\) and \(+4\) oxidation states. Therefore, when tin reacts with fluorine, two products are possible. Write balanced equations for the production of the two tin halide compounds and name them.
All the tools & learning materials you need for study success - in one app.
Get started for freeWhy are the tin(IV) halides more volatile than the \(\operatorname{tin}(\mathrm{II})\) halides?
Write balanced equations describing the reaction of lithium metal with each of the following: $\mathrm{O}_{2}, \mathrm{S}, \mathrm{Cl}_{2}, \mathrm{P}_{2}, \mathrm{H}_{2}, \mathrm{H}_{2} \mathrm{O},\( and \)\mathrm{HCl} .$
Slaked lime, \(\mathrm{Ca}(\mathrm{OH})_{2},\) is used to soften hard water by removing calcium ions from hard water through the reaction $$ \begin{array}{r}{\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{Ca}^{2+}(a q)+2 \mathrm{HCO}_{3}^{-}(a q) \rightarrow} \\ {2 \mathrm{CaCO}_{3}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)}\end{array} $$ Although \(\mathrm{CaCO}_{3}(s)\) is considered insoluble, some of it does dissolve in aqueous solutions. Calculate the molar solubility of \(\mathrm{CaCO}_{3}\) in water $\left(K_{\mathrm{sp}}=8.7 \times 10^{-9}\right) .$
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 Te(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 $$ \mathrm{TeF}_{6}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Te}(\mathrm{OH})_{6}(a q)+6 \mathrm{HF}(a q) $$ Tellurium hexafluoride can be prepared by the reaction of elemental tellurium with fluorine gas: $$ \mathrm{Te}(s)+3 \mathrm{F}_{2}(g) \longrightarrow \mathrm{TeF}_{6}(g) $$ 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 \(\mathrm{L}\) fluorine gas at 1.06 \(\mathrm{atm}\) and \(25^{\circ} \mathrm{C},\) what is the pH of a solution of \(\mathrm{Te}(\mathrm{OH})_{6}\) formed by dissolving the isolated TeF \(_{6}(g)\) in 115 \(\mathrm{mL}\) solution? Assume 100\(\%\) yield in all reactions.
There are two forms of solid sulfur: rhombic and monoclinic. The stable form of sulfur at \(25^{\circ} \mathrm{C}\) is the rhombic form. Upon heating, the rhombic form converts to the monoclinic form, which is the stable form of sulfur at high temperatures. Consider the process: $$ \mathrm{S}_{\text { rhombic }}(s) \longrightarrow \mathrm{S}_{\text { monoclinic }}(s) $$ Predict the signs of \(\Delta H\) and \(\Delta S\) for this process. Which form of sulfur has the more ordered structure (has the smaller positional probability)?
What do you think about this solution?
We value your feedback to improve our textbook solutions.