Chapter 8: Problem 15
How does Dalton's law of partial pressures help us with our model of ideal gases? That is, what postulates of the kinetic molecular theory does it support?
Chapter 8: Problem 15
How does Dalton's law of partial pressures help us with our model of ideal gases? That is, what postulates of the kinetic molecular theory does it support?
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Get started for freeConsider the unbalanced chemical equation below: $$\mathrm{CaSiO}_{3}(s)+\mathrm{HF}(g) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Suppose a \(32.9-\mathrm{g}\) sample of \(\mathrm{CaSiO}_{3}\) is reacted with \(31.8 \mathrm{L}\) of \(\mathrm{FH}\) at \(27.0^{\circ} \mathrm{C}\) and \(1.00\) atm. Assuming the reaction goes to completion, calculate the mass of the \(\mathrm{SiF}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced in the reaction.
Metallic molybdenum can be produced from the mineral moIybdenite, MoS \(_{2}\). The mineral is first oxidized in air to molybdenum trioxide and sulfur dioxide. Molybdenum trioxide is then reduced to metallic molybdenum using hydrogen gas. The balanced equations are $$\begin{array}{l}\operatorname{MoS}_{2}(s)+\frac{7}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{MoO}_{3}(s)+2 \mathrm{SO}_{2}(g) \\\\\mathrm{MoO}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{Mo}(s)+3 \mathrm{H}_{2} \mathrm{O}(l)\end{array}$$ Calculate the volumes of air and hydrogen gas at \(17^{\circ} \mathrm{C}\) and \(1.00\) atm that are necessary to produce \(1.00 \times 10^{3} \mathrm{kg}\) pure molybdenum from MoS \(_{2}\). Assume air contains \(21 \%\) oxygen by volume, and assume \(100 \%\) yield for each reaction.
Helium is collected over water at \(25^{\circ} \mathrm{C}\) and 1.00 atm total pressure. What total volume of gas must be collected to obtain \(0.586 \mathrm{g}\) helium? (At \(25^{\circ} \mathrm{C}\) the vapor pressure of water is 23.8 torr.)
In the presence of nitric acid, \(UO\) \(^{2+}\) undergoes a redox process. It is converted to \(\mathrm{UO}_{2}^{2+}\) and nitric oxide (NO) gas is produced according to the following unbalanced equation: $$\begin{aligned}\mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{UO}^{2 *}(a q) & \longrightarrow \\\\\mathrm{NO}(g)+& \mathrm{UO}_{2}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \end{aligned}$$ If \(2.55 \times 10^{2} \mathrm{mL} \mathrm{NO}(g)\) is isolated at \(29^{\circ} \mathrm{C}\) and \(1.5 \mathrm{atm}\), what amount (moles) of \(UO\) \(^{2+}\) was used in the reaction? (Hint: Balance the reaction by the oxidation states method.)
Xenon and fluorine will react to form binary compounds when a mixture of these two gases is heated to \(400^{\circ} \mathrm{C}\) in a nickel reaction vessel. A \(100.0\) -\(\mathrm{mL}\) nickel container is filled with xenon and fluorine, giving partial pressures of \(1.24\) atm and \(10.10\) atm, respectively, at a temperature of \(25^{\circ} \mathrm{C}\). The reaction vessel is heated to \(400^{\circ} \mathrm{C}\) to cause a reaction to occur and then cooled to a temperature at which \(\mathrm{F}_{2}\) is a gas and the xenon fluoride compound produced is a nonvolatile solid. The remaining \(\mathrm{F}_{2}\) gas is transferred to another A \(100.0\) -\(\mathrm{mL}\) nickel container, where the pressure of \(\mathrm{F}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(7.62\) atm. Assuming all of the xenon has reacted, what is the formula of the product?
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