Chapter 6: Problem 20
Write the chemical equation for the formation reaction of \(\mathrm{H}_{2} \mathrm{~S}(g)\).
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Chapter 6: Problem 20
Write the chemical equation for the formation reaction of \(\mathrm{H}_{2} \mathrm{~S}(g)\).
These are the key concepts you need to understand to accurately answer the question.
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Is the following reaction the appropriate one to use in determining the enthalpy of formation of methane, \(\mathrm{CH}_{4}(g) ?\) Why or why not? $$ \mathrm{C}(g)+4 \mathrm{H}(g) \longrightarrow \mathrm{CH}_{4}(g) $$
How fast (in meters per second) must an iron ball with a mass of \(56.6 \mathrm{~g}\) be traveling in order to have a kinetic energy of \(15.75 \mathrm{~J} ?\) The density of iron is \(7.87 \mathrm{~g} / \mathrm{cm}^{3} .\)
Hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\), is a poisonous gas with the odor of rotten eggs. The reaction for the formation of \(\mathrm{H}_{2} \mathrm{~S}\) from the elements is $$ \mathrm{H}_{2}(g)+\frac{1}{8} \mathrm{~S}_{8}(\text { rhombic }) \longrightarrow \mathrm{H}_{2} \mathrm{~S}(g) $$ Use Hess's law to obtain the enthalpy change for this reaction from the following enthalpy changes: $$ \begin{gathered} \mathrm{H}_{2} \mathrm{~S}(g)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{SO}_{2}(g) ; \Delta H=-518 \mathrm{~kJ} \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) ; \Delta H=-242 \mathrm{~kJ} \\ \frac{1}{8} \mathrm{~S}_{8}(\text { rhombic })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) ; \Delta H=-297 \mathrm{~kJ} \end{gathered} $$
A bullet weighing 245 grains is moving at a speed of \(2.52 \times 10^{3} \mathrm{ft} / \mathrm{s} .\) Calculate the kinetic energy of the bullet in joules and in calories. One grain equals \(0.0648 \mathrm{~g}\).
Phosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4}\), can be prepared by the reactio of phosphorus(V) oxide, \(\mathrm{P}_{4} \mathrm{O}_{10}\), with water. $$ \frac{1}{4} \mathrm{P}_{4} \mathrm{O}_{10}(s)+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q) ; \Delta H=-96.2 \mathrm{~kJ} $$ What is \(\Delta H\) for the reaction involving \(1 \mathrm{~mol}\) of \(\mathrm{P}_{4} \mathrm{O}_{10}\) ? $$ \mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(a q) $$
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