Chapter 6: Problem 18
Explain why oceanfront areas generally have smaller temperature fluctuations than inland areas.
Chapter 6: Problem 18
Explain why oceanfront areas generally have smaller temperature fluctuations than inland areas.
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Get started for freeCalculate \(w\) and \(\Delta E\) when 1 mole of a liquid is vaporized at its boiling point \(\left(80 .^{\circ} \mathrm{C}\right)\) and 1.00 atm pressure. \(\Delta H_{\text { vap }}\) for the liquid is 30.7 \(\mathrm{kJ} / \mathrm{mol}\) at \(80 .^{\circ} \mathrm{C} .\)
A swimming pool, 10.0 \(\mathrm{m}\) by \(4.0 \mathrm{m},\) is filled with water to a depth of 3.0 \(\mathrm{m}\) at a temperature of \(20.2^{\circ} \mathrm{C}\) . How much energy is required to raise the temperature of the water to \(24.6^{\circ} \mathrm{C} ?\)
You have a 1.00 -mole sample of water at \(-30 .^{\circ} \mathrm{C}\) and you heat it until you have gaseous water at \(140 .^{\circ} \mathrm{C}\) . Calculate \(q\) for the entire process. Use the following data. $$ \begin{aligned} \text { Specific heat capacity of ice } &=2.03 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g} \\ \text { Specific heat capacity of water } &=4.18 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g} \\ \text { Specific heat capacity of steam } &=2.02 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g} \end{aligned} $$ $$ \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H_{\mathrm{fision}}=6.02 \mathrm{kJ} / \mathrm{mol}\left(\mathrm{at} 0^{\circ} \mathrm{C}\right) $$ $$ \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H_{\mathrm{vaporization}}=40.7 \mathrm{kJ} / \mathrm{mol}\left(\mathrm{at} 100 .^{\circ} \mathrm{C}\right) $$
A gas absorbs 45 kJ of heat and does 29 kJ of work. Calculate \(\Delta E .\)
Calculate \(\Delta H\) for the reaction: $$ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) $$ given the following data: $$ 2 \mathrm{NH}_{3}(g)+3 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-1010 . \mathrm{kJ}\) $$ \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-317 \mathrm{kJ}\) $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-623 \mathrm{kJ}\) $$ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-286 \mathrm{kJ}\)
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