Chapter 7: Problem 9
Hess's law is really just another statement of the first law of thermodynamics. Explain.
Chapter 7: Problem 9
Hess's law is really just another statement of the first law of thermodynamics. Explain.
All the tools & learning materials you need for study success - in one app.
Get started for freeA sample of nickel is heated to \(99.8^{\circ} \mathrm{C}\) and placed in a coffeecup calorimeter containing \(150.0 \mathrm{g}\) water at \(23.5^{\circ} \mathrm{C}\). After the metal cools, the final temperature of metal and water mixture is \(25.0^{\circ} \mathrm{C} .\) If the specific heat capacity of nickel is \(0.444 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) what mass of nickel was originally heated? Assume no heat loss to the surroundings.
Calculate \(\Delta H\) for the reaction $$\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ given the following data: $$\begin{array}{lr}\text { Equation } & \Delta H(\mathrm{kJ}) \\ 2 \mathrm{NH}_{3}(g)+3 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) & -1010 \\ \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & -317 \\ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & -143 \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & -286 \end{array}$$
As a system increases in volume, it absorbs \(52.5 \mathrm{J}\) of energy in the form of heat from the surroundings. The piston is working against a pressure of \(0.500 \mathrm{atm}\). The final volume of the system is \(58.0 \mathrm{L}\). What was the initial volume of the system if the internal energy of the system decreased by \(102.5 \mathrm{J} ?\)
The enthalpy of neutralization for the reaction of a strong acid with a strong base is \(-56 \mathrm{kJ} / \mathrm{mol}\) water produced. How much energy will be released when \(200.0 \mathrm{mL}\) of \(0.400 \mathrm{M} \mathrm{HNO}_{3}\) is mixed with \(150.0 \mathrm{mL}\) of \(0.500 \mathrm{M}\) KOH?
For the process \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) at \(298 \mathrm{K}\) and 1.0 atm, \(\Delta H\) is more positive than \(\Delta E\) by 2.5 kJ/mol. What does the \(2.5 \mathrm{kJ} / \mathrm{mol}\) quantity represent?
What do you think about this solution?
We value your feedback to improve our textbook solutions.