Extra Credit 2
- Page ID
- 96902
Q5.3.1
A 500-mL bottle of water at room temperature and a 2-L bottle of water at the same temperature were placed in a refrigerator. After 30 minutes, the 500-mL bottle of water had cooled to the temperature of the refrigerator. An hour later, the 2-L of water had cooled to the same temperature. When asked which sample of water lost the most heat, one student replied that both bottles lost the same amount of heat because they started at the same temperature and finished at the same temperature. A second student thought that the 2-L bottle of water lost more heat because there was more water. A third student believed that the 500-mL bottle of water lost more heat because it cooled more quickly. A fourth student thought that it was not possible to tell because we do not know the initial temperature and the final temperature of the water. Indicate which of these answers is correct and describe the error in each of the other answers.
Answer: The second student was correct. The error that the first student made is that he ignored the mass of water. For the third student, it should be 2L bottle of water lost more heat since it takes longer time to reach the temperature of the fridge. The mistake the fourth student make is that even though we do not know the initial T and the final T, we could still answer this question; because the \(\Delta T\) are the same in two systems and it would not affect the comparison.
Q10.1.6
Open the PhET States of Matter Simulation to answer the following questions:
- Select the Solid, Liquid, Gas tab. Explore by selecting different substances, heating and cooling the systems, and changing the state. What similarities do you notice between the four substances for each phase (solid, liquid, gas)? What differences do you notice?
- For each substance, select each of the states and record the given temperatures. How do the given temperatures for each state correlate with the strengths of their intermolecular attractions? Explain.
- Select the Interaction Potential tab, and use the default neon atoms. Move the Ne atom on the right and observe how the potential energy changes. Select the Total Force button, and move the Ne atom as before. When is the total force on each atom attractive and large enough to matter? Then select the Component Forces button, and move the Ne atom. When do the attractive (van der Waals) and repulsive (electron overlap) forces balance? How does this relate to the potential energy versus the distance between atoms graph? Explain
Answer:
1. Similarities: Solid and liquid are more dense compared to gas. The higher temperature the system has, the faster the molecular move.
Differences: For water, the liquid state is denser than solid state; while other solids are denser than their liquid state. Different substances have different temperatures.
2.
| Temperature | Neon | Argon | Oxygen | Water |
| Solid | \(-260^{\circ}C\) | \(-230^{\circ}C\) | \(-242^{\circ}C\) | \(-116^{\circ}C\) |
| Liquid | \(-247^{\circ}C\) | \(-187^{\circ}C\) | \(-204^{\circ}C\) | \(55^{\circ}C\) |
| Gas | \(-218^{\circ}C\) | \(-84^{\circ}C\) | \(-79^{\circ}C\) | \(536^{\circ}C\) |
The higher temperature they have, the stronger IMF the substances have.
3. The potential energy first decreases in the negative region, and then keeps increasing.
When the distance between two atoms is greater than \(\varepsilon\), the total force on each atom attractive and large enough to matter.
When the distance between two atoms is \(\varepsilon\), the attractive (van der Waals) and repulsive (electron overlap) forces balance. At this point, the whole system has the lowest potential energy.
Q13.2.2
Explain why an equilibrium between Br2(l) and Br2(g) would not be established if the container were not a closed vessel shown below:
Answer:
The reason why the equilibrium cannot be established is that the pressure of \(Br_2\) will change in a not closed system, which will make the concentration of \(Br_2\) hard to determine.
Q14.1.4
Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
- \(HNO_3\)
\(PH^{4+}\) - \(H_2S\)
- \(CH_3CH_2COOH\)
\(H_2PO^{4-}\) - \(HS^{-}\)
Answer:
a.
\[ HNO_3 + H_2O \leftrightharpoons NO_3^- + H_3O^+ \]
b.
\[PH^{4+} + H_2O\leftrightharpoons PH_3 + H_3O^+ \]
c.
\[H_2S + H_2O\leftrightharpoons HS^- + H_3O^+\]
d.
\[CH_3CH_2COOH + H_2O\leftrightharpoons CH_3CH_2COO^- + H_3O^+\]
e.
\[H_2PO^{4-} +H_2O\leftrightharpoons HPO^{5-} + H_3O^+\]
f.
\[HS^- +H_2O\leftrightharpoons S^{2-} +H_3O^+\]
Q15.1.2
Complete the changes in concentrations for each of the following reactions:
\(\begin{alignat}{3}
&\ce{BaSO4}(s)⟶&&\ce{Ba^2+}(aq)\,+\,&&\ce{SO4^2-}(aq)\\
& &&x &&\underline{\hspace{45px}}
\end{alignat}\)
\(\begin{alignat}{3}
&\ce{Ag2SO4}(s)⟶&&\ce{2Ag+}(aq)\,+\,&&\ce{SO4^2-}(aq)\\
& &&\underline{\hspace{45px}} &&x
\end{alignat}\)
\(\begin{alignat}{3}
&\ce{Al(OH)3}(s)⟶&&\ce{Al^3+}(aq)\,+\,&&\ce{3OH-}(aq)\\
& &&x &&\underline{\hspace{45px}}
\end{alignat}\)
\(\begin{alignat}{3}
&\ce{Pb(OH)Cl}(s)⟶&&\ce{Pb^2+}(aq)\,+\,&&\ce{OH-}(aq)\,+\,&&\ce{Cl-}(aq)\\
& &&\underline{\hspace{45px}} &&x &&\underline{\hspace{45px}}
\end{alignat}\)
Answer: \(x;x\)
\(\begin{alignat}{3}
&\ce{Ca3(AsO4)2}(s)⟶&&\ce{3Ca^2+}(aq)\,+\,&&\ce{2AsO4^3-}(aq)\\
& &&3x &&\underline{\hspace{45px}}
\end{alignat}\)
Answer: \(2x\)