Hi,

Question 101

Which of the following explains the discrepancy between the observations of the two students?

Student A: Phase diagram of CO2, positive slope for the solid liquid phase boundary.
----This means that at a high P along the liq/solid boundary, CO2 prefers to be solid

Student B: Phase diagram of H2O, negative slope for the solid liquid phase boundary
----This means that at a high P along the liq/solid boundary, H2O prefers to be liquid

This boundary is to do with liquid and solid, so we can elminate D. We can also eliminate C because the different temperatures are needed to observe phase changes in the different elements. This gives you a 50/50 shot between A and B.

Water expands when it freezes, which eliminates B as an explanation.

This leaves you with A. A seems reasonable because from what we know, water is more dense than ice, and so it makes sense that at higher pressures, water would predominate more.

Question 104

For this reaction we mix 0.5 M of NaOH with 0.5 M of HCl in the passage.

We consider replacing NaOH with NH4OH, which is a weaker base.

Because NH4OH is a weaker base, it is less likely to dissociate and act as a base than NaOH is. One way we can interpret this is that it will release less energy than NaOH upon reaction.

This previous statement can eliminate C and D, leaving A and B as a 50/50 shot.

From Table 1, we know this acid base reaction is exothermic as it heats the water. When we react with NH4OH, we would expect a less exothermic reaction because NH4OH is less reactive than NaOH. This would heat the water less and result in a smaller temperature change.

This eliminates A and leaves B as the correct answer.

Question 108

We look at a lung setup wth a lower balloon (balloon 1) attached to the bottom of a soda bottle, making a sealed chamber (Chamber 1), which contains a 2nd balloon (balloon 2) that is open to the atmosphere (chamber 2).

Because chamber 1 is not open to the atmosphere, we know that it cannot equilibrate with atmospheric pressure, so any change in the volume of chamber 1 must be accompanied with a change in its internal pressure.

When you push in the lower balloon, you will decrease the volume of chamber 1, increasing its pressure. This will result in it pushing against balloon 2, which has pressure of Patm as it is open to the atmosphere.

Because of this pressure differential, balloon 2 will contract to attempt to equilibrate the pressure in chamber 1 with the Patm. This leaves us with B.

Please let me know if you would like clarification on any of this.

Katt