The normal boiling point of c2cl3f3 is 47.6°c and its molar enthalpy of vaporization is 27.49 kj/mol. what is the change in entr

Answer:

The adjustable legs along with the sand table.

Note: The question is incomplete. The full question is presented below.

Using Models to Address Questions Regarding Systems

Armando’s class was examining images of rivers shaped by flowing water. Most rivers appeared wide and shallow, except for one, which was narrow and deep. The students theorized that this river's narrowness and depth are due to:

• the steepness of the hill from which the water descends, or
• the diminutive size of the sand grains the water flows through.

To explore the answer to the question of why this river is so narrow and deep, Armando created the model outlined below.

Explanation:

The model constructed by Armando will facilitate addressing the question due to specific features:

1. Adjustable leg - as one theory proposed by the class suggests that the steep hill affecting the water's path could be the reason for the river's dimensions, the adjustable legs are designed to be raised or lowered to alter the slope, allowing testing of this theory.

2. Sand table - this acts as the streambed. By modifying the size of the sand grains, students can examine the second hypothesis that smaller sand grains contribute to the river's narrowness and depth.

The outcomes of their experimentation will lead them to a conclusion.

The vapor pressure of a substance varies with temperature. To identify which compound has a vapor pressure of 58 kilopascals at 65 degrees Celsius, we consult standard tables that list temperature against vapor pressure.

According to the tables or graphs, the answer is:

<span>2. ethanol</span>

Answer:

7.3

Explanation:

Using the Henderson Hasselbalch equation, one can determine the pH or pOH of a solution via its pKa. Remember, , and pKa = -logKa, where Ka denotes the acid's equilibrium constant.

The Henderson Hasselbalch formula:

In this context, acid X possesses two ionic forms: the carboxyl group and an alternative form. Initially, we have 0.1 mol/L of acid in 100 mL, which gives:

n1 = (0.1 mol/L)×(0.1 L) = 0.01 mol

Upon dissociation, it yields 0.005 mol of the carboxyl form and 0.005 mol of the other form with stoichiometry assumed constant.

Introducing NaOH at a concentration of 0.1 mol/L and 75 mL, the moles of become:

n2 = (0.1 mol/L)×(0.075 L) = 0.0075 mol

Thus, 0.0075 mol of reacts with 0.005 mol of the carboxyl form, leading to 0.0025 mol of , which in turn reacts with 0.005 mol of the alternating group, leaving 0.0025 mol of the latter.

The new solution’s volume is 175 mL, but the concentrations of both forms remain unchanged in volume, so we can utilize the moles in the equation.

Thus, we arrive at:

6.72 = pKa - log 4

pKa - log 4 = 6.72

pKa = 6.72 + log 4

pKa = 6.72 + 0.6

pKa = 7.3