Answer:
A. Substance E
A. Substance C
A. Substance A
Explanation:
Given that:
At 4 °C, Substance E has a vapor pressure of 86. torr and Substance F has a vapor pressure of 136. torr
Which has a higher boiling point?
A. Substance E
B. Substance F
C. Neither,EandF have the same boiling point
The vapor pressure varies inversely proportional to the boiling point.

Therefore, the lower the vapor pressure, the higher the boiling point.
At 4°C, Substance E with a lower vapor pressure of 86. torr will have a higher boiling point from the given information.
2.
Recall that :

therefore, the lower the enthalpy of vaporization, the higher the vapor pressure at any given temperature.
Given that:
Substance C has an enthalpy of vaporization smaller than that of substance D. Then, substance C has a higher vapor pressure.
3.
We've earlier said that:
The vapor pressure varies inversely proportional to the boiling point.

Therefore, the lower the vapor pressure, the higher the boiling point.
As such, Substance A will have a higher boiling point.
Ca=40
C=12
O=16
1 mole of CaCO3 has 100 grams
So 50 grams is 0.5 mole
Answer:
Molar mass X = 18.2 g/mol
Explanation:
Step 1: Data given
Mass of compound X = 231 mg = 0.231 grams
Mass of benzene = 65.0 grams
The freezing point of the solution is measured to be 4.5 °C.
Freezing point of pure benzene = 5.5 °C
The freezing point constant for benzene is 5.12 °C/m
Step 2: Calculate molality
ΔT = i*Kf*m
⇒ΔT = the freezing point depression = 5.5 °C - 4.5 °C = 1.0 °C
⇒i = the Van't hoff factor = 1
⇒Kf = the freezing point depression constant for benzene = 5.12 °C/m
⇒m = the molality = moles X / mass benzene
m = 1.0 / 5.12 °C/m
m = 0.1953 molal
Step 3: Calculate moles X
Moles X = molality * mass benzene
Moles X = 0.1953 molal * 0.065 kg
Moles X = 0.0127 moles
Step 4: Calculate molar mass X
Molar mass X = mass / moles
Molar mass X = 0.231 grams / 0.0127 moles
Molar mass X = 18.2 g/mol
The number of protons and the electron configuration of each