Answer:
(a) W
(b) X
(c) Y
Explanation:
Let's consider the following table with melting and boiling points for 4 substances.
Substance Melting Point (°C) Boiling Point (°C)
W -7 60
X 660 2500
Y 180 1330
Z 115 445
Let's consider that:
- Below the melting point, a substance is solid.
- Between the melting and the boiling point, a substance is liquid.
- Above the boiling point, a substance is gas.
(a) Which substance is a gas at 100°C?
At 100 °C, W is above the boiling point
(b) Which substance is a liquid for the largest range of temperature?
The largest difference between the melting point and the boiling point is that of X: 2500 - 660 = 1840.
(c) Which substance is liquid at 1000 °C and a gas at 2000°C?
Y is between the melting and the boiling point at 1000 °C and above the boiling point at 2000 °C.
Answer: The energy required to vaporize 12.5 g of liquid water is 28.2 kJ
Explanation:
Latent heat of vaporization is the amount of heat required to convert 1 gram of liquid into its vapor state without change in its temperature.
Given : The enthalpy of vaporization of water is 40.65 kJ/mol.
n = number of moles =
Thus 1 mole of water requires heat = 40.65 kJ
0.694 moles of water requires heat =
Thus the energy required to vaporize 12.5 g of liquid water is 28.2 kJ
Answer:
286 J/K
Explanation:
The molar Gibbs free energy for the vaporization (ΔGvap) is:
ΔGvap = ΔHvap - T.ΔSvap
where,
ΔHvap: molar enthalpy of vaporization
T: absolute temperature
ΔSvap: molar entropy of the vaporization
When T = Tb = 64.7 °C = 337.9 K, the reaction is at equilibrium and ΔGvap = 0.
ΔHvap - Tb . ΔSvap = 0
ΔSvap = ΔHvap/Tb = (71.8 × 10³ J/K.mol)/ 337.9 K = 212 J/K.mol
When 1.35 mol of methanol vaporizes, the change in the entropy is:
2 m/sec.
his average velocity divided by the travel time. is this helpful?
The statement is false.
This is because of the Law of the Conservation of Energy, which states that energy cannot be created or destroyed, only transferred.