Answer:
(a) The normal freezing point of water (J·K−1·mol−1) is
(b) The normal boiling point of water (J·K−1·mol−1) is 
(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole
Explanation:
Lets calculate
(a) - General equation -
=
= 
→ phases
ΔH → enthalpy of transition
T → temperature transition
=
=
(
is the enthalpy of fusion of water)
= 
(b) 
=
(
is the enthalpy of vaporization)
= 
(c)
=
°
°
=
°
°![C)]](https://tex.z-dn.net/?f=C%29%5D)
ΔT
°
°

= 109J/mole
Answer:
Chromosomes and I think its too many
Explanation:
Answer:
Explanation:
T1 = 150°C = (150 + 273.15)K = 423.15K
T2 = 45°C = (45 + 273.15)K = 318K
V1 = 693mL = 693cm³
Applying Charle's law, the volume of a given gas is directly proportional to is temperature provided that pressure remains constant.
V = kT
V1 / T1 = V2 / T2
693 / 423.15 = V2 / 318
V2 = (693 * 318) / 423.15 = 520.79cm³
The new volume of the gas is 520.79cm³