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:
.08 L or 80 ml
Explanation:
Use the equation V/t = V/t.
.04L / 150K = V / 300K
.04 / 150 * 300 = V
.08 L or 80 ml
Answer:
I got 3/8, hope this helps.
Explanation:
Answer:
Damian here! (ノ◕ヮ◕)ノ*:・゚✧
The newly hatched larva is in its first instar, a developmental stage that occurs between molts. It feeds until it grows too big for its cuticle, or soft shell, and then it molts. After molting, the larva is in the second instar. Ladybug larvae usually molt through four instars, or larval stages, before preparing to pupate.
Explanation:
hope this helps? :))