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:
0.238 M
Explanation:
A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.
C₁.V₁ = C₂.V₂
C₁ × 24.00 mL = 0.220 M × 52.00 mL
C₁ = 0.477 M
The concentration of Pb(ClO₃)₂ is:

<span>The answer is synthesis. This is a kind of reaction in which numerous reactants mix to make a single product. Synthesis responses discharge energy in the way of heat and light, so they are exothermic. An instance of a synthesis reaction is the creation of water from hydrogen and oxygen.</span>
Answer:
1. 0.00040 calories
2. 8.57 calories
3. 0.196 calories
4. 68 calories
5. 243 calories
6. 83680 joules
7. 1,054,368 joules
8. 2.45 calories
9. 556 (it says calories to calories so it wouldn't change)
10. 28367.52 joules
11. 59.6 calories
12. 449.6 joules
13. 0.00234 calories
14. 23292.328 joules
15. 22877693.6 joules
Hope this helps!
Explanation: