Answer:
Earth's far distance from the sun
Explanation:
We have hot summers and cold winters because of the tilt of the Earth's axis. The tilt of the Earth means the Earth will lean towards the Sun (Summer) or lean away from the Sun (Winter) 6 months later.
Answer: The entire water/ice solution is at the melting/freezing point, 32°F (0°C). Adding rock salt — or any substance that dissolves in water — disrupts this equilibrium.
Explanation: Hope this helps! Have a great day :)
1. B
2. C
3. A
That should do it
<span>London dispersion forces
is the weakest intermolecular force. It is a temporary force that happens when
electrons of two adjacent atoms occupy positions that make atoms form dipoles
which are temporary dipoles. This is also referred as dipole-dipole attraction.</span>
Answer:
A. ΔG° = 132.5 kJ
B. ΔG° = 13.69 kJ
C. ΔG° = -58.59 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.
ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)
where,
n: moles
ΔH°f: standard enthalpy of formation
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))
ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)
ΔH° = 178.3 kJ
We can calculate the standard entropy of the reaction (ΔS°) using the following expression.
ΔS° = ∑np . S°p - ∑nr . S°r
where,
S: standard entropy
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))
ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)
ΔS° = 160.6 J/K. = 0.1606 kJ/K.
We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.
ΔG° = ΔH° - T.ΔS°
where,
T: absolute temperature
<h3>A. 285 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ
<h3>B. 1025 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ
<h3>C. 1475 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ