Answer:
131.5 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
First, we will calculate the standard enthalpy of the reaction (ΔH°).
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)
) - 1 mol × ΔH°f(CaCO₃(s)
)
ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)
ΔH° = 179.2 kJ
Then, we calculate the standard entropy of the reaction (ΔS°).
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)
) - 1 mol × S°(CaCO₃(s)
)
ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)
ΔS° = 160.2 J/K = 0.1602 kJ/K
Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.
ΔG° = ΔH° - T × ΔS°
ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K
ΔG° = 131.5 kJ
Answer:
a. the maximum number of σ bonds that the atom can form is 4
b. the maximum number of p-p bonds that the atom can form is 2
Explanation:
Hybridization is the mixing of at least two nonequivalent orbitals, in this case, we have the mixing of one <em>s, 3 p </em> and <em> 2 d </em> orbitals. In hybridization the number of hybrid orbitals generated is equal to the number of pure atomic orbital, so we have 6 hybrid orbital.
The shape of this hybrid orbital is octahedral (look the attached image) , it has 4 orbital located in the plane and 2 orbital perpendicular to it.
This shape allows the formation of maximum 4 σ bond, because σ bonds are formed by orbitals overlapping end to end.
And maximum 2 p-p bonds, because p-p bonds are formed by sideways overlapping orbitals. The atom can form one with each one of the orbitals located perpendicular to the plane.
Mass = m = 40 grams
Volume = V = 9 cm³
Density = d = ?
Density is defined as the ratio of mass and volume.
So,
d = m/V
Using the values, we get
d = 40/9 = 4.44 g/cm³
This means the density of material would be 4.44 g/cm³
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