Answer:
D
Explanation:
The SI unit for density is the kilogram per cubic meter (kg/m3 ).
Answer:
The molecule has a bent geometry
Explanation:
Let us look again at the principles of VSEPR theory. The shape of a molecule depends on the number of electron pairs that surround the valence shell of the central atom in the molecule.
Lone pairs distort the molecular geometry away from what is expected on the basis of VSEPR theory.
The molecule described in the question has the form AEX2. Two substituents and one lone pair form three electron domains around the central atom. The expected geometry is trigonal planar but the observed molecular geometry is bent because of the lone pairs present.
1 carbon and 2 oxygen atoms CO2
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
