It's only a small difference (103 degrees versus 104 degrees in water),
and I believe the usual rationalization is that since F is more
electronegative than H, the electrons in the O-F bond spend more time
away from the O (and close to the F) than the electrons in the O-H bond.
That shifts the effective center of the repulsive force between the
bonding pairs away from the O, and hence away from each other. So the
repulsion between the bonding pairs is slightly less, while the
repulsion between the lone pairs on the O is the same -- the result is
the angle between the bonds is a little less.
Hope this helps!
This is your perfect answer.
Other Identifying Characteristics
Calcite is easy to identify even without testing the reaction to HCl, by its hardness, luster and cleavage. Another special property is magnetism
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
Answer:
A Graduated Cylinder has a limited round and hollow shape with each stamped line demonstrating the volume of fluid being estimated. While they are commonly more exact and exact than lab flagons and recepticles, they ought not be utilized to perform volumetric examination.
Molar mass Argon = 39.948 g/mol
1 mol ------ 39.948 g
mol ----- 20.0 g
mol = 20.0 * 1 / 39.948
= 0.5006 moles
1 mol --------------------- 22.4 L ( at STP )
0.5006 moles ------------- L
L = 0.5006 * 22.4
= 11.21 L
hope this helps!
