Answer:
Explanation:
Molar heat capacity at constant volume Cv of a gas = n x .5 R where n is degree of freedom of the gas molecules
CO₂ is a linear molecule , so number of degree of freedom = 3 + 2 = 5
3 is translational and 2 is rotational degree of freedom . There is no vibrational degree of freedom given .
So Cv = 5 / 2 R
= 2.5 R .
Answer:
To convert among units in the metric system, identify the unit that you have, the unit that you want to convert to, and then count the number of units between them. If you are going from a larger unit to a smaller unit, you multiply by 10 successively. If you are going from a smaller unit to a larger unit, you divide by 10 successively
Explanation:
hope it helps
Answer:
The reaction is spontaneous when T > 2170 K.
Explanation:
The spontaneity of a reaction is related to the standard Gibbs free energy (ΔG°). When ΔG° < 0 the reaction is spontaneous. ΔG° is related to the standard enthalpy (ΔH°) and the standard entropy (ΔS°) through the following expression:
ΔG° = ΔH° - T.ΔS°
where,
T is the absolute temperature
The reaction is spontaneous when ΔG° < 0, that is,
ΔH° - T.ΔS° < 0
ΔH° < T.ΔS°
T > ΔH°/ΔS° = (169.4 × 10³ J)/(78.1 J/K) = 2170 K
The reaction is spontaneous when T > 2170 K.
Answer:
A crystal or crystalline solid is a solid material whose constituents (such as atoms, molecules, or ions) are arranged in a highly ordered microscopic structure, forming a crystal lattice that extends in all directions. The scientific study of crystals and crystal formation is known as crystallography.
Explanation:
If we abbreviate the formula for nicotine as Nic, then the equations for two different equilibria of Nic in water are
Nic + H2O ---> NicH+ + OH-
NicH+ + H2O ---> NicH2 2+ + OH-
We can write the Kb1 expression for the first equation as
Kb1 = 1.0×10^-6 = [NicH+][OH-] / [Nic]
1.0×10^-6 = x^2 / 1.85×10^-3 - x
Approximating that x is negligible compared to 1.85×10^-3 simplifies the equation to
1.0×10^-6 = x^2 / 1.85×10^-3
x = 0.0000430
x = [OH-] = 4.30×10^-5 M
From the Kb2 expression
Kb2 = 1.3×10-11 = [NicH2 2+][OH-] / [NicH+]
1.1×10^-10 = x^2 / 4.30×10^-5 - x
Approximating that x is negligible compared to 4.30×10^-5 simplifies the equation to
1.1×10^-10 = x^2 / 4.30×10^-5
x = [OH-] = 6.88×10^-8
The concentration [OH-] can be computed as
[OH-] = 4.30×10^-5 M + 6.88×10^-8 M = 4.30×10^-5 M
This shows that the second equilibrium has a negligible effect on the pH.
We can now calculate for pH:
pOH = -log [OH-] = -log (4.30×10^-5 M) = 4.37
pH = 14 - pOH = 14 - 4.37 = 9.63
