Answer:
Keqq = 310
Note: Some parts of the question were missing. The missing values are used in the explanation below.
Explanation:
<em>Given values: ΔH° = -178.8 kJ/mol = -178800 J/mol; T = 25°C = 298.15 K; ΔS° = -552 J/mol.K; R = 8.3145 J/mol.K</em>
Using the formula ΔG° = -RT㏑Keq
㏑Keq = ΔG°/(-RT)
where ΔG° = ΔH° - TΔS°
㏑Keq = ΔH° - TΔS°/(-RT)
㏑Keq = {-178800 - (-552 * 298.15)} / -(8.3145 * 298.15)
㏑Keq = -14221.2/-2478.968175
㏑Keq = 5.73674166
Keq = e⁵°⁷³⁶⁷⁴¹⁶⁶
Keq = 310.05
The kinetic energy of gas particles depends on temperature. Greater the temperature higher will be the average kinetic energy
Kinetic energy is related to the temperature as:
KE = 3/2 kT
where k = Boltzmann constant
T = temperature
In the given example, since the temperature of O2 gas is maintained at room temperature, the average KE will also remain constant.
The fifth postulate of the kinetic molecular theory which states that the temperature of the gas depends on the average KE of the particles of the gas explains the above observation.
The Percent error in the calculation of densities of the metal during experiment is 0.03 %
<h3>What is Percentage error?</h3>
Percent error is the difference between estimated value and the actual value in comparison to the actual value and is expressed as a percentage.
Percent error = [Experimental value - Theoretical value] / Theoretical value x 100%
Percent error = 7.83 -7.86 / 7.86 x 100
= 0.03 %
Learn more about percentage error here ;
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Answer:
The “I” and “II” in copper oxide represents the number of electrons that the metal has provided when copper oxide is brought into contact with metal. Some uses for copper oxide are: Building copper-based structures. These structures gradually change color due to oxidation.
