Answer:
The chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.
Explanation:
The chemical potential of 2-propanol in solution relative to that of pure 2-propanol can be calculated using the following equation:
<u>Where:</u>
<em>μ (l): is the chemical potential of 2-propanol in solution </em>
<em>μ° (l): is the chemical potential of pure 2-propanol </em>
<em>R: is the gas constant = 8.314 J K⁻¹ mol⁻¹ </em>
<em>T: is the temperature = 82.3 °C = 355.3 K </em>
<em>x: is the mole fraction of 2-propanol = 0.41 </em>
Therefore, the chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.
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Answer:
1. C+ ---- O-
2. O+ ---- Cl-
3. O+ ----- F-
4. C+ ----- N-
5. Cl- ----- C+
6. S- ----- H+
7. S+ ----- Cl -
Explanation:
Electronegativity determines the polarity . There may be two atoms in a bond with high electronegativity, in such cases the positive charge is given to atom with comparatively lower electronegativity. Electronegativity determines the easiness with which an atom attract electrons in a chemical bond. A polar bond is formed when the difference in the electronegativity of two combining atoms is between 0.4 and 1.7. The correct direction is
1. C+ ---- O-
2. O+ ---- Cl-
3. O+ ----- F-
4. C+ ----- N-
5. Cl- ----- C+
6. S- ----- H+
7. S+ ----- Cl -
Answer:
Q was < K. Partial pressure of hydrogen decreased, iodine increased
Explanation:
<em>After iodine was added the Q was [Select] K so the reaction shifted toward the Products [Select] ,The partial pressure of hydrogen [Select], Iodine [Select] |,and hydrogen iodide Decreased</em>
Based on the equilibrium:
H2(g) + I2(g) ⇄ 2HI(g)
K of equilibrium is:
K = [HI]² / [H2] [I2]
<em>Where [] are concentrations at equilibrium</em>
And Q is:
Q = [HI]² / [H2] [I2]
<em>Where [] are actual concentrations of the reactants.</em>
<em />
When the reaction is in equilibrium, K=Q.
But as [I2] is increased, Q decreases and Q was < K
The only concentration that increases is [I2], doing partial pressure of hydrogen decreased, iodine increased
The percent of O in Cr₂O₃ : 31.58%
<h3>Further explanation</h3>
Given
Cr = 52.00 amu, O = 16.00 amu
Required
The percent of O
Solution
MW Cr₂O₃ = 2 x Ar Cr + 3 x Ar O
MW Cr₂O₃ = 2.52+3.16
MW Cr₂O₃ =152 amu
Biogas is not a fossil fuel
