How do the prefixes of the –saccharide words relate to the structure of the sugar?
1 answer:
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Answer:
A)
1. Reaction will shift rightwards towards the products.
2. It will turn green.
3. The solution will be cooler..
B) It will turn green.
Explanation:
Hello,
In this case, for the stated equilibrium:
In such a way, by thinking out the Le Chatelier's principle, we can answer to each question:
A)
1. If potassium bromide, which adds bromide ions, is added more reactant is being added to the solution, therefore, the reaction will shift rightwards towards the products.
2. The formation of the green complex is favored, therefore, it will turn green.
3. The solution will be cooler as heat is converted into "cold" in order to reestablish equilibrium.
B) In this case, as the heat is a reactant, if more heat is added, more products will be formed, which implies that it will turn green.
Regards.
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.
I hope it helps you!