Answer:
Pyridine solution has a greater concentration of hydroxide ions.
Explanation:
The pOH of the solution is defined as negative logarithm of hydroxide ion concentration in a solution.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)
- Higher the value of pOH lessor will be the hydroxide ion concentration and higher the concentration of hydrogen ions in the solution .
- Lower the value of pOH higher will be the hydroxide ion concentration and lower the concentration of hydrogen ions in the solution.
1) The pOH of the methylamine = 6.8
![6.8=-\log[OH^-]](https://tex.z-dn.net/?f=6.8%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=1.5848\times 10^{-7} M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.5848%5Ctimes%2010%5E%7B-7%7D%20M)
2) The pOH of the pyridine = 6.0
![6.0=-\log[OH^-]](https://tex.z-dn.net/?f=6.0%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=0.000001 M=1.0\times 10^{-6} M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.000001%20M%3D1.0%5Ctimes%2010%5E%7B-6%7D%20M)
Pyridine solution has a greater concentration of hydroxide ions than the solution of methylamine.
According to the information in the graph, it can be inferred that the amount of solute that will precipitate out of solution at 20°C is 130 grams.
<h3>How to calculate the amount of solute that precipitates out of solution?</h3>
To calculate the amount of solute that precipitates out of solution we must identify the solute data at 80°C and 20°C and identify the difference as shown below:
- Quantity of solute at 80°C: 170 grams.
- Quantity of solute at 20°C: 40 grams.
- 170 grams - 40 grams = 130 grams
According to the above, the amount of solute that will precipitate out of solution due to the change in temperature is 130 grams of KNO3.
Note: This question is incomplete because the graph is missing. Here is the graph
Learn more about solute in: brainly.com/question/7932885
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Answer:
C. Precipitate the crystals as fast as possible
Explanation:
C. Precipitate the crystals as fast as possible
Precipitation if done fast , can lead to the formation of impure crystals , as impurities also get stuck inside the crystals. So it must be done slowly to obtain the pure crystals . Slower the crystals form , purer they are .
All the other options can lead to a successful recrystallization.
The question is incorrect, nitration of methylbenzoate occurs at 3 position to get, methyl 3-nitrobenzoate.
This is because, in present reaction the reactant i.e. methylbenzoate, has low electron density at m-position.
Due to this, NO2+ prefentially attaches m-position to form methyl 3-nitro benzoate as the product.
A detailed reaction mechanism for formation of product is provided below.