The a mixture of 4-tert-butylphenol and 2- chlorobenzoic acid is separate by the by using bicarbonate solvent.
When solution of bicarbonate is added in mixture of 4-tert-butylphenol and 2- chlorobenzoic acid then 2- chlorobenzoic acid form a carboxylate ion whereas 4-tert-butylphenol is underacted and filtered out.
Since, only 2- chlorobenzoic acid which is acid is convert into its conjugate base by solution of bicarbonate in mixture of 4-tert-butylphenol and 2- chlorobenzoic acid .
However, phenol is less acidic than carboxylic acid. Both phenol and carboxylic acid is soluble in organic solvent . At that point as the phenol isolates as an oil, one needs to cool the blend in an ice shower to encourage crystallization
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<h3><u>Answer;</u></h3>
Nitrogen (N)
<h3><u>Explanation;</u></h3>
- Atoms may lose or gain electrons to achieve stability. Metal atoms lose electrons to attain stable configuration, and as a result they form positively charged ions called cations.
- <u>Non-metals on the other hand gain electrons to attain stable configuration, and as a result form negatively charged ions called anions.</u>
- <u>Among the elements given above nitrogen (N) gains electrons to form a negatively charged ion called anion. </u>
Answer:
The pH of the solution is 11.48.
Explanation:
The reaction between NaOH and HCl is:
NaOH + HCl → H₂O + NaCl
From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:
Now, we need to find the concentration of the OH⁻ ions.
![[OH^{-}] = \frac{n_{NaOH}}{V}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20)
Where V is the volume of the solution = 1.00 L
![[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20%3D%20%5Cfrac%7B3.01%20%5Ccdot%2010%5E%7B-3%7D%20moles%7D%7B1.00%20L%7D%20%3D%203.01%20%5Ccdot%2010%5E%7B-3%7D%20mol%2FL%20)
Finally, we can calculate the pH of the solution as follows:
![pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52](https://tex.z-dn.net/?f=%20pOH%20%3D%20-log%28%5BOH%5E%7B-%7D%5D%29%20%3D%20-log%283.01%20%5Ccdot%2010%5E%7B-3%7D%29%20%3D%202.52%20)
Therefore, the pH of the solution is 11.48.
I hope it helps you!
Answer is c. salt and water
