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saw5 [17]
3 years ago
14

To separate a mixture of benzoic acid and fluorene, we are going to use an acid-base extraction technique. Most carboxylic acids

are insoluble or slightly soluble in water, but they are highly soluble in dilute aqueous sodium hydroxide because the organic acid reacts with the base and produces carboxylate ions, a polar species. In a mixture of a carboxylic acid and an organic compound, the acid may be selectively removed from an organic solvent by extracting with a dilute NaOH solution. After separating the organic and aqueous layers, the carboxylic acid may be recovered from the aqueous solution by adding HCl, which causes precipitation of the carboxylic acid. Draw two structures below.

Chemistry
1 answer:
pentagon [3]3 years ago
8 0

Answer: please find attached to see the structure.

1. THE STRUCTURE OF BENZOIC ACID AND FLUORENE, soluble in ether and insoluble in water.

2. THE STRUCTURE OF CARBOXYLIC ACID BEEN EXTRACTED.

Explanation: the mixture of benzoic acid and fluorene are the first diagrams which shows the carboxylic acid attached to the benzene ring, which are soluble in ether and insoluble in water. When dissolved in NaOH(aq) is the carboxy ion becomes soluble in water but insoluble in ether, this is seen in the second diagram.

The third diagram shows the carboxylic acid been precipitated and soluble in ether but insoluble in water.

NOTE THE TWO MAIN DIAGRAM IS THE FIRST AND THE LAST DIAGRAM, WHERE CARBOXYLIC ACID DISSOLVES IN AQUEOUS SODIUM HYDROXIDE, AND WHEN THE ACID IS BEEN PRECIPITATED IN AQUEOUS HCl.

ALSO NOTE THE CHANGE IN BENZOIC RING MIXED WITH FLUORENE TO THAT OF THE ACID BEEN EXTRACTED.

Hope together with the picture, this has helped you.

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6 0
3 years ago
How does the structure of this umbrella relate to its function?
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45 Three samples of the same solution are tested, each with a different indicator. All three indicators, bromthymol blue, bromcr
ANEK [815]

<u>Answer:</u> The correct answer is Option 4.

<u>Explanation:</u>

Bromothymol blue, Bromocresol green and Thymol blue are the indicators which change their color according to the change in pH of the solution.

The pH range and color change of these indicators are:

  1. Bromothymol Blue: The pH range for this indicator is 6.0 to 7.5 and color change is from yellow to blue. It appears yellow below pH 6.0 and blue above pH 7.5
  2. Bromocresol green: The pH range for this indicator is 3.5 to 6.0 and color change is from yellow to blue. It appears yellow below pH 3.5 and blue above pH 6.0
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As, the highest pH of all the indicators is 9.6, so every indicator will appear blue above pH 9.6.

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3 0
3 years ago
Read 2 more answers
Phosphorus-32 (32P) is an isotope that is commonly used for medical and biological research. Phosphorus-32 has a half-life of 14
Pachacha [2.7K]

The activity of the sample when it was shipped from the manufacturer is 4.54 mCi

<h3>How to determine the number of half-lives that has elapsed </h3>

From the question given above, the following data were obtained:

  • Time (t) = 48 hours
  • Half-life (t½) = 14.28 days = 14.28 × 24 = 342.72 hours
  • Number of half-lives (n) =?

n = t / t½

n = 48 / 342.72

n = 0.14

<h3>How to determine the activity of the sample during shipping </h3>
  • Number of half-lives (n) = 0.14
  • Original activity (N₀) = 5.0 mCi
  • Activity remaining (N) =?

N = N₀ / 2ⁿ

N = 5 / 2^0.14

N = 4.54 mCi

Thus, the activity of the sample during shipping is 4.54 mCi

Learn more about half life:

brainly.com/question/2674699

5 0
3 years ago
A 52.0 mL portion of a 1.20 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a
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Answer:

C_3=0.125M

Explanation:

Hello!

In this case, we can divide the problem in two steps:

1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

V_1C_1=V_2C_2

So we solve for C2:

C_2=\frac{C_1V_1}{V_2}=\frac{52.0mL*1.20M}{278mL}\\\\C_2=0.224M

2. Now, since 111 mL of water is added, we compute the final volume, V3:

V_3=139+111=250mL

So, the final concentration of the 139 mL portion is:

C_3=\frac{139 mL*0.224M}{250mL}\\\\C_3=0.125M

Best regards!

8 0
3 years ago
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