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bezimeni [28]
2 years ago
10

You are separating anthracene from benzoic acid via an extraction between ethyl acetate and a basic aqueous solution in a separa

tory funnel. How would you recover the benzoic acid?.
Chemistry
1 answer:
anygoal [31]2 years ago
4 0

The benzoic acid Collect the top layer dry with Na2SO4 and evaporate the solvent.

Benzoic acid is sparingly soluble in water and forms a precipitate. The benzoic acid can then be separated by suction filtration through a Buchner funnel. Washing with sodium bicarbonate converts the benzoic acid to the more water-soluble form of sodium benzoate, which is extracted into the aqueous layer. Additionally, sodium bicarbonate neutralizes the catalytic acid in this reaction.

Benzoic acid containing the naphthalene impurity can be purified by crystallization in water. The reaction between benzoic acid and NaOH is an acid-base neutralization reaction. An acid reacts with a base to form salt and water. In aqueous solutions, NaOH increases the solubility of benzoic acid. Benzoic acid cannot move from one layer to another because ethanol is immiscible with water.

Learn more about Benzoic acid here:-brainly.com/question/28299797

#SPJ4

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Can you create or destroy energy? If yes why, if no why?
Step2247 [10]
The law of conservation of energy states that energy can neither be created nor destroyed.

It can only be converted from one form of energy to another.

I hope this helps! ❤️
4 0
2 years ago
Read 2 more answers
Calculate by (a)% weight and (b) %mole each of the elements present in sugar
Musya8 [376]

Explanation:

Molecular mass of sugar = C_{12}H_{22}O_{11} : = 432 g/mol

Atomic mass of carbon atom = 12 g/mol

Atomic mass of hydrogen atom = 1 g/mol

Atomic mass of oxygen atom = 16 g/mol

a) Percentage of an element in a compound:

\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of compound}}\times 100

Percentage of carbon by weight in C_{12}H_{22}O_{11}:

\frac{12\times 12 g/mol}{342 g/mol}\times 100=42.10\%

Percentage of hydrogen by weight in C_{12}H_{22}O_{11}:

\frac{22\times 1g/mol}{342 g/mol}\times 100=6.43\%

Percentage of oxygen by weight in C_{12}H_{22}O_{11}:

\frac{11\times 16g/mol}{342 g/mol}\times 100=51.46\%

b) Percentage of mole each of the elements present in sugar:

=\frac{\text{Moles of atoms of an element}}{\text{total moles of all types of atoms}}\times 100

In mole of sugar we have 12 moles of carbon atom , 22 moles of hydrogen atoms and 11 moles of oxygen atoms.

Percentage of carbon by mole in C_{12}H_{22}O_{11}:

\frac{12 mol}{45 mol}\times 100=26.66\%

Percentage of hydrogen by mole in C_{12}H_{22}O_{11}:

\frac{22 mol}{45 mol}\times 100=48.88\%

Percentage of oxygen by mole in C_{12}H_{22}O_{11}:

\frac{11 mol}{45 mol}\times 100=24.44\%

7 0
4 years ago
Sunflower oil contains 0.080 mol palmitic acid (C16H32O2)/mol, 0.060 mol stearic acid (C18H36O2)/mol, 0.27 mol oleic acid (C18H3
Cerrena [4.2K]

Answer:

x_H=0.882

x_N=0.118

Explanation:

In this reactor, oleic and linoleic acid react with hydrogen to form stearic acid. This reactions can be represented by:

Oleic: C_{18}H_{34}O_2 (l) + H_2 (g) \longrightarrow C_{18}H_{36}O_2 (l)

Linoleic: C_{18}H_{32}O_2 (l) + 2 H_2 (g) \longrightarrow C_{18}H_{36}O_2 (l)

Having this reactions in mind, the first thing is to determine the moles of hydrogen required:

<u>Base of caculation: 1 mol of sunflower oil</u>

For oleic acid: n_{Holeic}=\frac{1 mol H_2}{1 mol oleic}*\frac{0.27 mol oleic}{1 mol oil}*\frac{335 mol oil}{hr}

n_{Holeic}=frac{90.45 mol H_2}{hr}

For linoleic acid: n_{Hlinoleic}=\frac{2 mol H_2}{1 mol linoleic}*\frac{0.59 mol linoleic}{1 mol oil}*\frac{335 mol oil}{hr}

n_{Holeic}=frac{395.3 mol H_2}{hr}

n_{Htotal}=\frac{90.45 mol H_2}{hr}+\frac{395.3 mol H_2}{hr}

n_{Htotal}=frac{485.75 mol H_2}{hr}

Applying the excess:

n_{Htotal}=frac{485.75 mol H_2}{hr}*1.65=801.48 mol

Nitrogen: n_N= 801.48 mol*\frac{0.05 mol N}{0.95 mol}

n_N= 42.2 mol N

<u>After the reactions</u>:

n_H=801.48 mol-485.75mol=315.73 mol

and the nitrogen is inert.

Purge stream:

n_total=42.2+315.73 mol=357.93 mol

x_H=\frac{315.73mol}{357.93mol}=0.882

x_N=\frac{42.2mol}{357.93mol}=0.118

4 0
3 years ago
Consider the E2 elimination of 3‑bromopentane with hydroxide. The starting material consists of a chiral carbon with an in plane
Tanzania [10]

Answer:

Scheme is attached

Explanation:

When 3‑bromopentane reacts with hydroxide, (Z)-pent-2-ene will produce through E 2 (Z)-pent-2-ene.

Mechanism

Hydroxide ion (OH⁻ ) is a strong nucleophile so it abstract the proton from carbon next to the carbon attached with bromine.

The the carbon next to carbon of bromine gets -ve charge, mean while it shares its electrons with the carbon having bromine and make a double bond.

As bromine  is a good leaving group so it easily gets detached from carbon, so that carbon comes to its normal state.

As a result (Z)-pent-2-ene will produce. we call it Z because mostly trans products will form.

7 0
3 years ago
In the water molecule, the H-O-H bond angle is 105º. Which distribution of electrons around the central atom provides the best e
masha68 [24]

Answer:

The correct option is: 2 shared pairs, 2 lone pairs

Explanation:

Water is an inorganic molecule that is composed of one central oxygen atom and two hydrogen atoms bonded by covalent bonds. The molecular formula of water is H₂O and the bond angle is 109.47°.

The <em>oxygen is sp³ hybridized</em> and there are<em> two bond (or shared) pairs and two lone pairs </em>of electrons in a water molecule. The<em> structure of H₂O is bent </em>due to the <u>repulsive forces between the lone pairs of electrons</u>.

<em>Due to these repulsive forces, the H-O-H bond angle decreases from 109.47°, which is the ideal bond angle to 104.45°.</em>

6 0
4 years ago
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