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Arlecino [84]
2 years ago
10

Why does benzocaine precipitate during neutralization?.

Chemistry
1 answer:
Natalija [7]2 years ago
5 0

Explain why benzocaine precipitates during the neutralization.

With the addition of an acid, the amine group on the benzocaine will become protonated. The protonation creates a positive charge on the benzocaine molecule. Because of this, benzocaine will become soluble in a polar solvent. However, when the solution is returned to a more basic pH level, the amine will no longer be protonated. Thus, the benzocaine will no longer be soluble and will precipitate during neutralization.

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The name of the galaxy we live in is ______________.
Leni [432]

Answer:

Milky way

Explanation:

It has the shape of spilled milk

3 0
3 years ago
The question is below
rodikova [14]
The anode is the electrode where the oxidation occurs.

Cathode is the electrode where the reducction occurs.

Equations:

  Mn(2+) + 2e-  --->  Mn(s)                      Eo = - 1.18 V
2Fe(3+) + 2e-  ----> 2 Fe(2+)                2Eo = + 1.54 V

The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.

Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72

Answer: 2.72 V

3 0
3 years ago
What is the kernel form of a ferric ion? (Electron Configuration)
AnnZ [28]
<span>1s2, 2s2, 2p6, 3s2, 3p6, 3d5</span>
3 0
3 years ago
?Al + ?H2SO4 → ?Al2(SO4)3 +3H2
AfilCa [17]

The answer is: 27 grams of aluminium.

Balanced chemical reaction: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂.

n(H₂) = 1.5 mol; amount of hydrogen.

Form chemical reaction: n(Al) : n(H₂) = 2 : 3.

n(Al) = 2 · 1.5 mol ÷ 3.

n(Al) = 1.0 mol; amount of aluminium.

m(Al) = n(Al) · M(Al).

m(Al) = 1 mol · 27 g/mol.

m(Al) = 27 g; mass of aluminium.

5 0
3 years ago
A concentration cell consisting of two hydrogen electrodes (PH2 = 1 atm), where the cathode is a standard hydrogen electrode and
Lunna [17]

The pH of the unknown solution is 3.07.

<u>Explanation:</u>

<u>1.Find the cell potential as a function of pH</u>

From the Nernst Equation:

Ecell=E∘cell−RT /zF × lnQ

where

R denotes the Universal Gas Constant

T denotes the temperature

z denotes the moles of electrons transferred per mole of hydrogen

F denotes the Faraday constant

Q denotes the reaction quotient

Substitute the values,

E∘cell=0   lnQ=2.303logQ

E0cell=−2.30/RT /zF × log Q

Solving the equation,

<u>2. Find the Q  value</u>

Q=[H+]2prod pH₂, product/ [H+]2reactpH₂, reactant

Q=[H+]^2×1/1×1=[H+]2

Taking the log

logQ= log[H+]^2=2log[H+]=-2pH

From the formula,

Ecell=−2.303RT /zF× logQ

E cell= 2.303 × 8.314 CK mol (inverse)  × 298.15

K × 2pH /2×96 485  C⋅mol

( inverse)

E cell= 0.0592 V × pH

<u>3. Finding the pH value</u>

E cell= 0.0592 V × pH

pH = E cell/ 0.0592 V= 0.182V/ 0.0592V

pH=3.07

The pH of the unknown solution is 3.07.

7 0
3 years ago
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