<u>Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM</u>
Given:
Acetic Acid/Sodium Acetate buffer of pH = 5.0
Let HA = acetic acid
A- = sodium acetate
Total concentration [HA] + [A-] = 250 mM ------(1)
pKa(acetic acid) = 4.75
Based on Henderson-Hasselbalch equation
pH = pKa + log[A-]/[HA]
[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77
[A-] = 1.77[HA] -----(2)
From (1) and (2)
[HA] + 1.77[HA] = 250 mM
[HA] = 250/2.77 = 90.25 mM
[A-] = 1.77(90.25) = 159.74 mM
Silver from the anode gets dissolved to reach the cathode, where the spoon will be plated.
<h3>What is electroplating?</h3>
Electroplating is a way of electrolysis, where a thin layer of metal is used to plate a determined object. It is a kinda process to purify a material you want.
The anode contains the metal you want to plate with, in this case, the Ag.
The anode contains the half-reaction of oxidation so:
Ag(s) → Ag ⁺ (aq) + e⁻
In the cathode, you have the spoon, which it takes place the half-reaction of reduction:
Ag ⁺ (aq) + e⁻ → Ag(s)
The electrolytic cell, where the redox reaction takes place, must be filled with a AgNO₃ solution.
Silver from the anode gets dissolved to reach the cathode, where the spoon will be plated.
Learn more about the electroplating here:
brainly.com/question/20112817
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<h3><u>Full Question:</u></h3>
The following compound has been found effective in treating pain and inflammation (J. Med. Chem. 2007, 4222). Which sequence correctly ranks each carbonyl group in order of increasing reactivity toward nucleophilic addition?
A) 1 < 2 < 3
B) 2 < 3 < 1
C) 3 < 1 < 2
D) 1 < 3 < 2
<h3><u>Answer: </u></h3>
The rate of nucleophilic attack of carbonyl compounds is 2<3 <1.
Option B
<h3><u>Explanation. </u></h3>
Nucleophilic attack is explained as the attack of an electron rich radical to a carbonyl compound like aldehyde or a ketone. A nucleophile has a high electron density, so it searches for a electropositive atom where it can donate a portion of its electron density and become stable.
A carbonyl compound is a
hybridized carbon atom with a double bonded oxygen atom in it. The oxygen atom pulls a huge portion of electron density from carbon being very electropositive.
In a ketone, there are two factors that make it less likely to undergo a nucleophilic attack than aldehyde. Firstly, the steric hindrance of two carbon groups being attached with the carbonyl carbon makes it harder for the nucleophile to approach. Secondly, the electron push by the carbon groups attached makes the carbonyl carbon a bit less electropositive than the aldehyde one. So aldehydes are more reactive towards a nucleophilic addition reaction.
Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol