______ exhibits the highest

PLEASE PLEASE PLEASE HELP ME HAS TO BE TURNED IN AT 9:00

Kay [80]

Answer:

Option D will be the answer.

8 0

3 years ago

List and explain the four most important factors that control the relative reactivity of a carbonyl-containing functional group

ioda

Resonance, leaving group, carbonyl carbon delta+, and steric effect is the most crucial variables that affect the relative reactivity of a functional group containing a carbonyl in an addition or substitution process.

Discussion:

1. Carbonyl Carbon Delta+: The carbonyl group becomes more electrophilic and accelerates nucleophilic assault when the carbonyl carbon delta+ is bigger.

2. Resonance: When the carbonyl is transformed into the tetrahedral adduct, it may be lost. Loss of resonance increases the energy of the transition state for this nucleophilic assault because resonance has the function of stabilizing. Therefore, a carbonyl functional group's resistance to nucleophilic attack increases as resonance in the group increases in importance.

3. Leaving group: Tetrahedral adduct fragmentation is encouraged by a better LG.

4. Steric effects: The nucleophilic attack on carbonyl carbon is delayed when sterically impeded.

Learn more about carbonyl here:

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8 0

1 year ago

HELPPP how many Hg grams is .25 moles of CaO??

Eva8 [605]

15.5 grams
is your anwser

6 0

4 years ago

PLEASE HELP ME I WILL BRAINLIST YOU AND GIVE YOU POINTS

MissTica

Answer:

Carbon 12 and carbon 13

Carbon isotope 12 exist as 98.9%

Explanation:

8 0

3 years ago

Read 2 more answers

Calculate the pH of a 1.00 L buffer of 0.97 M CH3COONa / 1.02 M CH3COOH before and after the addition of the following species.

ivanzaharov [21]

Answer:

a) 4.73

b) 4.78

c) 4.66 (further addition) or 4.60 (starting from the original buffer solution)

Explanation:

Step 1: Data given

volume of the buffer = 1.00 L

Buffer = 0.97 M CH3COONa / 1.02 M CH3COOH

pKa CH3COOH = 4.75

Step 2: pH = pKa + log [CH3COONa]/[CH3COOH]

pH = 4.75 + log (0.97/1.02)

pH = 4.73

(b) pH after addition of 0.065 mol NaOH

Adding 0.065 mol NaOH will reduce the acid by that amount leaving 1.02 - 0.065 = 0.955 moles HA in 1 L so [HA] = 0.955; the neutralized acid produces A- in the same amount, increasing [A-] to 0.97 +0.065 = 1.035

pH = pKa + log[CH3COONa]/[CH3COOH]

pH = 4.75 + log(1.035/0.955)

pH = 4.78

c) pH after further addition of 0.144 mol HCl

The reverse will happen after the addition of HCl:

[HA] = 0.955 + 0.144 = 1.099

[A-] = 1.035 - 0.144 = 0.891

pH = 4.75 + log(0.891/1.099)

pH = 4.66

If we add 0.144 mol of HCl to the original buffer we will get:

[HA] = 1.02 + 0.144 = 1.164

[A-] = 0.97 - 0.144 = 0.826

pH = 4.75 + log(0.826/1.164)

pH = 4.60

3 0

3 years ago