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qaws [65]
2 years ago
6

3. What molecule or ion acts as the nucleophile in the base-catalyzed reaction between acetic anhydride and vanillin? A. Vanilli

n as a neutral phenol B. Vanillin as a phenoxide ion (conjugate base) C. Acetic anhydride D. Sodium hydroxide 4. According to the class notes, the first step in the acid-catalyzed reaction between acetic anhydride and vanillin is: A. Loss of a proton from the alcohol group in vanillin B. Protonation of acetic anhydride C. Protonation of the alcohol group in vanillin D. Departure of water as a leaving group in vanillin FOR QUESTIONS 5 AND 6 USE THE FOLLOWING INFORMATION: • Vanillin (solid) MW = 152.15 g/mol • Acetic anhydride (liquid) MW = 102.04 g/mol • Acetic anhydride density = 1.082 g/mL 5. How many mmoles of vanillin and how many mmoles of acetic anhydride are being reacted in the BASE CATALYZED reaction as described in the textbook? A. 1.97 mmol of vanillin and 8.45 mmol acetic anhydride B. 45.6 mmol of vanillin and 82 mmol acetic anhydride C. 8.45 mmol of vanillin and 2 mmol acetic anhydride D. 2 mmol of vanillin and 2 mmol acetic anhydride 6. How many mmoles of vanillin and how many mmoles of acetic anhydride are being reacted in the ACID CATALYZED reaction as described in the textbook? A. 23 mmol of vanillin and 102 mmol acetic anhydride B. 11 mmol of vanillin and 1 mmol acetic anhydride C. 1 mmol of vanillin and 10.6 mmol acetic anhydride D. 1 mmol of vanillin and 9 mmol acetic anhydride
Chemistry
1 answer:
Effectus [21]2 years ago
8 0

Answer:

B

B

A

C

Explanation:

3

B. Vanillin as a phenoxide ion (conjugate base)

4

B. Protonation of acetic anhydride

5

A. 1.97 mmol of vanillin and 8.45 mmol acetic anhydride

6

C. 1 mmol of vanillin and 10.6 mmol acetic anhydride

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If .758 moles of gas occupy a volume of 80.6L, how many moles will occupy a volume of 270.9L?
egoroff_w [7]

Answer:

n₂ = 2.55 mol

Explanation:

Given data:

Initial number of moles = 0.758 mol

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Final number of moles = ?

Solution:

Formula:

V₁/n₁ = V₂/n₂

V₁ = Initial volume

n₁ = initial number of moles

V₂ = Final volume

n₂ =  Final number of moles

now we will put the values in formula.

V₁/n₁ = V₂/n₂

80.6 L / 0.758 mol = 270.9 L/ n₂

n₂ = 270.9 L× 0.758 mol / 80.6 L

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4 0
3 years ago
A solution of hydrocloric acid has a molarity of 2.25 M HCl. If a reaction requires 5.80 g of HCI, what
Natali [406]

Answer:

0.071L

Explanation:

From the question given, we obtained the following data:

Molarity of HCl = 2.25 M

Mass of HCl = 5.80g

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Number of mole of HCl =?

Number of mole = Mass /Molar Mass

Number of mole of HCl = 5.8/36.45 = 0.159mole

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