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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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A graduated cylinder contains 18.0ml of water. what is the new water level after 35.6g of silver metal with a density of 10.5 g/
AlekseyPX

Answer:

The answer to your question is

1.- Volume = 3.4 ml

2.- Volume = 0.61 ml

3.- Mass = 2872.8 pounds

Explanation:

Problem 1

Volume = 18 ml

mass = 35.6 g

density = 10.5 g/ml

Process

1.- Calculate the volume of silver

Formula

density = \frac{mass}{volume}

solve for volume

volume = \frac{mass}{density}

Substitution

volume = \frac{35.6}{10.5}

<u>volume = 3.4 ml</u>

2.- Problem 2

Total volume = ?

Volume = 18 + 3.4

Volume = 21.4 ml

Data

mass = 8.3 g

density = 13.6 g(ml

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Formula

density = \frac{mass}{volume}

Solve for volume

volume = \frac{mass}{density}

Substitution

volume = \frac{8.3}{13.6}

Result

<u>volume = 0.61 ml</u>

3.- Problem 3

Data

volume = 345 gal

density = 1 g/ml

mass = ?

Formula

density = \frac{mass}{volume}

Solve for mass

mass = density x volume

Covert gal to ml

                            1 gal --------------- 3785 ml

                         345 gal -------------  x

                            x = (345 x 3785) / 1

                            x = 1305825 ml

Substitution

mass = 1 x 1305825

mass = 1305825 g

Convert g to pounds

                        1 g ------------------- 0.0022 pounds

              1305825 g ----------------   x

                        x = (1305825 x 0.0022)

                       <u> x = 2872.8 pounds</u>

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Check the attached file for the answer.

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Answer:

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Explanation:

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