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qaws [65]
3 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]3 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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Answer

thus, 4 moles of oxygen gas (O2) would have a mass of 128 g.

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What is the mole ratio of NH3 to N2?
Ksenya-84 [330]
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From this equation, one mole of nitrogen react with 3 moles of hydrogen to give 2 moles of ammonia.
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8 0
3 years ago
Liquid hexane
maks197457 [2]

<u>Answer:</u> The mass of H_2O produced is 2.52 g

<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

  • <u>For hexane:</u>

Given mass of hexane = 1.72 g

Molar mass of hexane = 86.18 g/mol

Putting values in equation 1, we get:

\text{Moles of hexane}=\frac{1.72g}{86.18g/mol}=0.020mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 8.0 g

Molar mass of oxygen gas= 32 g/mol

Putting values in equation 1, we get:

\text{Moles of water}=\frac{8.0g}{32g/mol}=0.25mol

The chemical equation for the combustion of hexane follows:

2C_6H_{14}+19O_2\rightarrow 12CO_2+14H_2O

By stoichiometry of the reaction:

If 2 moles of hexane reacts with 19 moles of oxygen gas

So, 0.020 moles of hexane will react with = \frac{19}{2}\times 0.020=0.19mol of oxygen gas

As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, hexane is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 2 moles of hexane produces 14 moles of H_2O

So, 0.020 moles of hexane will produce = \frac{14}{2}\times 0.020=0.14mol of H_2O

We know, molar mass of H_2O = 18 g/mol

Putting values in above equation, we get:

\text{Mass of }H_2O=(0.14mol\times 18g/mol)=2.52g

Hence, the mass of H_2O produced is 2.52 g

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Which of the following is NOT the same?
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All are the same. It equals to the same thing.

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