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professor190 [17]
3 years ago
5

Write the mechanism for the acid catalyzed hydrolysis of methyl benzoate

Chemistry
1 answer:
Elis [28]3 years ago
7 0
Hydrolysis of Methyl Benzoate yields Benzoic Acid and Methanol. This reaction is also called as Tranesterification (reverse of esterification). Acid in this reaction protonates the carbonyl oxygen, resulting in increasing electrophillic character of carbonyl carbon. Water acts as a nucleophile and methoxide leaves as a good leaving group. Mechanism is shown below,

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If an insect population increases and then decreases as the available supply of food changes, what does this demonstrate
Semenov [28]
D.

This is self-regulation because when the population of the insects becomes too large, it regulates itself and starts to decrease due to a shortage of resources.
7 0
3 years ago
FIRST GETS BRAINLIEST
lisov135 [29]

Answer:

0.07172 L = 7.172 mL.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT. </em>

where, P is the pressure of the gas in atm (P = 1.0 atm, Standard P).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 3.2 x 10⁻³ mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 273 K, Standard T).

<em>∴ V = nRT/P =</em> (3.2 x 10⁻³ mol)(0.0821 L.atm/mol.K)(273 K)/(1.0 atm) = <em>0.07172 L = 7.172 mL.</em>

8 0
3 years ago
Bases in solution produce what type of ions?
Natali [406]
Bases produce hydroxide ions, while acids produce hydrogen ions. 

Bases have a pH of above 7, and are bitter and slippery. 

Answer: <span>c. hydroxide ions</span>
4 0
3 years ago
5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf
mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and pK_{a} is as follows.

               pH = pK_{a} + log \frac{base}{acid}

where,     pH = 7.4 and pK_{a} = 7.21

As here, we can use the pK_{a} nearest to the desired pH.

So,      7.4 = 7.21 + log \frac{base}{acid}

             0.19 = log \frac{base}{acid}

            \frac{base}{acid} = 1.55

1 mM phosphate buffer means [HPO_{4}] + [H_{2}PO_{4}] = 1 mM

Therefore, the two equations will be as follows.

           \frac{HPO_{4}}{H_{2}PO_{4}} = 1.55 ............. (1)

  [HPO_{4}] + [H_{2}PO_{4}] = 1 mM ........... (2)        

Now, putting the value of [HPO_{4}] from equation (1) into equation (2) as follows.

             1.55[H_{2}PO_{4}] + [tex][H_{2}PO_{4}] = 1 mM

                        2.55 [H_{2}PO_{4}] = 1 mM

                             [H_{2}PO_{4}] = 0.392 mM

Putting the value of [H_{2}PO_{4}] in equation (1) we get the following.

                     0.392 mM + [HPO_{4}] = 1 mM

                          [HPO_{4}] = (1 - 0.392) mM

                              [HPO_{4}] = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0
3 years ago
The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?​
Natasha_Volkova [10]

Answer:

3.13%.

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 0.15

Half life (t½) = 2 mins

Number of half-life (n) = 5

Fraction of sample remaining =.?

Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:

Amount remaining (N) = 1/2ⁿ × original amount (N₀)

NOTE: n is the number of half-life.

N = 1/2ⁿ × N₀

N = 1/2⁵ × 0.15

N = 1/32 × 0.15

N = 0.15/32

N = 4.69×10¯³

Therefore, 4.69×10¯³ is remaining after 5 half-life.

Finally, we shall the fraction of the sample remaining after 5 half-life as follow:

Original amount (N₀) = 0.15

Amount remaining (N) = 4.69×10¯³

Fraction remaining = N/N₀ × 100

Fraction remaining = 4.69×10¯³/0.15 × 100

Fraction remaining = 3.13%

3 0
3 years ago
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