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lord [1]
3 years ago
8

Propose a mechanism for the reaction of acetic anhydride with water.

Chemistry
1 answer:
BabaBlast [244]3 years ago
3 0
Mechanism. 1) Nucleophilic Attack by the Alcohol. 4a.jpg. 2) Deprotonation by pyridine. 5a.jpg. 3) Leaving group removal.The esterification reaction is both slow and reversible. The equation for the reaction between an acid RCOOH and an alcohol R' OH.
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The best option is <span>D. The orientation of the reactants is critical. 
</span><span>Since chlorine and hydrogen both participate in the reaction and speed is not essential.</span>
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What is decarboxylation??<br><br>Gimme one reaction of it..​
kotegsom [21]

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Explanation:

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How does the number of coupons you use affect the amount of money you save?
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7 0
3 years ago
Ammonia and oxygen react to produce nitric oxide and water. Which of the following chemical equations describes this reaction an
aksik [14]

Answer:

Option B. 4NH3(g) + 5O2(g) —> 4NO(g) + 6H2O(g)

Explanation:

The reaction between ammonia, NH3 and oxygen, O2, will produce nitric oxide and water as shown below:

NH3(g) + O2(g) —> NO(g) + H2O(g)

Now, let us balance the equation.

This is illustrated below:

NH3(g) + O2(g) —> NO(g) + H2O(g)

There are 3 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of NH3 and 6 in front of H2O as shown below:

4NH3(g) + O2(g) —> NO(g) + 6H2O(g)

There are 4 atoms of N on the left side and 1 atom on the right side. It can be balance by putting 4 in front of NO as shown below:

4NH3(g) + O2(g) —> 4NO(g) + 6H2O(g)

Now, the are 2 atoms of O on the left side and a total of 10 atoms on the right side. It can be balance by putting 5 in front of O2 as shown below:

4NH3(g) + 5O2(g) —> 4NO(g) + 6H2O(g)

Now, we can see that the equation is balanced.

Therefore, option B gives the balanced chemical equation for the reaction.

5 0
4 years ago
Is 13.0 mv at 25 °c. calculate the concentration of the zn2 (aq) ion at the cathode?
Andrews [41]
Answer: 
 Zn =⇒ Zn+2(0.10) + 2e- (anode)
 Zn+2(?M) + 2e- === Zn(s) (cathode)
 
 Zn + Zn+2(?M) ===⇒ Zn+2(0.10) + Zn 
 E = E^o -0.0592 log Q; in this case E^o is zero. 
 E = - 0.0592 /n logQ where n is the number of electrons transferred, in this
case n = 2 
 23 mV x 1 volt/1000mv = 0.023 Volts 
 0.023 = -0.0592 / 2 log(0.10) / [Zn+2]
 0.023 = -0.0296 { log 0.10 – log [Zn+2] }
 0.023 = -0.0296{ -1 - log[Zn+2] }
 0.023 = +0.0296 + 0.0296log[Zn+2]
 -0.0066 = 0.0296log[Zn+2]
 -0.22= log[Zn+2]
 [Zn+2] = 10^-0.22 = 0.603 Molar

6 0
3 years ago
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