<span>Nucleophilic acyl substitution describe a class of substitution reactions involving nucleophiles and acyl compounds. In this type of reaction, a nucleophile – such as an alcohol, amine, or enolate – displaces the leaving group of an acyl derivative – such as an acid halide, anhydride, or ester. The resulting product is a carbonyl-containing compound in which the nucleophile has taken the place of the leaving group present in the original acyl derivative. Because acyl derivatives react with a wide variety of nucleophiles, and because the product can depend on the particular type of acyl derivative and nucleophile involved, nucleophilic acyl substitution reactions can be used to synthesize a variety of different products.</span>
Answer:
If the sugar concentration level of the must becomes too high at any given point--either at the beginning or during the fermentation--it starts to have an inhibiting effect on the yeast's ability to produce alcohol. In other words, the higher sugar concentration starts to act as a preservative effecting the fermentation in a negative way.
Explanation:
Fermentation is the process by which yeast converts the glucose in the wort to ethyl alcohol and carbon dioxide gas -- giving the beer both its alcohol content and its carbonation.
Answer:
Equilibrium concentration of 
is 12.5 M
Explanation:
Given reaction: 
Here, ![K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC_%7B2%7DH_%7B5%7DOH%5D%7D%7B%5BC_%7B2%7DH_%7B4%7D%5D%5BH_%7B2%7DO%5D%7D)
where 
represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations
Here, ![[C_{2}H_{4}]=0.015M](https://tex.z-dn.net/?f=%5BC_%7B2%7DH_%7B4%7D%5D%3D0.015M)
, ![[C_{2}H_{5}OH]=1.69M](https://tex.z-dn.net/?f=%5BC_%7B2%7DH_%7B5%7DOH%5D%3D1.69M)
and 
So, ![[H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M](https://tex.z-dn.net/?f=%5BH_%7B2%7DO%5D%3D%5Cfrac%7B%5BC_%7B2%7DH_%7B5%7DOH%5D%7D%7B%5BC_%7B2%7DH_%7B4%7D%5D%5Ctimes%20K_%7Bc%7D%7D%3D%5Cfrac%7B1.69%7D%7B0.015%5Ctimes%209.0%7D%3D12.5M)
Hence equilibrium concentration of is 12.5 M
1:) <span>Zn + 2 HCl = ZnCl2 + H<span>2
</span></span>2:) 4 Fe + 3 O2 = 2 Fe2O3
