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zavuch27 [327]
3 years ago
6

Write a mechanism for the bromination of acetanilide with bromine in acetic acid. Be as complete as possible and include all rel

evant resonance structures.

Chemistry
1 answer:
Karolina [17]3 years ago
4 0

Answer:

See figure 1 and 2

Explanation:

In this case, we can start with the general reaction. Acetanilide with bromine in acetic acid we will produce <u>p-bromo-acetanilide</u> (See figure 1).

In the reaction mechanism, we have as first step the attack of the double bond in the benzene ring to one of the Br atoms in bromine (Br_2) and a positive carbon will be produced in benzene. This positive charge can be moved inside of the benzne ring producing the r<u>esonance structures A, B and D</u>. Resonance structure C is produced when a double bond is generating between the "N" atom and the carbon in the benzene ring.

The final step in the mechanism is when we remove the "H" atom in structure A to produce again the double bond.

See figure 2

I hope it helps!

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What is the volume of 1,000 kg of ice? Please explain.
mars1129 [50]

Answer:

Volume of ice is 1.09 m³.

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Symbol:

The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.

Given data:

Mass of ice = 1000 Kg

Volume = ?

Solution:

The density of ice is 919 Kg/m³

d = m/v

v = m/d

v = 1000 Kg / 919 Kg/m³

v = 1.09 m³

6 0
3 years ago
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
3 years ago
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vichka [17]

Answer:

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Explanation:

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