The hydrocarbon is used in excess.
<h3><u>Explanation</u>:</h3>
The bromination of an arene is not simple as bromination of an alkane. This is because the carbocation or free radicle formation in benzene is a very energy consuming process. This is why a lewis base like aluminium bromide or ferric bromide is used. The ferric bromide takes in the bromine radicle and forms the brominium cation which helps in the formation of electrophile. Now this electrophile brominium cation attacks the benzene ring and forms a temporary sp3 hybrid carbon intermediate. Then the hydrogen is taken by the FeBr4- forming HBr and regenerating the FeBr3 as well as Aromaticity of the arene species at the same time. Here hydrocarbon is used in excess just to prevent the chances of multiple substitution in the same arene molecule.
102 grams of ammonia is formed when 3 moles of nitrogen and 6.7 moles of hydrogen reacts.
Explanation:
The equation given is of Haeber's process in which the nitrogen is limiting factor in the ammonia formation and hydrogen if in excess gets delimited.
We know that 1 mole of Nitrogen gives 2 moles of ammonia.
We have 3 moles of nitrogen here,
So, 6 moles of ammonia will be form
so from the formula
no of moles=mass/atomic mass
mass= no. of moles*atomic mass
= 6*17
= 102 grams of ammonia will be formed.
So, 6 moles or 102 grams of ammonia is formed when 3 mole of nitrogen and 6.7 mole of hydrogen reacts.
Answer:
V₂ = 15.3
Explanation:
Given data:
Initial volume = 12.0 L
Initial temperature = 20°C
Final temperature =100°C
Final volume = ?
Solution:
First of all we will convert the temperature into kelvin.
20°C + 273 = 293 K
100°C + 273 = 373 K
Formula:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 12.0 L × 373 K / 293 k
V₂ = 4476 L.K /293 k
V₂ = 15.3
V₂ = 1566 L.K / 298 K
V₂ = 5.3 L
I think it is D
It’s D I think
