Answer:
See explanation and picture below
Explanation:
First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.
In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.
In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.
Answer:
salt will dislove sand wont
Explanation:
Answer:
Option A. It has stayed the same.
Explanation:
To answer the question given above, we assumed:
Initial volume (V₁) = V
Initial temperature (T₁) = T
Initial pressure (P₁) = P
From the question given above, the following data were:
Final volume (V₂) = 2V
Final temperature (T₂) = 2T
Final pressure (P₂) =?
The final pressure of the gas can be obtained as follow:
P₁V₁/T₁ = P₂V₂/T₂
PV/T = P₂ × 2V / 2T
Cross multiply
P₂ × 2V × T = PV × 2T
Divide both side by 2V × T
P₂ = PV × 2T / 2V × T
P₂ = P
Thus, the final pressure is the same as the initial pressure.
Option A gives the correct answer to the question.
Answer:
The new temperature will be 565.83 K.
Explanation:
Gay Lussac's law establishes the relationship between the temperature and the pressure of a gas when the volume is constant. This law says that the pressure of the gas is directly proportional to its temperature. This means that if the temperature increases, the pressure will increase; or if the temperature decreases, the pressure will decrease.
In other words, Gay-Lussac's law states that when a gas undergoes a constant volume transformation, the ratio of the pressure exerted by the gas temperature remains constant:

When an ideal gas goes from a state 1 to a state 2, it is true:

In this case:
- P1= 180 kPa
- T1= 291 K
- P2= 350 kPa
- T2= ?
Replacing:

Solving:

T2= 565.83 K
<u><em>The new temperature will be 565.83 K.</em></u>
Answer:
The first question is 4
The second one is 1
Explanation:
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