Answer:
See explanation below
Explanation:
You are not providing the starting material, however, I manage to find a similar question to this, so I'm gonna use it as a basis to help you answer yours.
Now let's analyze what is happening in the reaction so we can predict the final product.
We have a ketone here, reacting at first with LDA. This is a very strong base that is commonly used in reactions with ketones and aldehydes to promove a condensation. To do this, as LDA is a strong base it will occur firts an acid base reaction, substracting the most acidic hydrogen in the molecule (Which in this case, is the Beta hydrogen of the carbonile). This will cause an enolate formation.
Then, this enolate will react with the CH3I and form a new product. The final result would be a ketone with a methyl group now attached. In the picture 2, you have the mechanism and final product.
Hope this helps
Answer:
The pressure inside the container will be 3.3 atmospheres
Explanation:
The relationship between the temperature and pressure of a gas occupying a fixed volume is given by Gay-Lussac's law which states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is kept constant.
Mathematically, it expressed as: P₁/T₁ = P₂/T₂
where P₁ is initial pressure, T₁ is initial temperature, P₂ is final pressure, T₂ is final temperature.
The above expression shows that the ratio of the pressure and temperature is always constant.
In the given question, the gas in the can attains the temperature of its environment.
P₁ = 3 atm,
T₁ = 25 °C = (273.15 + 25) K = 298.15 K,
P₂ = ?
T₂ = (55 °C = 273.15 + 55) K = 328.15 K
Substituting the values in the equation
3/298.15 = P₂/328.15
P₂ = 3 × 328.15/298.15
P₂ = 3.3 atm
Therefore, the pressure inside the container will be 3.3 atmospheres
Answer:
c. ΔH° is positive and ΔS° is positive.
Explanation:
Hello,
In this case, as the Gibbs free energy for a reaction is defined in terms of the change in the enthalpy and entropy as shown below:

Thus, as the reaction becomes spontaneous (ΔG°<0) at temperatures above 1100K (high temperatures), it necessary that c. ΔH° is positive and ΔS° is positive as the entropy will drive the spontaneousness as it becomes smaller than TΔS°.
Best regards.