Determine the molarity and mole fraction of a 1.15 m solution of acetone (CH3COCH3) dissolved in ethanol (C2H5OH). (Density of a
1 answer:
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Answer:
"A", "water changes from a gas to a solid to a liquid", according to this phase diagram, at at 0°C, as pressure is increased from 0atm to 10atm.
Explanation:
The question asks what happens at 0°C, as pressure is increased from 0atm to 10atm.
According to the question, the temperature is held constant. The pressure changes. In the phase diagram, we find the temperature 0°C on the horizontal axis, and all points where the temperature are 0°C are along that vertical line.
Since the pressure starts at 0atm and increases to 10atm, we start at the bottom, and move upward along that line, to see what phases of matter the substance changes to.
At the bottom, it is initially in a "gas" phase. As it moves up, it transitions to a "solid" phase. Later, as it continues moving up, it changes again into a "liquid" phase.
Thus, the answer would be "A", "water changes from a gas to a solid to a liquid", according to this phase diagram, at at 0°C, as pressure is increased from 0atm to 10atm.
Answer:
See explanation below
Explanation:
The question is incomplete. However in picture 1, you have the starting materials and the structure of the product, which you miss in this part.
Now, in picture 2, you have the starting reactant and the product, and the mechanism that is taking place here.
First, all what we have here is an acid base reaction. In the first step, we are using the acid medium to convert the reactant into an alcohol. The bromine there, is not leaving the molecule yet, because it's neccesary for the next step. The starting reactant is an alkene, in that way, we can convert the reactant in the first step into a secondary alcohol. In other words, the first reaction is a alkene hydration.
In the second step, we use a strong base. You may say this is a strong nucleophile and will do a Sn2 reaction to form another alcohol there, but it's not the case, because, before any kind of reaction happens, the priority here is always the acid base, so the base will react with the acidic hydrogen. In this case, it will substract an hydrogen from the OH. When this happens, the lone pair will do an auto condensation here, and attacks the bromine in the molecule. In this way, the molecule will become a cyclomolecule, and that way it form the final product.
See picture 2, for mechanism
