A chemical process dissolves 500 milligrams of iron oxide every 20 minutes. How long would it take this reaction to dissolve 2 l
1 answer:
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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
Answer:
Explanation:
Given that:
number of moles of super cooled liquid water = 1
Melting enthalpy of ice = 6020 J/mol
Freezing point =0 °C = (0 + 273 K)= 273 K
The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice ) = - 6020 J/mol × 1 mol /273 K = -22.051 J/K
Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)
Δ S = -1.4 J/K
Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K
Entropy change of universe = entropy change of the system+ entropy change of the surrounding
According to the second law of thermodynamics
Entropy change of universe >0
SO,
Entropy change of the system + entropy change in the surrounding > 0
Entropy change in the surrounding > - entropy change of the system
Entropy change in the surrounding > - (- 23.53 J/K)
Entropy change in the surrounding > 23.53 J/K
b) Make some comments on entropy changes from the obtained data.
From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.