what??? i need more information
Answer:
The final product of the reaction is (<em>2S,3S</em>)-2-ethoxy-3-methylpentane.
Explanation:
The given reaction undergoes mechanism in which the nucleophile attacks the backside and it is substituted by the elimination of bromine.
Due to the backside attack of nucleophile , the inverse in stereo-chemistry is observed.
After the substitution of ethoxy group, the configuration is assigned according to the priority it shows clock wise direction(R) - configuration.
When hydrogen faces the front side , it results shows inverse configuration i.e, S- configuration.
The chemical reaction is as follows.
Answer:
Explanation:
T1 = 150°C = (150 + 273.15)K = 423.15K
T2 = 45°C = (45 + 273.15)K = 318K
V1 = 693mL = 693cm³
Applying Charle's law, the volume of a given gas is directly proportional to is temperature provided that pressure remains constant.
V = kT
V1 / T1 = V2 / T2
693 / 423.15 = V2 / 318
V2 = (693 * 318) / 423.15 = 520.79cm³
The new volume of the gas is 520.79cm³
A + BC ---> AB + C
So here one reactant (A) is accepting a group which is being given by another compound (BC) however the A is not giving any group / element or ion
So this single displacement
Similarly in the given reaction
the anion OH- is only being replaced
The element Ca accepts OH- and H2O loses the same group to form new element H2
So the correct answer is
Single replacement also known as single displacement.
Explanation:
As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
..........(1)
..............(2)
The final reaction is as follows:
.............(3)
Therefore, adding (1) and (2) we get the final equation (3) and value of at 298 K will be as follows.
= +
= -314 kJ + (-80) kJ
= -394 kJ
Thus, we can conclude that at 298 K for the given process is -394 kJ.