Here we have to get the correct statements among the given, applicable for Diels-Alder reaction.
The true statements in case Diels-Alder reaction are-
1. An excess of Maleic anhydride is used.
2. The I.R. of the products are indistinguishable.
The Diels-Alder reaction is the most is the most important cyclo-addition reaction in organic chemistry. These are addition reactions in which ring systems are formed without eliminating any compounds.
There remains one diene and one dienophile. The reaction is reversible in nature and requires elevated temperature to obtain its transition state. The reaction rate become faster in certain condition like using of polar solvents.
Among the given statements the following statements are true-
1. An excess of maleic anhydride (the most effective di-enophile) is used to process the reaction in forward direction.
2. The products obtain in this reaction are stereoisomers thus are indistinguishable by infrared spectroscopy (IR).
The statements which are not true for the Diels-Alder reaction:
3. The re-crystallization of the products by any polar solvent like methanol is not feasible as it will cause the retro reaction due to stability of the transition state in polar solvent.
4. Cleaning of glassware are compulsory for any reaction it is not specifically true for Diels-Alder reaction.
5. The reaction occurs at elevated temperature thus flame is required.
Answer:
Pb(NO3)2
Cd(NO3)2
Na2SO4
Explanation:
In the first part, addition of HCl leads to the formation of PbCl2 which is poorly soluble in water. This is the first precipitate that is filtered off.
When the pH is adjusted to 1 and H2S is bubbled in, CdS is formed. This is the second precipitate that is filtered off.
After this precipitate has been filtered off and the pH is adjusted to 8, addition of H2S and (NH4)2HPO4 does not lead to the formation of any other precipitate.
The yellow flame colour indicates the presence of Na^+ which must come from the presence of Na2SO4.
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Answer:
483.27 minutes
Explanation:
using second faradays law of electrolysis
