Answer: please find attached to see the structure.
1. THE STRUCTURE OF BENZOIC ACID AND FLUORENE, soluble in ether and insoluble in water.
2. THE STRUCTURE OF CARBOXYLIC ACID BEEN EXTRACTED.
Explanation: the mixture of benzoic acid and fluorene are the first diagrams which shows the carboxylic acid attached to the benzene ring, which are soluble in ether and insoluble in water. When dissolved in NaOH(aq) is the carboxy ion becomes soluble in water but insoluble in ether, this is seen in the second diagram.
The third diagram shows the carboxylic acid been precipitated and soluble in ether but insoluble in water.
NOTE THE TWO MAIN DIAGRAM IS THE FIRST AND THE LAST DIAGRAM, WHERE CARBOXYLIC ACID DISSOLVES IN AQUEOUS SODIUM HYDROXIDE, AND WHEN THE ACID IS BEEN PRECIPITATED IN AQUEOUS HCl.
ALSO NOTE THE CHANGE IN BENZOIC RING MIXED WITH FLUORENE TO THAT OF THE ACID BEEN EXTRACTED.
Hope together with the picture, this has helped you.
Answer: 2.54g
Explanation:
Molar Mass of H2O2 = (2x1) + (2x16) = 34g/mol
1mole (34g) of H2O2 contains 6.02x10^23 molecules
Therefore Xg of H2O2 will contain 4.5x10^22 molecules i.e
Xg of H2O2 = (34x4.5x10^22)/6.02x10^23 = 2.54g
Answer:
Due to presence of a triple bond between the two N−atoms, the bond dissociation enthalpy (941.4 kJ mol
−1
) is very high. Hence, N
2
is the least reactive.
Answer:
At end point there will a transition from pink to colorless.
Explanation:
As the student put the vinegar in the titrator and NaOH in the beaker, it means that he has poured phenolphthalein in the NaOH solution.
The pH range of phenolphthalein is 8.3-10 (approx), it means it will show pink color in basic medium.
So on addition of phenolphthalein in NaOH the solution will become pink in color.
When we start pouring vinegar from titrator neutralization of NaOH will begin.
On complete neutralization , on addition of single drop of vinegar the solution will become acidic and there will be complete disappearance of pink color solution in the beaker.
Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄
Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.
For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:
Kp = 
where:
P(N₂O₄) and P(NO₂) are the partial pressure of each gas.
Calculating constant:
Kp = 
Kp = 0.0104
After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.
P(N₂O₄) + P(NO₂) = 200
P(N₂O₄) = 200 - P(NO₂)
Kp =
0.0104 = ![\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B200%20-%20P%28NO_%7B2%7D%29%20%20%7D%7B%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D%7D)
0.0104![[P(NO_{2} )]^{2}](https://tex.z-dn.net/?f=%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D)
+ 
- 200 = 0
Resolving the second degree equation:
= 
= 98.7
Find partial pressure of N₂O₄:
P(N₂O₄) = 200 - P(NO₂)
P(N₂O₄) = 200 - 98.7
P(N₂O₄) = 101.3
The partial pressures are = 98.7 MPa and P(N₂O₄) = 101.3 MPa
