Part 1 : Answer is only B substance is soluble in water.
In this experiment undissolved mass of each substance was measured. According to the given data, undissolved mass of substance B at 20 °C is 10 g while A is 50 g. Since, the initial added mass of each substance is 50 g, we can see that substance A is not soluble in water since the undissolved mass is 50 g.
Part 2 : Substance A is not soluble in water and substance B is soluble in water.
According to the given data, the undissolved mass of substance A remains as same as initial added mass, 50 g throughout the temperature range from 20 ° to 80 °C. Hence, we can conclude that substance A is not soluble in water.
But, according to the data, undissolved mass of substance B at 20 °C is 10 g. That means, 40 g of substance B was dissolved in water. When the temperature increases the undissolved mass of substance B decreases. Hence, we can conclude that substance B is soluble in water and solubility increases with temperature.
Option B. carboxylic acid
For example
CH3 - COOH, is the acetic acid or etanoic acid.
where - COOH is C bonded with O (trhough doble bond) and with OH (with a single bond)
The cell notation is 
║
and the cell potential is 0.464
The reaction occurred while losing of hydron is known as oxidation reaction
We can also tell that the reaction occurred while gaining of oxygen atom is known as oxidation reaction.
The reaction occurred while gaining of hydrogen is known as reduction reaction or we can say that the reaction occurred while losing oxygen atom is known as reduction reaction
An electrochemical cell's cell potential is defined as the difference in potential between two half cells. The electrons' capacity to go from one half cell to the other is what causes the potential difference. As a result of the chemical reaction being a redox reaction, electrons can travel across electrodes.
Calculating the Cell potential
E°cell = E°(reduction) - E°(oxidation)
= 0.34 - (-0.0124)
= 0.464
Hence the cell potential is 0.464
Learn more about Cell potential here
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From the combined gas law
that is P1V1/T1=P2V2/T2
P2 is therefore T2P1V1/TIV2
A gas at STP has a temperature of 273k and a pressure of 760torr
P1=760torr
T1=273K
V1=6ml
P2=?
T2=525+273=798K
V2=500ml
P2 ={(798k x760 torr x 6ml) /(273k x500ml)} = 26.7 torr
