Answer:
Answer: The solubility of B is high than the solubility of A.
Explanation:
The solubility is defined as the amount of substance dissolved in a given amount of solvent. More the solute gets dissolved, high will be the solubility and less the solute dissolved, low will be the solubility.
Mass of undissolved substance of substance A is more than Substance B at every temperature. This implies that less amount of solute gets dissolved in the given amount of solvent.
Therefore, B has high solubility than substance A.
Quantitative means to count how many of something. The correct answer is B, It has two eyes. Hope this helps
the compound with the smaller lattice energy is potassium sulphide here the size of the molecule play a important role
The quantity of energy released by the electrostatic attraction of oppositely charged ions is known as lattice energy (L.E). The ion's size and charge affect the lattice energy.
lattice energy is inversely proportional to size of ion and directly proportional to charge of the ion. They are each charged equally that is plus two and minus two However, because the Sulphur is larger in size and the oxygen is lesser in this case, The lattice energy of potassium oxide is larger the lattice energy of potassium sulphide is smaller.
To learn more about lattice energy :
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Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
