If 15 grams of an organic solid is dissolved in 80 ml of water, how many grams remain in the water after extraction with 50 ml of a solvent which has a partition (distribution) coefficient with water of 8? show calculations.

2.5 g

Explanation:

The partial coefficient (K) is the ratio of the solubility (S) of a solute in two different solvents. For the given problem:

K = Ssolvent/Swater

The solubility is the mass divided by the volume. If the fraction of the solute in water is called q, the fraction of it in the solvent is 1-q. Let's call m the initial mass of the solute, so, after the extraction:

mass at the solvent = (1-q)m

mass at water = qm

Ssolvent = (1-q)m/50

Swater = qm/80

8 = [(1-q)m/50]/qm/80

8 = 80*(1-q)/50q

400q = 80 - 80q

480q = 80

q = 0.167

The mass that remains at the water is:

0.167*15 = 2.5 g

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :