An organic chemist is planning to extract 1.00 g of an organic compound dissolved in 200.0 mL of water into a diethyl ether solvent. The partition coefficient for this system is 9.5, favoring the diethyl ether solvent. What percentage of the organic compound remains in the aqueous solvent if you perform three successive extractions with 50.0 mL of diethyl ether for each extraction?(a) 29.63% (b) 2.60% (c) 0.00169%

(b) 2.60%

Explanation:

Partition coefficient of the organic compound in ether is:

9.5 = Concentration in ether / Concentration in water

In the first extraction, X is the amount of organic compound extracted:

9.5 = X/50mL / (1-X)/200mL

9.5 = 200X / (50-50X)

475 - 475X = 200X

475 = 675X

X = 0.7037g are extracted.

Remains = 1g - 0.7037g = 0.2963g

Second extraction:

9.5 = X/50mL / (0.2963-X)/200mL

9.5 = 200X / (14.815-50X)

140.74 - 475X = 200X

140.74 = 675X

X = 0.2085g are extracted.

Remains = 0.2963g - 0.2085g = 0.0878g

Third extraction:

9.5 = X/50mL / (0.0878g-X)/200mL

9.5 = 200X / (4.3899-50X)

41.70 - 475X = 200X

41.70 = 675X

X = 0.2085g are extracted.

Remains = 0.0878g - 0.0618g = 0.026g remains

Percentage: 0.026g / 1g * 100 =

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-