Fe3+(aq)+6H2O(l)⇌Fe(H2O)63+(aq) : F e 3 + ( a q ) + 6 H 2 O ( l ) ⇌ F e ( H 2 O ) 6 3 + ( a q ) : blank is the Lewis acid and bl
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Answer:
159 mg caffeine is being extracted in 60 mL dichloromethane
Explanation:
Given that:
mass of caffeine in 100 mL of water = 600 mg
Volume of the water = 100 mL
Partition co-efficient (K) = 4.6
mass of caffeine extracted = ??? (unknown)
The portion of the DCM = 60 mL
Partial co-efficient (K) =
where; solubility of compound in the organic solvent and
= solubility in aqueous water.
So; we can represent our data as:
÷
Since one part of the portion is A and the other part is B
A+B = 60 mL
A+B = 0.60
A= 0.60 - B
4.6= ÷
4.6 =
4.6 × =
4.6 B = 0.6 - B
2.76 B = 0.6 - B
2.76 + B = 0.6
3.76 B = 0.6
B =
B = 0.159 g
B = 159 mg
∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.