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.
Hi! :)
Ba(OH)2 + 2 HNO3 → Ba(NO3)2 + 2 H2O
(18.2 mL) x (0.45 M Ba(OH)2) x (2 mol HNO3 / 1 mol Ba(OH)2) / (38.5 mL HNO3) = 0.43 M HNO3
D.
Blue litmus paper turns red when placed in a base.
An element with 8 valence electrons is Neon (Ne).
Ps: Any element under group 18 has 8 valence electrons except for Helium (He).
40.28 sorry if it is wrong:) but I tried