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.
Answer:
0.085 moles of N₂O₅ are needed
Explanation:
Given data:
Mass of NO₂ produces = 7.90 g
Moles of N₂O₅ needed = ?
Solution:
2N₂O₅ → 4NO₂ + O₂
Number of moles of NO₂ produced :
Number of moles = mass/ molar mass
Number of moles = 7.90 g/ 46 g/mol
Number of moles = 0.17 mol
now we will compare the moles of NO₂ with N₂O₅.
NO₂ : N₂O₅
4 : 2
0.17 : 2/4×0.17 = 0.085 mol
Thus, 0.085 moles of N₂O₅ are needed.
When solid <span>iron (iii) hydroxide is dissolved into water, it ionizes or it dissociates into ions. These ions are the iron (iii) ions and the hydroxide ions. Iron(III) oxide is classified as a base when in aqueous solution since it produces hydroxide ions. It is a weak base so it does not completely dissociate into the solution. The dissociation equation would be:
Fe(OH)3 <-----> Fe3+ + OH-
To write a complete reaction, the reaction should be balanced wherein the number of atoms of each element in the reactant side and the product side should be equal. Also, the phases of the substances should be written. We do as follows:
</span>
Fe(OH)3 (s) <-----> Fe3+ (aq) + 3OH- (aq)
Answer:
Answer: pH = 2.72
Explanation:
Calculate the pH of 0.010 M HNO2 solution. The K, for HNO2 is 4.6 x 104
