Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Answer:
<h3>The answer is 30 cm³</h3>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula
From the question
mass = 180 g
density = 6 g/cm³
We have
We have the final answer as
<h3>30 cm³</h3>
Hope this helps you
8.4 moles equals 469.14g Fe
a. 30 moles of H₂O
b. 2.33 moles of N₂
<h3>Further explanation</h3>
Given
a. 20 moles of NH₃
b. 3.5 moles of O₂
Required
a. moles of H₂O
b. moles of N₂
Solution
Reaction
4NH₃+3O₂⇒2N₂+6H₂O
a. From the equation, mol ratio NH₃ : H₂O = 4 : 6, so mol H₂O :
=6/4 x mol NH₃
= 6/4 x 20 moles
= 30 moles
b. From the equation, mol ratio N₂ : O₂ = 2 : 3, so mol N₂ :
=2/3 x mol O₂
= 2/3 x 3.5 moles
= 2.33 moles
