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:
C-Rb20
Explanation:
The chemical formula for Rubidium oxide is Rb20.
Hope this helps!
Answer:
Humans started to use pesticides dating back to before 2000 BC.
Hope this helps ^-^
Explanation:
Answer:
X= -Y + 2/2 Y=-X + 3/2
Explanation:
I don't know how to simplify it anymore. You would get x=-y+3/2 and y=-x+3/2
Answer:
The correct answer is 0.033 M
Explanation:
We have a solution of NaClO with a concentration of 5%w/w:
5% w/w= 5 g NaClO/100 g solution
The first dilution is 10 ml of solution in 100 ml. That is a 1/10 dilution (10ml/100 ml= 1/10). That means we are diluting 10 times the solution. We can calculate the resulting concentration after this first dilution as follows:
5%w/w x 10 ml/100 ml = 5% w/w/10= 0.5%w/w
Then, we take 6 ml of 0.5% w/w solution and we add 6 ml of dye in a reaction vessel. The total volume of the solution in the reaction vessel is 6 ml + 6 ml= 12 ml, and we are diluting twice the solution because 6 ml/12 ml= 1/2. We can calculate the resulting concentration of the solution after this second dilution as follows:
0.5% w/w x 6 ml/12 ml= 0.5% w/w/2= 0.25%w/w
Finally, we need to convert the concentration from %w/w to M (mol solution/1L solution). For this, we assume a density of the solution close to the density of water (1.00 g/ml) and we use the molecular weight of NaClO (74.44 g/mol):
0.25 g NaClO/100 g solution x 1 mol NaClO/74.44 g x 1.00 g solution/1 ml x 100 ml/0.1 L= 0.033 mol/L
= 0.033 M
