The name is Aluminum Cloride
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
1.)b
2.)true
3.)false
are the answer don't take me on my word
Answer:
D) winds that blow in the same direction at a consistent speed
Explanation:
i took the quiz got it right so i know the answer please trust me i know this is right i promise with all my heart
Answer: 9.9 grams
Explanation:
To calculate the moles, we use the equation:

a) moles of 

b) moles of 


According to stoichiometry :
1 mole of
combine with 1 mole of
Thus 0.33 mole of
will combine with =
mole of
Thus
is the limiting reagent as it limits the formation of product.
As 1 mole of
give = 1 mole of 
Thus 0.33 moles of
give =
of 
Mass of 
Thus theoretical yield (g) of
produced by the reaction is 9.9 grams