Lipids are hydrophobic; They would be insoluble, group together, and float to the top
Answer:
63.53% yield
Explanation:
The balanced equation for this reaction is 2NaCl + H2O -> 2NaOH +Cl2
First we must find the limiting reactant
From NaCl we can only produce 6.06 grams of Cl2 in <u>theory</u>
From H20 we can only produce 38.995 grams in theory
so we know NaCl is the limiting
% yield is (Actual/Theoretical) x100 so
(3.85/6.06)x100= 63.53% yield
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:
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Explanation:
Answer:
KCl is the answers for the question
Explanation:
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