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
It would be D
Because a covelant compound forms when 2 non metal atoms bond
The answer is four hundred
Answer:
HOAc is stronger acid than HClO
ClO⁻ is stronger conjugate base than OAc⁻
Kb(OAc⁻) = 5.5 x 10⁻¹⁰
Kb(ClO⁻) = 3.3 x 10⁻⁷
Explanation:
Assume 0.10M HOAc => H⁺ + OAc⁻ with Ka = 1.8 x 10⁻⁵
=> [H⁺] = √Ka·[Acid] =√(1.8 x 10⁻⁵)(0.10) M = 1.3 x 10⁻³M H⁺
Assume 0.10M HClO => H⁺ + ClO⁻ with Ka = 3 x 10⁻⁸
=> [H⁺] = √(3 x 10⁻⁸)(0.10)M = 5.47 x 10⁻⁵M H⁺
HOAc delivers more H⁺ than HClO and is more acidic.
Kb = Kw/Ka, Kw = 1 x 10⁻¹⁴
Kb(OAc⁻) = 5.5 x 10⁻¹⁰
Kb(ClO⁻) = 3.3 x 10⁻⁷
Answer:
Density of the object is 8759.494 grams/L
Explanation:
As we know density of an object is mass of the object divided by its volume
Given
Volume of the object is equal to the change in volume of water with in the cylinder when the object was immersed in water.
Hence, volume of object is equal to
mL
Mass of the object is 
grams
Density
grams/L
Density of the object is 8759.494 grams/L
