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
the gas it has in it. it’s moving because of the gas
Climate change from the use of metals i think sorry if i’m wrong that one seems the most ligit to me
Answer:
T₂ = 669.2 K
Explanation:
Given data:
Initial pressure = 660 torr
Initial temperature = 26 °C (26 +273 = 299 K)
Final volume = 280 mL ( 280/1000 = 0.28 L)
Final pressure = 940.0 torr
Final volume = 0.44 L
Final temperature = ?
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₁V₁/T₁ = P₂V₂/T₂
T₂ = P₂V₂ T₁ /P₁V₁
T₂ = 940 torr × 0.44 L × 299 K / 660 torr × 0.28 L
T₂ = 123666. 4 torr. L. K / 184.8 torr. L
T₂ = 669.2 K
