Answer:
tama yon sagot nya gayahin mo nalang
Answer:
are a gas at very low volumes, when gas particles are very close together
a gas at very low temperatures, when gas particles have very little kinetic energy
a gas with highly polar molecules that have very strong intermolecular forces
Explanation:
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:
0.01144L or 1.144x10^-2L
Explanation:
Data obtained from the question include:
V1 (initial volume) = 20.352 mL
P1 (initial pressure) = 680mmHg
P2 (final pressure) = 1210mmHg
V2 (final volume) =.?
Using the Boyle's law equation P1V1 = P2V2, the volume of the container can be obtained as follow:
P1V1 = P2V2
680 x 20.352 = 1210 x V2
Divide both side by 1210
V2 = (680 x 20.352)/1210
V2 = 11.44mL
Now we need to convert 11.44mL to L in order to obtain the desired result. This is illustrated below:
1000mL = 1 L
11.44mL = 11.44/1000 = 0.01144L
Therefore the volume of the container is 0.01144L or 1.144x10^-2L
