A dissolving liquid composed of polar molecules is a polar solvent.
The distinction of polar and non-polar liquids is important because the like dissolves like rule. This rule states that the solubility is greater when the polarity of the liquid is similar to the polarity of the solute.
So, to dissolve polar compounds (e.g. ionic compounds) you should use polar solvents (e.g. water).
Answer: polar solvent
Answer:
The molar mass of the gas is 44 g/mol
Explanation:
It is possible to solve this problem using Graham's law that says: Rates of effusion are inversely dependent on the square of the mass of each gas. That is:
If rate of effusion of nitrogen is Xdistance / 48s and for the unknown gas is X distance / 60s and mass of nitrogen gas is 28g/mol (N₂):
6,61 = √M₂
44g/mol = M₂
<em>The molar mass of the gas is 44 g/mol</em>
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I hope it helps!
Answer:
b 13kg/m³ is the answer to this question
Answer:
If there reacted 1.5 moles of O2, there will be produced 1.0 mol of Fe2O3
Explanation:
Step 1: Data given
Number of moles oxygen reacted = 1.5 moles
Step 2: The balanced equation
4Fe + 3O2 → 2Fe2O3
Step 3: Calculate moles of Fe2O3
For 4 moles Fe consumed, we need 3 moles of O2 to produce 2 moles of Fe2O3
For 1.5 moles O2 consumed, we'll have 2/3 * 1.5 = 1.0 mol of Fe2O3
If there reacted 1.5 moles of O2, there will be produced 1.0 mol of Fe2O3
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
