Answer: CoBr3 < K2SO4 < NH4 Cl
Justification:
1) The depression of the freezing point of a solution is a colligative property, which means that it depends on the number of particles of solute dissolved.
2) The formula for the depression of freezing point is:
ΔTf = i * Kf * m
Where i is the van't Hoof factor which accounts for the dissociation of the solute.
Kf is the freezing molal constant and only depends on the solvent
m is the molality (molal concentration).
3) Since, you are assuming equal concentrations and complete dissociation of the given solutes, the solute with more ions in the molecular formula will result in the solution with higher depression of the freezing point (lower freezing point).
4) These are the dissociations of the given solutes:
a) NH4 Cl (s) --> NH4(+)(aq) + Cl(-) (aq) => 1 mol --> 2 moles
b) Co Br3 (s) --> Co(3+) (aq) + 3Br(-)(aq) => 1 mol --> 4 moles
c) K2SO4 (s) --> 2K(+) (aq) + SO4 (2-) (aq) => 1 mol --> 3 moles
5) So, the rank of solutions by their freezing points is:
CoBr3 < K2SO4 < NH4 Cl
Molecule is a chemical substance that cannot be broken down into another chemical substance.
Capillary action is defined as the ability of a liquid to go up a narrow space without the help or opposition of external forces. One of the most important factors affecting capillary action is the intermolecular forces within a substance. The higher the IMF, the greater the capillary action. The H-bonding in water gives it greater IMF than acetone, so water has greater capillary action.
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Answer:</h3>
382.63 K
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Explanation:</h3>
We are given;
- Volume of Iodine as 71.4 mL
- Mass of Iodine as 0.276 g
- Pressure of Iodine as 0.478 atm
We are required to calculate the temperature of Iodine
- We are going to use the ideal gas equation;
- According to the ideal gas equation; PV = nRT, where R is the ideal gas constant, 0.082057 L.atm/mol.K.
T = PV ÷ nR
But, n, the number of moles = Mass ÷ Molar mass
Molar mass of iodine = 253.8089 g/mol
Thus, n = 0.276 g ÷ 253.8089 g/mol
= 0.001087 moles
Therefore;
T = (0.478 atm × 0.0714 L) ÷ (0.001087 moles × 0.082057)
= 382.63 K
Thus, the temperature of Iodine in Kelvin is 382.63 K
