For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:
ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3
Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol
m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg
For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf
Tf = -1.59 celsius
Answer:
Explanation:
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Sorry if I'm wrong but I think that it is B.
Answer:
4.0 moles
Explanation:
The following data were obtained from the question:
Volume (V) = 12L
Pressure = 5.6 atm
Temperature (T) = 205K
Gas constant (R) = 0.08206 atm.L/Kmol
Number of mole (n) =?
Using the ideal gas equation: PV = nRT, the number of mole of the gas can be obtained as follow
PV = nRT
5.6 x 12 = n x 0.08206 x 205
Divide both side by 0.08206 x 205
n = (5.6 x 12)/(0.08206 x 205)
n = 4.0 moles
Therefore, the number of mole of the gas is 4.0 moles
