Answer: sharing
Reason: They do this to gain stability. The reason they don’t actually transfer is because the difference in electronegativity values are above a certain value.
Answer:
2
b= they are grouped differently, but all the atoms are still there.
Beef and cheddar I believe!!!
Answer is: V<span>an't Hoff factor (i) for this solution is 2,26.
</span>Change in freezing point
from pure solvent to solution: ΔT =i · Kf · m.
<span>Kf - molal freezing-point depression constant for water is 1,86°C/m.
</span>m - molality, moles of solute per kilogram of solvent.
n(K₂SO₄) = 16,8 g ÷ 174,25 g/mol
n(K₂SO₄) = 0,096 mol.
m(K₂SO₄) = 0,096 mol/kg.
ΔT = 0,405°C.
i = 0,405 ÷ (1,86 · 0,096)
i = 2,26.
Answer:
548 g/mol
Explanation:
The freezing point depression of a solvent occurs when a nonvolatile solute is added to it. Because of the interactions between solute-solvent, it is more difficult to break the bonds, so the phase change will need more energy, and the freezing point will drop, which is called cryoscopy.
The drop in temperature can be calculated by:
ΔT = Kf*W*i
Where Kf is the cryoscopy constant of the solvent, W is the molality, and i is the van't Hoff factor, which indicates the fraction of the solute that dissolves.
The molality represents how much moles (n) of the solute is presented in each kg of the solvent (m2), thus
W = n/m2
The number of moles is the mass of the solute (m1) in g, divided by the molar mass (M1) of it:
W = m1/(M1*m2)
So, by the data:
0.2214 = 0.632/(M1*0.00521)
0.00115M1 = 0.632
M1 = 548 g/mol
