<h2>.An oily cell membrane.</h2>
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When dT = Kf * molality * i
= Kf*m*i
and when molality = (no of moles of solute) / Kg of solvent
= 2.5g /250g x 1 mol /85 g x1000g/kg
=0.1176 molal
and Kf for water = - 1.86 and dT = -0.255
by substitution
0.255 = 1.86* 0.1176 * i
∴ i = 1.166
when the degree of dissociation formula is: when n=2 and i = 1.166
a= i-1/n-1 = (1.166-1)/(2-1) = 0.359 by substitution by a and c(molality) in K formula
∴K = Ca^2/(1-a)
= (0.1176 * 0.359)^2 / (1-0.359)
= 2.8x10^-3
Answer:
D 4,3,2
Explanation:
4 Co + 3 O2 ----> 2 Co2O3
By Tri chlorination, three chlorine atoms can replace the 3 hydrogen atoms in propane molecule to give tri-chlorinated products according to the attached picture.
Answer:
The boiling point elevation is 3.53 °C
Explanation:
∆Tb = Kb × m
∆Tb is the boiling point elevation of the solution
Kb is the molal boiling point elevation constant of CCl4 = 5.03 °C/m
m is the molality of the solution is given by moles of solute (C9H8O) divided by mass of solvent (CCl4) in kilogram
Moles of solute = mass/MW =
mass = 92.7 mg = 92.7/1000 = 0.0927 g
MW = 132 g/mol
Moles of solute = 0.0927/132 = 7.02×10^-4 mol
Mass of solvent = 1 g = 1/1000 = 0.001 kg
m = 7.02×10^-4 mol ÷ 0.001 kg = 0.702 mol/kg
∆Tb = 5.03 × 0.702 = 3.53 °C (to 2 decimal places)
