N-Octanol and water are chosen because the connection between a substance's hydrophilicity and lipophilicity is measured by
(n-Octanol/Water partition coefficient). When a chemical is more dissolves in fat-like solvents like n-octanol, the value is more significant than one, when it's more dissolved in water, the value is lower.
What is the partition coefficient?
- The partition coefficient for the two-phase network comprising n-octanol and water is known as the
value. N-Octanol-Water Partition Ratio is another name for it.
- The connection between a substance's hydrophilicity (its ability to dissolve in water) and lipophilicity (its ability to dissolve in fat) is measured by
. The value is bigger if a drug is more accessible in fat-like liquids like n-octanol and less if a compound seems more water-soluble.
- Owing to linkage or fragmentation, substances that are involved in the octanol-water combination as multiple synthetic entities are each given a unique
ratio.
So, N-Octanol is chosen because it has a carbon/oxygen ratio that is comparable to that of lipids and because it shows both hydrophobic and hydrophilic properties. N-octanol, therefore, resembles the makeup and characteristics of cells and other living things.
Learn more about octanol here:
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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:
C = 9460 Kj
Explanation:
Given data:
Mass of copper = 2kg
Latent heat of vaporization = 4730 Kj/Kg
Energy required to vaporize 2kg copper = ?
Solution:
Equation
Q= mLvap
by putting values,
Q= 2kg × 4730 Kj/Kg
Q = 9460 Kj
I think it’s the one that has a Br
132 g of C , 22 g of H , 176 g of O
132 + 22 + 176 => 330 g <span>of the substance
</span>Now convert the masses in <span>moles :
</span>
C = 12.0 u H = 1.0 u O = 16.0 u
C = 132 / 12.0 => 11 moles
H = 22 / 1.0 => 22 moles
O = 176 / 16.0 => 11 moles
Using the values obtained the lowest proportion in mols of elements present, simply divide the values found for the least of them<span>:
</span>
C = 11 / 11 => 1
H = 22 / 11 => 2
O = 11 / 11 => 1
formula empirically <span>is : CH</span>₂O
hope this helps!