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FrozenT [24]
10 months ago
8

Octanol is slightly soluble in water, and water is very soluble in octanol. why is it important to presaturate octanol with wate

r and water with octanol when measuring ow?
Chemistry
1 answer:
notsponge [240]10 months ago
8 0

N-Octanol and water are chosen because the connection between a substance's hydrophilicity and lipophilicity is measured by K_{OW} (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 K_{OW} 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 K_{OW}. 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 K_{OW} 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:

brainly.com/question/7768749

#SPJ4

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We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r:                      32.00

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2. Mass of O₂

\text{Mass of C$_{8}$H}_{18} = \text{16.52 mol O}_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol O}_{2}} = \textbf{528.6 g O}_{2}\\\text{The reaction requires $\large \boxed{\textbf{528.67 g O}_{2}}$}

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Calculation:

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(b) <em>Thickness of foil </em>

The foil is a rectangular solid.

V  = lwh                            Divide each side by lw

h = V/(lw)

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   = 1.50 × 10⁻³ cm           Convert to millimetres

   = 0.015 mm                  Convert to micrometres

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