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sleet_krkn [62]
3 years ago
14

Below are the structures of three fatty acids, palmitic acid, stearic acid, and oleic acid. what combination of these fatty acid

s could you use to make a fat that is liquid at room temperature?
Chemistry
2 answers:
irina1246 [14]3 years ago
7 0
I think the fatty acids that may be used to make a fat that is liquid at room temperature are One plamitic acid, one stearic acid, and one oleic acid.
Fatty acids are composed of linked chains of carbon atoms with an organic acid group at the end of the chain. Liquid fats or Oils are mainly obtained from plants or fish sources, and have high percentages of unsaturated fatty acids.

12345 [234]3 years ago
3 0

Answer:

A combination of oleic acid & palmitic acid.

Explanation:

Oleic acid is an unsaturated fatty acid and unsaturated fatty acids are liquid at room temperature because of the presence of double bond in them.

Between palmitic acid and stearic acid, we will choose palmitic acid because palmitic acid (16 carbon chain) has less number of carbon atoms as compared to stearic acid (18 carbon chain) that is why it is liquid at room temperature. The fatty acids which have more number of carbon atoms have more Tm (transition temperature) so they will require more temperature for changing phase from solid to liquid. So, it simply means they will be solid at room temperature that is why we will not choose stearic acid.

Tm ∝ No. of carbon atoms in the lipid ∝  hydrophobic interactions.

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Calculate the new boiling point of a solution if 10.00 g of a non-ionizing compound (C3H5(OH)3) is dissolved in 90.00 g of H2O.
erma4kov [3.2K]

Answer:

Boiling T° of solution = 100.6

Explanation:

Formula for elevation of boiling point is:

ΔT = Kb . m . i

where ΔT means Boiling T° of solution - Boiling T° of pure solvent

Our solute is a non ionizing compound.

i = 1, because it is a non ionizing compound. i, indicates the ions dissolved in solution.

m = molality (moles of solute dissolved in 1 kg of solvent)

90 g of solvent = 0.09 kg of solvent

We convert mass of solute to moles (by the molar mass):

10 g . 1 mol /92.09 g = 0.108 moles

m = 0.108 mol /0.09 kg = 1.21 m

Let's replace data: Boiling T° of solution - 100°C = 0.51 °C/m . 1.21 m . 1

Boiling T° of solution = 0.51 °C/m . 1.21 m . 1 + 100°C

Boiling T° of solution = 100.6

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s344n2d4d5 [400]
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Explanation:

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rjkz [21]

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Hope this helped!

7 0
3 years ago
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