If two fatty acids, glycerol, and a phosphate group undergo a dehydration reaction, what molecule forms
1 answer:
<h3>Answer:</h3>
A Phospholipid is formed.
<h3>Explanation:</h3>The reaction of fatty acids, glycerol and phosphoric acid is given below. In this reaction the carboxylic groups of fatty acids and the hydrogyl groups of glycerol undergo an Esterification Reaction with the elimination of two water molecules and formation of Ester functional group. Also, the third hydroxyl group of glycerol undergo dehydration reaction with phosphoric acid. The resulting molecule is known as phospholipid.
Biologically, Phospholipids are found in the cell membranes. The long hydrocarbon chains of fatty acids acts as hydrophobic (water fearing) tails and the phosphate group acts as hydrophilic (water loving) head. They form a barrier protecting the internal environment of cell from external surrounding.
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The pH of 0.35 M of CH₃NH₂ is 12.09 and the for methylamine is 4.4 x 10⁻⁴.
<h3>What is pH?</h3>
pHis the quantitative measure of acidity and basicity of an aqueous or other liquid solution. The scale range goes from 0 to 14. Water has a pH of 7 and is neutral in nature.
<h3>What is dissociation constant?</h3>The dissociation constant is an equilibrium constant that describes the dissociation or ionization of a base or an acid
describes the dissociation of an acid
describes the dissociation of a base
For methylamine,
Initial concentration of methylamine = 0.35 M
Initial concentration of products = 0
Let, at equilibrium concentration of CH₃NH₂ = 0.35 - x
Then, concentration of CH₃NH₃⁺ and OH⁻ is x and x respectively
The dissociation constant for methylamine, = 4.4 x 10⁻⁴
pOH = -log[OH] = -log(0.0124) = 1.91
pH + pOH = 14
pH = 14 - 1.91 = 12.09
Thus, the pH of methylamine is 12.09 and is 4.4 x 10⁻⁴
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