(a) Interpretation: The category to which the given lipid structure belongs should be identified. Concept Introduction: Lipids are a type of chemical substances which is not soluble in water but soluble in non-polar solvents. Lipids are of two type simple lipids and complex lipids. Simple lipids are those where Glycerol attached with fatty acid. In complex lipids there are two major lipids i, e, Phospholipids and Glycolipids.

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5. A solution of sucrose is fermented in a vessel until the evolution of CO2 ceases. Then, the product solution is analyzed and found to contain, 45% ethanol; 5% acetic acid; and 15% glycerin by weight. If the original charge is 500 kg, evaluate; e. The ratio of sucrose to water in the original charge (wt/wt). f. Moles of CO2 evolved. g. Maximum possible amount of ethanol that could be formed. h. Conversion efficiency. i. Per cent excess of excess reactant. Reactions: Inversion reaction: C12H22O11 + H2O →2C6H12O6 Fermentation reaction: C6H12O6 →→2C2H5OH + 2CO2 Formation of acetic acid and glycerin: C6H12O6 + C2H5OH + H₂O→ CH3COOH + 2C3H8O3

Answer 1

Question

Chapter 20.9, Problem 20.6QC

Interpretation Introduction

(a)

Interpretation:

The category to which the given lipid structure belongs should be identified.

Concept Introduction:

Lipids are a type of chemical substances which is not soluble in water but soluble in non-polar solvents. Lipids are of two type simple lipids and complex lipids. Simple lipids are those where Glycerol attached with fatty acid. In complex lipids there are two major lipids i, e, Phospholipids and Glycolipids.

Interpretation Introduction

(b)

Interpretation:

The components of the compound should be determined.

Concept Introduction:

Lipids are a type of chemical substances which is not soluble in water but soluble in non-polar solvents. Lipids are of two type simple lipids and complex lipids. Simple lipids are those where Glycerol attached with fatty acid. In complex lipids there are two major lipids i,e, Phospholipids and Glycolipids.

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5. A solution of sucrose is fermented in a vessel until the evolution of CO2 ceases. Then, the product solution is analyzed and found to contain, 45% ethanol; 5% acetic acid; and 15% glycerin by weight. If the original charge is 500 kg, evaluate; e. The ratio of sucrose to water in the original charge (wt/wt). f. Moles of CO2 evolved. g. Maximum possible amount of ethanol that could be formed. h. Conversion efficiency. i. Per cent excess of excess reactant. Reactions: Inversion reaction: C12H22O11 + H2O →2C6H12O6 Fermentation reaction: C6H12O6 →→2C2H5OH + 2CO2 Formation of acetic acid and glycerin: C6H12O6 + C2H5OH + H₂O→ CH3COOH + 2C3H8O3

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