Besides glycerol what 3 other molecules make up a triglyceride?

1 answer:

6 0

Triglyceride is a molecule constituted by one one molecule of glycerol and 3 molecules of fatty acids. The structure of a glycerol is like a letter "E", where the vertical line ( | ) is the glycerol and the three horizontal lines are long chains of organic acids, each with one COOH group, called fatty acids. The answer is that the other 3 molecules that make up a triglyceride are fatty acids.

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Which term describes a chemical equation in which all highly soluble compounds are written as dissociated ions

BartSMP [9]

Answer: IONIC EQUATION.

Explanation:

A chemical equation is defined as the form by which a chemical reaction is represented mathematically. These are written in the form of symbols and chemical formulas of reactants and products which are taking part in the chemical reaction. A chemical equation can be written in two forms, these include:

--> MOLECULAR EQUATION: in this type of equations, the compounds are written and represented in a molecular form. This is sometimes referred to as a balanced equation.

--> IONIC EQUATION: This is a type of chemical equation in which the electrolytes in aqueous solution are expressed as dissociated ions. A typical illustrated example is seen in the reaction between AgNO3(aq) and NaCl(aq) :

Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) → AgCl(s) + Na+(aq) + NO3-(aq)

The (aq) written in the above equation signifies they are in aqueous solution.

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the elements helium (he), neon (ne), and xenon (xe) are all part of the same on the periodic table plz answer correctly gets a b

bagirrra123 [75]

Noble gas. Noble gas, any of the seven chemical elements that make up Group 18 (VIIIa) of the periodic table. The elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og)

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Read 2 more answers

The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be

Mrac [35]

Answer: The freezing point of solution is 5.35°C

Explanation:

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 4.90°C/m

$m_{solute}$ = Given mass of solute (naphthalene) = 2.60 g

$M_{solute}$ = Molar mass of solute (naphthalene) = 128.2 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC$