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Dmitrij [34]
3 years ago
9

What term is used to describe alkaline hydrolysis of fats and oils​

Chemistry
1 answer:
Gnesinka [82]3 years ago
8 0

Answer:

Saponification

Explanation:

When treated with a base such as NaOH, the ester is transformed back to alcohol and the sodium salt of carboxylic acid. Saponification is the name <u>given</u> to the reaction since it is utilized in the production of soap.

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Upper n subscript 2 (g) plus 3 upper H subscript 2 (g) double-headed arrow 2 upper N upper H subscript 3 (g). At equilibrium, th
artcher [175]

Answer:

The <u>equilibrium constant</u> is:

              k_c=0.0030M^{-2}

Explanation:

The correct equation is:

  •   N₂(g)    +    3H₂(g)    ⇄    2NH₃(g)

Thus, with the equilibrium concentrations you can calculate the equilibrium constant, Kc.

The equation for the equilibrium constant is:

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Substituting:

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         k_c=0.0030M^{-2}

6 0
4 years ago
Suppose you wanted to dissolve 40.0 g NaOH in enough H2O to make 6.00 dm3 of solution
dezoksy [38]

Molarity of solution = 1.6 M

<h3>Further explanation</h3>

Given

40 g NaOH

6 L solution

Required

Steps to solve the problem of molarity

Solution

No additional information about the question.

If you want to make the solution above, then we just need to put the existing NaOH (40 g) into 6 L of water, then do the stirring (in a warm temperature above the hot plate will speed up the NaOH dissolving process)

But if you want to know the molarity of a solution, then

  • 1. we calculate the moles of NaOH

\tt mol=\dfrac{mass}{MW}

MW(molecular weight) of NaOH=

Ar Na+ Ar O + Ar H

23 + 16 + 1 = 40 g/mol

so mol NaOH :

\tt mol=\dfrac{40~g}{40~g/mol}=1~mol

  • 2. Molarity(M)

\tt M=\dfrac{n}{V}\\\\M=\dfrac{1}{6}\\\\M=0.16

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