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3 years ago

20. Sudan III is used to test for

Chemistry

2 answers:

monitta3 years ago

6 0

Answer:

Lipid tests

Sudan III is a red fat-soluble dye that is utilized in the identification of the presence of lipids, triglycerides and lipoproteins. The Reaction: Sudan III reacts with the lipids or triglycerides to stain red in colour.

notka56 [123]3 years ago

6 0

Answer:

I think soo fats and oils

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<h3>Further explanation</h3>

Given

0.2 moles of Potassium Oxide

500 ml of water = 0.5 L

Required

The molarity

Solution

Molarity is a way to express the concentration of the solution

Molarity shows the number of moles of solute in every 1 liter of solution or mmol in each ml of solution

$\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}$

Assume volume of solution = volume of water(addition of solution volume from volume Potassium Oxide is negligible

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The pH of a 0.78M solution of 4-pyridinecarboxylic acid HC6H4NO2is measured to be 2.53.Calculate the acid dissociation constant

nika2105 [10]

Answer : The value of acid dissociation constant is, $1.1\times 10^{-5}$

Solution : Given,

Concentration pyridinecarboxylic acid = 0.78 M

pH = 2.53

First we have to calculate the hydrogen ion concentration.

$pH=-\log [H^+]$

$2.53=-\log [H^+]$

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Now we have to calculate the acid dissociation constant.

The equilibrium reaction for dissociation of (weak acid) is,

$HC_6H_4NO_2\rightleftharpoons C_6H_4NO_2^-+H^+$

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At eqm. (0.78-x) x x

The expression of acid dissociation constant for acid is:

$k_a=\frac{[C_6H_4NO_2^-][H^+]}{[C_6H_4NO_2]}$

As, $[H^+]=[C_6H_4NO_2^-]=x$

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$k_a=\frac{(x)\times (x)}{(0.78-x)}$

$k_a=\frac{(2.95\times 106{-3})\times (2.95\times 106{-3})}{(0.78-2.95\times 106{-3})}$

$K_a=1.1\times 10^{-5}$

Therefore, the value of acid dissociation constant is, $1.1\times 10^{-5}$

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