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3 years ago

What is the method used to prepare emulsion

Chemistry

1 answer:

olasank [31]3 years ago

8 0

Answer:

The methods commonly used to prepare emulsions can be divided into two categories

A) Dry gum method

B) Wet gum method

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45) George is making spaghetti for dinner. He places 4.01 kg of water in a pan and brings it to a boil.

lesya692 [45]

On adding salt.....The boiling temperature increases.....

So ∆t= KB * molality
=O.52*(58/58)/4
= O.52*1/4
= 0.13
So increase is 100+.13=100.13°c

5 0

2 years ago

Calculate the pH and fraction of dissociation ( α ) for each of the acetic acid ( CH 3 COOH , p K a = 4.756 ) solutions. A 0.002

marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The $pK_a=4.756$

The value of the dissociation constant = $K_a$

$pK_a=-\log[K_a]$

$K_a=10^{-4.756}=1.754\times 10^{-5}$

Initial concentration of the acetic acid = [HAc] =c = 0.00225

Degree of dissociation = α

$HAc\rightleftharpoons H^++Ac^-$

Initially

At equilibrium ;

(c-cα) cα cα

The expression of dissociation constant is given as:

$K_a=\frac{[H^+][Ac^-]}{[HAc]}$

$1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}$

$1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}$

$1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}$

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

$[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M$

The pH of the solution ;

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

$=-\log[0.0001901 M]=3.72$

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3 years ago

Write an equation for the reaction that takes place during the laboratory preparation of dry hydrogen

olya-2409 [2.1K]

Answer:

Zn+2HCl→ZnCl2+H2

Explanation:

Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc.

8 0

3 years ago

3. What mass of copper could be deposited from a copper(II) Sulphate solution using a current of 5.0 A over 100 seconds? ( F =96

arsen [322]

Mass of copper : 0.165 g

<h3>Further explanation</h3>

Given

5.0 A over 100 seconds

Required

Mass of copper

Solution

Faraday's law:

<em>The mass of the substance formed at each electrode is proportional to the electric current flowing in the electrolysis</em>

<em /> $\tt W=\dfrac{e.i.t}{96500}$ <em />

e = Ar / valence = eqivalent weight

i = current

t = time

W = weight

CuSO₄ ----> Cu²⁺ + SO₄²⁻

Cu ----> Cu²⁺ + 2e

e = Ar/2

= 63,5/2 = 31,75

$\tt W=\dfrac{31.75\times 5\times 100}{96500}=0.165~g$

8 0

2 years ago

??????????????? why ????????????????

stealth61 [152]

Answer:

because

Explanation:

4 0

2 years ago

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What is the method used to prepare emulsion​

What is the method used to prepare emulsion