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Brrunno [24]
3 years ago
7

Soap is a very interesting chemical. We even discussed it on the discussion board. How does it work, exactly?

Engineering
1 answer:
soldi70 [24.7K]3 years ago
4 0

Answer:

Saponification is a process in which soap is formed from mixtures of  sodium or potassium salts of fatty acids. These fatty acids are reacted high temperature of At 80°C-100°C with alkali to extract salt. These alkali can be sodium hydroxide or potassium hydroxide.

Soap has both polar (ionic) and non polar molecules due to which it has characteristics of both hydrophilic substance (having tendency to mix with water) and hydrophobic substance (have tendency to mix with oils) and due to this nature it can act as an emulsifier.

An emulsifier has tendency to diffuse one liquid into another   liquid which is incapable of mixing with homogeneous liquid like water.

Cleansing action takes place due to presence of ionic and non-polar properties at same time, in combination with solubility principles. The ionic end of soap molecule is the salt end. It is hydrophilic (water soluble) in nature. The non-polar end cotains long hydrocarbon chains and is hydrophobic (water repellent).

When immiscible liquids like grease or oil mixed with soap water, non polar end (hydrophobic end) absorbs the dirt which means the soap will form the micelles and trap the dirt in it. As micelles is soluble in water it will remove the dirt with it.

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marishachu [46]

Answer:

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Explanation:

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8 0
3 years ago
Read 2 more answers
A well is located in a 20.1-m thick confined aquifer with a conductivity of 14.9 m/day and a storativity of 0.0051. If the well
ahrayia [7]

Answer:

S = 5.7209 M

Explanation:

Given data:

B = 20.1 m

conductivity ( K ) = 14.9 m/day

Storativity  ( s ) = 0.0051

1 gpm = 5.451 m^3/day

calculate the Transmissibility ( T ) = K * B

                                                       = 14.9 * 20.1 = 299.5  m^2/day

Note :

t = 1

U = ( r^2* S ) / (4*T*<em> t </em>)

  = ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4

Applying the thesis method

W(u) = -0.5772 - In(U)

       = 7.9

next we calculate the pumping rate from well ( Q ) in m^3/day

= 500 * 5.451 m^3 /day

= 2725.5 m^3 /day

Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping

S = \frac{Q}{4\pi T} * W (u)

 where : Q = 2725.5

               T = 299.5

               W(u)  = 7.9

substitute the given values into equation above

S = 5.7209 M

4 0
3 years ago
A square silicon chip (k = 152 W/m·K) is of width 7 mm on a side and of thickness 3 mm. The chip is mounted in a substrate such
Harrizon [31]

Answer:

The steady-state temperature difference is 2.42 K

Explanation:

Rate of heat transfer = kA∆T/t

Rate of heat transfer = 6 W

k is the heat transfer coefficient = 152 W/m.K

A is the area of the square silicon = width^2 = (7/1000)^2 = 4.9×10^-5 m^2

t is the thickness of the silicon = 3 mm = 3/1000 = 0.003 m

6 = 152×4.9×10^-5×∆T/0.003

∆T = 6×0.003/152×4.9×10^-5 = 2.42 K

7 0
3 years ago
Engine oil flows through a 25‐mm‐diameter tube at a rate of 0.5 kg/s. The oil enters the tube at a temperature of 25°C, while th
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Answer:

a) the log mean temperature difference (Approx. 64.5 deg C)

b) the rate of heat addition into the oil.

The above have been solved for in the below workings

Explanation:

3 0
4 years ago
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