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3 years ago

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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A PMMA plate with a 25 mm (width) x 6.5 mm (thickness) cross-section has a contained crack of length 2c = 0.5 mm in the center o

victus00 [196]

Answer:

LAOD = 6669.86 N

Explanation:

Given data:

width $= 25 mm = 25\times 10^{-3} m$

thickness $= 6.5 mm = 6.5\times 10^{-3} m$

crack length 2c = 0.5 mm at centre of specimen

$\sigma _{applied} = 1000 N/cross sectional area$

stress intensity factor = k will be

$\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}$

$= 6.154\times 10^{6} Pa$

we know that

$k =\sigma_{applied} (\sqrt{\pi C})$

$=6.154\sqrt{\pi (2.5\times 10^{-04})}$ [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

if $K_C = 1.15 Mpa m^{1/2}$ then load will be

$Kc = \sigma _{frac}(\sqrt{\pi C})$

$1.15 MPa = \sigma _{frac}\times \sqrt{\pi (2.5\times 10^{-04})}$

$\sigma _{frac} = 41.04 MPa$

$load = \sigma _{frac}\times Area$

$load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N$

LAOD = 6669.86 N

3 0

2 years ago

After testing a model of a fuel-efficient vehicle, scientists build a full-sized vehicle with improved fuel efficiency. Which st

Mazyrski [523]

B evaluate the solution. Sorry if I'm wrong.

4 0

3 years ago

Casein, a dairy product used in making cheese, contains 25% moisture when wet. A dairy sells this product for $40/100 kg. If req

Nataly_w [17]

Based on the percent moisture content of the dried product, the mass of dried casein produced os 852.3 kg.

<h3>What is the mass of casein in wet casein?</h3>

The mass of casein in 1000 Kg of wet casein is 75% 1000 kg = 750 Kg

Mass of water 250 kg

The mass of casein is constant while the moisture content can be changed.

At 12% moisture content;

750 kg = 88%%

100 % = 100 ×750/88 = 852.27 kg

Therefore, the mass of dried casein produced os 852.3 kg.

Learn more about mass at: brainly.com/question/24658038

#SPJ1

3 0

2 years ago

Metal and dirt are not considered contaminants to oll.<br> A) O True<br> B) O False

Likurg_2 [28]

Answer:

true

Explanation:

6 0

2 years ago

Read 2 more answers

An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown

barxatty [35]

Answer:

The specific weight of unknown liquid is found to be 15 KN/m³

Explanation:

The total pressure in tank is measured to be 65 KPa in the tank. But, the total pressure will be equal to the sum of pressures due to both oil and unknown liquid.

Total Pressure = Pressure of oil + Pressure of unknown liquid

65 KPa = (Specific Weight of oil)(depth of oil) + (Specific Weight of unknown liquid)(depth of unknown liquid)

65 KN/m² = (8.5 KN/m³)(5 m) + (Specific Weight of Unknown Liquid)(1.5 m)

(Specific Weight of Unknown Liquid)(1.5 m) = 65 KN/m² - 42.5 KN/m²

(Specific Weight of Unknown Liquid) = (22.5 KN/m²)/1.5 m

<u>Specific Weight of Unknown Liquid = 15 KN/m³</u>

4 0

3 years ago