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aalyn [17]
3 years ago
12

What purpose does ammonia lauryl sulfate serve as a surfactor

Chemistry
1 answer:
Alborosie3 years ago
8 0

Answer:

See below  

Step-by-step explanation:

Ammonium lauryl sulfate has the structural formula CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂OSO₂O⁻ NH₄⁺.

The long nonpolar hydrocarbon chain and the ionic sulfate end group make it a surfactant.  

The ionic end tends to dissolve in water, but the nonpolar chain does not. This makes the compound an excellent <em>foaming agent,</em> so it is used in many shampoos and toothpastes.

The molecules form <em>micelles</em> in water, small spherical shapes with the polar heads outside, facing the water, and the nonpolar tails are inside.

They reduce the surface tension or the water so that, when you brush your teeth or shampoo your hair, the air bubbles are stable and do not break.

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it would be an element because its an element

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Which of the following electronic transitions would emit the lowest energy photon?
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i will think c-n=3 to NM=2

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Which two body systems are primarily responsible for locomotion
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smooth and skeleton muscle

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2 years ago
Compute the percent ionic character of the interatomic bonds for the following compounds : a. TiO2 b. ZnTe c. CsCld. InSb e. MgC
sesenic [268]

Answer:

a. 63.2%

b. 11.7%

c. 73.3%

d. 0.995%

e. 55.5%

Explanation:

An ionic compound is a compound that is formed by ions, so one of the elements must donate electrons (which is the cation, the positive ion), and the other will receive these electrons (which is the anion, the negative ion).

The power of an element has to attract the electrons is called electronegativity, and so, as higher is the difference of electronegative of the elements, it is more probable that one of them will "still" the electrons and will form an ionic compound. The percent of this ionic character can be found by the Pauling's equation:

%IC = (1 - e^{-0.25*(x_A - x_B)^2}) *100%

Where x_A - x_B is the electronegativity difference of the elements. Thus, consulting an electronegativity table:

a. x_{Ti} = 1.5

x_{O} = 3.5

%IC = (1 - e^{-0.25*(3.5 - 1.5)^2})*100%

%IC = 63.2%

b. x_{Zn} = 1.6

x_{Te} = 2.1

%IC = (1 - e^{-0.25*(2.1 - 1.6)^2})*100%

%IC = 11.7%

c. x_{Cs} = 0.7

x_{Cl} = 3.0

%IC = (1 - e^{-0.25*(3.0 - 0.7)^2})*100%

%IC = 73.3%

d. x_{In} = 1.7

x_{Sb} = 1.9

%IC = (1 - e^{-0.25*(1.9 - 1.7)^2})*100%

%IC = 0.995 %

e. x_{Mg} = 1.2

x_{Cl} = 3.0

%IC = (1 - e^{-0.25*(3.0 - 1.2)^2})*100%

%IC = 55.5%

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3 years ago
Answer True or False for each of the following statements. (a) The carburization surface was maintained at slightly less than 0.
fomenos

Answer:

verdadero/a

falso/b

verdadero/c

Explanation:

6 0
3 years ago
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