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jekas [21]
2 years ago
12

Describes two uses of surfactants at least one must be something that was not described in the passage

Chemistry
1 answer:
Morgarella [4.7K]2 years ago
4 0

Answer:

We don't have the passage.  A random sampling of surfactant uses includes:

  • removal of oily materials from objects (clothes and dishes)
  • forms remarkable structures called bubbles
  • Assists in forming emulsions (e.g., mayonaise and paints)

Explanation:

The structure of a surfactant makes one end of a molecule hydrophilic and the other end hydrophobic.  In water, they self-assemble into micelles, an arrangement in which the hydrophobic ends align towards the center, and the hydrophilic ends are pointed outwards to the water.  This self-assembly is apparant when bubbles are made.  The  molecules quickly align themselves such that the hyrophilic ends are oriented inwards towards a thin layer of water and the hydrophobic ends are pointed outward to the air.  This arrangement allows a mono-molecular sphere of water molecules to remain stable enough to float, reflect light, and please.  These same properties allow the inverse to occur.  Soap molecules surround a hydrophobic mass (e.g., the hamburger grease on your shirt) and solubilize it into small micelles which are then carried away in the surrounding water.

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aleksklad [387]
<h2>H_2O  + H_2 + O_2</h2>

Explanation:

1. Water decomposition

  • Decomposition reactions are represented by-

       The general equation: AB → A + B.

  • Various methods used in the decomposition of water are -
  1. Electrolysis
  2. Photoelectrochemical water splitting
  3. Thermal decomposition of water
  4. Photocatalytic water splitting
  • Water decomposition is the chemical reaction in which water is broken down giving oxygen and hydrogen.
  • The chemical equation will be -

        H_2O  + H_2 + O_2

Hence, balancing the equation we need to add a coefficient of 2 in front of H_2O on right-hand-side of the equation and  2 in front of H_2 on left-hand-side of the equation.

     ∴The balanced equation is -

       2 H_2O → 2 H_2 + O_2

2. Formation of ammonia

  • The formation of ammonia is by reacting nitrogen gas and hydrogen gas.

      N_2 + H → NH_3

Hence, for balancing equation we need to add a coefficient of 3 in front of hydrogen and 2 in front of ammonia.

   ∴The balanced chemical equation for the formation of ammonia gas is as  follows -

     N_2+3H→ 2NH_3.

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5 0
3 years ago
How much 10.0 M HNO must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the pH
Naya [18.7K]

Explanation:

It is given that molarity of acetic acid = 0.0100 M

Therefore, moles of acetic acid = molarity of acetic acid × volume of buffer

            Moles of acetic acid = 0.0100 M × 1.00 L

                                              = 0.0100 mol

Similarly, moles of acetate = molarity of sodium acetat × volume of buffer

                                           = 0.100 mol

When HNO_{3} is added, it will convert acetate to acetic acid.

Hence, new moles acetic acid = (initial moles acetic acid) + (moles HNO_{3})

                                                = 0.0100 mol + x

New moles of sodium acetate = (initial moles acetate) - (moles HNO_{3})

                                        = 0.100 mol - x

According to Henderson - Hasselbalch equation,

           pH = pK_{a} + log\frac{[conjugate base]}{[weak acid]}

             pH = pKa + log\frac{(new moles of sodium acetate)}{(new moles of acetic acid)}

           4.95 = 4.75 + log\frac{(0.100 mol - x)}{(0.0100 mol + x)}

       log\frac{(0.100 mol - x)}{(0.0100 mol + x)}  = 4.95 - 4.75

                                            = 0.20

\frac{(0.100 mol - x)}{(0.0100 mol + x)}  = antilog (0.20)

                                           = 1.6

Hence,    x = 0.032555 mol

Therefore, moles of HNO_{3} = 0.032555 mol

volume of HNO_{3} = \frac{moles HNO_{3}}{molarity of HNO_{3}}

                                = \frac{0.032555 mol}{10.0 M}

                                 = 0.0032555 L

or,                             = 3.25           (as 1 L = 1000 mL)

Thus, we can conclude that volume of HNO_{3} added is 3.26 mL.

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3 years ago
The electron-domain geometry of a carbon-centered compound ch4 is tetrahedral. the hybridization of the central carbon atom is _
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The  hybridization  of  the   central  carbon  compound  CH4  tetrahedral  is
SP^3 hybridization 

In   tetrahedral  molecular  geometry    a central atom  is  located  at the  center  with  4  substituent  that  are located at  the  corners  of  tetrahedron. Example  in methane  molecules  is made  up of  b equally  spaced sp^3  hybrid  orbital  forming  bond  angle  of   109.5
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3 years ago
Is the following statement about our solar system true or false?
Archy [21]

Answer:

true

Explanation:

5 0
3 years ago
Is a solution that has a pH of 7.52 acidic or basic?
Luda [366]

Answer:

on the PH scale anything with a ph of 1-7 is acidic, anything 7-14 is basic, in this case the solution mentioned in the question is basic.

8 0
3 years ago
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