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GuDViN [60]
3 years ago
5

Identify one type of noncovalent bond present in each solid.

Chemistry
1 answer:
NeX [460]3 years ago
7 0

Answer:

Explanation:

The question is not complete, the cmplete question is:

Identify one type of noncovalent bond present in each solid.

1) Table salt (NaCl)                2) Graphite (repeating)

a. hydrogen bonds

b. ionic interactions

c. van der Waals interactions

d. hydrophobic interactions

Answer:

1) Table salt

b. ionic interactions

Ionic bond are formed between atoms with incomplete outermost shell. Some atoms add electrons to their outermost shell to make the shell complete hence making it a negative ion while some atoms loses their electron to make the outermost shell complete becoming a positive ion. In NaCl, sodium (Na) has 1 electron in its outermost shell which it transfers to Cl which has 7 electrons in the outermost shell. Hence after the bonding the outermost shell of the atoms become complete.

2) Graphite

c. Van Der Waals interaction

Van der waal  forces are weak interaction between molecules that exist between close atoms. Carbon atoms in graphite planes have covalent bond, these graphite planes are known as graphenes. Bonds between graphenes are very weak and are van der waals forces.

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Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81
Gnom [1K]

Answer:

Explanation:

Given parameters;

pH  = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

      pH = -log[H₃O⁺]

      pOH = -log[OH⁻]

 pH + pOH = 14

Now let us solve;

   pH = 8.74

             since  pH = -log[H₃O⁺]

                           8.74 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{8.74}

                             [H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

       pH + pOH = 14

                 pOH = 14 - 8.74

                  pOH = 5.26

                  pOH = -log[OH⁻]

                     5.26  = -log[OH⁻]

                     [OH⁻] = 10^{-5.26}

                      [OH⁻] = 5.5 x 10⁻⁶mol dm³

2.  pH = 11.38

             since  pH = -log[H₃O⁺]

                           11.38 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{11.38}

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           pH + pOH = 14

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                  pOH = 2.62

                  pOH = -log[OH⁻]

                     2.62  = -log[OH⁻]

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3. pH = 2.81

             since  pH = -log[H₃O⁺]

                           2.81 =  -log[H₃O⁺]

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                             [H₃O⁺]  = 1.55 x 10⁻³ mol dm³

           pH + pOH = 14

                 pOH = 14 - 2.81

                  pOH = 11.19

                  pOH = -log[OH⁻]

                     11.19  = -log[OH⁻]

                     [OH⁻] = 10^{-11.19}

                      [OH⁻] =6.46 x 10⁻¹²mol dm³

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