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Mashcka [7]
3 years ago
5

Which allotrope of carbon is used in lubricants, sporting goods, and aircraft?

Chemistry
1 answer:
Morgarella [4.7K]3 years ago
4 0

The allotrope of carbon that is used in lubricants, sporting goods, and aircraft is graphite. The answer is letter A. Graphite is a good conductor of heat and is very soft. The layers of lattice of each are bound to the adjacent layers by weak Van der Waals forces which allow the layers to slide over one another easily.

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Consider the titration of a 20.0-mL sample of 0.105 M HC2H3O2 with 0.125 M NaOH. Determine each quantity. a. the initial pH b. t
Oksi-84 [34.3K]

Answer:

Explanation:

Given that:

Concentration of HC_2H_3O_2 \  (M_1) = 0.105 M

Volume of  HC_2H_3O_2 \  (V_1) = 20.0 mL

Concentration of NaOH (M_2) = 0.125 M

The  chemical reaction can be expressed as:

HC_2H_3O_2_{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O_{(l)}

Using the ICE Table to determine the equilibrium concentrations.

          HC_2 H_3 O_2 _{(aq)} + H_2O _{(l) } \to C_2 H_3O_2^- _{(aq)} + H_3O^+_{ (aq)}

I            0.105                                     0                  0

C              -x                                         +x                +x

E            0.105 - x                                  x                  x

K_a = \dfrac{[C_2H_5O^-_2][H_3O^+]}{[HC_2H_3O_2]}

K_a = \dfrac{(x)(x)}{(0.105-x)}

Recall that the ka for HC_2H_3O_2= 1.8 \times 10^{-5}

Then;

1.8 \times 10^{-5} = \dfrac{(x)(x)}{(0.105 -x)}

1.8 \times 10^{-5} = \dfrac{x^2}{(0.105 -x)}

By solving the above mathematical expression;

x = 0.00137 M

H_3O^+ = x = 0.00137  \ M \\ \\  pH = - log [H_3O^+]  \\ \\  pH = - log ( 0.00137 )

pH = 2.86

Hence, the initial pH = 2.86

b)  To determine the volume of the added base needed to reach the equivalence point by using the formula:

M_1 V_1 = M_2 V_2

V_2= \dfrac{M_1V_1}{M_2}

V_2= \dfrac{0.105 \ M \times 20.0 \ mL }{0.125 \ M}

V_2 = 16.8 mL

Thus, the volume of the added base needed to reach the equivalence point = 16.8 mL

c) when pH of 5.0 mL of the base is added.

The Initial moles of HC_2H_3O_2 = molarity × volume

= 0.105  \ M \times 20.0 \times 10^{-3} \ L

= 2.1 \times 10^{-3}

number of moles of 5.0 NaOH = molarity × volume

number of moles of 5.0 NaOH = 0.625 \times 10^{-3}

After reacting with 5.0 mL NaOH, the number of moles is as follows:

                    HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}

Initial moles   2.1*10^{-3}       0.625 * 10^{-3}           0                      0

F(moles) (2.1*10^{-3} - 0.625 \times 10^{-3})    0      0.625 \times 10^{-3}         0.625 \times 10^{-3}

The pH of the solution is then calculated as follows:

pH = pKa + log \dfrac{[base]} {[acid]}

Recall that:

pKa for HC_2H_3O_2=4.74

Then; we replace the concentration with the number of moles since the volume of acid and base are equal

∴

pH = 4.74 + log \dfrac{0.625 \times 10^{-3}}{1.475 \times 10^{-3}}

pH = 4.37

Thus, the pH of the solution after the addition of 5.0 mL of NaOH = 4.37

d)

We need to understand that the pH at 1/2 of the equivalence point is equal to the concentration of the base and the acid.

Therefore;

pH = pKa = 4.74

e) pH at the equivalence point.

Here, the pH of the solution is the result of the reaction in the (C_2H_3O^-_2) with H_2O

The total volume(V) of the solution = V(acid) + V(of the base added to reach equivalence point)

The total volume(V) of the solution = 20.0 mL + 16.8 mL

The total volume(V) of the solution = 36.8 mL

Concentration of (C_2H_3O^-_2) = moles/volume

= \dfrac{2.1 \times 10^{-3} \ moles}{0.0368 \ L}

= 0.0571 M

Now, using the ICE table to determine the concentration of H_3O^+;

             C_2H_5O^-_2 _{(aq)} + H_2O_{(l)} \to HC_2H_3O_2_{(aq)} + OH^-_{(aq)}

I              0.0571                                0                      0

C              -x                                       +x                     +x

E             0.0571 - x                             x                       x

Recall that the Ka for HC_2H_3O_2 = 1.8 \times 10^{-5}

K_b = \dfrac{K_w}{K_a} = \dfrac{1.0\times 10^{-14}}{1.8 \times 10^{-5} }  \\ \\ K_b = 5.6 \times 10^{-10}

k_b = \dfrac{[ HC_2H_3O_2] [OH^-]}{[C_2H_3O^-_2]}

5.6 \times 10^{-10} = \dfrac{x *x }{0.0571 -x}

x = [OH^-] = 5.6 \times 10^{-6} \ M

[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{5.6 \times 10^{-6} }

[H_3O^+] =1.77 \times 10^{-9}

pH =-log  [H_3O^+]   \\ \\  pH =-log (1.77 \times 10^{-9}) \\ \\ \mathbf{pH = 8.75 }

Hence, the pH of the solution at equivalence point = 8.75

f) The pH after 5.09 mL base is added beyond (E) point.

             HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}

Before                             0.0021              0.002725         0

After                                   0                     0.000625        0.0021

[OH^-] = \dfrac{0.000625 \ moles}{(0.02 + 0.0218 )  \ L}

[OH^-] = \dfrac{0.000625 \ moles}{0.0418 \ L}

[OH^-] =  0.0149 \ M

From above; we can determine the concentration of H_3O^+ by using the following method:

[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{0.0149}

[H_3O^+] = 6.7 \times 10^{-13}

pH = - log [H_3O^+]

pH = -log (6.7 \times 10^{-13} )

pH = 12.17

Finally, the pH of the solution after adding 5.0 mL of NaOH beyond (E) point = 12.17

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When various components of a mixture do not combine chemically it is a Combination of substances in which individual components do not combine chemically but retain their individual properties. Mixture in which one or more substances are distributed evenly in another substance.

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