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2 years ago

Can I get help with this, please!!

Chemistry

1 answer:

kondor19780726 [428]2 years ago

8 0

Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa

Explanation:

Pressure in the submarine = 108.9 kPa

Volume, V = 2.4 * 10^5 L

Pressure, P = 116k Pa

Temperature, T = 312 K

Ideal gas law: PV = nRT or n = PV / RT

So, moles of gas, n =116 KPa * 2.4 * 10 ^5L / 8.314 LK Pa K^-1 *312 K

= 1.073 *10^4 mol

when temperature is changed to 293K,

PV = nRT or P = nRT / V

=1.073 *10^4 mol *8.314 LK Pa mol^-1 K^-1 *293 K / 2.4*10^5L

=108.9 K Pa

Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa

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The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate

Tems11 [23]

Answer:

6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

pKa =.?

Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

Finally, we shall determine the pka for the reaction as follow:

Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

6 0

3 years ago

When a aqueous solution of a certain acid is prepared, the acid is dissociated. Calculate the acid dissociation constant of the

stepan [7]

<h2> $K_a$ = $\dfrac{[H^{+}] [A^{-}]}{[HA]}$ </h2>

Explanation:

When an aqueous solution of a certain acid is prepared it is dissociated is as follows-

${\displaystyle {\ce {HA$ ⇄ ${H^+}+{A^{-}}} }}$

Here HA is a protonic acid such as acetic acid, $CH_3COOH$

The double arrow signifies that it is an equilibrium process, which means the dissociation and recombination of the acid occur simultaneously.

The acid dissociation constant can be given by -

$K_a$ = $\dfrac{[H^{+}] [A^{-}]}{[HA]}$

The reaction is can also be represented by Bronsted and lowry -

$\\{\displaystyle {\ce {{HA}+ H_2O}$ ⇄ $[H_3O^+] [A^-]$

Then the dissociation constant will be

$K_a$ = $\dfrac{[H_3O^{+}] [A^{-}]}{[HA]}$

Here, $K_a$ is the dissociation constant of an acid.

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3 years ago

To understand the relation between the strength of an acid or a base and its pKa and pKb values. The degree to which a weak acid

elena-14-01-66 [18.8K]

Answer:

pKa = 3.675

Explanation:

pKa = - Log Ka

∴ <em>C</em> X-281 = 0.079 M

∴ pH = 2.40

let X-281 a weak acid ( HA ):

∴ HA ↔ H+ + A-

⇒ Ka = [H+] * [A-] / [HA]

mass balance:

⇒<em> C</em> HA = 0.079 M = [HA] + [A-]

⇒ [HA] = 0.079 - [A-]

charge balance:

⇒ [H+] = [A-] + [OH-]... [OH-] is negligible; it comes from to water

⇒ [H+] = [A-]

∴ pH = - log [H+] = 2.40

⇒ [H+] = 3.981 E-3 M

replacing in Ka:

⇒ Ka = [H+]² / ( 0.079 - [H+] )

⇒ Ka = ( 3.981 E-3 )² / ( 0.079 - 3.981 E-3 )

⇒ Ka = 2.113 E-4

⇒ pKa = - Log ( 2.113 E-4 )

⇒ pKa = 3.675

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