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olganol [36]
3 years ago
14

The p K a pKa of acetic acid is 4.76 . 4.76. The p K a pKa of trichloroacetic acid is 0.7 . 0.7. Calculate the equilibrium disso

ciation constant, K a , Ka, of each acid.
Chemistry
1 answer:
Andrei [34K]3 years ago
3 0

Answer:

Acetic acid Ka = 1.74 × 10⁻⁵

Trichloroacetic acid Ka = 2 × 10⁻¹

Explanation:

Let's consider the acid dissociation of acetic acid.

CH₃COOH(aq) ⇄ CH₃COO⁻(aq) + H⁺(aq)

The pKa of acetic acid is 4.76. The acid dissociation constant (Ka) is:

pKa = -log Ka

- pKa = log Ka

Ka = anti log (-pKa)

Ka = anti log (-4.76)

Ka = 1.74 × 10⁻⁵

Let's consider the acid dissociation of trichloroacetic acid.

CCl₃COOH(aq) ⇄ CCl₃COO⁻(aq) + H⁺(aq)

The pKa of trichloroacetic acid is 0.7. The acid dissociation constant (Ka) is:

pKa = -log Ka

- pKa = log Ka

Ka = anti log (-pKa)

Ka = anti log (-0.7)

Ka = 2 × 10⁻¹

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The answer is false.
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You are given a bottle of solid X and three aqueous solutions of X—one saturated, one unsaturated, and one supersaturated. How w
quester [9]

In a saturated solution, extra solid X would remain solid, dissolve in an unsaturated solution, and crystallize in a supersaturated one.

A solution is said to be saturated when there is a maximum amount of solute present that has been dissolved in the solvent. As a result, the system is in an equilibrium between the dissolved and undissolved solutes: A solution is considered to be unsaturated if the solute concentration is less than the equilibrium solubility. A supersaturated solution is one that has more solute than is necessary to generate a saturated solution at a given temperature.

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5 0
2 years ago
The frequency light of 3.52*10-7
zimovet [89]

Answer:

The speed of light, c, equals the wavelength, λ (pronounced lambda), times the frequency, ν, (pronounced noo).

c=λν

c is a constant. It is usually given as 3.00×108 m/s or 3.00×1010 cm/s rounded to three significant figures.❤

7 0
3 years ago
the element carbon has two common isotopes: C-12 (12 U) AND C-13 (13.003355 U). IF THE AVERAGE ATOMIC MASS OF CARBON IS 12.0107
oksian1 [2.3K]

Answer:

The percent isotopic abundance of C- 12 is 98.93 %

The percent isotopic abundance of C- 13 is 1.07 %

Explanation:

we know there are two naturally occurring isotopes of carbon, C-12 (12u)  and C-13 (13.003355)

First of all we will set the fraction for both isotopes

X for the isotopes having mass 13.003355

1-x for isotopes having mass 12

The average atomic mass of carbon is 12.0107

we will use the following equation,

13.003355x + 12 (1-x) = 12.0107

13.003355x + 12 - 12x = 12.0107

13.003355x- 12x = 12.0107 -12

1.003355x = 0.0107

x= 0.0107/1.003355

x= 0.0107

0.0107 × 100 = 1.07 %

1.07 % is abundance of C-13 because we solve the fraction x.

now we will calculate the abundance of C-12.

(1-x)

1-0.0107 =0.9893

0.9893 × 100= 98.93 %

98.93 % for C-12.

7 0
3 years ago
Copper has two naturally occurring isotopes. Cu−63 has a mass of 62.939 amu and relative abundance of 69.17%.
fiasKO [112]
The answer is 64.907 amu.

The atomic mass of an element is the average of the atomic masses of its isotopes. The relative abundance of isotopes must be taken into consideration, therefore:
atomic mass of copper = atomic mass of isotope 1 * abundance 1 + atomic mass of isotope 2 * abundance 2

We know:
atomic mass of copper = 63.546 amu
The atomic mass of isotope 1 is: 62.939 amu
The abundance of isotope 1 is: 69.17% = 0.6917
The atomic mass of isotope 1 is: x
The abundance of isotope 2: 100% - 69.17% = 30.83% = 0.3083

Thus:
63.546 amu = 62.939 amu * 0.6917 + x * 0.3083
63.546 <span>amu = 43.535 amu + 0.3083x
</span>⇒ 63.546 amu - 43.535 amu = 0.3083x
⇒ 20.011 amu = 0.3083x
   ⇒ x = 20.011 amu ÷ 0.3083 = 64.907 amu
7 0
3 years ago
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