What is the best safety practice recommended before leaving the chemistry laboratory?

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

2 years ago

The mass and volume of each box is given in the chart below. Calculate the density of each box. Density = Mass/ volume which box

stira [4]

Box C will have the greatest density.

All boxes have the same volume.

Explanation:

We calculate the density using the following formula:

density = mass / volume

density of Box A = 10 g / 20 cm³ = 0.5 g/cm³

density of Box B = 30 g / 20 cm³ = 1.5 g/cm³

density of Box C = 170 g / 20 cm³ = 8.5 g/cm³

Box C will have the greatest density.

All boxes have the same volume.

Learn more about:

density

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7 0

2 years ago

has a standard free‑energy change of − 3.59 kJ / mol at 25 °C. What are the concentrations of A , B , and C at equilibrium if, a

Mamont248 [21]

Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.

Explanation:

The reaction equation is as follows.

$A + B \rightarrow C$

Initial : 0.3 0.4 0

Change: -x -x x

Equilbm: (0.3 - x) (0.4 - x) x

We know that, relation between standard free energy and equilibrium constant is as follows.

$\Delta G = -RT ln K$

Putting the given values into the above formula as follows.

$\Delta G = -RT ln K$

$-3.59 kJ/mol = -8.314 \times 10^{-3} kJ/mol K ln (\frac{x}{(0.3 - x)(0.4 - x)})$

x = 0.1417

Hence, at equilibrium

[A] = 0.3 - 0.1417

= 0.1583 M

[B] = 0.4 - 0.1417

= 0.2583 M

[C] = 0.1417 M

5 0

2 years ago

The strength of gravitational force depends on ___.

MrRissso [65]

Answer:

c.mass and distance

Explanation:

gravitation=mass/distance ²

3 0

2 years ago

Neptunium-237 was the first isotope of a transuranium element to be discovered. the decay constant is /s. what is the half-life

kari74 [83]

The half-life in years of Neptunium-237 which was the first isotope is 2.1 $10^{6}$ years.

Neptunium is most stable and Neptunium-237 is undergoes alpha decay, it means Neptunium-237 is decays by the emission of alpha particles . Seven alpha particles is emitted during decay of Neptunium-237. Neptunium-237 is radioactive actinide elements and first transuranium element.

The transuranium synthesis process involves creating a transuranium element through the transmutation process . The transmutation process is the process of creating heavy elements from light elements. Hence the process is the transmutation of light elements. There are two types: artificial and natural transmutation.

to learn more about transuranium element.

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6 0

1 year ago