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Dmitrij [34]
3 years ago
14

Which of the following can be considered evidence?

Chemistry
1 answer:
melomori [17]3 years ago
5 0

Answer:

Shiny like a limousine

You're spending like a cash machine

Smile, show your golden teeth

That's how you cover up your cavities

You're glowing like a diamond ring

I saw you bought some other things

You're spending like a cash machine

To cover up your insecurities

Don't front, no needExplanation:

3455

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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
3 years ago
g Write a molecular equation for the precipitation reaction that occurs (if any) when the following solutions are mixed. If no r
Mrac [35]

Answer:

CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓

Explanation:

We identify the reactants:

CuBr₂ and Pb(CH₃COO)₂

The products will be: Cu(CH₃COO)₂ and PbBr₂

You may know these information:

Salts from acetate are soluble.

Bromide can make solid salts with these cations: Ag⁺, Pb²⁺, Hg₂²⁺, Cu⁺

PbBr₂ is formed, so this will be our precipitate

The equation is:

CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓

8 0
3 years ago
If you need to measure the volume of liquid in a bottle of eye drops, what unit would be the most practical?
aniked [119]
The volume of eye drop fluid in a bottle is specified in milliliter, because in medical 1 drop=12ml
7 0
3 years ago
What is true of all eukaryotic organisms?
Basile [38]

Answer:

They are multicelled

Explanation:

I just did a quiz on it UwU

4 0
3 years ago
What is the mass in grams of a single formula unit of silver chloride, AgCI? A) 4.21 x 1021 g B) 8.61 x 10258 C) 1.66 x 10-248 D
bekas [8.4K]

Answer: D) 2.38*10^-^2^2 g

Explanation: The question asks to convert formula unit to grams. It is a unit conversion problem.

1 mole equals to Avogadro number of formula units. So, to convert the given number of formula units to moles, we need to divide by the Avogadro number. After this, we do moles to grams conversion and for this the moles are multiplied by the molar mass of the compound. Molar mass of AgCl is 143.32 gram per mol.

1FormulaUnitAgCl(\frac{1mol}{6.022*10^2^3formulaUnits})(\frac{143.32g}{1mol})

= 2.38*10^-^2^2 g

So, the correct option is D) 2.38*10^-^2^2 g

7 0
3 years ago
Read 2 more answers
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