When [H+] = 4.0 × 10–9 M in water at 25°C, then
Kw =1.0×10–14.
b. pOH = 7.
c. [H3O+] = [OH–].
d. pH= 7
e. A-D are all correct

Answers

Answer 1

Answer:

a. Kw = 1.0 × 10⁻¹⁴

Explanation:

a. Let's consider the self-ionization of water.

2 H₂O(l) ⇄ H₃O⁺(aq) + OH⁻

The ion-product of water (Kw) at 25 °C is:

Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴

c. Considering [H⁺] = [H₃O⁺] = 4.0 × 10⁻⁹ M, the concentration of OH⁻ is:

[OH⁻] = 1.0 × 10⁻¹⁴/[H₃O⁺] = 2.5 × 10⁻⁶ M

b. We can calculate the pOH using the following expression.

pOH = -log [OH⁻] = -log 2.5 × 10⁻⁶ = 5.6

d. We can calculate the pH using the following expression.

pH = -log [H⁺] = -log 4.0 × 10⁻⁹ = 8.4

Answer 2

When [H+] = 4.0 × 10–9 M in water at 25°C, then the [tex]Kw =1.0\times 10^-^1^4.[/tex]

What is dissociation constant?

Kw is the dissociation constant or ionization constant of water.

The self-ionization of water will be

[tex]\rm 2 H_2O(l) = H_3O^+(aq) + OH^-[/tex]

The dissociation constant of the water at 25°C will be

[tex]\rm Kw = [H_3O^+][OH^-] = 1.0 \times 10^-^1^4[/tex]

Thus, the correct option is a. [tex]Kw =1.0\times 10^-^1^4.[/tex]

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...

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from google

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I just took the quiz, have a nice day!

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Question 5
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Answer:

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Hello!

In this case, since percent compositions are used to identify the empirical formula of an unknown compound, we can assume we have 54.0 g of sodium, 8.50 g of boron and 37.5 g of oxygen, and we compute the moles of each one:

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Answers

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