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3 years ago

What is the pH of a solution with an H3O+ concentration of 5.67 x 10-4 M?

Chemistry

1 answer:

Valentin [98]3 years ago

3 0

Answer:

Explanation:

The pH of a solution can be found by using the formula

$pH = - log [ { H_3O}^{+}]$

From the question we have

$ph = - log(5.67 \times {10}^{ - 4} ) \\ = 3.2464$

We have the final answer as

Hope this helps you

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If a muriatic acid solution has a pH of 2.50, what is the hydronium ion concentration (hydronium ion concentration is also the m

Irina18 [472]

Answer:

[H₃O⁺] = 0.032 M

Explanation:

The acidity of a solution is measured by its pH, which is the logarithm of the inverse of the molar concentration of hydronium (H₃O⁺) ions:

pH = log 1 / [H₃O⁺] = - log [H₃O⁺]

When you know the pH value you can find hydronium concentration using the antilogaritm function:

$pH=-log[H_3O^{+}]\\ \\ {[H_3O^+]}=10^{-pH}\\ \\ {[H_3O^+]}=10^{-2.50}\\ \\ {[H_3O^+]}=0.0032$

The unit of molar concentration is M.

To prove your answer you can take the logarithm of 0.0316:

log (0.0032) ≈ 2.500

5 0

3 years ago

Calculate the standard potential for the following galvanic cell: Fe(s) | Fe2+(aq) || Ag+(aq) | Ag(s) which has the overall bala

Nataly [62]

Answer:

E° = 1.24 V

Explanation:

Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)

According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:

Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻

Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an

E° = 0.80 V - (-0.44 V) = 1.24 V

6 0

3 years ago

NaOH + HCl → NaCl + H2O

bearhunter [10]

Answer:

0.00496 g H2O

Explanation:

7 0

3 years ago

Naphthalene, commonly found in moth balls, is composed of 93.7% carbon and 6.3% hydrogen. The molar mass of naphthalene is 128 g

Vikki [24]

Answer:

Molecular formula naphthalene → C₁₀H₈

Empirical formula naphthalene → C₅H₄

Explanation:

Centesimal composition means that in 100 g of compound we have x g of the element. Therefore in 100 g of naphthalene we have:

93.7 g of C

6.3 g of H

Let's make a rule of three:

In 100 g of naphthalene we have 93.7 g of C and 6.3 g of H

In 128 g of naphthalene we would have:

128 . 93.7 / 100 = 120 g of C

128. 6.3 / 100 = 8 g of H

We convert the mass to moles, by molar mass:

120 g . 1mol / 12 g = 10 moles C

8 g . 1mol/ 1g = 8 moles H

Molecular formula naphthalene → C₁₀H₈

Empirical formula naphthalene → C₅H₄

(The sub-index of each element is divided by the largest possible number)

6 0

3 years ago

Read 2 more answers

WILL MARK BRAINLIEST!!!!

Lyrx [107]

Answer:

5.702 mol K₂SO₄

General Formulas and Concepts:

<u>Atomic Structure</u>

Reading a Periodic Table
Compounds
Moles

<u>Stoichiometry</u>

Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 993.6 g K₂SO₄

[Solve] moles K₂SO₄

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of K: 39.10 g/mol

[PT] Molar Mass of S: 32.07 g/mol

[PT] Molar mass of O: 16.00 g/mol

Molar Mass of K₂SO₄: 2(39.10) + 32.07 + 4(16.00) = 174.27 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 993.6 \ g \ K_2SO_4(\frac{1 \ mol \ K_2SO_4}{174.27 \ g \ K_2SO_4})$
[DA] Divide [Cancel out units]: $\displaystyle 5.7015 \ mol \ K_2SO_4$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

5.7015 mol K₂SO₄ ≈ 5.702 mol K₂SO₄

7 0

3 years ago