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3 years ago

Percentage composition of HBr

Chemistry

1 answer:

Radda [10]3 years ago

5 0

Answer:

Percentage composition H in HBr = 1.236%

Percentage composition Br in HBr = 98.764%

Explanation:

H = 1g/mol

Br = 79.904g/mol

HBr = Molar Mass 80.904 g/mol

Percentage composition of X = (X / Molar Mass) x 100

H = (1/80.904) x 100 = 1.236%

Br = (79.904/80.904) x 100 = 98.764%

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Explain whether hydrogen bonding is present between hydrogen fluoride (HF) molecules

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A 50.00-mL solution of 0.0350 M aniline ( Kb = 3.8 × 10–10) is titrated with a 0.0113 M solution of hydrochloric acid as the tit

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Answer:

pH = 3.70

Explanation:

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0.0500L × (0.0350mol / 1L) = 1.750x10⁻³ mol aniline

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C₆H₅NH₂ + H₂O ⇄ C₆H₅NH₃⁺ + OH⁻ Kb = 3.8x10⁻¹⁰

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At equivalence point, all aniline reacts producing C₆H₅NH₃⁺. C₆H₅NH₃⁺ has its own equilibrium with water thus:

C₆H₅NH₃⁺(aq) + H₂O(l) ⇄ C₆H₅NH₂(aq) + H₃O⁺(aq)

Where Ka is defined as:

Ka = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺] = Kw / Kb = 1.0x10⁻¹⁴ / 3.8x10⁻¹⁰ = 2.63x10⁻⁵ (1)

As all aniline reacts producing C₆H₅NH₃⁺, moles in equilibrium are:

[C₆H₅NH₃⁺] = 1.750x10⁻³ mol - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in (1):

2.63x10⁻⁵ = [X] [X] / [1.750x10⁻³ mol - X]

4.6x10⁻⁸ - 2.63x10⁻⁵X = X²

0 = X² + 2.63x10⁻⁵X - 4.6x10⁻⁸

Solving for X:

X = -2.3x10⁻⁴ → False answer, there is no negative concentrations

X = 2.0x10⁻⁴ → Right answer

As [H₃O⁺] = X, [H₃O⁺] = 2.0x10⁻⁴.

Now, pH = -log[H₃O⁺]. Thus, pH at equivalence point is:

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Given the following equation:

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Mass of KMnO4 required : 31.607 g

<h3>Further explanation </h3>

The reaction equation is the chemical formula of reagents and product substances

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products

Reaction

2KMnO₄ → K₂MnO₄ + MnO₂ + O₂

mol of O₂ :

$\tt \dfrac{1.6}{16}=0.1$