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2 years ago

Examples of pure and impure substances

Chemistry

2 answers:

erastova [34]2 years ago

8 0

Pure- table salt
Impure- vegetable oil

Sladkaya [172]2 years ago

8 0

Examples of pure substances are water, diamond, table salt

Examples of impure substances are muddy water, air, sugar in water

Explanation:

A pure substance is one that is made up of only one element or one chemical compound. It has no impurities. They can be elements or compounds.
It has definite properties and composition.
Examples are water, diamond, table salt, etc.
An impure substance is one that is made up of two or more elements or compounds that are not chemically bonded. It is a mixture.
It has no definite composition or properties.
Examples are muddy water, sugar in water, air, etc.

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2 years ago

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14. All the following are necessary parts of a neutralization reaction except A. an indicator. B. a salt. C. water. D. an acid.

ValentinkaMS [17]

The correct answer is B. a Salt because The reaction of an acid and a base is called a neutralization reaction because the properties of both the acid and base are diminished or neutralized when they react. A neutralization reaction is a reaction of an acid with a base in aqueous solution to produce water and a salt, as shown by the following equation:

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3 years ago

Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid

Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

$35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol$

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

$\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M$

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb) where Cb is the initial concentration of the base

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

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Tpy6a [65]

Answer:

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Galina-37 [17]

Hi,

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