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vova2212 [387]
3 years ago
15

Which of the following is considered applied research?

Chemistry
1 answer:
dexar [7]3 years ago
3 0

The correct answer is option B.

Research focused on how to cure cancer is an example of applied research.  

The examples of basic research are :

A.) Research focused on how the human brain works

C.) Research focused on detecting life on other planets

D.) Research focused on human behavior

Applied research targets to solve specific human problems that have commercial value in the present day. Basic research is about understanding the fundamentals of science and nature, and it is essential to laying down the foundation of applied research.  


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In a solution of pure water, the dissociation of water can be expressed by the following: H2O(l) + H2O(l) ⇌ H3O+(aq) + OH−(aq) T
kakasveta [241]

Answer:

  • [H₃O⁺] = 2.90 × 10⁻¹⁰ M

Explanation:

1)<u><em> Ionization equilibrium equation: given</em></u>

  • H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)

2) <em><u>Ionization equilibrium constant, at 25°C, Kw: given</u></em>

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As from the ionization equilibrium equation, as from the fact it is stated, the concentration of both ions, at 25°C, are equal:

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  • ⇒ Kw = [H3O⁺] [OH⁻] = 1.0 × 10⁻⁷  × 1.0 × 10⁻⁷  = 1.0 × 10⁻¹⁴ M

<u><em>4) A solution has a [OH⁻] = 3.4 × 10⁻⁵ M at 25 °C </em></u><em><u>and you need to calculate what the [H₃O⁺(aq)] is.</u></em>

Since the temperature is 25°, yet the value of Kw is the same, andy you can use these conditions:

  • Kw = 1.0 × 10⁻¹⁴ M², and

  • Kw = [H3O⁺] [OH⁻]

Then you can substitute the known values and solve for the unknown:

  • 1.0 × 10⁻¹⁴ M² = [H₃O⁺] × 3.4 × 10⁻⁵ M

  • ⇒ [H₃O⁺]  = 1.0 × 10⁻¹⁴ M² / ( 3.4 × 10⁻⁵ M ) = 2.9⁻¹⁰ M

As you see, the increase in the molar concentration of the ion [OH⁻] has caused the decrease in the molar concentration of the ion [H₃O⁺], to keep the equilibrium law valid.

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Answer: The component has a higher boiling point

Explanation:

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Answer:

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Explanation:

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