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goblinko [34]
3 years ago
7

A compound accepts electrons from another substance to form a covalent bond. Which term best describes this compound’s behavior?

Lewis acid Arrhenius base Bronsted-Lowry acid Bronsted-Lowry base
Chemistry
1 answer:
Cloud [144]3 years ago
5 0

A compound accepts electrons from another substance to form a covalent bond. The compound acts as a Lewis base.

<h3>What are the most common acid-base theories?</h3>
  • Arrhenius: acids release H⁺ and bases release OH⁻.
  • Bronsted-Lowry: acids donate H⁺ and bases accept H⁺.
  • Lewis: acids accept electrons and bases donate electrons.

A compound accepts electrons from another substance to form a covalent bond. Which term best describes this compound’s behavior?

  • Lewis acid. YES.
  • Arrhenius base. NO, because OH⁻ is not involved.
  • Bronsted-Lowry acid. NO, because H⁺ is not involved.
  • Bronsted-Lowry base. NO, because H⁺ is not involved.

A compound accepts electrons from another substance to form a covalent bond. The compound acts as a Lewis base.

Learn more about Lewis acid-base theory here: brainly.com/question/7031920

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Explanation:

Cubic decimeter is the same unit as liter; so, mole per cubic decimeter is mole per liter, and that is the unit of concentration of molarity. Thus, what is asked is the molarity of the solution. This is how you find it.

1. <u>Take a basis</u>: 1 dm³ = 1 liter = 1,000 ml

2. <u>Calculate the mass of 1 lite</u>r (1,000 ml) of solution:

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Here, the density is given through the specific gravity

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Take density of water as 1.00 g/ml.

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3. <u>Calculate the mass of solute</u> (pure acid)

  • % m/m = (mass of solute / mass of solution) × 100

  • 56 = mass of solute / 1,250 g × 100

  • mass of solute = 56 × 1,250g / 100 = 700 g

4. <u>Calculate the number of moles of solute</u>:

  • moles = mass in grams / molar mass = 700 g / 70 g/mol = 10 mol

5. <u>Calculate molarity (mol / dm³)</u>

  • M = number of moles of solute / liter of solution = 10 mol / 1 liter = 10 mol/liter.
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Your question is missing some information.

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