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mezya [45]
2 years ago
9

Which term refers to the formation of ions as a result of a chemical reaction?

Chemistry
2 answers:
levacccp [35]2 years ago
8 0

Answer: Option (d) is the correct answer.

Explanation:

A process in which a compound in a chemical reaction results in the formation of ions is known as ionization.

Whereas dissociation is a process in which a compound  divides into atoms, ions or radicals.

A chemical compound which when dissolved in a solution forms ions and thus, conducts electricity is known as electrolyte.

A chemical reaction in which a compound reacts or combines with water is known as hydration.

Therefore, we can conclude that ionization is the term which refers to the formation of ions as a result of a chemical reaction.

tangare [24]2 years ago
5 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Ionization is the term <span>refers to the formation of ions as a result of a chemical reaction.

</span>Ionization<span>, in general, occurs whenever sufficiently energetic charged particles or radiant energy travel through gases, liquids, or solids. Charged particles, such as alpha particles and electrons from radioactive materials, cause extensive </span>ionization<span>along their paths.</span>
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A bottle of wine contains 9.81 grams of C2H5OH, dissolved in 87.5 grams of water. The final volume of the solution is 100.0 mL.
dimulka [17.4K]

Answer:

[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)

Explanation:

1. Molarity = moles solute / Volume solution in Liters

=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH

=> volume of solution (assuming density of final solution is 1.0g/ml) ...

volume solution =  9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution

Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH

2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)

From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln

= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.

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