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melomori [17]
3 years ago
7

In an aqueous solution, 42% of a substance dissociates to release hydronium ions. Which of the following statements is true for

the substance?
It is a weak acid.

It is a weak base.

It is a strong acid.

It is a strong base.
Chemistry
1 answer:
const2013 [10]3 years ago
5 0

Answer: The statement it is a weak acid is true for the substance.

Explanation:

An acid that dissociates completely when dissolved in water to give hydrogen (H^{+}) or hydronium (H_{3}O^{+}) ions is called a strong acid.

For example, HCl is a strong acid.

HCl + H_{2}O \rightarrow H_{3}O^{+} + Cl^{-}

An acid that dissociates partially or weakly when dissolved in water to given hydrogen or hydronium ions is called a weak acid.

For example, CH_{3}COOH is a weak acid.

CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H_{3}O^{+}

A strong base is a base which when dissolved in water then it dissociates completely to give hydroxide ions.

For example, NaOH is a strong acid.

A weak base is a base which when dissolved in water then it dissociates partially or weakly to give hydroxide ions.

For example, NH_{3} is a weak base.

Hence, in an aqueous solution where 42% of a substance dissociates to release hydronium ions shows that the dissociation is less than 50%. This means that substance is dissociating weakly so, it is a weak acid.

Thus, we can conclude that the statement it is a weak acid is true for the substance.

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<h3>Answer:</h3>

0.6 g NaCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

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[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

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<u>Step 2: Identify Conversions</u>

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Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

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<u>Step 3: Stoichiometry</u>

  1. Set up:                    \displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})
  2. Multiply/Divide:                                                                                               \displaystyle 0.551373 \ g \ NaCl

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

0.551373 g NaCl ≈ 0.6 g NaCl

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