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prisoha [69]
2 years ago
11

Identify each definition that applies to the compound in red. Check all that apply. Upper F superscript minus in red plus upper

H upper S upper O subscript 4 superscript minus right arrow upper H upper F plus upper S upper O subscript 4 superscript 2 minus. Arrhenius acid Bronsted-Lowry acid Arrhenius base Bronsted-Lowry base.
Chemistry
1 answer:
likoan [24]2 years ago
4 0

The fluoride ion in the reaction has been the acceptor of the hydrogen. Thus, the fluoride ion has been the Brønsted Lowry base.

<h3 /><h3>What is the Brønsted Lowry acid and base concept?</h3>

The Brønsted Lowry acid and base concept can be described based on the proton acceptor and donor. The Brønsted Lowry acid has been the proton donor, and the Brønsted Lowry base has been the proton acceptor.

The Arrhenius concept is based on hydrogen and hydroxide donors. The Arrhenius base has been the hydroxide donor, and the Arrhenius acid has been the hydrogen donor.

Since the fluoride ion has been the proton acceptor in the reaction, it has been the Brønsted Lowry base.

Lear more about Brønsted Lowry acid and base, here:

brainly.com/question/21736327

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Answer:

Here's what I get  

Explanation:

3. Molar concentration by formula.

\begin{array}{rcl}M_{\text{a}}V_{a} & = & M_{\text{b}}V_{b}\\M_{a} \times \text{0.025 00 L} & = & \text{0.3840 mol/L} \times \text{0.034 52 L}\\0.025 00M_{a}\text{ L} & = & \text{0.013 26 mol}\\M_{a}&= &\dfrac{\text{0.013 26 mol}}{\text{0.025 00 L}}\\\\& = &\textbf{0.5302 mol/L}\end{array}

(i) Comparison of molar concentrations

The formula gives a calculated value of 0.5302 mol·L⁻¹.

Dimensional analysis gives a calculated value of 0.1767 mol·L⁻¹.

The first value is three times the second.

It is wrong because the formula assumes that the acid supplies just enough moles of H⁺ to neutralize the OH⁻ from the NaOH.

Instead, I mol of H₃PO₄ provides 3 mol of H⁺, so your calculated concentration is three times the true value.

(ii) When is the formula acceptable?

The formula is acceptable only when the molar ratio of acid to base is 1:1.

Examples are

HCl + NaOH ⟶ NaCl + H₂O

H₂SO₄ + Ca(OH)₂ ⟶ CaSO₄ + 2H₂O

H₃PO₄ + Al(OH)₃ ⟶ AlPO₄ + 3H₂O

 

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