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Ugo [173]
2 years ago
6

Using the Brønsted-Lowry concept of acids and bases, identify the Brønsted-Lowry acid and base in each of the following reaction

s:
HSO3−(aq)+H2O(l)→H2SO3(aq)+OH−(aq)
(CH3)3N(g)+BCl3(g)→(CH3)3NBCl3(s)
Drag the appropriate items to their respective bins.
Chemistry
1 answer:
Luden [163]2 years ago
4 0

Answer:

1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)

<u>The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻</u>

<u />

2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)

<u>There are no Brønsted-Lowry acids and bases in this reaction.</u>

Explanation:

According to the Brønsted-Lowry concept, when an acid (HA) and a base (B) undergoes a chemical reaction, the acid (HA) loses a proton and forms its conjugate base (A⁻), whereas the base gains (B) the proton to form its conjugate acid (HB⁺).

<em>The chemical equation for this reaction is:</em>

HA  +  B  ⇌  A⁻  +  HB⁺

Given reactions:

1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)

<u>The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻</u>

Reason: In this reaction, the acid H₂O loses a proton and forms its conjugate base, OH⁻. Whereas, the base HSO³⁻ gains a proton to form its conjugate acid, H₂SO₃.

2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)

<u>There are no Brønsted-Lowry acids and bases in this reaction.</u>

Reason: In this reaction, there is no exchange of proton between the acid and the base.

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2 years ago
Which of the following statements is TRUE? Question 5 options: The emission spectrum of a particular element is always the same
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Answer:

All of the above are true

Explanation:

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3 years ago
I need part 1 and 2 please , just separate answers
Vladimir79 [104]

First, we have to remember the molarity formula:

M=\text{ }\frac{moles\text{ of solute}}{L\text{ solution}}

Part 1:

In this case, our solute is sodium nitrate (NaNO3), and we have the mass dissolved in water, then we have to convert grams to moles. For that, we need the molecular weight:

M.W_{NaNO_3}=\text{ 23+14+16*3= 85 g/mol}

Then, we calculate the moles present in the solution:

3.976\text{ g NaNO}_3\text{ * }\frac{1\text{ mol}}{85\text{ g}}=\text{ 0.04678 mol NaNO}_3

Now, we have the necessary data to calculate the molarity (with the solution volume of 200 mL):

M=\frac{0.04678\text{ mol}}{200\text{ mL*}\frac{1\text{ L}}{1000\text{ mL}}}=\text{ 0.2339 M}

The molarity of this solution equals 0.2339 M.

Part 2:

In this case, we have the same amount (in moles and mass) of sodium nitrate, but a different volume of solution, then we only have to change it:

M=\text{ }\frac{0.04678\text{ mol}}{275\text{ mL *}\frac{1\text{ L}}{1000\text{ mL}}}=\text{ 0.1701 M}

So, the molarity of this solution is 0.1701 M.

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