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Ad libitum [116K]
2 years ago
6

How many moles of sulfur dioxide are produced with 6.0 mols of carbon dioxide?

Chemistry
1 answer:
kodGreya [7K]2 years ago
6 0

Answer:

228.38g of CO2

Explanation:

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When discussing acids and bases, any substance that donates a proton, by definition, is considered a(n)
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Bronsted-Lowry acid is your answer
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3 years ago
Using the Brønsted-Lowry concept of acids and bases, identify the Brønsted-Lowry acid and base in each of the following reaction
Luden [163]

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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3 years ago
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ivanzaharov [21]
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3 years ago
PLEASE CAN SOMEONE EXPLAIN this FOR ME WHAT IS RADIOACTIVE ISOTOPES AND ANSWER THIS QUETION? A nitrogen atom has 7 protons and m
zalisa [80]

Answer:

\frac{15}{7}N

Explanation:

We already know that the mass number of an atom is the sum of the number of protons and the number of neutrons.

So, the mass number of this isotope is;

Number of protons = 7

Number of neutrons = 8

Mass number = 7 + 8 = 15

Hence, the isotope is;

\frac{15}{7}N

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