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Ainat [17]
3 years ago
12

List the number of each type of atom on the right side of the equation hbr(aq)+2naoh(aq)→2nabr(s)+h2o(l)

Chemistry
2 answers:
ser-zykov [4K]3 years ago
5 0

On the basis of the given unbalanced equation, that is:

HBr (aq) + 2NaOH (aq) → 2NaBr (s) + H2O (l)

On the right side of the equation, there are 2 atoms of sodium (Na), 2 atoms of bromine (Br), 2 atoms of hydrogen (H), and 1 atom of oxygen (O₂).

After balancing the equation correctly we get:

HBr (aq) + NaOH (aq) → NaBr (s) + H2O (l)

On the right side, one atom of Na, 1 atom of Br, 1 atom of H and one atom of O₂.

AlexFokin [52]3 years ago
4 0

Answer:

As per the given unbalanced equation:-

HBr(aq) + 2NaOH(aq) --> 2NaBr(s) + H2O(l)

Na:-2

Br:-2

H:- 2

O:- 1

If correctly balanced then

HBr(aq) + NaOH(aq) --> NaBr(s) + H2O(l)

Na:-1

Br:-1

H:- 2

O:- 1

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Answer:

1. C₃H₆O₃

2. C₆H₁₂

3. C₆H₂₄O₆

4. C₆H₆

5. N₂O₄

Explanation:

1. Determination of the molecular formula.

Empirical formula => CH₂O

Mass of compound = 90 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂O]ₙ = 90

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Divide both side by 30

n = 90/30

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Molecular formula = [CH₂O]ₙ

Molecular formula = [CH₂O]₃

Molecular formula = C₃H₆O₃

2. Determination of the molecular formula.

Empirical formula => CH₂

Mass of compound = 84 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂]ₙ = 84

[12 + (2×1)]n = 84

[12 + 2]n = 84

14n = 84

Divide both side by 14

n = 84/14

n = 6

Molecular formula = [CH₂]ₙ

Molecular formula = [CH₂]₆

Molecular formula = C₆H₁₂

3. Determination of the molecular formula.

Empirical formula => CH₄O

Mass of compound = 192 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₄O]ₙ = 192

[12 + (4×1) + 16]n = 192

[12 + 4 + 16]n = 192

32n = 192

Divide both side by 32

n = 192/32

n = 6

Molecular formula = [CH₄O]ₙ

Molecular formula = [CH₄O]₆

Molecular formula = C₆H₂₄O₆

4. Determination of the molecular formula.

Empirical formula => CH

Mass of compound = 78 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH]ₙ = 78

[12 + 1]n = 78

13n = 78

Divide both side by 13

n = 78/13

n = 6

Molecular formula = [CH]ₙ

Molecular formula = [CH]₆

Molecular formula = C₆H₆

5. Determination of the molecular formula.

Empirical formula => NO₂

Mass of compound = 92 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[NO₂]ₙ = 92

[14 + (2×16)]n = 92

[14 + 32]n = 92

46n = 92

Divide both side by 46

n = 92/46

n = 2

Molecular formula = [NO₂]ₙ

Molecular formula = [NO₂]₂

Molecular formula = N₂O₄

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