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RoseWind [281]
3 years ago
10

When a 14.2-g sample of mercury (II) oxide is decomposed into its elements by heating, 13.2 g Hg is obtained. What is the percen

t composition of the compound?
Chemistry
1 answer:
nevsk [136]3 years ago
4 0
Mercury (ii) oxide is made up of mercury and oxygen. The total mass of mercury (ii) oxide is 14.2 g, after decomposition 13.2 g of mercury were formed, therefore the mass of oxygen 1 g (14.2 g -13.2 g).
Percentage of oxygen = (1/14.2)×100 = 7.04%
Percentage of mercury = (13.2/14.2) × 100 = 92.96%
Therefore, percentage composition of the compound, oxygen is 7.04% and mercury is 92.96%.
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EXTRA CREDIT
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Answer:

\huge\boxed{\sf 36\ H\ atoms}

Explanation:

<u>Molecular formula from Glucose:</u>

C₆H₁₂O₆

<u>3 moles of Glucose:</u>

3C₆H₁₂O₆

In 1 mole of Glucose, there are 12 hydrogen atoms.

<u>In 3 moles:</u>

= 12 × 3

= 36 H atoms

\rule[225]{225}{2}

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3. Given 20g of Barium Hydroxide, how many grams of
anastassius [24]

The number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

First, we will write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

Ba(OH)₂ + 2NH₄NO₃ → 2NH₄OH + Ba(NO₃)₂

This means, 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

Now, we will calculate the number of moles of barium hydroxide present.

Mass of barium hydroxide (Ba(OH)₂) = 20 g

Using the formula

Number\ of\ moles = \frac{Mass}{Molar\ mass}

Molar mass of Ba(OH)₂ = 171.34 g/mol

∴ Number of moles of Ba(OH)₂ present =\frac{20}{171.34}

Number of moles of Ba(OH)₂ present = 0.116727 mole

Now,

Since 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

Then,

0.116727 mole of barium hydroxide will react with 2 × 0.116727 mole of ammonium nitrate

2 × 0.116727 = 0.233454 mole

∴ Number of moles of NH₄NO₃ required is 0.233454 mole

Now, for the mass of ammonium nitrate (NH₄NO₃) required

From the formula

Mass = Number of moles × Molar mass

Molar mass of NH₄NO₃ = 80.043 g/mol

∴ Mass of NH₄NO₃ required = 0.233454 × 80.043

Mass of NH₄NO₃ required = 18.68636 g

Mass of NH₄NO₃ required ≅ 18.7g

Hence, the number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

Learn more on determining mass of reactant required here: brainly.com/question/11232389

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