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lora16 [44]
3 years ago
15

When heated, mercury oxide produces oxygen plus mercury. What would be the combined mass of oxygen and mercury if 20g of mercury

oxide were heated?
Chemistry
1 answer:
vivado [14]3 years ago
7 0
<span>Releasing the oxygen from the mercury does not change the mass. If you started with 20 g total you would still have 20 g total after the heating</span>
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What is an ionic compound? How is electrical neutrality maintained in an ionic compound?
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<u>Answer:</u> When oppositely charged species interact.

<u>Explanation:</u>

Ionic compound is defined as the compound which is formed by the complete transfer of electrons from one element to another element.

In these compounds, oppositely charged ions are attracted towards each other to form a compound.

<u>For Example:</u>  NaCl is formed by the attraction of Na^+\text{ and }Cl^- ions. So, same number of oppositely charges maintain electrical neutrality.

Electrical neutrality in a compound is achieved when atoms forming a compound have same number of opposite charges. Basically the charges on cation is balanced by the charges on anion.

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For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox
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Answer:

1) Maximun ammount of nitrogen gas: m_{N2}=10.682 g N_2

2) Limiting reagent: NO

3) Ammount of excess reagent: m_{N2}=4.274 g

Explanation:

<u>The reaction </u>

2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)

Moles of nitrogen monoxide

Molecular weight: M_{NO}=30 g/mol

n_{NO}=\frac{m_{NO}}{M_{NO}}

n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol

Moles of hydrogen

Molecular weight: M_{H2}=2 g/mol

n_{H2}=\frac{m_{H2}}{M_{H2}}

n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted

m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}

m_{N2}=10.682 g N_2

2) <u>Limiting reagent</u>: NO

3) <u>Ammount of excess reagent</u>:

m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}

m_{N2}=4.274 g

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