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____ [38]
3 years ago
14

Burning magnesium gives us magnesium oxide. This is an example of a reaction.

Chemistry
2 answers:
frez [133]3 years ago
5 0

Answer: -

Oxidation, Oxygen

Explanation: -

Burning of magnesium gives us magnesium oxide.

The chemical symbol of magnesium is Mg.

The chemical symbol of magnesium oxide is MgO.

So, the balanced chemical equation is

2Mg + O2 – 2 MgO.

We see that the element added is O Oxygen. So, the reaction is oxidation.

In oxidation always, oxygen is the common element involved in all chemical reactions of this kind.

VikaD [51]3 years ago
3 0

The given reaction can be represented as,

2Mg (s) + O_{2}(g) ---> 2MgO (s)

In this reaction, two elements (Mg and O_{2}) combine to form a single compound, MgO. Hence, this reaction is an example of chemical combination or synthesis. Here as the element Mg is oxidized in the presence of O_{2} to form MgO. This is referred to as combustion reaction. In combustion reaction, elements or compounds react with oxygen to give their respective oxides.

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How many liters of oxygen gas, at standard
Karo-lina-s [1.5K]

Answer:

Explanation:

  • For the balanced reaction:

<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)​.</em>

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

  • Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

  • Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

<em><u>Using cross multiplication:</u></em>

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

  • Finally, we can get the volume of oxygen using the information:

<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>

<em></em>

<em><u>Using cross multiplication:</u></em>

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

5 0
3 years ago
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An alloy of bronze is manufactured by melting 51.2 g of copper with 6.84 g of tin. What is the percent copper in the bronze?
Harlamova29_29 [7]

Answer:

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Given data:

Mass of copper = 51.2 g

Mass of tin = 6.84 g

Percentage of copper = ?

Solution:

Formula:

Percentage of copper = mass of copper / total mass × 100

Now we will determine the total mass:

Total mass = mass of copper + mass of  tin

Total mass = 51.2 g + 6.84 g

Total mass = 58.04 g

Now we will calculate the percentage of copper.

Percentage of copper = 51.2 g / 58.04 g × 100

Percentage of copper = 0.88 × 100

Percentage of copper = 88%

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