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scZoUnD [109]
2 years ago
13

The compound iron oxide can exist with either iron(II) ions or iron(III) ions. Conduct Internet research to learn about the diff

erences between iron(II) oxide and iron(III) oxide. Give the chemical formula for each compound. Describe their appearance and uses. Based on your findings, are these two forms of the same compound, or are they two completely different compounds?
(What goes in the 6 empty spaces?)

Chemistry
1 answer:
melomori [17]2 years ago
7 0

Based on our finding these two compounds are not same they are completely different from each other as Formula of both compounds are different, their appearance is also different from each other.

<h3>What is Iron (II) Oxide ? </h3>

The formula of the Iron II oxide is FeO. Common name of Iron (II) Oxide is Ferrous Oxide. Iron (II) Oxide is a black colored powder. The mineral form of Iron (II) oxide is known as Wustite. Iron (II) Oxide is used as a pigment. It is also used to make dyes.

<h3>What is Iron (III) Oxide ? </h3>

The formula of the Iron (III) Oxide is Fe₂O₃. Common name of Iron (III) Oxide is Ferric oxide. Iron (III) Oxide appears as Red-Brown solid. It is also known as Hematite. Iron (III) oxide is used as pigments. It is used in dental composites , cosmetics. It is also used to apply the final polish on metallic jewellery.

           

Thus from the above conclusion we can say that Based on our finding these two compounds are not same they are completely different from each other as Formula of both compounds are different, their appearance is also different from each other.

Learn more about the Iron (II) Oxide here: brainly.com/question/14143857
#SPJ1

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The average atomic mass of your mixture is 1.03 u .

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A 11.97g sample of NaBr contains 22.34 % Na by mass. Considering the law of constant composition (definite proportions), how man
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<h3>Answer:</h3>

2.125 g

<h3>Explanation:</h3>

We have;

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We are required to determine the mass of 9.51 g of a NaBr sample.

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In this case,

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A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass

  • But;

% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100

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Mass of the element = (% composition of an element × mass of the compound) ÷ 100

Therefore;

Mass of sodium = (22.34% × 9.51 g) ÷ 100

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