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Iron (II) oxide, FcO, crystal has a cubic structure and each edge of the unit cell is 5.0 Å. Taking density of the oxide as 4.0 g cm–3. Calculate the no. of Fe2+ and O2– ions present in each unit cell.
- a)Fe2+: 4; O2–: 4
- b)Fe2+: 3; O2–: 3
- c)Fe2+: 1; O2–: 1
- d)Fe2+: 3; O2–: 4
Correct answer is option 'A'. Can you explain this answer?
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Iron (II) oxide, FcO, crystal has a cubic structure and each edge of t...
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Iron (II) oxide, FcO, crystal has a cubic structure and each edge of t...
Angstroms long. The density of FeO is 5.7 g/cm3. Calculate the number of Fe and O atoms in one unit cell.
First, we need to calculate the volume of one unit cell:
Volume = (edge length)3 = (5.0 angstroms)3 = 125 angstroms3
Next, we need to convert the volume to cubic centimeters (cm3) so that we can use the density to calculate the mass of the unit cell:
1 angstrom = 10^-8 cm
Volume = 125 angstroms3 x (10^-8 cm/angstrom)^3 = 1.25 x 10^-22 cm3
Now we can calculate the mass of the unit cell:
Density = mass/volume
mass = density x volume = (5.7 g/cm3) x (1.25 x 10^-22 cm3) = 7.125 x 10^-22 g
Next, we need to calculate the number of atoms in the unit cell. Iron has a molar mass of 55.85 g/mol, and oxygen has a molar mass of 16.00 g/mol. The formula for FeO tells us that there is one iron atom and one oxygen atom in each unit of FeO.
Number of Fe atoms = (mass of Fe/molar mass of Fe) = (55.85 g/mol)/(7.125 x 10^-22 g) = 7.84 x 10^21
Number of O atoms = (mass of O/molar mass of O) = (16.00 g/mol)/(7.125 x 10^-22 g) = 2.25 x 10^22
Therefore, there are approximately 7.84 x 10^21 iron atoms and 2.25 x 10^22 oxygen atoms in one unit cell of FeO.
First, we need to calculate the volume of one unit cell:
Volume = (edge length)3 = (5.0 angstroms)3 = 125 angstroms3
Next, we need to convert the volume to cubic centimeters (cm3) so that we can use the density to calculate the mass of the unit cell:
1 angstrom = 10^-8 cm
Volume = 125 angstroms3 x (10^-8 cm/angstrom)^3 = 1.25 x 10^-22 cm3
Now we can calculate the mass of the unit cell:
Density = mass/volume
mass = density x volume = (5.7 g/cm3) x (1.25 x 10^-22 cm3) = 7.125 x 10^-22 g
Next, we need to calculate the number of atoms in the unit cell. Iron has a molar mass of 55.85 g/mol, and oxygen has a molar mass of 16.00 g/mol. The formula for FeO tells us that there is one iron atom and one oxygen atom in each unit of FeO.
Number of Fe atoms = (mass of Fe/molar mass of Fe) = (55.85 g/mol)/(7.125 x 10^-22 g) = 7.84 x 10^21
Number of O atoms = (mass of O/molar mass of O) = (16.00 g/mol)/(7.125 x 10^-22 g) = 2.25 x 10^22
Therefore, there are approximately 7.84 x 10^21 iron atoms and 2.25 x 10^22 oxygen atoms in one unit cell of FeO.
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Iron (II) oxide, FcO, crystal has a cubic structure and each edge of the unit cell is 5.0 Å. Taking density of the oxide as 4.0 g cm–3. Calculate the no. of Fe2+ and O2– ions present in each unit cell.a)Fe2+: 4; O2–: 4b)Fe2+: 3; O2–: 3c)Fe2+: 1; O2–: 1d)Fe2+: 3; O2–: 4Correct answer is option 'A'. Can you explain this answer?
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Iron (II) oxide, FcO, crystal has a cubic structure and each edge of the unit cell is 5.0 Å. Taking density of the oxide as 4.0 g cm–3. Calculate the no. of Fe2+ and O2– ions present in each unit cell.a)Fe2+: 4; O2–: 4b)Fe2+: 3; O2–: 3c)Fe2+: 1; O2–: 1d)Fe2+: 3; O2–: 4Correct answer is option 'A'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Iron (II) oxide, FcO, crystal has a cubic structure and each edge of the unit cell is 5.0 Å. Taking density of the oxide as 4.0 g cm–3. Calculate the no. of Fe2+ and O2– ions present in each unit cell.a)Fe2+: 4; O2–: 4b)Fe2+: 3; O2–: 3c)Fe2+: 1; O2–: 1d)Fe2+: 3; O2–: 4Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Iron (II) oxide, FcO, crystal has a cubic structure and each edge of the unit cell is 5.0 Å. Taking density of the oxide as 4.0 g cm–3. Calculate the no. of Fe2+ and O2– ions present in each unit cell.a)Fe2+: 4; O2–: 4b)Fe2+: 3; O2–: 3c)Fe2+: 1; O2–: 1d)Fe2+: 3; O2–: 4Correct answer is option 'A'. Can you explain this answer?.
Iron (II) oxide, FcO, crystal has a cubic structure and each edge of the unit cell is 5.0 Å. Taking density of the oxide as 4.0 g cm–3. Calculate the no. of Fe2+ and O2– ions present in each unit cell.a)Fe2+: 4; O2–: 4b)Fe2+: 3; O2–: 3c)Fe2+: 1; O2–: 1d)Fe2+: 3; O2–: 4Correct answer is option 'A'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Iron (II) oxide, FcO, crystal has a cubic structure and each edge of the unit cell is 5.0 Å. Taking density of the oxide as 4.0 g cm–3. Calculate the no. of Fe2+ and O2– ions present in each unit cell.a)Fe2+: 4; O2–: 4b)Fe2+: 3; O2–: 3c)Fe2+: 1; O2–: 1d)Fe2+: 3; O2–: 4Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Iron (II) oxide, FcO, crystal has a cubic structure and each edge of the unit cell is 5.0 Å. Taking density of the oxide as 4.0 g cm–3. Calculate the no. of Fe2+ and O2– ions present in each unit cell.a)Fe2+: 4; O2–: 4b)Fe2+: 3; O2–: 3c)Fe2+: 1; O2–: 1d)Fe2+: 3; O2–: 4Correct answer is option 'A'. Can you explain this answer?.
Solutions for Iron (II) oxide, FcO, crystal has a cubic structure and each edge of the unit cell is 5.0 Å. Taking density of the oxide as 4.0 g cm–3. Calculate the no. of Fe2+ and O2– ions present in each unit cell.a)Fe2+: 4; O2–: 4b)Fe2+: 3; O2–: 3c)Fe2+: 1; O2–: 1d)Fe2+: 3; O2–: 4Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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Iron (II) oxide, FcO, crystal has a cubic structure and each edge of the unit cell is 5.0 Å. Taking density of the oxide as 4.0 g cm–3. Calculate the no. of Fe2+ and O2– ions present in each unit cell.a)Fe2+: 4; O2–: 4b)Fe2+: 3; O2–: 3c)Fe2+: 1; O2–: 1d)Fe2+: 3; O2–: 4Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Iron (II) oxide, FcO, crystal has a cubic structure and each edge of the unit cell is 5.0 Å. Taking density of the oxide as 4.0 g cm–3. Calculate the no. of Fe2+ and O2– ions present in each unit cell.a)Fe2+: 4; O2–: 4b)Fe2+: 3; O2–: 3c)Fe2+: 1; O2–: 1d)Fe2+: 3; O2–: 4Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Iron (II) oxide, FcO, crystal has a cubic structure and each edge of the unit cell is 5.0 Å. Taking density of the oxide as 4.0 g cm–3. Calculate the no. of Fe2+ and O2– ions present in each unit cell.a)Fe2+: 4; O2–: 4b)Fe2+: 3; O2–: 3c)Fe2+: 1; O2–: 1d)Fe2+: 3; O2–: 4Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Iron (II) oxide, FcO, crystal has a cubic structure and each edge of the unit cell is 5.0 Å. Taking density of the oxide as 4.0 g cm–3. Calculate the no. of Fe2+ and O2– ions present in each unit cell.a)Fe2+: 4; O2–: 4b)Fe2+: 3; O2–: 3c)Fe2+: 1; O2–: 1d)Fe2+: 3; O2–: 4Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Chemistry tests.
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