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3.0 g of metal oxide on reduction gives 1.40 g of metal. No of oxygen atoms in3.0 g of that metal oxide is (N is Avagadro's number)?
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3.0 g of metal oxide on reduction gives 1.40 g of metal. No of oxygen ...
Given data:
- Mass of metal oxide = 3.0 g
- Mass of metal produced = 1.40 g

Calculating the mass of oxygen:
- Mass of oxygen = Mass of metal oxide - Mass of metal
- Mass of oxygen = 3.0 g - 1.40 g
- Mass of oxygen = 1.60 g

Calculating the number of moles of oxygen:
- Molar mass of oxygen = 16 g/mol
- Number of moles of oxygen = Mass of oxygen / Molar mass of oxygen
- Number of moles of oxygen = 1.60 g / 16 g/mol
- Number of moles of oxygen = 0.10 mol

Calculating the number of oxygen atoms:
- Number of oxygen atoms in 1 mole = Avogadro's number = 6.022 x 10^23 atoms/mol
- Number of oxygen atoms in 0.10 mol = 0.10 mol x 6.022 x 10^23 atoms/mol
- Number of oxygen atoms in 0.10 mol = 6.022 x 10^22 atoms
Therefore, there are 6.022 x 10^22 oxygen atoms in 3.0 g of the metal oxide.
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3.0 g of metal oxide on reduction gives 1.40 g of metal. No of oxygen atoms in3.0 g of that metal oxide is (N is Avagadro's number)?
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3.0 g of metal oxide on reduction gives 1.40 g of metal. No of oxygen atoms in3.0 g of that metal oxide is (N is Avagadro's number)? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 3.0 g of metal oxide on reduction gives 1.40 g of metal. No of oxygen atoms in3.0 g of that metal oxide is (N is Avagadro's number)? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3.0 g of metal oxide on reduction gives 1.40 g of metal. No of oxygen atoms in3.0 g of that metal oxide is (N is Avagadro's number)?.
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