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When 3.0 g of carbon is burnt in 8.00 g ...

When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

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Our answer will be governed by the law of constant proportions. Now, since carbon and oxygen combine in the fixed proportion of `3 : 8` by mass to produce 11 g of carbon dioxide, therefore, the extra mass of carbon dioxide (11g) will be obtained even if we burn 3 g of carbon in 50 g of oxygen. The extra oxygen (`50 - 8 = 42 g` oxygen) will remains unreacted.
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LAKHMIR SINGH & MANJIT KAUR-ATOMS AND MOLECULES-Exercise
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  2. Write the full form of IUPAC.

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  3. Name the scientist who gave : (a) law of conservation of mass. ...

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  4. Name the law of chemical combination : (a) Which was given by Lavo...

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  5. Name the scientist who gave atomic theory of matter.

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  6. Which postulate of Dalton’s atomic theory is the result of the law o...

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  7. Which postulate of Dalton’s atomic theory can explain the law of def...

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  8. Which ancient India philosopher suggested that all matter is composed...

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  9. Name any two laws of chemical combination.

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  10. If 100 grams of pure water taken from different sources is decomposed ...

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  11. If 100 grams of calcium carbonate (whether in the form of marble or ...

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  12. What are the builiding blocks of matter ?

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  13. How is the size of an atom indicated ?

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  14. Name the unit in which the radius of an atom is usually expressed.

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  15. Write the relation between nanometre and metre.

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  16. The radius of an oxygen atom is 0.073 nm. What does the symbol 'nm' ...

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  17. State whether the following statement is true or false : The symbol...

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  18. Define 'molecular mass' of a substance.

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  19. What is meant by saying that 'the molecular mass of oxygen is 32'...

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  20. Fill in the following blanks with suitable words : (a) In water , t...

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  21. (a) Name the element used as a standard for atomic mass scale. (b) W...

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