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in Some Basic Concepts of Chemistry by (30.3k points)
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(a) How can you illustrate the law of multiple proportions by using oxides of metals containing 78.7% and 64.5% of the metal? 

(b) Match the following:

1/12th the mass of C12 atom - 1 mole 

1 g of hydrogen atom – amu 

22.4 L O at NTP – gram mole 

180 g of glucose – gram atom 

6.022 × 1023 particles – molar volume

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(a) In 100 g samples of the two oxides, the masses of the metal are 78.7 g and 64.5 g respectively. 

First Oxide : 

Mass of oxygen = 100 – 78.7 = 21.3 g 

No. of parts by mass of oxygen combining with one part by mass of metal = \(\frac{78.7}{21.3}\) = 3.7g

Second oxide: 

Mass of oxygen = 100 – 64.5 = 35.5 g 

No. of parts by mass of oxygen combining with one part by mass of metal = \(\frac{64.5}{35.5}\) = 1.9g

The ratio of masses of oxygen combining with a fixed mass of metal = 3.7 : 1.9 = 2: 1, a simple whole number ratio.

(b) 1/12th the mass C12 atom – amu 

1 g of hydrogen atom – gram atom 

22.4 L O2 at NTP – molar volume

180 g of glucose – gram mole 

6.022 × 1023 particles – 1 mole

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