The following data are obtained when dinitrogen and dioxygen react together to form different compounds:

 Mass of dioxygen Mass of dinitrogen (i) 16 g 14 g (ii) 32 g 14 g (iii) 32 g 28 g (iv) 80 g 28 g

(a) Which law of chemical combination is obeyed by the above experimental data? Give its statement.

(b) Fill in the blanks in the following conversions:

(i) 1 km = …………………. mm = …………………. pm

(ii) 1 mg = …………………. kg = …………………. ng

(iii) 1 mL = …………………. L = …………………. dm3

Asked by Abhisek | 1 year ago |  75

##### Solution :-

(a) If we fix the mass of N2 at 28 g, the masses of O2 that will combine with the fixed mass of N2 are 32 grams, 64 grams, 32 grams and 80 grams.

The mass of O2 bear whole no. ratio of 1: 2: 1: 5. Therefore, the given information obeys the law of multiple proportions.

The law of multiple proportions states, “If 2 elements combine to form more than 1 compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers.”

(b) Convert:

(a) 1 km = ____ mm = ____ pm

• 1 km = 1 km * $$\dfrac{ 1000 \; m }{ 1 \; km }× \dfrac{ 100 \; cm }{ 1 \; m }* \dfrac{ 10 \; mm }{ 1 \; cm }$$

$$∴1 km = 10^{ 6 } mm$$

• 1 km = $$1 km * \dfrac{ 1000 \; m }{ 1 \; km } * \dfrac{1 \; pm}{10^{ -12 } \; m}$$

$$∴ 1 km = 10^{ 15 } pm$$

Therefore, 1 km = $$0^{ 6 }mm = 10^{ 15 }pm$$

(b) 1 mg = ____ kg = ____ ng

• 1 mg = $$1 mg * \dfrac{ 1 \; g }{ 1000 \; mg }* \dfrac{ 1 \; kg }{ 1000 \; g } 1 mg =10^{ -6 }kg$$

• 1 mg = $$1 mg *{ 1 \; g }{ 1000 \; mg }* \dfrac{ 1 \; ng }{ 10^{ -9 } \; g }$$

1 mg =$$10^{ 6 } ng$$

Therefore, 1 mg =$$10^{ -6 }kg = 10^{ 6 } ng$$

(c) 1 mL = ____ L = ____ $$dm^{ 3 }$$

• 1 mL = 1 mL * $$\dfrac{1 \; L}{1000 \; mL}$$

1 mL =$$10^{ -3 } L$$

• 1 mL = $$1cm^{ 3 } = 1 *\dfrac{1 \; dm \; \times \; 1 \; dm \; \times \;1 \; dm }{10 \; cm \; \times \; 10 \; cm \; \times \; 10 \; cm } cm^{ 3 }$$

1 mL =$$10^{ -3 } dm^{ 3 }$$

Therefore, 1 mL = $$10^{ -3 }L = 10^{ -3 }dm^{ 3 }$$

Answered by Pragya Singh | 1 year ago

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