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

Solution:

(a)

If we fix the mass of dinitrogen at 28 g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 32 g, 64 g, 32 g, and 80 g

The masses of dioxygen bear a whole number ratio of 1:2:2:5. Hence, the given experimental data obeys the law of multiple proportions. The law states that if two elements combine to form more than one 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) (i) $1 \mathrm{~km}=1 \mathrm{~km} \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} \times \frac{100 \mathrm{~cm}}{1 \mathrm{~m}} \times \frac{10 \mathrm{~mm}}{1 \mathrm{~cm}}$

$\therefore 1 \mathrm{~km}=10^{6} \mathrm{~mm}$

$1 \mathrm{~km}=1 \mathrm{~km} \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} \times \frac{1 \mathrm{pm}}{10^{-12} \mathrm{~m}}$

$\therefore 1 \mathrm{~km}=10^{15} \mathrm{pm}$

Hence, $1 \mathrm{~km}=10^{6} \mathrm{~mm}=10^{15} \mathrm{pm}$

(ii) $1 \mathrm{mg}=1 \mathrm{mg} \frac{1 \mathrm{~g}}{1000 \mathrm{mg}} \times \frac{1 \mathrm{~kg}}{1000 \mathrm{~g}} \times$

$\Rightarrow 1 \mathrm{mg}=10^{-6} \mathrm{~kg}$

$1 \mathrm{mg}=1 \mathrm{mg} \times \frac{1 \mathrm{~g}}{1000 \mathrm{mg}} \times \frac{1 \mathrm{ng}}{10^{-9} \mathrm{~g}}$

$\Rightarrow 1 \mathrm{mg}=10^{6} \mathrm{ng}$

$\therefore 1 \mathrm{mg}=10^{-6} \mathrm{~kg}=10^{6} \mathrm{ng}$

(iii) $1 \mathrm{~mL}=1 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}$

$\Rightarrow 1 \mathrm{~mL}=10^{-3} \mathrm{~L}$

$1 \mathrm{~mL}=1 \mathrm{~cm}^{3}=1 \frac{1 \mathrm{dm} \times 1 \mathrm{dm} \times 1 \mathrm{dm}}{10 \mathrm{~cm} \times 10 \mathrm{~cm} \times 10 \mathrm{~cm}} \mathrm{~cm}^{3}$

$\Rightarrow 1 \mathrm{~mL}=10^{-3} \mathrm{dm}^{3}$

$\therefore 1 \mathrm{~mL}=10^{-3} \mathrm{~L}=10^{-3} \mathrm{dm}^{3}$