CHECK YOUR PROGRESS

1.179 g of copper on heating in air gave 1.476 g of a black oxide of copper(ll). 1.375 g of black copper(ll) oxide on reduction with hydrogen gave 1.098 g of copper. Show that these results illustrate the law of constant proportions.

In one experiment, hydrogen gas was passed through 1.06 g of copper(ll) oxide when 0.85 g of copper was obtained. In another experiment, 3.19 g of copper was dissolved in concentrated nitric acid. The solution was evaporated to dryness and the residue was ignited to get 3.98 g of copper(ll) oxide. Show that these data are in accordance with the law of constant proportions. 

In one experiment, 6.10 g of iron was heated and treated with steam to obtain 8.72 g of iron(lll) oxide. In another experiment, 12.84 g of iron(lll) oxide was reduced with hydrogen to get 8.98 g of iron. Show that these results illustrate the law of constant proportions. 

Hydrogen gas and oxygen gas combine in the ratio $1:8$ by mass to give water. What mass of oxygen gas will react completely with $6.0 \mathrm{g}$ of hydrogen gas? 

Phosphorus forms two oxides. The phosphorus content in one oxide is $56.36\%$ while that in the other oxide it is $43.66\%$. Show that these data are in agreement with the law of multiple proportions. 

Phosphorus forms two chlorides. The phosphorus content in one chloride is 22.54% while that in the other oxide it is 14.87%. Show that these data are in agreement with the law of multiple proportions.

Note- The question given in the book seems to be incorrect. The corrected question is as follows:

Phosphorus forms two oxides. The phosphorus content in one oxide is $22.54\%$ while that in the other oxide it is $14.87\%$. Show that these data are in agreement with the law of multiple proportions.

Calculate the number of water molecules contained in a drop of water weighing 0.05 g.

Calculate the mass percentage of oxygen present in the following compounds and state the law of chemical combination associated. Given: $\mathrm{H} = 1, \mathrm{O} = 16$. a. water $\left({\mathrm{H}}_2\mathrm{O}\right)$b. Hydrogen peroxide $\left({\mathrm{H}}_2{\mathrm{O}}_2\right)$.