Two experiments were conducted: 4.125 g of copper(II) oxide, CuO2, was reduced by heating at high pressure with excess hydrogen to yield 3.294 g of copper Cu. 1.179 g of copper Cu was dissolved in nitric acid to yield copper(II) nitrate, which was converted to 1.476 g copper(II) oxide, CuO2, on ignition. Show that the results illustrate the law of definite proportions by calculating the percentage of oxygen in copper(II) oxide in experiments 1 and 2.

Explanation:

4.125 g of copper oxide contains 3.294 g of Cu . Rest is oxygen

oxygen is 4.125 - 3.294 = .831 g

Ratio of oxygen and copper = .831 : 3.294

.831 / .831 : 3.294 / .831

= 1 : 3.96

1 : 4 ( Approx )

percentage of oxygen in copper oxide = 20 % approx

in the second experiment ,

1.476 g of copper oxide is formed from 1.179 g of Cu

oxygen in this copper oxide = 1.476 - 1.179 = .297 g

Ratio of oxygen and copper

.297 : 1.179

.297/.297 : 1.179/ .297

1 : 3.96

1 : 4 ( approx )

percentage of oxygen in copper oxide = 20 % approx

So it follows the law of definite proportions .

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.