Black oxide of copper and red oxide of copper | vedclass

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Question

(C) In black oxide,$79.9 \ g$ of copper combines with $(100 - 79.9) = 20.1 \ g$ of oxygen.In red oxide,$88.8 \ g$ of copper combines with $(100 - 88.8

Vedclass · Accepted Answer

Multiple Proportions

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(C) In black oxide,$79.9 \ g$ of copper combines with $(100 - 79.9) = 20.1 \ g$ of oxygen.In red oxide,$88.8 \ g$ of copper combines with $(100 - 88.8) = 11.2 \ g$ of oxygen.For $79.9 \ g$ of copper in red oxide,the amount of oxygen is $\frac{11.2 	imes 79.9}{88.8} \approx 10.08 \ g$.The ratio of oxygen masses combining with a fixed mass of copper $(79.9 \ g)$ is $20.1 : 10.08 \approx 2 : 1$.Since the ratio is a simple whole number,this illustrates the Law of Multiple Proportions.

Black oxide of copper and red oxide of copper contain $79.9\%$ and $88.8\%$ of metal respectively. This data illustrates the law of:

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