Chemistry, asked by isaacraja, 8 months ago

0.8567g of cupric oxide contains 0.6842 g ofcopper. what is the eqivalent weight of copper?​

Answers

Answered by S10305
1

Answer:

In first experiment:

Copper oxide = 1.375 g

Copper left = 1.098 g

Therefore oxygen present = 1.375 – 1.098g = 0.277 g

begin mathsize 11px style Hence space % space of space oxygen space present space in space CuO space equals fraction numerator 0.277 space cross times space 100 over denominator 1.375 end fraction space equals space 20.14 end style

Second experiment:

Copper taken = 1.179 g

Copper oxide formed = 1.476 g

Therefore oxygen present = 1.476-1.179g = 0.297 g

begin mathsize 11px style Hence space % space of space oxygen space of space CuO space equals fraction numerator 0.297 space cross times space 100 over denominator 1.476 end fraction space equals space 20.12 end style

Since percentage of oxygen is the same in both the above cases, so the law of constant composition is illustrated.

Explanation:

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Answered by kaustumbh3136
0

Answer:In first experiment:

Copper oxide = 1.375 g

Copper left = 1.098 g

Therefore oxygen present = 1.375 – 1.098g = 0.277 g

begin mathsize 11px style Hence space % space of space oxygen space present space in space CuO space equals fraction numerator 0.277 space cross times space 100 over denominator 1.375 end fraction space equals space 20.14 end style

Second experiment:

Copper taken = 1.179 g

Copper oxide formed = 1.476 g

Therefore oxygen present = 1.476-1.179g = 0.297 g

begin mathsize 11px style Hence space % space of space oxygen space of space CuO space equals fraction numerator 0.297 space cross times space 100 over denominator 1.476 end fraction space equals space 20.12 end style

Since percentage of oxygen is the same in both the above cases, so the law of constant composition is illustrated.

Explanation:

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