"Unveiling The Secret Of Conservation" The Law Of Mass And Definite Proportion Explained!"

In this video, we delve into the intriguing realms of the Law of Conservation of Mass and the Law of Definite Proportion.
Whether you're a student, educator, or just someone curious about the wonders of chemistry, this video is tailored for you! Gain a solid understanding of these laws and appreciate their significance in the broader scientific landscape.
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[@giftchristopher8543]

Pls sir ..when are you starting tutorial for jupeb

[@AfolabiAyomide19]

1.percentage of Cu in sample1 =2.01÷2.25 ×100= 89.3%
Percentage of O in sample 1= 0.24÷2.25 × 100= 10.7%
2. Percentage of Cu in sample 2= 2.18× 100 ÷2.44 = 89.3%
Percentage of o in sample 2= 100% - 89.3% = 10.7 %
3.percentage of Cu in sample 3= 0.24×100÷2.25 = 89.3%
Percentage of o in sample 3= 100%- 89.3%= 10.7%

[@askchemistryhangout]

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[@AgroviewFarmservices]

Percentage composition of Cu in sample 1 is 89.3%
Percentage composition of O in sample 1.is 10.6%
Percentage composition Cu sample 2 is 89.3%
Percentage composition O sample 2 10.6%
Percentage composition of Cu 3 89.3%
Percentage composition of O sample 3. 10.6%

[@askchemistryhangout]

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[@alaekpereTemple-tw6bo]

In sample 1 the percentage composition by mass of cu in cuo is :
2.01÷ 2.25 ×100÷1= 89.3%
Percentage comp.by mass in oxygen cuo is :
0.24÷2.25×100÷1= 10.7%
In sample 2
The percentage composition by mass of cu in cuo is :
1.92÷2.15×100÷1 = 89.3%
Percentage composition of oxygen in cuo is:
0.23÷2.15×100÷1=10.7%
Sample 3 the percentage composition of mass of cu in cuo is:
2.18÷2.44×100÷1= 89.3%
Percentage composition of oxygen in cuo is:= 0.26÷2.44×100×1=10.7%

[@askchemistryhangout]

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