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In an experiment 1.288 g of copper oxide was obtained from 10.3 g of Cu. In another experiment 3.672 g of copper oxide gave on reduction 2.938 g of copper. Which law of chemical combination can be illustrated by this example ?
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In an experiment 1.288 g of copper oxide was obtained from 10.3 g of C...
**Law of Conservation of Mass**
The law of conservation of mass, also known as the law of definite proportions, states that in a chemical reaction, matter is neither created nor destroyed. This means that the total mass of the reactants must be equal to the total mass of the products.
**Example 1:**
In the first experiment, 10.3 g of copper (Cu) reacted with an unknown amount of oxygen (O) to form 1.288 g of copper oxide (CuO).
**Reactants:**
Cu (copper) = 10.3 g
O (oxygen) = unknown
**Product:**
CuO (copper oxide) = 1.288 g
To apply the law of conservation of mass, we need to calculate the total mass of the reactants and compare it with the total mass of the product.
**Total mass of reactants:**
Cu = 10.3 g
O = unknown
**Total mass of product:**
CuO = 1.288 g
Since the mass of the oxygen is unknown, we can assume it as 'x' for calculation purposes.
Therefore, the total mass of the reactants is 10.3 g + x g, and the total mass of the product is 1.288 g.
According to the law of conservation of mass, the total mass of the reactants should be equal to the total mass of the products.
10.3 g + x g = 1.288 g
Solving for 'x', we get:
x g = 1.288 g - 10.3 g
x g = -9.012 g
Since mass cannot be negative, this indicates an error in the experiment. However, if we assume that the mass of oxygen is negligible compared to copper, we can approximate the mass of oxygen as zero.
Therefore, the law of conservation of mass is illustrated by this experiment, as the total mass of the reactants (Cu) is approximately equal to the total mass of the product (CuO).
The law of conservation of mass, also known as the law of definite proportions, states that in a chemical reaction, matter is neither created nor destroyed. This means that the total mass of the reactants must be equal to the total mass of the products.
**Example 1:**
In the first experiment, 10.3 g of copper (Cu) reacted with an unknown amount of oxygen (O) to form 1.288 g of copper oxide (CuO).
**Reactants:**
Cu (copper) = 10.3 g
O (oxygen) = unknown
**Product:**
CuO (copper oxide) = 1.288 g
To apply the law of conservation of mass, we need to calculate the total mass of the reactants and compare it with the total mass of the product.
**Total mass of reactants:**
Cu = 10.3 g
O = unknown
**Total mass of product:**
CuO = 1.288 g
Since the mass of the oxygen is unknown, we can assume it as 'x' for calculation purposes.
Therefore, the total mass of the reactants is 10.3 g + x g, and the total mass of the product is 1.288 g.
According to the law of conservation of mass, the total mass of the reactants should be equal to the total mass of the products.
10.3 g + x g = 1.288 g
Solving for 'x', we get:
x g = 1.288 g - 10.3 g
x g = -9.012 g
Since mass cannot be negative, this indicates an error in the experiment. However, if we assume that the mass of oxygen is negligible compared to copper, we can approximate the mass of oxygen as zero.
Therefore, the law of conservation of mass is illustrated by this experiment, as the total mass of the reactants (Cu) is approximately equal to the total mass of the product (CuO).
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In an experiment 1.288 g of copper oxide was obtained from 10.3 g of C...
Ok i understand
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In an experiment 1.288 g of copper oxide was obtained from 10.3 g of Cu. In another experiment 3.672 g of copper oxide gave on reduction 2.938 g of copper. Which law of chemical combination can be illustrated by this example ?
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In an experiment 1.288 g of copper oxide was obtained from 10.3 g of Cu. In another experiment 3.672 g of copper oxide gave on reduction 2.938 g of copper. Which law of chemical combination can be illustrated by this example ? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about In an experiment 1.288 g of copper oxide was obtained from 10.3 g of Cu. In another experiment 3.672 g of copper oxide gave on reduction 2.938 g of copper. Which law of chemical combination can be illustrated by this example ? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In an experiment 1.288 g of copper oxide was obtained from 10.3 g of Cu. In another experiment 3.672 g of copper oxide gave on reduction 2.938 g of copper. Which law of chemical combination can be illustrated by this example ?.
In an experiment 1.288 g of copper oxide was obtained from 10.3 g of Cu. In another experiment 3.672 g of copper oxide gave on reduction 2.938 g of copper. Which law of chemical combination can be illustrated by this example ? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about In an experiment 1.288 g of copper oxide was obtained from 10.3 g of Cu. In another experiment 3.672 g of copper oxide gave on reduction 2.938 g of copper. Which law of chemical combination can be illustrated by this example ? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In an experiment 1.288 g of copper oxide was obtained from 10.3 g of Cu. In another experiment 3.672 g of copper oxide gave on reduction 2.938 g of copper. Which law of chemical combination can be illustrated by this example ?.
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