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5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO were produced in the process. Show that these results illustrate the law of constant proportions.
? Related: Laws of Chemical Combinations
Verified Answer
5.06 g of pure cupric oxide (CuO), on complete reduction by heating in...
Given that,
Mass of CuO, W1 = 5.06g
Mass of Cu, Wcu= 4.04g
Mass of O, Wo = W1 - Wcu = 5.06-4.04 = 1.02g
Now,
mass of Cu / Mass of O = Wcu/Wo = 4.01/1.02 = ~3.9
Mass of Cu, Wcu= 1.3g
Mass of CuO, W2 = 1.63g
Mass of O, Wo = W2 - Wcu = 1.63-1.3 = 0.33g
Now,
mass of Cu / Mass of O = Wcu/Wo = 1.33/0.33 = ~3.9
Hence, law of constant proportion is followed.
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5.06 g of pure cupric oxide (CuO), on complete reduction by heating in...
The Law of Constant Proportions
The law of constant proportions, also known as the law of definite proportions, states that in a chemical compound, the elements are always present in a definite proportion by mass. This means that the ratio of the mass of each element in a compound is always the same, regardless of the amount or source of the compound.
Explanation of the Experiment
In the given experiment, cupric oxide (CuO) was reduced by heating in a current of hydrogen to produce metallic copper. The mass of the CuO and the resulting metallic copper were measured, and it was found that 5.06 g of CuO gave 4.04 g of copper. This implies that the ratio of the mass of copper to the mass of oxygen in CuO is 4.04:0.98 (5.06-4.04), which is approximately 4:1.
In the second part of the experiment, metallic copper was dissolved in nitric acid, and the resulting solution was carefully dried and ignited. The mass of the resulting CuO was measured to be 1.63 g. This implies that the mass ratio of copper to oxygen in CuO is 1.63:0.39 (1.63-1.24, where 1.24 g is the mass of oxygen in the CuO produced). This ratio is also approximately 4:1, which is the same as in the first part of the experiment.
Conclusion
The fact that the mass ratio of copper to oxygen in CuO is constant in both parts of the experiment illustrates the law of constant proportions. This law is a fundamental principle in chemistry, and it allows scientists to predict the composition of compounds based on the masses of their constituent elements.
The law of constant proportions, also known as the law of definite proportions, states that in a chemical compound, the elements are always present in a definite proportion by mass. This means that the ratio of the mass of each element in a compound is always the same, regardless of the amount or source of the compound.
Explanation of the Experiment
In the given experiment, cupric oxide (CuO) was reduced by heating in a current of hydrogen to produce metallic copper. The mass of the CuO and the resulting metallic copper were measured, and it was found that 5.06 g of CuO gave 4.04 g of copper. This implies that the ratio of the mass of copper to the mass of oxygen in CuO is 4.04:0.98 (5.06-4.04), which is approximately 4:1.
In the second part of the experiment, metallic copper was dissolved in nitric acid, and the resulting solution was carefully dried and ignited. The mass of the resulting CuO was measured to be 1.63 g. This implies that the mass ratio of copper to oxygen in CuO is 1.63:0.39 (1.63-1.24, where 1.24 g is the mass of oxygen in the CuO produced). This ratio is also approximately 4:1, which is the same as in the first part of the experiment.
Conclusion
The fact that the mass ratio of copper to oxygen in CuO is constant in both parts of the experiment illustrates the law of constant proportions. This law is a fundamental principle in chemistry, and it allows scientists to predict the composition of compounds based on the masses of their constituent elements.
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5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO were produced in the process. Show that these results illustrate the law of constant proportions. Related: Laws of Chemical Combinations?
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5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO were produced in the process. Show that these results illustrate the law of constant proportions. Related: Laws of Chemical Combinations? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO were produced in the process. Show that these results illustrate the law of constant proportions. Related: Laws of Chemical Combinations? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO were produced in the process. Show that these results illustrate the law of constant proportions. Related: Laws of Chemical Combinations?.
5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO were produced in the process. Show that these results illustrate the law of constant proportions. Related: Laws of Chemical Combinations? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO were produced in the process. Show that these results illustrate the law of constant proportions. Related: Laws of Chemical Combinations? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO were produced in the process. Show that these results illustrate the law of constant proportions. Related: Laws of Chemical Combinations?.
Solutions for 5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO were produced in the process. Show that these results illustrate the law of constant proportions. Related: Laws of Chemical Combinations? in English & in Hindi are available as part of our courses for JEE.
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5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO were produced in the process. Show that these results illustrate the law of constant proportions. Related: Laws of Chemical Combinations?, a detailed solution for 5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO were produced in the process. Show that these results illustrate the law of constant proportions. Related: Laws of Chemical Combinations? has been provided alongside types of 5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO were produced in the process. Show that these results illustrate the law of constant proportions. Related: Laws of Chemical Combinations? theory, EduRev gives you an
ample number of questions to practice 5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO were produced in the process. Show that these results illustrate the law of constant proportions. Related: Laws of Chemical Combinations? tests, examples and also practice JEE tests.
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