CBSE Study Materials
2013-14
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Class IX
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SI NO.
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Subject
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Title
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1
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Chemistry
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2
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Physics
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3
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Biology
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Class VI
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SI NO.
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Subject
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Title
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1
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Science
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Class VII
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SI NO.
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Subject
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Title
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1
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Science
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Class VIII
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SI NO.
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Subject
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Title
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1
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Science
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Class X
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SI NO.
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Subject
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Title
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1
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Chemistry
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2
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Physics
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3
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Biology
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Courtesy: KVS ZIET
Bhubaneswar
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Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination Law of conservation of mass Law of constant proportion Empirical formula 1. If 10 grams of CaCO 3 on heating gave 4.4g of CO 2 and 5.6g of CaO, show that these observations are in agreement with the law of conservation of mass.(Based on Law of conservation of mass) Solution: Mass of the reactants = 10g ; Mass of the products = 4.4 + 6.6g = 10g Since the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass. 2. 1.375 g of cupric oxide was reduced by heating and the weight of copper that remained was 1.098g. In another experiment 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate was converted into cupric oxide by ignition . The weight of cupric oxide formed was 1.476 g. which law of chemical combinations does this data state? Solution: in first experiment: Copper oxide = 1....
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