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2,75 g of cupric oxide was reduced by he...

2,75 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper that remained was 2.196 g. Another experiment, 2.358 g of copper was dissolved in nitric acid and the resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed was 2.952 g. Show that these results illustrate law of constant composition.

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To demonstrate that the results illustrate the law of constant composition, we will analyze the two experiments step by step. ### Step 1: Analyze the First Experiment In the first experiment, we have: - Weight of cupric oxide (CuO) = 2.75 g - Weight of copper (Cu) after reduction = 2.196 g The reaction can be represented as: ...
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1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper that remained was 1.098 g In another experiment, 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed was 1.476 g . Show that these result illustrate the law of constant composition.

In an experiment, 2.4 g of Iron oxide on reduction with Hydrogen yields 1.68 g of Iron. In another experiment, 2.9 g of Iron oxide give 2.03 g of Iron on reduction with Hydrogen. Show that the above data illustrates the law of constant proportion.

1.80g of a certain metal burnt in oxygen gave 3.0g of its oxide 1.50g of the same metal heated in steam gave 2.50g of its oxide. Show that these illustrate the law of constant proportion .

Weight of copper oxide obtained by heating 2.16 g of metallic copper with HNO_(3) and subsequent ingnition was 2.70 g In another experient, 1.15 g of copper oxide on reduction yielded 0.92 g of copper. Show that the results illustrance the law of definite proportions.

The same current if passed through solution of silver nitrate and cupric salt connected in series. If the weights of silver deposited is 1.08g . Calculate the weight of copper deposited

In an experiment 5.0 g of CaCO_(3) on heating gave 2.8 g of CaO and 2.2 g of CO_(2) . Show that these results are in accordance to the law of conservation of mass.

12.976 g of lead combines with 2.004 g of oxygen to form PbO_(2). PbO_(2) can also be produced by heating lead nitrate and it was found that the percentage of oxygen present in PbO_(2) is 13.38 %. With the help of given informations, illustrate the law of constant composition.

0.7 g of iron combine directly with 0.4 g of sulphur to form FeS. If 2.8 g of Fe are dissolved in dilute HCl and excess of Na_2S solution is added, 4.4 g of FeS are precipitated. Show that these data prove law of constant composition.

112 mL of hydrogen combines with 56 mL of oxygen of form water. When 224 mL of hydrogen is passes over hand cupric oxide, the cupric oxide loses. 0.160 g of weight. All volumes are measured at STP . Show that the result agrees with the law of constant composition (22.4 L hydrogen and oxygen at STP weigh, respectively, 2g and 32 g )

In an experiment 2.4g of iron oxide in reduction with hydrogen gave 1.68 g of iron. In another experimet, 2.7 g of iron oxide gave 1.89 g of iron on reduction. Which law is illustrated from the above data?

AAKASH INSTITUTE ENGLISH-SOME BASIC CONCEPT OF CHEMISTRY-TRY YOURSELF
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