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I only need number 4\'s answer. Thanks in advance Jaffrey Zagnut couldn’t find a

ID: 880066 • Letter: I

Question

I only need number 4's answer.

Thanks in advance

Jaffrey Zagnut couldn’t find any nitric acid so he tried to dissolve his copper sample in hydrochloric acid instead. Unfortunately his copper wouldn’t dissolve in HCl. Why will copper dissolve in nitric acid but not in hydrochloric acid (after all, HCl is a stronger acid than HNO3).

1. Copper (II) hydroxide is converted into copper (II) oxide by heating the test tube containing Cu(OH)2 in a hot water bath. Is it necessary to use distilled water in this water bath? Why or why not?

2. Copper metal doesn’t “rust” in the presence of oxygen at room temperature. However, it will react with O2 at elevated temperatures. Write a balanced chemical equation describing the formation of copper (II) oxide when copper metal is heated in air.

3. When zinc is dissolved in sulfuric acid a gas is produced. What is the chemical identity of this gas? How is it produced?

4. Jaffrey Zagnut started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment. At the end of the experiment he obtained 0.038 g of a black product. What was his percent yield? What is the most likely source of the error in his experiment? (Hint: consider question 2 above)

Explanation / Answer

Solution :- starting mass of Cu = 0.032 g

Experimental mass of Cu = 0.038 g

Percent yield =( experimental value /theoretical value )*100 %

                        = (0.038 g / 0.032 g)*100%

                       = 118.8 %

Percent yield never exceed 100 %

The source of error is may be Jaffrey Zagnut has heated the copper metal at higher temperature in presence air this will cause the formation of the copper oxide as shown by the following reaction equation

2Cu + O2 ------ > 2CuO

Therefore in the product Cu is combined with the oxygen

This added oxygen increases the mass of the final product therefore the percent yield goes beyond 100% which is incorrect.