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Investigating the reaction that takes place during the thermal decomposition of copper carbonate.

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Investigating the reaction that takes place during the thermal decomposition of copper carbonate. In this investigation I have been given the following equations for the thermal decomposition of copper carbonate: 1. 2CuCO Cu O + 2CO + O (s) (s) (g) (g) 2. CuCO CuO + CO (s) (s) (g) The aim of the investigation is to find out which reaction actually takes place during the thermal decomposition of copper carbonate, and which is false. Each reaction has different products, and by studying these products, specifically the gas produced, I will be able to find out which reaction has taken place. The two reactions produce different volumes of gas in a ration 1.25:1. Therefore, by working out what volume of gas would be produced in each reaction, by the decomposition of a certain amount of copper carbonate, when the experiment is carried out I will be able to see which reaction has taken place by measuring the volume of gas produced. In order to produce an accurately measurable volume of gas I plan to heat 1.5 grams of copper carbonate. If reaction 1 takes place, the thermal decomposition of 1.5 grams of copper carbonate will produce 364.56 dm of gas. But, if reaction 2 takes place, 291.6 dm of gas will be produced. ...read more.


When doing this make sure your eyes are level with the water level, so that you can clearly see the volume markings, and make an accurate reading. The volume should indicate which equation is correct for the reaction that has taken place. I have chosen to use this method, as I think that it is the most reliable way within my means to find out which equation is correct. If carried out carefully and safely, according the above method, all of the gas produced should be obtained. And, allowing for a small margin of error, due to air in the rubber tubing and a small amount of air which may get into the measuring cylinder when it is turned over, the volume of gas measured should be equal or close to one of the volumes I have predicted on page 1, therefore indicating which equation is correct for the reaction. Results Concentration of H2O2 / % Repeat number Volume on burette 20 seconds At stated times 30 seconds 20 1 48.7 46.6 20 2 49.3 47.2 20 3 49.1 47.2 40 1 47.9 44.0 40 2 48.0 43.9 40 3 48.0 44.2 60 1 47.0 40.8 60 2 46.8 40.6 60 3 47.9 45.9 80 1 46.9 38.8 80 2 46.6 39.0 80 3 ...read more.


Oxygen dissolving into the water. Oxygen dissolves into water and this means that the results were inaccurate as the actual reading on the measuring cylinder may not have been realistic. This would not have affected my conclusion as it takes a considerable amount of time for oxygen to dissolve into water therefore it would be minimal factor. No buffer solution was added. As there was no buffer solution added to the liver homogenate there could have been different amounts of enzyme in the samples. This would not have affected my conclusion as my results were very consistent but this could have been the contributing factor to the anomaly. The apparatus used. Some of the apparatus that was used may have been new to the user which would make it difficult for them to use the apparatus. As it may be difficult to use the apparatus this could lead to human error or if the apparatus was set-up incorrect could lead inaccuracy in the measurements. I believe it was a number of factors such as the bung being placed onto the conical flask after the reaction had already started and the time not being measured accurately as well as not being familiar to the apparatus resulted in the anomalous result I had. Overall I was able to determine that the different amount of gas produced and the concentration of the hydrogen peroxide was proportional. ...read more.

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