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Copper carbonate thermal decomposition
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Decomposition of copper carbonate
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Determining the Correct Equation for the Decomposition of Copper Carbonate Introduction and background information: Important points to note: ‘At room temperature, 25°C and atmospheric pressure at 1 atmosphere, I mole of any gas will occupy a volume of 24 dm³.’ We will need this to work out how much copper carbonate to decompose to obtain a sufficient amount of carbon dioxide gas. To work out the amount of copper carbonate to use I will need to use the following equations: Number of moles = Mass / Mr PV = nRT P = Pressure V = Volume n = Number of moles R = Gas constant T = Temperature We can substitute n (number of moles) with the first equation to get: PV = (mass/Mr) x RT When rearranged this gives: Mass = (Mr x PV) / RT This will allow me to work out the mass needed. Aim: The aim of this experiment is to determine which of the following equations is correct: [IMAGE]Equation 1: 2CuCO3 (s) Cu2O (s) + 2CO2 (g) + ½ O2 (g) [IMAGE]Equation 2: CuCO3 (s) CuO (s) + CO2 (g) I will do this by decomposing the copper carbonate. I will need to calculate if the volume of carbon dioxide produced is equal to what the equation suggests. Equipment: * Heatproof mat * Bunsen burner * Boiling tube * Bung * Delivery tube * Water bath * Measuring cylinder (250 cm³) * Digital weighing scales * Spatula * Clamp and stand * One molar copper carbonate (1 gram) * Two molar copper carbonate (1 gram)
Mix 50ml of WATER with 15ml (1 tbsp) of BAKING SODA in one beaker or cup
4 "That amount of any gas that occupies a volume of 22414 mL in normal conditions is called one mole [eine solche Menge irgendeines Gases, welche das Volum von 22412 ccm im Normalzustand einnimt nennt man ein Mol]"
(CNG). That's because the gas is confined to a pressure of approximately 3,600 pounds p...
However, the mass of copper carbonate equals the total mass of the copper oxide and carbon dioxide because the same atoms are present but in different ways. It is possible to determine which equation is correct by measuring the volume of... ... middle of paper ... ... ume produce by equation 1, i.e. 18.5cm3) this would indicate that less gas has been produced, and thus suggest equation 2 is correct.
Substance Mr/Ar - and hence, mass of 1 mole (g). Copper Carbonate (CuCO3) 123.5 Copper Oxide (Cu2O) 143 Copper Oxide (CuO) 79.5 Carbon Dioxide (CO2) 44 Oxygen (O2) 32 --------------------------------------------------------------------------------------------------------------------- So using the equation of a reaction, it is possible to predict the masses of products that will be made by a given mass of reactants. In this experiment, from the two equations given, it is possible to calculate how much gas would be given off by each.
Avogadro is well known for his hypothesis known as Avogadro's Law. His law states that at a given temperature, equal volumes of gas contain the same number of molecules equal to about 6.0221367 x 10 to the 23rd power.A Mole of a substance is the quantity of the substance that weights the same as its molecular mass. One mole of any substance is Equal to Avogadro's number. Therefore Avogadro's law can be stated in terms of moles, namely that equal volumes of gases at the same temperature and pressure contain the same number of moles.
Obtaining Zinc Oxide from Calamine Introduction Calamine is a mineral containing zinc carbonate (ZnCO₃) On heating it decomposes as: [IMAGE]ZnCO₃ ZnO + CO₂ (C = 12, 0 = 16, Zn = 65) This equation allows you to calculate a theoretical conversion of calamine into zinc oxide. As when using the theoretical conversion; [IMAGE]ZnCO₃ ZnO + CO₂ [IMAGE]65+12+48 65+16 + 12+32 [IMAGE]125 81 + 44 This means that one mole of calamine weighs 125g and when heated it produces 81g of zinc oxide and 44g of carbon dioxide. Therefore to work out how much zinc oxide is produced from 1g of calamine we divide 81 by 125.
== § Test tubes X 11 § 0.10 molar dm -3 Copper (II) Sulphate solution § distilled water § egg albumen from 3 eggs. § Syringe X 12 § colorimeter § tripod § 100ml beaker § Bunsen burner § test tube holder § safety glasses § gloves § test tube pen § test tube method = == = =
I will not add a catalyst to my solution and I will not stir my solution. · I will use 25cm3 of hydrochloric acid. · I will use 1g of calcium carbonate.
Moles Volume HCl Volume Water 2 M 10 cm 3 0 cm 3 1.5 M 7.5 cm 3 2.5 cm 3 1 M 5 cm 3 5 cm 3 0.5 M 2.5 cm 3 7.5 cm 3
First we pour 10mL of infiltration solution and 30mL of sodium bicarbonate in a small beaker.
the bulk to ordinary matter; the volume of an atom is nearly all occupied by the
remaining 20 percent is due to other gasses that are present in very small amounts? (Murck,
of Copper Sulphate. To do this I plan to work out the amount of water