Chemistry Aim: I am trying to find out whether the equation 2CuCO3 ® Cu2O + 2CO2 + ½O2 or the equation CuCO3 ® CuO + CO2 is correct for the decomposition of copper carbonate.
Introduction: This is my background research, this information may help me when I am trying to see which equation is correct. I have used the book Chemistry 1 by OCR as my reference.
Relative Atomic Mass: This is the mass of an atom of the element relative to the mass of an atom of carbon-12 which has a mass of exactly 12.
Relative Molecular Mass: This is the mass of a molecule of a compound relative to an atom of carbon-12.
Mole: This is the unit of an atom of
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Equation 1 produces the most gas.
Number of moles = Volume of Gas (cm3) = 80cm3
Molar Volume (cm3) 24000cm3
= 0.0033 moles of gas
1.25 moles of gas is produced from 1 mole of copper carbonate (CuCO3)
» 0.0033 moles of gas is produced from (0.0033)
1.25
= 0.00264 moles of Copper Carbonate
m = n x M \ M[CuCO3] = [63.5 + 12 + (3 x 16)] = 123.5g mol 1
\ 0.00264 x 123.5g mol 1
= 0.33g of CuCO3
I can now work out how much gas 0.33g of CuCO3 will give in equation 2:
n = m = 0.33g = 0.00267 moles of Copper Carbonate
M 123.5g mol 1
1 mole of CuCO3 gives 1 mole of gas
\ 0.00267 moles of CuCO3 gives 0.00267 moles of gas.
Volume of gas (cm3) = number of moles x molar volume (cm3)
= 0.00267 x 24000 cm3
= 64cm3
I can now carry out the experiment and compare the actual value with the two predicted values I have just calculated. Which ever value is closest to the actual value is the correct value and so its equation will be the correct equation.
Method
I will now carry out a risk assessment of the
Finally, we measured the mass of the remaining copper in the test tube. This lead to the discovery that the mass of the copper in relation to our original 1.08 grams of copper oxide was 0.96 grams. From this data, we could find that the measurement of oxygen in our amount of copper oxide was 0.12 grams. With these significant weights, we were then able to convert those results to the amount of moles each element had in contribution to our total quantity of copper oxide. Specifically, we found that there was 0.015 moles of copper in our copper oxide, and 0.0075 moles of oxygen. So, that made the calculated ratio between moles of copper and moles of oxygen to be 2.0, or a 2:1 ratio of copper to oxygen. Thus, the formula of the red copper oxide can be expressed as . The lab worked the way it did primarily because of the chemical characteristics of copper oxide. Since the oxygen could separate from the copper when presented with a heat source, it allowed for the amount of each element in our quantity of copper oxide to be measured and then calculated into a ratio, which determined the formula for the
This experiment is based on determining the chemical formula for a hydrated compound containing copper, chloride, and water molecules in the crystal structure of the solid compound, using law of definite proportion. The general formula of the compound is CuxCly•zH2O, and aim is to determine chemical formula of this compound.
Example problem: How many grams of carbon dioxide are produced when 20.0 g acetylene (C2H2) is burned?
Purpose: The purpose of this experiment is to observe a variety of chemical reactions and to identify patterns in the conversion of reactants into products.
Co 2+ (aq) + 2 NO3- (aq) + 3 Na+ (aq) + PO4 3- (aq) CoPO4 (s) + 3 Na+ (aq) + 2 NO3- (aq)
Mol of Cu = 0.035983 mol Fe * 1 mol Cu1 mol Fe = 0.035983 mol [III]
1. Suppose that you have an unlimited supply of copper (II) sulfate to react with iron. How many moles of copper would be
b. Using the newsvendor model, Cu = 1 - 0.2 = 0.8 and Co = .2. Cu /(Cu + Co) = .8.
= = == Calcium Carbonate + Hydrochloric Acid Calcium Chloride + Water + Carbon Dioxide Equipment ---------
= 0.1 △H (J)= -Q/1000 Molar Enthalpy (KJ/mol)= Q/1000/n Uncertainty Calculations 1 Error measuring solutions, +/- 1% 2 Time measurement error, +/- 1 second Combined Error is the square root of the sum of the squares of the relative uncertainties = √((0.01 x 2)²+ (1/25)²) = 0.0447 =44.7 x
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4g of Na CO x 1 mol of Na CO 4g of CaCl x 1mol of CaCl
cos2 x = sin2 x + cos2 x sin2 x cos2 x cos2 x cos2 x = 1 2 cos2 x = RHS
Changes in matter occur every day. There are two types of ways matter can be altered; physically and chemically. Physical changes do not change the composition of the matter while chemical changes occur when one or more substances turn into a completely new substance.
This is a simple equation that doesn’t properly prove the reaction. It is very complex and starts with this: