Find out the compounds that would get formed when heating copper carbonate.
In this experiment, I'll try to find out the compounds that would get formed when heating copper carbonate. The colour of CuO and Cu2O are black and red respectively. Heating copper carbonate strongly will produce copper (ll) oxide and carbon dioxide that will be given off so basically the equation that results from this is: CuCO3 (s) ? CuO (s) + CO2 (g). By heating for about 3g of the green powder of copper carbonate, I should obtain a new compound with the black colour proving the presence of copper (ll) oxide. The volume of the carbon dioxide that will result from heating copper carbonate depends on the mass of copper carbonate. Actually, it is proportional to it: the bigger the mass of copper carbonate the bigger the volume of gas given off and the bigger the mass of the product formed. The time of heating is very important as well because the copper carbonate isn't completely burnt, it will affect the quantity of gas and the mass of the compound formed. In the preliminary experiment, I just identified which compound that was formed knowing the colours. Using the same apparatus as in the proper experiment, I heated 1.00gram of a green powder of copper carbonate and obtained 0.30gram of copper (ll) oxide. That experiment was limited in the fact that I couldn't measure directly the volume of gas that was given of in the reaction and, considering the accuracy of the
Plan for rate of reaction for halogenoalkanes
Harry Denis Candidate No. 9038 Centre No. 10534 Plan for rate of reaction for halogenoalkanes Aim Aim of the experiment is to find out and compare the rate of displacement of the halide ion and how is varies with respect to the carbon halogen bond. This will occur through a nucleophilic attack. General equation: general formula: R-X +: Nu- ? R-Nu + X- CnH2n+1X (where X is the halogen) Background information halogenoalkanes In the reaction of halogenoalkanes bond polarity show us that C-F would be the most reactive then C-CL, C-Br and C-I would be the lest reactive. This is due to the electronegativity of the halogen atoms. Electronegativity is the measure of how strongly an atom in a compound attaches electron in a bond. The greater the difference in the electronegativity of two atoms, the more polar is the covalent bond between the two atoms. So in a carbon halogen bond the halogen is more electronegative than carbon. Consequently the bond between them is polarized so the halogen atom is slightly negative. Electronegativity values E.G. C 2.5 F 4.0 Cl 3.0 Br 2.8 I 2.5 H :Nu- H R C+ X - R C Nu +:X- H H According to bond enthalpy the reverse is true. Bond enthalpy shows use that C-I bond is the most reactive and the C-F bond is lest
The aim of the experiment is to identify 6 unknown colourless liquids by carrying out different chemical reactions on each of the liquids.
The aim of the experiment is to identify 6 unknown colourless liquids by carrying out different chemical reactions on each of the liquids. Flow chart showing tests/ reactions to identify the liquids. a) Bromoethane b) Butan-1- ol c) Cyclohexene d) Ethanoic Acid e) 2-Methylpropan-2-ol f) Water a b c d e f Forms CO2 gas and a white precipitate when reacting with sodium hydrogen carbonate. Yes No d- Ethanoic Acid a b c e f Decolourises bromine water Yes No c- Cyclohexene a b e f Forms a pale cream precipitate when reacting with silver nitrate solution. Yes No a- Bromoethane b e f Colour change occurs from orange to green when reacting with acidified potassium dichromate (vi) Yes No b- Butan-1-ol e f Turns blue cobalt chloride paper pink Yes No f- Water e Reacts with concentrated Hydrochloric acid, and Chlorine is substituted in place of the OH group. It then reacts with bromine water, decolourising the solution. Yes e- 2-Methylpropan-2-ol Background Information & Methods for Testing: Haloalkanes: Bromoethane is a haloalkane. Haloalkanes have the general formula of CnH2n+1X where the X represents the halogen. Bromoethane is a primary haloalkane. Carbon1 is attached to two hydrogen atoms and one alkyl group, hence a primary haloalkane. Molecular formula: C2H5Br = CH3CH3Br Structural formula of Bromoethane: Carbon
Physics : To find the center of mass of various shapes.
PHYSICS LAB CENTER OF MASS Aim: To find the center of mass of various shapes. Shapes: Center of Mass: Towards the middle, a little eastwards. Where all edges are furthest possible. Center of Gravity: In the middle of the width of the shape, since it's symmetrical. Towards the lower part of it vertically. Center of Mass: Totally symmetrical shape, therefore center of mass is exactly in the center. Center of Mass: Furthest possible away from each edge, towards the middle of the shape. Conclusion: All the shapes have the center of mass towards the middle. They all have the center of mass the furthest from each edge as possible. The shapes that are symmetrical have the center of mass in the middle, whereas the square shape has it exactly in its center, since its symmetrical anyway you look at it. The triangle, vertically symmetrical, also has a center of mass in its middle horizontally, although not vertically. Vertically its center of mass is nearer to the bottom, probably making the surface area before and after the center of mass equal. Since on top its thinner but longer and at the bottom its wider but lower. The fact that the center of mass of the object is towards the middle tells us that the pull of gravity is focused in the middle of the object. Since the gravity is focused on the center of mass, this means that all of the forces acting upon the body cancel out.
Iron Oxalate Lab
Marina Horta Iron Oxalate Lab 10/22/09 Abstract; The purpose of this lab was to synthesize, isolate, and determine Fe content of a complex; and calculate the formula of the Potassium Oxalate Iron(III) Complex using chemical analyses. This lab used filtration, titration, acid base reactions, redox , synthesis, dilutions , beers law and red tide computer system to establish that the final formula for the complex was K3[Fe(C2O4)3] •2 H2O. Introduction; Part 1; The purpose of part 1 to synthesize, purify and, mass crystals of potassium oxalate ferrate (III) complex. Potassium oxalate ferrate (III) is a complex, an ion consisting of a ligand and a metal covalently bonded together. A ligand is an ion with the tendency to bond with metals because it typically donates electron pairs. This relationship between metals and ligands is only one of the key components for the synthesis of the crystals. For the precipitation of the complex ion alcohol was used to reduce solubility, Part: 2 Part two determines the percent oxalate in the complex produced in the previous lab. Using volumetric analysis a method (similar to gravimetric analyses) in which the volume of a known
Thermal Decomposition of copper carbonate
Thermal Decomposition of copper carbonate Aim: Copper has two oxides, Cu2O, and CuO. Copper carbonate, CuCO3 decomposes on heating to form one of these oxides and an equation can be written for each possible reaction Equation 1: 2CuCO3 (s) ? Cu2O (s) + 2CO2 (g) + 1/2 O2 (g) Equation 2: CuCO3 (s) ? CuO (s) + CO2 (g) The aim of this experiment is to prove which equation is correct. This can be done by volumetric analysis i.e. calculating the volume of gas produced. This is then compared to the calculated volume of gas produced in each equation and the equation with the nearest volume of gas is correct. This is a thermal decomposition reaction and when an element like copper can form two oxides, which one forms is based on the stability of the compound formed. The more stable the compound, the more likely it is to form. The stability of a compound with respect to its elements can be predicted by the ?Hf (molar heat of formation). This is the energy change when 1 mole of a compound is formed from its elements. If it is exothermic (negative), then the compound is stable with respect to its elements. If it is endothermic (positive), then the compound is unstable with respect to its elements. In general, the lower the value of ?Hf, the greater the energetic stability of the compound with respect to its elements. The molar heat of formation of CuO is -155.2, and the molar
To investigate the rate at which different metal carbonates decompose (thermally) by measuring the amount of carbon dioxide produce when each metal carbonate is heated in a certain amount of time.
Gail Wingham Chemistry Sc1 - Heating Metal Carbonates Aim: To investigate the rate at which different metal carbonates decompose (thermally) by measuring the amount of carbon dioxide produce when each metal carbonate is heated in a certain amount of time. When a metal carbonate is heated the bonds that join the carbon to the metal element are broken and so the amount of carbon dioxide produced certifies the rate of thermal decomposition and the reactivity of the metal carbonated. Background Information: Thermal decomposition is the breaking up of compounds using heat. When metal carbonates are thermally decomposed (or heated) the bonds between the metal and the carbonates are broken and carbon dioxide and metal oxides are produced. An example of this reaction is when Calcium Carbonate (CaCO3) is heated; CaCO3 --> CO2+ CaO In this reaction from Calcium Carbonate, Carbon Dioxide (CO2) and Calcium Oxide (CaO) is produced. Relative Atomic Mass of an element (or Relative Formula Mass if a compound) has the same number of particles. eg. 100g of CaCO3 has the same amount of particles as 44g of CO2 and in 56g of CaO We know that this is true because of the relative atomic mass in each compound. The relative atom mass of an element or a compound is equal to the mass in one mole of the substance. For example: We know that one mole of CaCO3 has a mass of 100g. This is
Which oxide is formed when copper carbonate decomposes.
Introduction Aim Copper has two oxides, namely, CuO and Cu2O. Copper carbonate, CuCO3 decomposes on heating to form one of these oxides and an equation can be written for each possible reaction: Equation 1: 2CuCO3 (s) Cu2O (s) + 2CO2 (g) + 1/2O2 (g) Equation 2: CuCO3 (s) CuO (s) + CO2 (g) Our aim for the investigation is to find out which of the two equations above is correct and which oxide is formed when copper carbonate decomposes. Using the mole theory, an experiment is to be designed to measure the volume of gas, which will hence prove that which is the correct equation for the practical. The students also have to determine the quantities of reagents to be used in the experiment. Background Information Copper is a transition metal. It is found abundantly at many locations as a primary mineral in basaltic lavas and also as reduced from copper compounds. Copper pure is very malleable and ductile and can be rolled into sheets, hammered into thin leaves, and drawn into wire. It is an excellent conductor of heat and electricity. Some of its physical and chemical properties are given below: Symbol Cu Atomic Number 29 Relative Atomic Mass 63.54 Melting Point 083°C Boiling Point 2567°C Density 8.96 g cm3 Colour Reddish brown Group In Periodic Table Ib Isotopes Cu - 63 and Cu - 65 It is extracted by reduction on heating with carbon from its
Determine BaCl2.2 H2O -Gravimetric Analysis Lab
Mark Levi Chemistry Gravimetric Analysis Lab Data Collection Quantitative results for the weight of the crucible during the experiment What is Being Measured Mass (± 0.01 g) Crucible and Lid 26.64 Crucible, Lid and Barium Chloride Sample 28.65 Crucible, Lid and Barium Chloride Sample after 1st heating session 28.37 Crucible + Lid and Barium Chloride Sample after 2nd heating session 28.38 Crucible + Lid and Barium Chloride Sample after 3rd heating session 28.38 Observations Before, During and After the Experiment Heating Session Before During And After First Substance in crucible is chips of lucid, white, shiny solid. The solid does not seem to show any major appearance changes Second Solid, white, shiny small flecks. The solid seems to look like powder Third White, shiny, solid in semi-powdery form. No changes being viewed Data Processing and Presentation Calculations Initial Mass of Hydrated Barium Chloride: 28.65 ± 0.01g (Mass of Crucible and Lid and Barium Chloride Sample) - 26.64 ± 0.01g (Mass of Crucible and Lid) = 2.01 ± 0.02g (Mass of Hydrated Barium Chloride before any heating process) Mass of Barium Chloride after 1st Heating: 28.37 ± 0.01g (Mass of Crucible, Lid and Heated Barium Chloride after 1st heating) - 26.64 ± 0.01g (Mass of Crucible and Lid) = 1.67 ± 0.02g (Mass of Barium Chloride after 1st Heating) Mass of
Rate of reaction of different concentrations of sodium thiosulphate.
Science Coursework: Rate of Reaction The aim of the experiment: In this experiment I will investigate how different concentrations of sodium thiosulphate affects the time for the cross on a piece of paper to disappear. The sodium thiosulphate's concentration will vary in the stages of the experiment as proportions of water will be diluted with the chemical to see how it effects the time of the visibility of the cross. Prediction: I predict the higher the concentration of the sodium thiosulphate the faster the rate of reaction therefore shorter the time will be for the cross (x) to disappear. Background Information: Sodium thiosulphate is a liquid so its particles will be in constant motion. Before a reaction can occur between two liquids their particles must collide together in order to create a reaction. It is not enough for the particles to collide, the bonds between the atoms they have to be broken before new molecules can be produced. In liquids and gases their particles are continually moving at various speeds, which then collide in different ways. There are different ways of collisions occurring which have different effects such as 'head on' and 'glancing' collisions. The particles that contain a lot of kinetic energy (at a high temperature) collide with a large force therefore gain a heavy contact with another particle (head on). This is known to be successful
