Copper is an element that occurs in minor amounts in the earth’s crust. Estimates of average crustal prevalence are on the order of 0.0058 percent by weight. Deposits considered to be economically recoverable at current market prices may contain as little as 0.5 percent of copper or less‚ depending on the mining method‚ total reserves‚ and the geologic setting of the deposit. Here is the flow sheet on how to process copper from Zenith. Copper Process Flow Sheet: Copper ore is first quarried from
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Coursework 4 Copper Name of student: Alfiya Ospanova Tutorial group: G Copper is one of the earliest elements explored by humanity. Due to the special red shade it was easy to find in ancient times. The metal was named according to place where it was found‚ the Cyprus islands. Copper produced in massive stars and spread in the world as minerals like chalcocite and chalcopyrite (sulfide of copper)‚ azurite and malachite (carbonate of copper)‚ cuprite (oxide of copper). The atomic
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heat CuO + H2O 4) CuO + H2SO4 Cu2+(aq) + SO42-(aq) + H2O 5) Cu2+(aq) + Zn Cu + Zn2+(aq) Stage 1- Dissolution of Copper Mass of vial & copper = 3.537g Mass of vial = 2.505g Mass of Copper = 1.032g Moles of copper = mass (g) ÷ (63.5g/mol) = 0.0162 mol (3s.f) Copper description: Initially‚ the copper was a solid that was cut into thin strips. It had a shiny/metallic and opaque look‚ with an orange colour. HNO3 description: The nitric
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Research question/Aim Extract copper from copper oxide by reduction using butane gas Hypothesis The butane gas will reduce the copper oxide and leave behind pure copper. This will allow us to find the empirical formula of the copper oxide as well as the masses of oxygen and copper Variables Dependant variable The dependant variable will be the reduction of the copper oxide which will be calculated at different temperatures. Independent variable The independent variable here
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Percent Copper and Formula of a Copper Compound Sofia Economides Date Preformed: 3/19/2012 Date Submitted: 3/26/2012 Lab Partners: Anita Smith‚ Taylor Bell‚ Derek Bourgeois 1. Purpose a. To determine the percent of copper and the formula weight of a copper compound. 2. Introduction a. Magnesium is a very active metal which can reduce copper (II) ions in a weight sample of a compound to metallic copper. The reduction of copper ions in a compound to a copper metal
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Modbury High School SACE Stage 1 Chemistry Topic 5 Mole Concept and Stoichiometry Assignment 5: Volumetric analysis (titrations)‚ stoichiometry SOLUTIONS Note: Write answers neatly and legibly in your exercise book or on pad paper. ALWAYS include a title and name for your work and clearly indicate each answer. 1. a) 23.08 and 23.00 mL are concordant titre values. Average titre = (23.08 + 23.00) = 23.04 mL 2 b) Ca(OH)2
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Copper Sun By Sharon M. Draper Ch. 1 (V) Betrothed: engaged to‚ be married Amari and Besa had been formally betrothed to each other last year. Kente: A colorful fabric of Ghanaian origin Their fingers flew as they chatted together‚ the pale fibers stretching and uncurling into threads for what would become kenet cloth. Ch. 1 (Q) “Be content with woman’s work child. It is enough.” I think this is a memorable quote in which got me thinking of how different women’s and men’s jobs where and
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Title: Stoichiometry Reaction Objectives: 1. To decompose sodium hydrogen carbonate (sodium bicarbonate) by heating. 2. To accurately measure the degree of completion of the reaction by analysing the solid sodium carbonate product. 3. To calculate amount of product with given amount of reactant. 4. To determine amount of heat release in the reaction. Results: Part 1: Thermal Decomposition of NaHCO3 Materials Mass (g) Clean and dry test tube 15.1632 Clean test tube + NaHCO3 17.1647
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Na2CO3(s)+CO2(g)+H2O(g). The was the only equation that matched up exactly with my data in terms of percentage. To start off with‚ when I balanced out the equation‚ I got 2 NaHCO3→ 1 Na2CO3(s)+ 1 CO2(g)+ 1 H2O(g). Therefore when I set up my stoichiometry problem I got 3.2 grams NaHCO3 over 1 x 1 mol NaHCO3 over 84.007g NaHCO3 x 1mol Na2CO3 over 2 mol NaHCO3 x 105.987g Na2CO3 x 1 mol Na2CO3. Hence‚ I multiplied 3.2 x 1 x 1 x 105.987 and got 339.1584. Afterwards‚ I divided 339.1584 by 84.007 and
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volume. In this case‚ experimenters were given aqueous solutions of NaOH and CaCl2 in known molarities and then had them react with one another to yield a precipitate of Ca(OH)2. The precipitate was filtered out of the remaining aqueous solution of stoichiometry. In our case‚ all four tests yielded more mass than should have occurred. Experimenters attribute this completely evaporated from the filtered precipitate‚ which would add excess mass. By dealing with such small quantities of reagents‚ any small
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