...mass. Then, we had to fill a 400-ml beaker with deionized water and test it to make sure that the ph was just basic. We then combined the deionized water and salt by adding about 150ml of the deionized water to the salt mixture. We then stirred the combined mixture for about 2-3 minutes and then let it sit so that the precipitate would settle. After it settled, we covered the beaker with a watch glass and warmed it up on a hot plate at 75C for about 15 minutes, periodically stirring the solution. After 15 minutes, we removed the beaker from the heat and allowed the precipitate to settle again. While we waited for the precipitate to settle, we prepared wash water by heating up about 30 ml of deionized water at 70-80C. We then placed filter paper in a filter funnel to set up gravity for filtering. On the side, we took some of the solution’s supernatant and half-filled two test tubes using a pipet. Thereafter, we took the rest of the solution started filtering it with the use of a rubber policeman. IF necessary, we were to use the wash water to wash away any precipitate into the flter. After the filtering was completed, we were to remove the filter, with the precipitate on top of it, and store it away so that it was able to air-dry until the next lab class. We took two trials of this. Meanwhile, we came back to the two test tubes we...
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...reactants is the limiting reactant and which is the excess reactant. • Determine the theoretical mass of precipitate that should form. • Compare the actual mass with the theoretical mass of precipitate and calculate the percent yield. Materials: Balance 0.70 M sodium carbonate solution, Na2CO3(aq) Graduated cylinder 0.50 M calcium chloride solution, CaCl2(aq) Beaker (250 mL) Wash Bottle (distilled H2O) Filter paper Funnel Iron ring Ring stand Procedure: Part I: The Precipitation Reaction (Day 1) 1. Obtain two clean, dry 25 mL graduated cylinders and one 250 mL beaker. 2. In one of the graduated cylinders, measure 25 mL of the Na2CO3 solution. In the other graduated cylinder, measure 25 mL of the CaCl2 solution. Record these volumes in your data table. 3. Pour the contents of both graduated cylinders into the 250 mL beaker and observe the results. Record these qualitative observations in your observations table. Allow the contents of the beaker to sit undisturbed for approximately 5 minutes to see what happens to the suspended solid particles. Meanwhile, proceed to step 4. 4. Obtain a piece of filter paper and put your initials and your partner’s initials on it using a pencil. Measure and record the mass of the filter paper, then use it to set up a filtering apparatus. 5. Gently swirl the beaker and its contents to suspend the precipitate in the solution, then pour it carefully and slowly into the filter funnel. It takes time to complete the...
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...precipitated on the optimum needed concentration and easily detected. 1st group of cations The group (I) cations form the insoluble chlorides. The dilute hydrochloric acid (generally 1-2 M concentration) is used as the group reagent. The Concentrated HCl is not used because it forms a soluble complex ion ([PbCl4]2-) with Pb2+ and would not be detected. The most important cations of 1st group are Ag+, Hg22+, and Pb2+. Color detection test is not suitable for their chlorides because all are white solid compounds. The analyzing of group is done by adding the respective salt into water and follow it with dilute HCl. This results in a white precipitate in which NH4OH is added. The insoluble precipitate shows the presence of Plumbous (Pb2+) and if precipitate is soluble, then it shows the presence of Argentous (Ag+).If the white precipitate turns black, it shows the presence of Hg22+. The obtained filtrate after first group analysis is used for the test of second group. Usually Pb2+ is also included in the 2nd analytical group because the solubility of PbCl2 is more than the chlorides of other two ions. Confirmation Test For lead Pb2+ + 2KI → PbI2 + 2 K+ Pb2+ + K2CrO4 → PbCrO4 + 2K+ For Silver Ag+ + KI → AgI + K+ 2Ag+ + K2CrO4 → Ag2CrO4 + 2K+ For mercury ion Hg22++ 2KI → Hg2I2 + 2K+ 2Hg22+ + 2NaOH → 2Hg2O + 2Na+ + H2O 2nd analytical group of cations The ions of this group form acid-insoluble sulfides. Cations of the 2nd group...
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...considered for aluminum alloys. In this work, the microstructural and corrosion properties of friction stir welded 7075 Al alloy were studied. The microstructures of the base metal, bore metal, thermo-mechanically affected zone (TMAZ) and weld region were characterized by optical microscopy and transmission electron microscopy. Micro-hardness profile was obtained across the weld. The pitting corrosion properties of the weldments were studied in 3.5% NaCl solution. Friction stir welding of this alloy resulted in fine recrystallized grains in weld nugget which has been attributed to frictional heating and plastic flow. The process also produced a softened region in the weld nugget, which may be due to the dissolution and growth of possible precipitates, identified as Mg32(Al,Zn)49. Corrosion resistance of weld metal has been found to be better than that of TMAZ and base metal. 1. INTRODUCTION High strength precipitation hardening 7XXX series aluminum alloys, such as 7075 are used extensively in aerospace industry. These alloys are difficult to join by conventional fusion welding techniques. Hence realizing a fusion-welded joint in such alloys without impairing the mechanical properties is a difficult task for the welding engineer. Consequently the welding engineer has to rely on rivets and fasteners with substantial increase in fabrication cost and structure weight. The process of Friction stir welding is described in...
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...are; hydroxide, chloride, nitrate, oxide, carbonate, and sulphate. Testing for anion includes, testing for carbonate it requires you adding a dilute acid to the carbonate sample, using a pipette take out the air in the sample and bubble it through limewater. It will go cloudy which mean that carbon dioxide is present. Testing for sulphate and Sulphide, if sulphate dissolve in dilute nitric acid, then a white precipitate forms with barium chloride solution. If you was to add hydrochloric acid, the sulphate will dissolve but if it doesn’t dissolve it is a Sulphide. Testing for sliver nitrate you will do a halogens, or halide ions test. You add sliver nitrate to a solution, this could be chloride, bromide or iodide, and if they react with the sliver they will form a ppt. A white, cream or yellow ppt. If chloride dissolve in dilute nitric acid, then a white precipitate forms with silver nitrate solution. The precipitate will dissolves in dilute ammonia solution. If bromide dissolve in dilute nitric acid, then a cream precipitate forms with silver nitrate solution. The precipitate is insoluble in dilute ammonia solution, but will dissolve in concentrated ammonia solution. If iodide...
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...the products for the following reactions, balance the equation, then classify the type of reaction: 1) ____ Na + ____ FeBr3 ( 2) ____ NaOH + ____ H2SO4 ( 3) ____ C2H4O2 + ____ O2 ( 4) ____ NH3 + ____ H2O ( 5) ____ PbSO4 + ____ AgNO3 ( 6) ____ PBr3 ( 7) ____ HBr + ____ Fe ( 8) ____ KMnO4 + ____ ZnCl2 ( 9) ____MnO2 + ____ Sn(OH)4 ( 10) ____ O2 + ____ C5H12O2 ( 11) ____ H2O2 ( 12) ____ PtCl4 + ____ Cl2 ( Predicting Reaction Products (cont.) For each of the following double replacement reactions, determine what the products of each reaction will be. When you have predicted the products, balance the equation and use a solubility to determine which of the products (if any) will precipitate (be insoluble). Assume all reactions take place in aqueous solutions (aq). 1) ____ Ca(OH)2 + ____ HF ( 2) ____ Pb(NO3)2 + ____ K2CrO4 ( 3) ____ NaC2H3O2 + ____ H2SO4 ( 4) ____ Cu(OH)2 + ____ H3PO4 ( 5) ____ AgNO3 + ____ Na2CO3 ( 6) ____ Pb(OH)2 + ____ Hg2S ( Predicting Reaction Products - Solutions Balance the equations and predict the products for the following reactions: 1) 3 Na + 1 FeBr3 ( 3 NaBr + 1 Fe 2) 2 NaOH + 1 H2SO4 ( 1 Na2SO4 + 2 H2O 3) 1 C2H4O2 + 2 O2 ( 2 CO2 + 2 H2O 4) 1 NH3 + 1 H2O ( 1 NH4OH 5) 1 PbSO4 + 2 AgNO3 ( 1 Ag2SO4 + 1 Pb(NO3)2 6) 4 PBr3 ( 1 P4 + 6 Br2 7) 2 HBr + 1 Fe ( 1 H2 + 1 FeBr2 OR 6 HBr + 2 Fe ( 3 H2 + 2 FeBr3 ...
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...organisms) and the actual concentration of oxygen is called the biological demand in oxygen. Principle: The Winkler test is used to determine the level of dissolved oxygen in water samples and to estimate the biological activity in the water sample. An excess of Manganese(II) salt, iodide (I-) and hydroxide (OH-) ions are added to a water sample causing a white precipitate of Mn(OH)2 to form. This precipitate is then oxidized by the dissolved oxygen in the water sample into a brown Manganese precipitate. In the next step, a strong acid (either hydrochloric acid or sulphuric acid) is added to acidify the solution. The brown precipitates then convert the iodide ion (I-) to Iodine. The amount of dissolved oxygen is directly proportional to the titration of Iodine with a thiosulphate solution. Method First Manganese(II) sulfate is added to an environmental water sample. Next, Potassium iodide is added to create a pinkish-brown precipitate. In the alkaline solution, dissolved oxygen will oxidize manganese(II) ions to the tetravalent state. 2 Mn(OH)2(s) + O2(aq) → 2 MnO(OH)2(s) MnO(OH)2 appears as a brown precipitate. There is some confusion about...
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...aldehydes and ketones were determined using tests such as oxidation with sodium nitrate and sodium hydroxide, 2,4-dinitrophenylhydrazones test, Semicarbones test, Tollens’ test, Fehling’s test, Iodoform test, and the chromic acid test. The objectives of the experiment were met since the physical properties of the aldehydes and ketones were observed using these tests. Give an Unknown Letter D sample was tested with 2,4-dinitrophenylhydrazones test, Semicarbones test, Tollens’ test, Fehling’s test, Iodoform test, and the chromic acid test. Unknown D presumed miscible with a color change of Tang orange precipitate in 2,4-Dinitrophenyldrazones test. Semicarbones Test Unknown D had a melting of point 123C. Tollen’s test for unknown D tested negative with no formation of black precipitate of reduced silver. In Fehling’s test for unknown D tested negative with no formation of reddish precipitate of copper(I) oxide metallic deposits. Iodoform test of unknown D tested positive and Chromic Acid test for unknown D tested negative. Once conducting all test unknown D tested positive for a Keytone. Comparing results to all known samples that are keytones and viewing there melting points and properties unknown D is concluded to be 2-Heptanone. Data/Results/Discussion Generally, the chromic acid test reagent (Jones reagent) and Tollens reagent was to identify aldehyde. Iodoform test reagent is to identify methyl ketone. A. 2,4-dinitrophenylhydrazine can react with both...
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...equations for each precipitate formed. The first part of this lab required us to combine four unknown solutions in well plates and see which react. Next, we had to create eight mixtures, in the well plates. Most of these mixtures created precipitates. The second part of the lab was essential in identifying which solution was in each of the unknown bottles. We were able to identify all four unknown solutions. Solution A was NaCl, Sodium chloride, solution B was AgNO3, Silver Nitrate, solution C was KI, potassium iodide, and solution D was KNO3, potassium nitrate. The task of identifying each of the unknown solutions was a little tricky, but with the right equations, it was not too hard. The first unknown bottle that we were able to identify was B. We were first able...
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...strength and hardness of some metal alloys may be enhanced by the formation of extremely small uniformly dispersed second-phase particles within the original phase matrix in a process known as precipitation or age hardening. The precipitate particles act as obstacles to dislocation movement and thereby strengthen the heat-treated alloys. Many aluminum based alloys, copper-tin, certain steels, nickel based super-alloys and titanium alloys can be strengthened by age hardening processes. In order for an alloy system to be able to be precipitation-strengthened, there must be a terminal solid solution that has a decreasing solid solubility as the temperature decreases. E.g: Al-4.5% Cu, Al-6% Zn-2.5%Mg, Cu-2%Be, Ni-17%Cu-8%Sn, Ti-6%Al-4%V. Stages: Solution treatment: A suitable alloy is heated to a temperature at which a second phase (Usually present in small quantities) dissolves in the more abundant phase. The metal is left at this temperature until a homogeneous solid solution is attained. Quenching is the second step where the solid is rapidly cooled forming a supersaturated solid solution of αSS which contains excess of second phase and is not an equilibrium structure. The atoms do not have time to diffuse to potential nucleation sites and thus θ precipitates do not form. Aging is the third step where the supersaturated is heated below...
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...the following: 1. Title 2. List of Materials 3. Safety that includes MSDS risk assessment for all the materials used ( instructions were given to you already and they are in Moodle) 4. Pre-lab questions 5. Data Tables with the results obtained ( observations and predictions) 6. Post Lab: Part I : Single Replacement Reactions a. For every reaction that took place you must write the balanced chemical equation b. Which metal reacted the most? c. Rank your metals from more to least active Part II: Double replacement Reactions a. For every reaction where you observed precipitate, write the complete balanced molecular equation, the complete ionic equation and the net ionic equation; use the solubility rules to identify the precipitate and the states of matter of each substance participating in the reaction. b. Which cation produced the most number of precipitates? c. Write general rules of solubility that you observed. 7. Final Conclusion and error analysis To do Lab Report for Single and Double Replacement reactions : This is a formal lab report. It must be typed or written with blue or black ink. Make sure that you include the following: 1. Title 2. List of Materials 3. Safety that includes MSDS risk assessment for all the materials used ( instructions were given to you already and they are in Moodle) 4. Pre-lab questions 5. Data Table with the results obtained ( observations and predictions0 6. Post Lab: Part I : Single...
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...Organic Chemistry Name: Rochelle Bailey Lab #: 7 Date: March 9, 2015 Introduction: Aldehydes and ketones are two of several types of compounds that contain the carbonyl group. Reactions that occur because of the presence of the carbonyl group include nucleophilic addition reactions and base catalysed condensations. Aldehydes are also easily oxidized, which provides a convenient means to distinguish them from ketones. The carbonyl group in aldehydes and ketones is highly polarized; the carbonyl carbon bears a substantial partial positive charge and is susceptible to nucleophilic attack. Further, since it is sp2 hybridized it is relatively open to attack. Because the carbonyl contains no good leaving group, addition occurs rather than substitution. Aldehydes and ketones are polar compounds; however, the pure compounds do not undergo hydrogen bonding as the alcohols do. Thus the boiling points of aledehydes and ketones are lower than alcohols, but higher than alkanes or ethers. Low molecular weight carbonyl compounds are water soluble. Tests used to classify aldehydes and ketones Question 1 Identify the tests used to classify aldehydes and ketones. For each test identify the reagent(s) used and tell what observations/ results are expected. Write equations to show reactions (where appropriate). There are various tests that can be used to identify aldehydes and ketones. Some tests can be used to identify the carbonyl group in compounds while others can be used to distinguish...
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...Based on the results of your laboratory experiment, would you be able to differentiate between phosphate and hydroxide based on the precipitates they formed? Please explain your answer for full marks. Answer: You are able to differentiate between phosphate and hydroxide only if you mix them with CO+2, Mg+2, as these are the only reactions where they had different reactions during our lab. 8. Write the remained of the discussion identifying the unknowns from the experiment (3 positive ions and 3 negative ions). For full marks you must explain how you determined the identity of each unknown. Answer: Positive (cation) ions Unknown 1: The first unknown is Cobalt. I think it is Cobalt because it matches the final previous reactions with chlorine, iodine, phosphate, sulfate, and oxalate. Unknown 2: The second unknown is Barium. I think it is Barium because it matches the...
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...Objective: To identify the components of the solution in its pure form with various food tests and state the justifications. Observations: Tests | Observations | Reducing sugar test a) Glucose | * White glucose solution changes colour to blue when Benedict’s solution is added. After the solution is heated, the blue solution forms moderate amount of brick red precipitate. | b) Fructose | * White fructose solution changes colour to blue when Benedict’s solution is added. After heating, blue solution forms more amount of brick red precipitate. | c) Lactose | * White lactose solution changes colour to blue when Benedict’s solution is added. After heating, blue solution forms least amount of brick red precipitate. Solution still have a slight blue colour. | Non-reducing sugar test a) Sucrose | * Sucrose solution is neutralized with hydrochloric acid and sodium hydroxide solution. No changes is observed. Blue colour forms when the solution is added with Benedict’s solution. After heating, brick red precipitate is formed. | Iodine test a) Starch | * Blue-black precipitate is formed after 3 drops of iodine is added. | | Sudan III test | * Oil and distilled water were immiscible forming two separate layer. When Sudan III is added, fats globules are stained red. Fat layers forms at the top layer. The solution was opaque and forms at the bottom layer. | Emulsion test | * After shaking- White, opaque solution was formed. * After...
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...Winkler test for dissolved oxygen The Winkler test is used to determine the concentration of dissolved oxygen in water samples. Dissolved oxygen (D.O.) is widely used in water quality studies and routine operation of water reclamation facilities. An excess of manganese(II) salt, iodide (I–) and hydroxide (OH–) ions is added to a water sample causing a white precipitate of Mn(OH)2 to form. This precipitate is then oxidized by the dissolved oxygen in the water sample into a brown manganese precipitate. In the next step, a strong acid (either hydrochloric acid or sulfuric acid) is added to acidify the solution. The brown precipitate then converts the iodide ion (I–) to iodine. The amount of dissolved oxygen is directly proportional to the titration of iodine with athiosulfate solution.[1] Today, the method is effectively used as its colorimetric modification, where the trivalent manganese produced on acidifying the brown suspension is directly reacted with EDTA to give a pink color.[2] As manganese is the only common metal giving a color reaction with EDTA, it has the added effect of masking other metals as colorless complexes. History The test was originally developed by Ludwig Wilhelm Winkler, in later literature referred to as Lajos Winkler, while working at Budapest University on his doctoral dissertation in 1888.[3] The amount of dissolved oxygen is a measure of the biological activity of the water masses. Phytoplankton and macroalgae present in the water mass-produce oxygen...
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