https://www.youtube.com/watch?v=NFTB0IVTLEI Sodium Hydroxide Solution, NaOH Metal Cation Observation (must include ionic equation that represents the observed reaction) Ca2+ Zn2+ Pb2+ Cu2+ Cr3+ Fe2+ Fe3+ Dilute Ammonia Solution, NH3 Metal Cation Observation (must include ionic equation that represents the observed reaction for the ions in bold) Ca2+ Zn2+ Pb2+ Cu2+ Cr3+ Fe2- Fes-
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the following experiment is reference from this video link, they are the first 2 experiments ;
reference
https://www.youtube.com/watch?v=NFTBOIvTLtI
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- While working in a metal processing facility, Letlen had accidentally mixed two metal vatstogether creating an alloy. One vat was labeled for cadmium, while the other was not. It canbe assumed that these are of pure metal composition. To identify this metal, Letlen took 1.000 g of the homogenous alloy sample composed ofcadmium and the unknown metal, dissolved, and diluted it to exactly 100.0 mL in avolumetric flask. A 20.00-mL aliquot was taken and titrated this using 22.82 mL of 0.05000M EDTA. In a second 20.00-mL aliquot, the Cd was masked through the addition of HCN/NaCN buffer.The titration of the unknown metal in the aliquot required 15.13 mL of EDTA.MW: Cd (112.411 g/mol) a. Calculate the moles of Cd and the moles of unknown metal in the 20.00-mL aliquot.b. Calculate the moles of Cd and the moles of unknown metal in the sample.A 600.0 mg sample consisting of only CaC2O4and Mg C2O4is heated at 500oC converting the two salts to CaCO3and MgCO3. The sample weighs 465.0 mg. If the sample had been heated at 900oC where the products are CaO and MgO, what would the mixture of oxides weigh? Show the clear and complete solutionContamination of precipitate by co-precipitation can be minimized by (tick all that may apply) A. Digestion B. Peptization C. Washing D. Re-precipitation
- Using basic conditions, MnO4- can be used as titrant for the analysis of Mn2+, with both the analyte and the titrant ending up as MnO2. In the analysis of a mineral sample for manganese, a 0.5165-g sample is dissolved, and the manganese is reduced to Mn2+. The solution is made basic and titrated with 0.03358 M KMnO4, requiring 34.88 mL to reach the endpoint. Calculate the %w/w Mn in the mineral sample. Answer: % Mn =In this activity, you will develop an experimental procedure to solve a problem. Consider the followingscenario:You are the manager of a chemical stockroom, and find a bottle containing approximately one liter of aclear and colorless solution of unknown identity and concentration. Your only clue to its identity is thatit was found between bottles of silver fluoride and sodium fluoride, so it is likely an aqueous solution ofone of those two compounds. You will need to develop a procedure to determine the following:a) The identity of the unknown solutionb) The concentration of the unknown solutionWrite out a precise procedure, which includes all glassware, reagents, and steps. You will also need towrite the calculations that you would need to determine the concentration of the solution. Assume thatyou have access to any reagent you might needWhich among these is/are properties for the type of precipitate required in gravimetric analysis? Hygroscopic Known composition Reactive with atmospheric constituents Large particles
- Prepare 100.00 mL of a solution with ALL the following chemicals into together; CHEMICALS PROVIDED solid FeCl3 6H2O, iron(II) chloride hexahydrate (source of FeCl3) dilute hydrochloric acid in the rack – 3.0 M solid sodium salicylate – 99.7% assay Concentrations needed in the 100mL solution FeCl3: 0.020 M HCl: 0.050 M Salicylate (mg/L): 40.25 PLEASE SHOW HOW TO CALUATE AND WORK. The only equations you will need to use are the equations for molarity and dilutions.The level of dissolved oxygen in a water sample can be determined by the Winkler method. In a typical analysis, a 100.0-mL sample is made basic, and treated with a solution of MnSO4, resulting in the formation of MnO2. An excess of KI is added, and the solution is acidified, resulting in the formation of Mn2+ and I2. The liberated I2 is titrated with a solution of 0.00870 M Na2S2O3, requiring 8.90 mL to reach the starch indicator end point. Calculate the concentration of dissolved oxygen as parts per million of O2.As part of a soil analysis on a plot of land, a scientist wants to determine the ammonium content using gravimetric analysis with sodium tetraphenylborate, Na+B(C6H5)4−. Unfortunately, the amount of potassium, which also precipitates with sodium tetraphenylborate, is non‑negligible and must be accounted for in the analysis. Assume that all potassium in the soil is present as K2CO3 and all ammonium is present as NH4Cl. A 5.095 g soil sample was dissolved to give 0.500 L of solution. A 150.0 mL aliquot was acidified and excess sodium tetraphenylborate was added to precipitate both K+ and NH4+ ions completely. B(C6H5)4-+K+⟶KB(C6H5)4(s) B(C6H5)4-+NH4+⟶NH4B(C6H5)4(s) The resulting precipitate amounted to 0.269 g. A new 300.0 mL aliquot of the original solution was made alkaline and heated to remove all of the NH4+ as NH3. The resulting solution was then acidified, and excess sodium tetraphenylborate was added to give 0.129 g of precipitate. Find the mass percentages of NH4Cl and…
- Calculations for Acetate Buffer Solutions Molarity of acetic in acetate buffer (below is the data to answer the answer) Deionized water Acetate Buffer Ammonia Buffer pH of solution 6.62 4.59 9.49 Ph of solution after addition of 1 mL of 0.6 M NaOH 12.41 4.71 9.50 pH of solution after addition of 1 mL of 0.6 M HCl 1.38 3.87 9.24 Preparation of Buffer solutions Mass of sodium acetate in acetate buffer ~ 1.1981 Volume of 3.0 M acetic acid in acetate buffrer ~ 3.81 Mass of ammonium chloride ~ 1.2721 Volume of 5.0 M NH4OH ~ 6.00M is a cation which exists as a nitrate compound in an aqueous solution. Part of this solution is separated andH2S gas is sent. Then a black precipitate is given. Next, another part of this M's nitrate solution is separated.When NaOH is added, white precipitate is given. With excess NaOH, white precipitate dissolve and givecolourless solution. What could be the cation?After reacting the nitrate with cadmium to produce nitrite, the nitrite is then reacting with sulfanilamide and N-(1-naphthyl)-ethylenediamine, to produce a purple dye molecule that can be quantified on a spectrophotometer. The N-(1-naphthyl)-ethylenediamine, called NNED for convenience, reagent is made by mixing 1 gram of NNED in 1 liter of water. However, we don't always want to make an entire liter of solution because the NNED solution only lasts about 1 month before going bad and turning brown. How many grams of NNED will need to be added to make 493.364 milliliters of solution? (round to 3 decimal places)