10-4 mol KIO3 4.675 x 10-4 mol KIO3 x (6 mol Na2S2O3 / 1 mol KIO3) = 0.002805 mol Na2S2O3 0.002805 mol Na2S2O3 / 0.03005 L Na2S2O3 = 0.093344426 M Na2S2O3 Determination of the Ascorbic Acid Titration Mass of the ascorbic acid sample‚ g Volume of Sodium Thiosulfate solution‚ mL Weight % of the Ascorbic Acid in the tablet 1 0.1018 19.50 85 % 2 0.1016 19.20 88 % 3 0.1003 19.20 89 % IO3- + 8 I- + 6 H+ → 3 I3- + 3 H2O I3- + 2 S2O32- → 3 I- + S4O62- C6H8O6 + I3- + H2O → C6H6O6 + 3 I- + 2 H+ TITRATION
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amount of heat absorbed or released during the dissolving of ammonium nitrate and of sodium acetate in water. 2.0 g of solid NaOH is dissolved in 100 mL of water. The initial and final temperatures are measured and recorded. The heat of solution is calculated (ΔH1) Reaction 2. Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → Na+(aq) + Cl-(aq) + H2O(l) 50 mL of 1.0 M hydrochloric acid solution is combined with 50 mL of 1.0 M sodium hydroxide solution. The initial and final temperatures are recorded‚ and the heat
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Introduction The purpose of this lab was to explore the properties of an unknown compound. An unknown was given and a cation flame test and anion test was performed to determine the identity of the compound. Once the identity was determined‚ the properties were explored. Experimental To determine the cation of the compound‚ a cation flame test was performed. A bunsen burner was lit until a medium blue flame was burning. The given unknown was scooped onto a nichrome wire loop. The wire was held
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such as produced on evaporation‚ but rather in the form of ions‚ the basic elements and the acid radicals being separated‚ nevertheless it is customary and convenient to consider them as combined into the form of salts. Among these‚ common salt‚ or sodium chloride‚ makes up the bulk of the material‚ being nearly 78 percent of the total mass of salt‚ or over 27 mille (thousand) of the salinity (which is taken as 35 in round numbers). In the accompanying table the composition of sea water salts is given
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a) Calculate the molar enthalpy change for the reaction b) Calculate the minimum quantity of magnesium required to ensure it is in excess. c) Calculate the temperature change if only 0.8 g of magnesium is added. 3. When 5.73 g of sodium chloride (NaCl) dissolves in 100 cm3 of water‚ the temperature of the water fell from 22.4 oC to 19.8 oC. Calculate the enthalpy change of the reaction. 4. When 2.3 g of magnesium chloride dissolves in 200 cm3 of water‚ the temperature rose by
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47-51 of Bio 203L Lab Manual. Results/ Discussion: The effect of the chloride ion on its rate of diffusion in agar means it will diffuse faster because it is more concentrated. We graphed the diffusion of chloride ions rather than sodium ions because while the sodium ions do diffuse in to the agar the chloride ions from the NaCl will combine with silver ions‚ from AgNO3 and create a white precipitate. The white precipitate allows us to measure the the white band through the agar at different points
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3.13 Test for Saponin glycosides 2g of dried plant powder add 5ml of distill water shake vigorously. Persistent of foam indicate the presence of saponin glycosides. 3.14 Test for Steroids (Salkowski’s test) 5ml of test solution add with 2ml of Chloroform and then Conc. H2SO4 sides of the test tubes. The appearance of color at the interface indicates the presence of steroids. 3.15 Experimental Procedure for preliminary Phytochemical studies (Brindha et al.‚ 1981) S. No Name of the Test
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max expansion‚ covers more surface area‚ lays flatter on surfaceWater: did not form even circle‚ reached max expansion very quickly‚ formed more of a bubble on surfaceBOTH: Clear and shiny | e) Making iodine in water: potassium iodide (KI)‚ and sodium hypochlorite (bleach‚ NaClO). | KI- ClearNaOCl- ClearWhen combined they solution immediately turned a bright yellow. |
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theoretical yield of the product if 1.20 moles of aluminium and 2.40 moles of iodine are used. (Ans: 489.218 g) (b) Calculate the percentage yield of the product if 450 g of AlI3 is obtained. (Ans: 91.98%) 4. A salt solution is produced when 2.9 g of sodium chloride‚ NaCl dissolved in 200 ml of water. Calculate the molality (m) of the NaCl solution‚ given that the density of water is 1.00 g ml⁻1. (Ans: 0.25 m) [Apr 2013] 1 5. Carbon and hydrogen atoms make up a compound which yields 0.345 g of carbon
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ion was dropped into the test solution containing the calcium ion‚ the calcium ion was no longer attracted to the water‚ but was instead attracted to the oxalate. This attraction resulted in a compound that is insoluble in water‚ a precipitate. The sodium and chloride ions are soluble in water‚ so they did not experience a change. In the Home 1 water sample‚ there were medium amounts of the calcium ion‚ medium to high amounts of the chloride ion‚ and a small amount of the sulfate ion. Comparatively
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