complex and bind more tightly to EBT than Ca2+ b. What is the color change at endpoint? The color change at endpoint is sky blue/light blue. 5) A 50.0 ml water sample requires 16.33 ml of 0.0109 M Na2H2Y to reach EBT endpoint. a. Calculate the moles of hardening ions in the water sample. 1 mol of Na2H2Y = 1 mol of CaCO3 If mol of Na2H2Y = 0.0109 M x 0.01633 L = 1.78 x 10-4 mol then there are 1.78 x 10-4 mol of CaCO3 b. Express the hardness concentration in mg CaCO3 / L sample. (1.78
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Making up Hydrogen Peroxide Volume required 250 cm3 and concentration required 0.1 moldm-3 Given concentration of H2O2 = 1.7 moldm¬-3 Number of moles (n¬¬¬¬¬¬¬¬¬¬¬) = Concentration (moldm-3) x Volume (dm-3) = 0.1 x 0.25 = 0.025 mol Volume (dm-3) = Number of Moles (n) X 1000 Concentration (moldm-3) = (0.025/1.7) x 1000 = 14.7 cm3 Distilled water required: 250 cm3 – 14.7
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smallest particle‚ mole. Measuring the mass was a primary difficulty at that time since one mole of a substance was unable to weigh without using developed technology. Even though‚ it was clear that everything was made out of a small unit‚ there was no evidence that could determine what it was and how to measure it. So‚ as most scientific ideas had to be created with indirect evidence‚ it was difficult for theories to develop further. Therefore‚ the discovery of the mass of one mole‚ Avogadro’s number
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KMnO4 dispensed 28.5ml 26.3ml Moles of KMnO4 5.7E-4 5.26E-4 Moles of C2O42- 1.425E-3 1.315E-3 Table 2: (Second Titration) C2O42- Analysis Sample 1 Sample 2 Molarity of KMnO4 0.02m 0.02m Weight of Sample 0.237g 0.225g Final Buret Reading 4.2ml 4.5ml Initial Buret Reading 0ml 0ml Volume of KMnO4 dispensed 4.2ml 4.5ml Moles of KMnO4 8.4E-5 9.0E-5 Moles of C2O42- 2.1E-4 2.25E-4 Calculations: 1.) 2 KMnO4 + 5 K2C2O4 + 8 H2SO4 = 2 MnSO4 + 10 CO2 + 8 H2O + 6 K2SO4 Find moles of C2O42- Trial 1: KMnO4 =
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Topic 1.2 AMOUNT OF SUBSTANCE The mole Reacting masses and atom economy Solutions and titrations The ideal gas equation Empirical and molecular formulae Ionic equations Mill Hill County High School THE MOLE Since atoms are so small‚ any sensible laboratory quantity of substance must contain a huge number of atoms: 1 litre of water contains 3.3 x 1025 molecules. 1 gram of magnesium contains 2.5 x 1022 atoms. 100 cm3 of oxygen contains 2.5 x
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Rutherford Gold-Foil Experiment Shot alpha particles at gold foil; the particles either went straight through‚ or bounced back at large angles Atoms‚ Molecules‚ Formula Units‚ Mole‚ Mole Conversions‚ Percent Composition Identifying atoms‚ molecules‚ and formula units Atoms Molecules Formula units Understanding one mole
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Experiment 5- Standardization of NaOH and determination of Molarity of an unknown Acid Objectives 1. Preparation and standardization of a 0.1M NaOH solution 2. To learn the technique of titration 3. Determination of the concentration of an unknown diprotic acid. Introduction Titration can be traced to the origins of volumetric analysis‚ which began in the late eighteenth century. Study of analytical chemistry began in France and the first burette was made by Francois Antoine Henri
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calcium in water (b) Adding silver nitrate solution to sodium chloride solution A36.2 (p.36-7) (a) Rate of formation of NO2(g) == 0.24 mol dm–3 s–1 (b) From the equation‚ mole ratio of NO2 to N2 = 2 : 1. Rate of consumption of N2(g) =× Rate of formation of NO2(g) =× 0.24 mol dm–3 s–1 = 0.12 mol dm–3 s–1 (c) From the equation‚ mole ratio of NO2 to O2 = 2 : 2 = 1 : 1. Rate of consumption of O2(g) = Rate of formation of NO2(g) = 0.24 mol dm–3 s–1 A36.3 (p.36-12) (a) Instantaneous rate of reaction
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Titration and Volumetric Analysis The purpose of this experiment is to determine the [NaOH] of a solution by titrating it with standard HCl solution‚ to neutralize a known mass of an unknown acid using the NaOH solution as a standard‚ to determine the moles of NaOH required to neutralize the unknown acid‚ and to calculate the molecular mass of the unknown acid. Procedure: Part A: Standarized 0.10M HCl solution and unknown NaOH solution were poured into two beakers. The burets were then filled with
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^-4 in HCl. Solution: pH = -log [H3o+] pH= -log[5.4 x 10 ^-4 ] = 3.27 pNa = -log [2.00 x 10^-3 ] pNa = 2.669 For pCl = -log[Cl NaCl] + -log[ClHCl] pCl = 2.595 Concentration of Solutions 2. Normality (N) is the ratio between the no. of moles of solute in one equivalent per liter of solution. • An equivalent is the ratio between Reacting units can be a: the molar mass of the substance •Base‚ Acid or salt and the Avogadro’s number of the reacting
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