FACULTY OF RESOURCE SCIENCE AND TECHNOLOGY DEPARTMENT OF CHEMISTRY STK 1094 – Analytical Chemistry 1 EXPERIMENT NO : 1 TITLE OF EXPERIMENT : ACID – BASE TITRATIONS DATE OF EXPERIMENT : 27 SEPTEMBER 2013 GROUP MEMBERS & MATRIX NUMBERS : 1. DELVINO DISONNEO ANAK DALIN (44903) 2. GOH CHIA HONG (44927) 3.LINDA CHONG (41889) 4.LIM WHYE KIT LEONARD(41874) 5. LAB FACILITATOR : REPORT DUE DATE : 4 OCTOBER 2013 INTRODUCTION The laboratory
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023 moles of nitrogen gas in a 4.2 L container at a temperature of 20.C? 2. Oxygen gas is collected at a pressure of 123 kPa in a container‚ which has a volume of 10.0 L. What temperature must be maintained on 0.500 moles of this gas in order to maintain this pressure? Express the temperature in degrees Celsius. 3. How many moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100. C? After determining the number of moles‚ calculate
Free Gas Pressure Carbon dioxide
(iii) How much excess reagent remains after the reaction is complete? Answers: (i) Solution for determining the limiting reagent Determine the moles of Al2S3 and H2O aluminum sulfide: 15.00 g ÷ 150.158 g/mol = 0.099895 mol water: 10.00 g ÷ 18.015 g/mol = 0.555093 mol Divide each mole amount by equation coefficient aluminum sulfide: 0.099895 mol ÷ 1 mol = 0.099895 water: 0.555093 mol ÷ 6 mol = 0.0925155 which is less than 0.099895 (for aluminum sulfide) Therefore
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Chapter 3: Stoichiometry 3: Stoichiometry 5: Thermochemistry 8: Covalent Bonding and Molecular Structure 15: Chemical Equilibrium 16: Acids and Bases 3.2 Stoichiometry and Compound Formulas 3.1 The Mole and Molar Mass 3.2 Stoichiometry and Compound Formulas 3.3 Stoichiometry and Chemical Reactions 3.4 Stoichiometry and Limiting Reactants 3.5 Chemical Analysis Chapter Summary Chapter Summary Assignment Reference Tools Periodic Table Molarity Calculator Molar Mass Calculator Unit
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Mass of empty bottle (g) 21.5924 Mass of ascorbic acid used (g) 0.1002 Number of moles: (mass of ascorbic acid used)/(molecular mass of ascorbic acid) = 0.1002g/(176.10gmol-1) = 5.69 x10-4 moles M1: concentration of stock ascorbic acid V1: volume in volumetric flask with stock solution M2: concentration of diluted ascorbic acid V2: volume in volumetric flask with diluted solution (M1 x 100)/1000 = 5.69 x10-4 moles M1: 5.69 x10-3 mol/L M1V1 = M2V2 5.69 x10-3 mol/L x (10/1000) = M2 x (100/1000)
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Heats of Reaction – Hess’s Law Overview Hess’s Law states that the energy change for a reaction depends on the enthalpy of the reactants and products and is independent of the pathway of the reaction. In this experiment you will use calorimetry to measure the heats of reaction for three reactions: Reaction 1 Reaction Equation NaOH(s) → Na+(aq) + OH-(aq) Solid NaOH is dissolved in water 2 Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → Na+(aq) + Cl-(aq) + H2O(l) Solutions of NaOH and HCl are mixed 3 NaOH(s)
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molecules Mass number & atomic number Atomic structure (neutrons‚ protons & electrons) Mole concept & conversion Avogadro’s concept Empirical & molecular formulae Isotopes FHSC1114 Physical Chemistry Centre for Foundation Studies‚ UTAR Objectives To define relative atomic masses of atoms & molecules To define & determine mass no. & atomic no. To determine no. of neutrons‚ protons & electrons To understand mole concept & Avogadro’s concept To determine the empirical & molecular formulae Atomic
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performed. The heat given off by these two reactions was used to determine the stoichiometric ratio and the limiting reactants in each experiment. Introduction: Coefficients in a balanced equations show how many moles of each reactant is needed to react with each other and how many moles of each product that will be formed. Stoichiometry allows us to calculate the amount of reactants needed and also the amount of product. The major basis of stoichiometry is formed by the law of definite proportions
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of NaOH used Average volume of NaOH Calculations 1. Moles of NaOH = 0.500 mol/L X 0.010525 L = 5.26 x 10-3 mol 2. NaOH + HCl --> H2O + NaCl NaOH: HCl 1:1 ratio Moles of HCl = Moles of NaOH = 5.26 X 10-3 mol 3. Molarity of HCl = 5.26 x 10-3 mol / 0.010 L = 0.526 M 4. Moles of NaOH = 0.500 mol/L x 0.016275 L = 8.14 X 10-3 mol 5. CH3COOH + NaOH --> H2O + CH3COONa CH3COOH: NaOH 1:1 ratio Moles of CH3COOH = Moles of NaOH = 8.14 X 10-3 mol 6. Molarity of CH3COOH = 8.14 x 10-3
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Studies 265747595250 Finding suspicious moles or skin cancer early is the key to treating skin cancer successfully.Examining yourself is usually the first step in detecting skin cancer. You are a family physician and you have a few patients coming in to have their moles observed. For each patient use the ABCD chart to identify whether the mole is suspicious. Write a summary in the space provided‚ answering the following: 1. Is there a possibility the mole is cancerous and why/why not? 2. What
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