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    post lab 6

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    Jacob Hernandez October 23‚ 2014 Hima Lahari Marneni CHEM 1411. 114 Post Lab Report 6 Introduction The purpose of this lab is to determine the concentration of HCl when titrated with a standardized NaOH solution‚ through means of the titration method. My hypothesis is When an unknown concentration of HCl is titrated with a known volume of standardized NaOH solution‚ one can gather enough data to determine the concentration of the unknown (HCl). My prediction is If an unknown concentration of HCl

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    Potentiometric Titration of Sodium Carbonate Otieno O. Victor University of Detroit Mercy Quantitative Analysis Lab CHM 3880 Fall 2011 Partner: Edwin Gay Abstract The PH at each point during the titration of sodium carbonate unknown sample was determined. An Unknown sample of Na2CO3 was titrated with a standard HCL solution. In addition to titration‚ the pH at each point of titration was measured using PH meter. The % of the unknown Na2CO3 was 25.83% Introduction1 The purpose of this experiment

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    The chemical equation for Thiopental Sodium is C11H17N2NaO2S. First synthesized in 1935‚ the proper name for the drug is Thiopental Sodium it also goes by Sodium Pentothal but that is actually a trademark of the drug ’s manufacturer‚ Abbott Laboratories. A yellow water-soluble crystal‚ it is classified as an ultra-short acting barbiturate‚ a type of drug that depresses the central nervous system‚ slowing the heart rate and lowering blood pressure. Widely used in medicine as an anesthetic during surgery

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    SCIENCE SA 2 2013 14

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    OUR OWN HIGH SCHOOL‚ Al WARQA’A‚ DUBAI SUMMATIVE EXAMINATION-2‚ MARCH -2014 Name: ______________________________________ Grade: VII Sec: ____ M Marks: 90 Time: 2 1/2 hours Subject: SCIENCE Date: 09/04/2014 General Instructions: 1. This question paper has 4 pages. 2. All questions are compulsory. 3. Question numbers 1 to 10 are questions of one mark each. 4. Questions 11 to 16 are very short answer questions of two marks each. 5. Question numbers 17 to 24 are short answer questions of 3 marks each

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    AQA GCSE Chemistry Unit 1 C 1: Fundamental ideas: C 1.1. Atoms‚ elements and compounds: * All substances are made up of atoms. * Elements contain only one atom. * Compounds contain more than one atom. * An atom has a tiny nucleus in its centre‚ surrounded by electrons. C 1.2. Atomic structure: * Atoms are made up of protons‚ neutrons and electrons. * Protons and electrons have equal and opposite electrical charges. Protons are positively charged‚ and electrons are negatively

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    Iodometric Titration Lab

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    molarity and mass percent of an analysts in an commercial product. In this experiment my partner and I prepared 500mL of 0.07M sodium thiosulfate pentahydrate. As well‚ we added 0.05 grams of sodium bicarbonate and mixed it together with DI water. We got 60 mL of sodium thiosulfate solution in a clean 100-mL beaker. Then filled the buret to just above the 0 mL mark with sodium thiosulfate solution. Allowed a few milliliters to pour through the buret tip so that any trapped air can be flushed through

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    to a green and yellow color‚ and NaCl mixed with HCl changed to hold a slightly yellow tint. CuCo3 with NaOH changed to a darker color‚ also irreversible. These are all signs that a chemical change occurred. Classify the following properties of sodium metal as physical or chemical: • Silver metallic color: physical • Turns gray in air: chemical • Melts at 98oC:

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    Stoichiometry Lab Report

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    The decomposition of sodium carbonate is definitely‚ Na2CO3(s)+CO2(g)+H2O(g). The was the only equation that matched up exactly with my data in terms of percentage. To start off with‚ when I balanced out the equation‚ I got 2 NaHCO3→ 1 Na2CO3(s)+ 1 CO2(g)+ 1 H2O(g). Therefore when I set up my stoichiometry problem I got 3.2 grams NaHCO3 over 1 x 1 mol NaHCO3 over 84.007g NaHCO3 x 1mol Na2CO3 over 2 mol NaHCO3 x 105.987g Na2CO3 x 1 mol Na2CO3. Hence‚ I multiplied 3.2 x 1 x 1 x 105.987 and got 339

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    Lab report

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    Acid Base Titration Purpose: The purpose is to calculate the molarity of a NaOH solution by titrating the base with 5mL of standard HCl solution in each trial. By adding the base with unknown molarity to the acid with 0.10M the molarity of NaOH can be calculated. The base‚ NaOH‚ helps bring the pH of the acid‚ HCl‚ closer to seven‚ which neutralizes it. When using the buret the amount of NaOH used is able to be determined. Then by writing a balanced chemical equation and using the titration

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    Experimental Part 1: Reaction First the glassware apparatus for the reaction was set up. 3.861-grams of isoborneol‚ 2.21-mL of glacial acetic acid‚ and 4.39-mL of 6% NaOCl solution were mixed in a 125-mL Erlenmeyer flask. Another 35-mL of 6% NaOCl solution was added to a separatory funnel and supported over the flask. The NaOCl in the sep funnel was slowly added into the Erlenmeyer flask with vigorous swirling‚ approximately 2-mL every 30 seconds until all the NaOCl from the sep funnel had been

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