|(Mg) | | |Trial 1 |Trial 2 |Trial 1 |Trial 2 | |Volume of 1.00 M HCl |50.0 mL ± 0.5 mL |50.0 mL ± 0.5 mL |50.0 mL ± 0.5 mL |50.0 mL ± 0.5 mL | |Final temperature‚ t2 |28.9(C ± 0.1(C |28.8(C ± 0.1(C |44.8(C ± 0.1(C |44.4(
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accomplish this the experiment was split into two parts; part A and Part B. in Part A of the experiment a standardized 0.05 M solution of HCl was titrated into a 25 mL solution of saturated Ca(OH)2 which contained 2 drops of orange methyl identifier. Once the titration began‚ the HCl was added until the methyl orange endpoint was reached‚ and as a result the volume of the HCl needed for the endpoint to be reached could be used in determining the moles and in turn the molar solubility and the solubility constant
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Summary The Project entitled “A Study on customer perception and effectiveness of student’s scheme on sale of laptop in Bangalore” with reference to HCL Infosystems was conducted in partial fulfillment of the Post Graduate Diploma in Management. The Study was undertaken with the objective of: Analyzing the customer perception regarding HCL Laptops and finding out the effectiveness of students scheme on sale of Laptops. The study has also taken into consideration the key demand drivers‚ and
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/ 63.54 g Cu = 0.00097 mol Cu 5. Write a balanced reaction of zinc with HCl. - Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g) 6. How many moles of HCl are needed to react completely with all of the zinc in a post 1982 penny? (2.43 g Zn x 1 mol) / 65.38 g Zn = 0.037 mol Zn (0.037 mol Zn x 2mol HCl) / 1 mol Zn = 0.074 mol HCl is needed 7. In a procedure developed to determine the percent zinc in post 1982 pennies‚ 50 ml of an HCl solution was used to react (dissolve) all of the zinc in the penny. To ensure
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If the temperature of an exothermic equilibrium system is increased‚ the equilibrium position shifts to use up the heat by producing more reactants. III. Materials: Reagents: Apparatus: * 0.4 M CoCl2 ő graduated cylinder * 0.5 M HCl ő
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Chemistry 104: Analysis of Commercial Antacid Tablets Hydrochloric acid (HCl) is one of the substances found in gastric juices secreted by the lining of the stomach. HCl is needed by the enzyme pepsin to catalyze the digestion of proteins in the food we eat. Heartburn is a symptom that results when the stomach produces too much acid (hyperacidity). Antacids are bases used to neutralize the acid that causes heartburn. Despite the many commercial brand‚ almost all antacids act on excess stomach
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Variables: Manipulated - Concentration of HCl Responding - Reaction time of HCl and Mg Controlled - Size of Mg piece‚ amount of HCl in test tube‚ room temperature‚ method of agitation. Materials List: 1. Test tube rack 2. Graduated Cylinder (10mL) 3. 3 Test Tubes 4. 50 mL beaker 5. 250 mL flask (for the HCl) 6. Dropper 7. Stopwatch 8. 17cm strip of Magnesium Ribbon 9. HCl (.5 molarity) 10. HCl (1 molarity) 11. HCl (2 molarity) 12. Goggles 13. Scissors/Knife (for cutting Mg)
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the reaction below. 2HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l) + HCl(aq) (in excess) (limiting reagent) (1) (unreacted) This reaction cannot be used directly to titrate the CaCO3 because it is very slow when the reaction is close to the endpoint. Instead the determination is achieved by adding an excess of hydrochloric acid to react with all of the CaCO3 and then titrating the remaining unreacted HCl with NaOH solution to determine the amount of acid which did not react with
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Concentrations of HCl on the Rate of Reaction Between HCl and Sodium Thiosulfate Chemistry Lab report Aim: The aim of this experiment is to determine how concentrations of HCl acid affect the rate of reaction when reacted with Sodium Thiosulfate (Na2S2O3). This experiment would require measuring the mass of the reaction over a period of time to be able to determine the rate of the reaction. Safety Precautions: In this experiment‚ SO2 gas is produced from the reaction between Sodium Thiosulfate and HCl‚ and this
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unrefined sodium carbonate‚ soda ash‚ a titration is done using a standardized solution of HCl. Aqueous HCl is a strong acid and therefore almost completely disassociates into H+ and CL-. Therefore‚ when HCl is used in a titration‚ the H+ is the titrant. Carbonate in aqueous solution is able to accept a proton‚ i.e. it acts as a base. When carbonate accepts the H+ a bicarbonate ion is formed. Na2CO3(aq) + HCl(aq) NaHCO3(aq) + NaCl(aq) This is not the complete reaction for the titration
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