2(22.99)+1(12.01)+3(16)g/mole = 0.2329g 105.99g/mole =2.197×10-3 moles From the equation ‚ 1 Mole Na2C03 react 2 mole HCl So‚ 2.197×10-3 moles Na2C03 = 2.197×10-3x 2 1 = 4.394x10-3 mole HCl Thus‚molarity HCl = n/v = 4.394x10-3 mole 42.87x10-3L = 0.1025 M. The suitable indicator is phenolphthalein. Primary standard
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acid‚ CH3COOH. • • • • • a strong acid: hydrochloric acid‚ HCl. an acidic commercial cleanser. a basic commercial cleanser. to use the titration curves to calculate the percent of the active ingredients in the commercial cleansers. to determine the Ka of a weak acid. Procedure Overview • • • • after calibration of the pH electrode and determination of the flow rate for the automatic titration‚ a NaOH solution is standardized against HCl. a pH titration curve for acetic acid is obtained and its pKa
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the coin using a triangular file. 2) Place the penny in a 50 mL of a pre-determined concentration of HCl solution overnight. 3) Then determine the percent copper from the mass of copper. 4) Find the percent of zinc by precipitation and titration 5) If there is a known amount of HCl added to the beaker that penny is placed in‚ then determine the amount of HCl that was used up in the reaction of HCl and Zinc. Calculate the % of zinc in penny. 6) If zinc reacts with NaOH in the titration experiment
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chloride (HCl) : Formula : HCl Nature : Covalent. Molecular Mass : 36·5u [i.e. HCl = 1 + 35·5 = 36·5] Occurrence : (i) Present in gastric juices. (ii) Present in volcanic gases. Laboratory preparation of HCl gas : Hydrogen chloride gas is prepared in laboratory by heating conc. H2SO4 with NaCl. NaCl + H2SO4 Heat → NaCl + NaHSO4 Heat → NaHSO4 + HCl Sodium hydrogen sulphate Na2SO4 + HCl Sodium sulphate (g) (g) ↑ ↑ Physical Properties of HCl : (i)
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a paper tray using an electronic scale. 2. Take (60 ml-80ml) of HCl and dissolve crushed eggshell in solution for eight to ten hours or until reaction between eggshell and acid is complete. 3. Rinse burette with NaOH solution and place in burette stand. 4. Rinse pipette with HCl. 5. Rinse conical flask with distilled water. 6. Filter acid and eggshell solution into beaker through funnel to remove unreacted HCl. (This step is necessary not for chemical reasons‚ but rather so that
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|(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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11.1 Washing with cold distilled water Washing process is not only remove soluble protein‚ fat‚ and pro-oxidant‚ but may also remove antioxidant. Therefore the oxidative stability of muscle ultimately depends upon the antioxidant/prooxidant balance that remains after the muscle is washed. It suggests that washing may promote oxidation in mince since it effectively removes antioxidants from the fish. In addition‚ the neutral lipids are more easily removed than polar lipids‚ phospholipids and free
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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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