/ 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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concentration of the components of a buffer. For a given addition of acid or base‚ the buffer component concentration ratio change less when the concentration are similar than what they are different. Materials: Acetic acid (MW=60 g/mol)‚ NaOH solution (0.5M)‚ HCl solution (0.001M)‚ calibration buffer (pH3.5 and pH 5.5)‚ 7-UP‚ 100 Plus. Apparatus: pH meter‚ pipettes (10mL)‚ volumetric flask (250mL)‚ beakers (150mL)‚ burettes‚ burette holder and stand‚ funnel‚ graduated
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The equilibrium constant was found by titrating a series of reactions containing H2O‚ HCl‚ and the unknown ester with only the last solution containing the unknown alcohol. Equilibrium constants can be easily found once the concentrations of all reactants are known using Keq= [ acid][alcohol]/[ester][water]. Procedure: Part A: Prepare six bottles with the following volumes in each: Bottle # 3M HCl in mL H2O in mL Ester in mL Alcohol in mL 1 5 5 0 0 1/A 5 5 0 0 2 5 0 5 0 3 5 1 4 0 4
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Irresistible? By Mikayla Messing 8/3/12 Section 623 Abstract To examine the effectiveness of buffers by titrating two sets of five different solutions using HCl and NaOH and monitoring the pH change of the various solutions. The data collected shows that the buffer systems made with sodium acetate and acetic acid were effect when titrated with the strong acid and the strong base. Comparison of all the solutions shows that the concepts of buffers holds true for the results from the experimentation
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standard solutions are used in determining the concentrations of other solutions to an extremely high accuracy. They are typically used in titrations and other analysis techniques as standardization solutions. A secondary standard solution‚ such as HCl solution‚ is a solution which must be standardized first against a primary standard‚ but afterwards‚ it will be stable enough for titrimetric work (Titration). Titration involves the gradual addition of a solution of accurately known concentration (standard
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gastric acid (HCl). In order to determine which antacid is the more effective‚ four different brands of antacids were chosen‚ and tested. The purpose of an antacid is to balance out the pH inside a person’s stomach. PH is the measurement of molar concentration of hydrogen ions that are present in the solution (N. Tro‚ 733-735). If this experiment were to be repeated‚ it would be to show which antacid would be more effective to buy as a consumer. In this experiment an antacid is dissolved in HCl‚ and then
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Heats of Reaction Lab Report Purpose: To measure the heats of reaction for three related exothermic reactions and to verify Hess’s Law of Heat Summation. NaOH(s) ( Na+(aq) + OH-(aq) ΔH = -10.6kcal/mol NaOH(s) + H+(aq) + Cl-(aq) ( H2O + Na+(aq) + Cl-(aq) ΔH = -23.9kcal/mol Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) ( H2O + Na+(aq) + Cl-(aq) ΔH = -13.3kcal/mol Background: Energy changes occur in all chemical reactions; energy is either absorbed or released. If energy is released in
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peroxide increased the rate of reaction increased as well. When the enzyme concentration increased the rate of catalase activity increased too. When catalase was subjected to an increase of temperature changes‚ the rate of reaction increased. Once the protein denatured around 100ºC the catalase activity decreased. Catalase operated with a high efficiency when the pH of the enzyme was 7. As the catalase pH was altered to acidic or basic‚ the rate of reaction decreased.
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solvents will be measured in order to explore the rate of reaction on different solvents’ effects Materials: Apparatus: 50.00 mL burette Chemicals: 0.04M NaOH 10.00 mL pipette t-butyl chloride 100 cm3 volumetric flask Acetone Conical flask Isopropanol Measuring cylinder Bromothymol blue Procedures: 1. 150 cm3 of 0.04M NaOH(aq) is placed in a beaker. 2. 100 cm3 of a 50/50 acetone/watermixture (by volume) is put into a stoppered flask or bottle and is mixed well. 3. 1.00 cm3 (±0
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9.5 – 5.0 = 4.5. Therefore 4.5 mL of NaOH was needed to neutralize HCl. Table 3: The amount of NaOH (mL) needed to neutralize HCl in Condition 1‚ 2 ‚ and 3 and their 3 trials. Trial 1 (mL) Trial 2 (mL) Trial 3 (mL) Condition 1 4.5 4.0 4.0 Condition 2 8.0 8.0 9.0 Condition 3 9.5 9.0 9.5 2. Sample Average Calculation of Condition 1: Table 4: The average volume (mL) of HCl & NaOH in each Condition Average Volume of HCl & NaOH Condition 1 9.2 Condition 2 16.5
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