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    PreLab assesment

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    of HCL that is neutralized by 30.5 mL of 0.50 M NaOH:  H30+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq) --> Na+ (aq) + Cl (aq) + H20 (l))  0.50 M means 0.50 mol/L --> 0.50m mol/mL you added 30.5 mL so:  0.50 *30.5 = 15.25 mmol NaOH.  This means there is also 15.25 mmol HCl in your original solution.  So there is 15.25 mmol / 25 mL --> 15.25/25‚0=0‚ mmol/mL --> 0‚61 mol/L  ANSWER: 0.61 M HCl  6. Calculate the molarity of 15 mL of NaOH that is neutralized by 38.8 ML of 0.20 M HCl 0.20 m HCL 0.20

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    5. Write a balanced reaction of zinc with HCl. 6. How many moles of HCl are needed to react completely with all of the zinc in a post 1982 penny? 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 complete reaction‚ the solution contains twice as many moles of HCl that is actually needed. What concentration of HCl should be used? 8. In the scenario

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    question: To what extent does the concentration of hydrochloric acid affect the rate of the following reaction: 2 HCl(aq) + CaCO3(s) → CaCl2(aq) + H2O(l) + CO2(g) Data Collection and Processing: Table1: Different volumes of Co2 gas produced by Different concentrations of HCL acid. Volume of CO2gas formed from 5 different concentrations of HCL acid ±0.5ml 5 different concentrations of HCL acid (Mol) ±0.5ml Time (sec) ±0.1 0.25mol 0.50mol 1.0mol 1.5mol 2.0mol 0.00 0.00 0.00 0.00 0.00 0.00 10 2.00 5.00

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    Antacid Stomach Lab

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    results among the four brands we tested‚ with 0.011 mols of HCl. Introduction: Stomach acid‚ or gastric acid‚ is the chief cause of the discomfort associated with heartburn.

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    Student

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    excess hydrochloric acid is neutralized by the antacid. Different antacids use different metal hydroxides‚ such as Al(OH)3 or Mg(OH)2. The general formula for this reaction is: M(OH)y + Y HCl [pic] Y H2O + M+y + Y Cl- Additionally‚ some brands of antacids use calcium carbonate as a neutralizing reagent. CaCO3 + 2 HCl (aq) [pic] H2CO3 (aq) + Ca2+ (aq) + 2 Cl-(aq) The carbonic acid formed in this reaction may undergo further reaction: H2CO3 (aq) [pic] H2O (l) + CO2 (g) Procedure Outline This

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    Water Analysis

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    of this experiment was to determine the pH level of an unknown HCl solution. While doing so‚ we would also be receiving a good and fundamental understanding of how to use modern lab equipment to determine the chemical and physical properties of water bodies. Method: Make sure safety gear is always on and equipment is always clean. Part 1- Direct Analysis by pH Test Kit We grab a 50ml beaker and obtain a sample of unknown HCl as said to do in the procedure section in page 3. A pH test kit

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    Standardization of Solution and Titration Lab Report‚ Preparing a Dilute HCl Solution from a Concentrated One Titrating NAOH Solution with HCl Solution (of Known Concentration) Chemistry lab report(by abdazino abdalla) International College Objective preparing a dilute HCl solution from a concentrated one titrating NAOH solution with HCl solution (of known concentration) Procedure Section A: Preparation of 100.0cm3 0.480 mol/dm3 HCl solution a)Determine the volume of the concentrated acid needed

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    Chemical Equilibrium

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    Chemical Equilibrium: Le Chatelier Principle By Sarah Ramos and Kristina Todorovic Chemistry 203 DEN Dr. Mohamed El-Maazawi Part A. Acid-Base Indicators Purpose In this part of the experiment‚ we will find a reagent that will shift the acid-base equilibrium reaction described by Equation (2) in one direction and then a second reagent that will cause the equilibrium position to shift back in the opposite direction. Introduction An acid–base indicator

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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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    Back Titration

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    8950g ± 0.002g = 1.657 ± 0.017 mol/dm3 0.0056dm3 ± 0.00005dm3 × 204.23g /mole Concentration of NaOH = 1.657 ± 0.017 mol/dm3 Part B : Standardization of 2M HCl Data Collection : Volume of HCl in a conical flask (ml) ± 0.05ml Burette Reading of NaOH(ml) ± 0.05ml Volume of NaOH solution for neutralization End point of titration - 10ml ± 0.05ml Initial final = final – initial = (10.3 ± 0.05ml) – (0.0 ml ± 0.05) Colorless

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