Chemistry Book Notes: Chapter 21: Buffers and the Titration of Acids and Bases 21-1 Henderson-Hasselbalch Equation THE HH EQUATION OFTEN CAN BE USED TO CALCULATE THE pH OF A BUFFER SOLUTION -buffer >a solution containing both a weak acid and its conjugate base can resist a change in pH by neutralizing either an added acid or an added base. Ex. acetic acid-acetate soln (acid with conj. Base) > Kc for a buffer reaction can = 1/Ka or 1/Kb if you add an acid or base because
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Experiment 12 Calorimetry and Heat of Reactions ____________________________________________________________________________________________________________ PERFORMANCE GOALS: 1. To learn how to use of a calorimeter 2. To learn how to collect and manipulate data in the computer 3. To calculate the calorimeter constant 4. To use Hess’ Law to find the heat or formation of magnesium oxide CHEMICAL OVERVIEW: • Enthalphy: (ΔH) : when chemical or physical changes occur at a constant
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dissolves in solution. A more a substance dissolves‚ the higher the Ksp value it has. In this experiment‚ a system of a sparingly soluble salt in water is studied. From the solubility information at various temperatures‚ the changes in standard enthalpy‚ standard entropy‚ and standard free energy were established. II. THEORETICAL BACKGROUND The reaction that is studied in this experiment is the dissolution of borax in water. “Borax” is a naturally occurring compound; it is in fact the most
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Law To find out the enthalpy change of Mg+ ½O2=MgO‚ we used a calorimeter‚ thermometer‚ 0.2 g of Mg‚ 0.2 g of MgO‚ and 2.0 M of HCl. We used a thermometer to measure the initial and final temperatures in Celcius. We recorded the initial temperature of the HCl. After we put the Mg or MgO into the calorimeter‚ we put in the HCl and covered it using a lid‚ mixed it around with the thermometer‚ and recorded the final temperature. We also used this method to find the enthalpies of the equations for the
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Sample Midterm Examination The midterm examination is two (2) hours long. In the actual examination you would not be allowed to consult your books or notes; however‚ you would be able to use a calculator. We strongly recommend that you sit this examination as if you were writing it in an invigilated setting. You will get a much better idea of your degree of preparedness for the actual examination. Part A: Short–answer Questions (1 mark each) 1. Calculate the density of bromine‚ given that 120.0
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Chemistry 105 A Final Exam 06/24/10 First Letter of last Name Dr. Jessica Parr Question 1 2 3 4 5 6 7 8 9 10 Maximum points 18 8 12 12 8 10 12 10 6 6 (Sub-T) PLEASE PRINT YOUR NAME IN BLOCK LETTERS Name: __________________________________________ Last 4 Digits of USC ID:_____ _____ _____ _____ Lab TA’s Name: _________________________________ Score (102) Grader Question 11 12 13 14 15 16 17 18 19 20 (Sub-T) TOTAL
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1. | | | What is the pressure‚ in atm‚ of a tank of gas with a regulator that reads 1250mmHg? | | | Student Response | Correct Answer | A. | 1.64 | | B. | 490 | | C. | 1.79 | | D. | 0.608 | | E. | 1.25 | | | Score: | 0/1 | | | 2. | | | A 147.9-L sample of dry air is cooled from 88.0°C to 22.1°C while the pressure is maintained at 2.85 atm. What is the final volume in L? (Do not type the units.) | | | Student Response | Correct
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| Experiment 2 | Calorimetry | | Chemistry 1310 | 7/21/2013 | | This experiment is to calibrate a constant pressure calorimeter to experimentally determine a series of heats of reaction that will be used to predict the enthalpy of reaction for another reaction using Hess’ Law and to determine heats of dissolution for a number of ionic salts that will be used to predict lattice energy again by using Hess’ Law. Heat may increase during experiment and undergo exothermic reaction
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Thermodynamics-Enthalpy of Reaction and Hess’s Law Purpose To demonstrate the principle of Hess’s Law and to find the heat capacity of the coffee cup calorimeter using three different reactions. Data Tave = (46.4-45.2)/2 = 45.8 qwater = -(100g)(4.184)(46.56-45.8) = -318 J Ccal = 318J/(46.56-21.2) = -12.53J/g*C Tinitial = (27.1+23.8)/2 = 25.45 qrxn = -(100g)(4.18)(38.43-24.45)+(-12.53x12.98) =-5400J/.1mol(1J/1000kJ) = -54.0 kJ/mol Tinitial = (26.0-24.5)/2 qrxn =-(100g)(4
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of the trend line for the other isomers to have a higher energy‚ while the less steep trend gradient means this switches over between 3 and 4 carbon‚ indicating that isomers with a carbon atom number greater than that will have increasingly lower enthalpy levels. This variance in results when performed under near identical experimental conditions‚ suggests that energy produced during the combustion reactions
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