test3s-1 - 4. We first compute the initial concentration of...

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Up: Back to the Chemistry 2000 test index Chemistry 2000, Section A Spring 1996 Test 3 Solutions 1. Denote the acid by HA. is the equilibrium constant for the dissociation . Since is small, very little of the lactic acid will dissociate so that . Also, one proton is produced for every lactate anion so we have 2. We first need to convert the pH and to an activity of and to a , respectively: We also want If we solve equation 1 for and substitute this result in equation 2 , we get
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3. Calcium fluoride is . is the equilibrium constant for the reaction Thus . Since two fluoride ions are liberated for every calcium ion, we have . Therefore
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Unformatted text preview: 4. We first compute the initial concentration of lactose: The equilibrium constant is quite large. This implies that almost all of the lactose will have dissociated at equilibrium, leaving us with 0.29mol/L each of glucose and galactose. As for the glucose, The equilibrium concentration of lactose is therefore . 5. The water starts out at its boiling point ( ) and cannot be heated beyond this point (provided it does not superheat). The piece of iron starts out at and ends up at . The heat balance equation is therefore where is the number of moles of water vaporized. Therefore Marc Roussel Wed Nov 6 07:49:00 MST 1996...
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This note was uploaded on 03/03/2012 for the course CHEM 2000 taught by Professor Roussel during the Fall '06 term at Lethbridge College.

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test3s-1 - 4. We first compute the initial concentration of...

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