In order to wash out the impurity, you looked up some sketchy method from 1952 which Hank Pym wrote during his Ph. D. under E. J
. Corey. The method requires you add HCl to the mixture. After you rid your sample of any impurities, you take some pH paper and find out your solution has a pH of 3.0. Assuming the pKa of Pym Particles is 4.5 and that the molar absorptivity of the protonated Pym Particle is 4.82 * 102 L/mol*cm and 8.54 * 103 L/mol*cm for 400 nm and 690 nm respectively, calculate the expected A400 and A690 of the solution.
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Answer: 502 Joules
Explanation:
To calculate the mass of water, we use the equation:
Density of water = 1 g/mL
Volume of water = 40.0 mL
Putting values in above equation, we get:
When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.
The equation used to calculate heat released or absorbed follows:
q = heat absorbed by water
= mass of water = 40.0 g
= final temperature of water = 20.0°C
= initial temperature of water = 17.0°C
= specific heat of water= 4.186 J/g°C
Putting values in equation 1, we get:
Hence, the joules of heat were re-leased by the lead is 502