Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is no
1 answer:
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Answer:
Explanation:
Hello,
In this case, for such formation of sulfur hexafluoride, the standard enthalpy of formation is -1220.47 kJ/mol (data extracted from NIST database). Next, we compute the moles in 10.0 grams of sulfur hexafluoride as shown below:
Next, for the given energy, we compute the total heat that is liberated:
Finally, we conclude such symbol has sense since negative heat is related with liberated heat.
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Answer:
A breakdown of the breaking buffer was first listed with its respective component and their corresponding value; then a table was made for the stock concentrations in which the volume that is being added was determined by using the formula . It was the addition of these volumes altogether that make up the 0.25 L (i.e 250 mL) with water
Explanation:
Given data includes:
Tris= 10mM
pH = 8.0
NaCl = 150 mM
Imidazole = 300 mM
In order to make 0.25 L solution buffer ; i.e (250 mL); we have the following component.
Stock Concentration Volume to be Final Concentration
added
1 M Tris 2.5 mL 10 mM
5 M NaCl 7.5 mL 150 mM
1 M Imidazole 75 mL 300 mM
. is the formula that is used to determine the corresponding volume that is added for each stock concentration
The stock concentration of Tris ( 1 M ) is as follows:
.
The stock concentration of NaCl (5 M ) is as follows:
.
The stock concentration of Imidazole (1 M ) is as follows:
.
Hence, it is the addition of all the volumes altogether that make up 0.25L (i.e 250 mL) with water.