How many moles of Ar gas are present in a container with a volume of 78.4 L at STP?
1 answer:
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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.
Explanation:
Expression for the speed is as follows.
where, = root mean square speed
k = Boltzmann constant
T = temperature
M = molecular mass
As the molecular weight of oxygen is 0.031 kg/mol and R = 8.314 J/mol K. Hence, we will calculate the value of as follows.
=
= 498.5 m/s
Hence, for oxygen atom is 498.5 m/s.
For nitrogen atom, the molecular weight is 0.028 kg/mol. Hence, we will calculate its speed as follows.
=
= 524.5 m/s
Therefore, speed for nitrogen is 524.5 m/s.