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.
In a polyatomic ion, the ate ending indicates one more oxygen than the ite ending. D. more.
First, we need to transform km to meters.
1 km = 1000 meters
2.6 * 1000 = 2600
Now, there are 7 days in a week.
Multiply.
2600 * 7 = 18,200
Final answer: 18,200 meters a week
Explanation:
The reaction equation will be as follows.
Hence, the expression for 
is as follows.
![K_{a} = \frac{[H_{2}SO^{-}_{4}][H^{+}]}{[H_{3}AsO_{4}]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BH_%7B2%7DSO%5E%7B-%7D_%7B4%7D%5D%5BH%5E%7B%2B%7D%5D%7D%7B%5BH_%7B3%7DAsO_%7B4%7D%5D%7D)
Let us assume that the concentration of both ![[H_{2}AsO^{-}_{4}]](https://tex.z-dn.net/?f=%5BH_%7B2%7DAsO%5E%7B-%7D_%7B4%7D%5D)
and ![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
is x.
x = 0.01118034
This means that the concentration of is 0.01118034.
Since, we know that the relation between pH and concentration of hydrogen ions is as follows.
pH = ![-log [H^{+}]](https://tex.z-dn.net/?f=-log%20%5BH%5E%7B%2B%7D%5D)
= 
= 1.958
Thus, we can conclude that the pH of a 0.500 M solution of arsenic acid is 1.958.
