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.
Answer:
Final concentration of NaOH = 0.75 M
Explanation:
For 
:-
Given mass = 90.0 g
Molar mass of NaOH = 39.997 g/mol
The formula for the calculation of moles is shown below:
Thus,
Molarity is defined as the number of moles present in one liter of the solution. It is basically the ratio of the moles of the solute to the liters of the solution.
The expression for the molarity, according to its definition is shown below as:
Where, Volume must be in Liter.
It is denoted by M.
Given, Volume = 3.00 L
So,
<u>Final concentration of NaOH = 0.75 M</u>
The answer is A. Water
Bronsted-Lowry base compounds are those that can accept protons
Bronsted-Lowry Acid Compounds are those that can recieve one
Water / H2O is an Amphoteric compund which mean that its molecul can act as a Base and Acid compound, so the answer is A.
Alloys are supposed to give greater strength to metals, which is why gold is mixed with others to make it harder. They have greater strength and are more resistant to erosion.
Answer:
5.25 moles.
Explanation:
The decomposition reaction of NaN₃ is as follows :
We need to find how many grams of N₂ produced in the process.
From the above balanced chemical reaction, we conclude that the ratio of moles of sodium azide and nitrogen gas are 2 : 3.
2 moles of sodium azide decomposes to give 3 moles of nitrogen gas. So,
3.5 moles of sodium azide decomposes to give 
moles of nitrogen gas.
Hence, the number of moles produced is 5.25 moles.
