Answer:
The answers are in the explanation
Explanation:
- Initial pH: An acid solution more dilute has a higher pH because concentration of H⁺ decreases.
- pH at the half‐equivalence point: In a titration curve. The pH at the half-equivalence point will be higher because the initial pH is higher and the equivalence point pH is the same.
- NaOH volume needed to reach the equivalence point: As the diulte solution has a higher pH, the NaOH volume you need is lower than original solution.
- pH at the equivalence point: The pH at the equivalence point will be always the same (pH = 7,0). Because is the pH where the total H⁺ of the acid were consumed.
I hope it helps!
P1v1=p2v2
47.5*125=66.2*v2
v2= p1v1/p2
=(47.5)(125)/(66.2)
=89.69ml
<span>The atomic number represents the equal number of protons and electrons in an element. Hope i helped
Cheers,
Belive1234
</span>
To be able to obtain the mass of the given volume of a substance, we need data on its density since it is the mass per volume of a substance. We multiply this value to the given volume. For water, it is approximately <span>10.02 lb/gal.
Mass of water = 200 gal ( 10.02 lb/gal ) = 200.4 lb
Hope this helped.</span>
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.
