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:
This question appears incomplete
Explanation:
This question appears incomplete because the data provided only makes it possible to calculate the certainty of the acetic acid content per total volume of the vinegar. Thus, the 4% means for every 100 mL of the vinegar, there is 4 mL of acetic acid present. To calculate the volume of acetic acid in any other volume of vinegar, the formula will be
volume of acetic acid = 4/100 × total volume of vinegar
Answer:
Explanation:
Hello there!
In this case, sine the solution of this problem require the application of the Raoult's law, assuming heptane is a nonvolatile solute, so we can write:
Thus, we first calculate the mole fraction of chloroform, by using the given masses and molar masses as shown below:
Therefore, the partial pressure of chloroform turns out to be:
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1 --> Jaws
2 --> Four limbs
4 --> Mammary & fur
5 --> Walking on two legs
