The phenomenon known as "salting-out" occurs at very high ionic strengths, when protein solubility declines as ionic strength rises. As a result, salting out may be used to segregate proteins according to how soluble they are in salt solutions.
Because large levels of sodium chloride disturb the bonds and structure of the active site, the rate of enzyme activity will gradually decrease as the concentration of sodium chloride rises. As a result, some of the active sites get denaturized and the starch loses its ability to attach to them. As more enzymes get denatured and eventually cease to function, enzyme activity will steadily wane.
Here, we have to get the number of atoms present in the 100 plane of the FCC crystal lattice.
There will be 2 atoms in 100 plane of FCC crystal lattice.
In the face centered crystal (FCC) lattice there are atoms at each corner of the cube and each are shared by 4 another atoms. And an atom is present at the face of the crystal.
For the 100 plane of the Miller indices the intercepts are a, ∞, ∞ or 2a, ∞, ∞.
Thus, for the 4 atoms of the corner at the cube shared by 4 other atoms will contribute, 4 × 
= 1 and the un-shared atoms at the face will contribute another 1, which make the total atom 1 + 1 = 2.
Answer:
1.7 mL
Explanation:
<em>A chemist must prepare 550.0 mL of hydrochloric acid solution with a pH of 1.60 at 25 °C. He will do this in three steps: Fill a 550.0 mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (8.0 M) stock hydrochloric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated hydrochloric acid that the chemist must measure out in the second step. Round your answer to 2 significant digits.</em>
Step 1: Calculate [H⁺] in the dilute solution
We will use the following expresion.
pH = -log [H⁺]
[H⁺] = antilog - pH = antilog -1.60 = 0.0251 M
Since HCl is a strong monoprotic acid, the concentration of HCl in the dilute solution is 0.0251 M.
Step 2: Calculate the volume of the concentrated HCl solution
We want to prepare 550.0 mL of a 0.0251 M HCl solution. We can calculate the volume of the 8.0 M solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂/C₁
V₁ = 0.0251 M × 550.0 mL/8.0 M = 1.7 mL
The reaction CFCl3 + UV Light -> CFCl2 + Cl does not need another reactant as with CFCl3 because the reaction itself is reactive to light. Note that there are reactions that are sensitive to light to form products and when this type of reaction are not exposed to light, no reaction occurs.
The answer is option number one
