First, we have to correct the equation in the question to b(g)⇆ 1/2 A(g)
at the first equation A(g)⇆ 2 B(g) so,
Kc = [B]^2 [ A] = 0.03
by reverse the equation 2B⇆ A
∴ Kc(original) = [A] / [B]^2
= 1/0.03 = 33 M^-1
and the new equation B⇆ (1/2) A
So, the new Kc = √Kc(original = √33
∴ KC = 5.7
Answer: His and Lys are deprotonated but Asp will be protonated.
Explanation:
As the pH is given as 7.4 and pK of His is given as 6.00. There will occur a positive charge on His when it's pH < pK therefore, it is neutral at the given pH.
As the pK value of Lys is 10.53 that is greater than the pH of 7.40. Therefore, charge on Lys is positive.
As the pK value of Asp is 3.65 which is less than the pH value of 7.40. Hence, Asp has a negative charge.
Therefore, we can conclude that His and Lys are deprotonated but Asp will be protonated.
Answer:
The cell membrane is the outer layer of the cell. The primary function of the cell membrane is to regulate what substances enter and leave the cell. The cell membrane is selectively permeable, or "semi-permeable", meaning that the membrane only allows certain substances to enter or leave the cell.
Explanation:
already had this question and this is the correct answer
The answer is; C
These lines that join areas of equal pressure during an air system are called isobars. The black lines in the diagram below are the isobars. They are good indicators of the strength of these systems. The innermost closed lines indicate the positions of relative maxima (high-pressure areas) and minima (low-pressure areas) in the pressure field.
