Answer:
Incomplete question
Complete question:
8. Compartments A and B are separated by a membrane that is permeable to K+ but not to Na+ or Cl-. At time zero, a solution of KCl is poured into compartment A and an equally concentrated solution of NaCl is poured into compartment B. Which would be true once equilibrium is reached?
A. The concentration of Na+ in A will be higher than it was at time zero.
B. Diffusion of K+ from A to B will be greater than the diffusion of K+ from B to A.
C. There will be a potential difference across the membrane, with side B negative relative to side A.
D. The electrical and diffusion potentials for K+ will be equal in magnitude and opposite in direction.
E. The concentration of Cl- will be higher in B than it was at time zero.
Answer: D. The electrical and diffusion potentials for K+ will be equal in magnitude and opposite in direction.
Explanation:
Diffusion is the movement of molecules from region of higher concentration to lower concentration through a semipermeable membrane.
Since the k+ is the permeable membrane, the k+ ion in the KCl would move in equal magnitude and direction in the solution.
Answer:
The correct answer is AMP+H2O→ Adenosine + pi
Explanation:
The above reaction is least energetic because there is no phosphoanhydride bond present with adenosine mono phosphate.Phospho anhydride bond is an energy rich bond.
As a result hydrolysis of AMP generates very little amount of energy in comparison to the hydrolysis of ATP and ADP.
Answer:
![[base]=0.28M](https://tex.z-dn.net/?f=%5Bbase%5D%3D0.28M)
Explanation:
Hello,
In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:
![pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D10%5E%7BpH-pKa%7D%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D10%5E%7B4.9-4.76%7D%5C%5C%5C%5C%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%3D1.38%5C%5C%5C%5C)
![[base]=1.38[acid]=1.38*0.20M=0.28M](https://tex.z-dn.net/?f=%5Bbase%5D%3D1.38%5Bacid%5D%3D1.38%2A0.20M%3D0.28M)
Regards.
The answer to your question is a
Ksp - solubility product constant is equivalent to equilibrium constant, except this constant is used to determine the solubility of ions of a solid in a solution.
ksp is the product of the soluble ions in the compound. Higher the ksp value, higher the degree of solubility.
ZnCO₃ (s) ---> Zn²⁺ (aq) + CO₃²⁻ (aq)
n n
ksp = [Zn²⁺][CO₃²⁻]
In the equation equal amounts of ions Zn²⁺ and CO₃²⁻ ions are soluble.
amount of ions soluble = n
ksp is therefore equal to;
ksp = n x n
ksp = n²
ksp = 1 * 10⁻¹⁰ M
therefore
1 * 10⁻¹⁰ M = n²
n = 1 x 10⁻⁵ M
therefore concentration of CO₃²⁻ = 1 x 10⁻⁵ M
