I’m not too sure I hope someone answers for you
Answer:
See explanation
Explanation:
Hello there!
In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:
Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:
![Ksp=[Ag^+][Cl^-]](https://tex.z-dn.net/?f=Ksp%3D%5BAg%5E%2B%5D%5BCl%5E-%5D)
And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:
I - 0 0
C - +x +x
E - x x
Which leads to the following modified equilibrium expression:
Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.
Regards!
Answer:
passively diffuses down its concentration gradient through the endothelial cell plasma membrane out of the cell and then passiveley diffuses through the plasma membrane into the cytoplasm of the smooth muscle cell, where it acts to decrease contraction.
Explanation:
Hello,
At first, we must consider that 
and 
, passively diffuses through membranes. As it is produced by an enzyme and accumulates in the endothelial cell cytosol, passively diffuses down its concentration gradient through the endothelial cell plasma membrane out of the cell and then passiveley diffuses through the plasma membrane into the cytoplasm of the smooth muscle cell, where it acts to decrease contraction.
Best regards
Answer:
0.259 kJ/mol ≅ 0.26 kJ/mol.
Explanation:
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = ??? J).
m is the mass of the ice (m = 100.0 g).
c is the specific heat of water (c of ice = 4.186 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 21.56°C - 25.0°C = -3.44°C).
<em>∵ Q = m.c.ΔT</em>
∴ Q = (100.0 g)(4.186 J/g.°C)(-3.44°C) = -1440 J = -1.44 kJ.
<em>∵ ΔH = Q/n</em>
n = mass/molar mass = (100.0 g)/(18.0 g/mol) = 5.556 mol.
∴ ΔH = (-1.44 kJ)/(5.556 mol) = 0.259 kJ/mol ≅ 0.26 kJ/mol.
Answer:
is capable of combining with oxygen to form iron oxide
