Answer: I am unsure of what you mean.
Explanation:
Please explain better.
I believe this property is vital since the sticky nature allows the wells to coat easily with the desired antigen by simply adding a small amount of it and allowing it time to incubate. Adhesion is the tendency of dissimilar particles or surfaces to cling to one another ( cohesion refers to the tendency of similar or identical particles/surfaces to cling to one another).
Answer:
The 
expression for the weak base equilibrium is:
![K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
Explanation:
The expression of the equilibrium constant of base 
can be given as:
![K_c=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N][H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%5BH_2O%5D%7D)
]![K_b=K_c\times [H_2O]=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3DK_c%5Ctimes%20%5BH_2O%5D%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
As we know, water is pure solvent, we can put ![[H_2O]=1](https://tex.z-dn.net/?f=%5BH_2O%5D%3D1)
![K_b=K_c\times 1=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3DK_c%5Ctimes%201%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
So, the the expression for the weak base equilibrium is:
A chemestretic equation equation which is formed by h20 mc square hydrogen peroxide and the equation of cf6c7bu7c
Hope it helped
The reaction is:
NH4 (NO3) (s) ⇄ N2O (g) + 2 H2O (g)
This means that 1 mol of NH4 (NO3)s produces 3 moles of gases.
Now find the number of moles in 1.71 kg of NH4 (NO3)
Molar mass = 2*14g/mol + 4 * 1g/mol + 3*16g/mol = 80 g/mol
# moles = mass / molar mass = 1710 g / 80 g/mol = 21.375 mol of NH4(NO3)
We already said that every mol of NH4(NO3) produces 3 moles of gases, then the number of moles of gases produced is 3 * 21.375 = 64.125 mol
Now use the equation for ideal gases to fin the volume
pV = nRT => V = nRT / p = (64.125 mol)(0.082atm*liter / K*mol) * (119 +273)K / (731mmHg *1 atm/760mmHg) =
V = 2143.01 liters
