Antibodies can destroy pathogens by (i) binding to and blocking the pathogen's receptors, thus causing neutralization of the pathogen, (ii) binding to the pathogen and activating complement, and (iii) binding to the pathogen and facilitating its opsonization and uptake by macrophages, which utilize their Fc receptors ...
pH of the acetyl choline solution before incubation = 7.65
![[H_{3}O^{+}]=10^{-7.65}=2.24*10^{-8}M](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%3D10%5E%7B-7.65%7D%3D2.24%2A10%5E%7B-8%7DM)
pH of the solution after incubation = 6.87
![[H_{3}O^{+}]=10^{-6.87}=1.35*10^{-7}M](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%3D10%5E%7B-6.87%7D%3D1.35%2A10%5E%7B-7%7DM)
The difference in concentration of hydronium ion before and after incubation
=
-
=
This difference in hydronium ion concentration can be attributed to the increase in the concentration of acetic acid, which is formed when acetylcholine is hydrolyzed by acetycholinesterase. The mole ratio of acetylcholine to acetic acid is 1:1.
Therefore the moles of acetylcholine = 
B.
Delta G is negative and Delta S is positive.
430 g of AgCl would be needed to make a 4.0m solution with a volume of 0.75 L.
<h3>What is Molarity?</h3>
- The amount of a substance in a specific volume of solution is known as its molarity (M).
- The number of moles of a solute per liter of a solution is known as molarity.
<h3>Calculation of Required amount of AgCl</h3>
Remember that mol/L is the unit of molarity (M).
We can compute the necessary number of moles of solute by multiplying the concentration by the liters of solution, according to dimensional analysis.
0.75L×4.0M=3.0mol
Then, using the periodic table's molar mass for AgCl, convert from moles to grams:
3.0mol×143.321gmol=429.963g
The final step is to round to the correct significant figure, which in this case is two: 430g.
Hence, 430 g of AgCl would be needed to make a 4.0m solution with a volume of 0.75 L.
Learn more about Molarity here:
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