Answer:
Explanation:
Let the tyrosine molecule be represented by TH . It will ionise in water as follows
TH ⇄ T⁻ + H⁺
Let C be the concentration of undissociated TH and α be the degree of dissociation
TH ⇄ T⁻ + H⁺
c 0 0 ( before )
c( 1-α ) αc αc ( after ionisation)
Ka = α²c² / c( 1-α )
= α²c ( neglect α in the denominator as it is very small )
pKa = 10
Ka = 10⁻¹⁰
pH = 8.5
H⁺ = 10⁻⁸°⁵
αc = 10⁻⁸°⁵
α²c =Ka = 10⁻¹⁰
α x10⁻⁸°⁵ = 10⁻¹⁰
α = 10⁻¹⁰⁺⁸°⁵
= 10⁻¹°⁵ = 1 / 31.62
Percentage of dissociation = 100 / 31.62
= 3.16 %
percent of tyrosine side chains deprotonated
Answer:
34.92 grams NaCl (in 1 L of solution)
Explanation:
The chemical formula of sodium chloride is NaCl. From the formula, we can calculate the molar mass of NaCl:
MM(NaCl)= MM(Na) + MM(Cl) = 23 g/mol + 35.4 g/mol = 58.4 g/mol
A solution of NaCl with a molarity of 0.598 M has 0.598 moles of NaCl per liter of solution. So, we multiply the moles by the molar mass of NaCl to calculate the mass we need:
mass of NaCl = 0.598 mol x 58.4 g/mol = 34.92 g NaCl
Therefore, we need 34.92 grams of NaCl to prepare 1 liter of a solution with a molarity of 0.598 M.
Answer:
The 13.76% of propionic acid is in the dissociated form in the solution
Explanation:
Concentration of propionic acid = c = 0.61mM = 
Degree of dissociation = α
At initial
c 0 0
At equilibrium
c - cα cα cα
The value of dissociation constant of propionic acid = 
The expression of dissociation constant of propionic acid is given by :
Solving the equation for 
:
The 13.76% of propionic acid is in the dissociated form in the solution
Answer:
i think the answer is a. if not then c
Mass of substance in grams 60.1
