Each of the Isotopic mass * its Abundance Sum together divide by 100.
(179.946706*0.12 + 181.948206*26.5 + 182.9502245*14.3 + 183.9509326*30.64 + 185.954362*28.43 ) / 100
You get 183.84 (5s.f. ) or (183.8417786)
_award brainliest if helped!
Answer is: 6.022·10²² molecules of glucose.
c(glucose) = 100 mM.
c(glucose) = 100 · 10⁻³ mol/L.
c(glucose) = 0.1 mol/L; concentration of glucose solution.
V(glucose) = 1 L; volume of glucose solution.
n(glucose) = c(glucose) · V(glucose).
n(glucose) = 0.1 mol/L · 1 L.
n(glucose) = 0.1 mol; amount of substance.
N(glucose) = n(glucose) · Na (Avogadro constant).
N(glucose) = 0.1 mol · 6.022·10²³ 1/mol.
N(glucose) = 6.022·10²².
Answer:
Ok:
Explanation:
So, you can use the Henderson-Hasselbalch equation for this:
pH = pKa + log(
) where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.
We can solve that
1 = log(
) and so 10 = or 10HA = A-. For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.
Hi there
Formula
M1V1=M2V2
(455*6)/2500
Answer
1.092 M
