Let us see the structure of ascorbic acid
As shown there is no COOH group however the OH group can lose a proton and forms conjugate base
The conjugate base formed is stabilized due to resonance
More the stability of conjugate base more the strength of acid
Hence ascorbic acid behaves as an acid
1.94 moles
I did 35/18.02 because 18.02 is the molar mass of water
Answer:
The answer to your question is 33.4 ml
Explanation:
Data
volume 1 = V1 = 42 ml
temperature 1 = T1 = 20°C
temperature 2 = T2 = -60°C
Volume 2 = V2 = x
Process
1.- Convert celsius to kelvin
T1 = 20 + 273 = 293°K
T2 = -60 + 273 = 233°K
2.- Use the Charles' law to solve this problem

Solve for V2
V2 = 
3.- Substitution
V2 = 
4.- Simplification
V2 = 
5.- Result
V2 = 33.4ml
<span>Conductor, and there you go, i hope this helped but if its wrong, i am extremly sorry</span>
Answer: 0.25 mol
Explanation:
Use the formula n=N/NA
n= number of mols
N = number of particles
Nᵃ = Avogadros constant = 6.02 x
So, n=
The 10 to the power of 23 cancels out and you are left with 1.505/6.02, which is approximately 1/4. This is the same as 0.25 mol.
Hope this helped :)