[Co(CN)₆]³⁻ → Yellow
[Co(NH₃)₆]³⁺ → Orange
[CoF₆]³⁻ → Blue
Explanation:
- All the given compounds have octahedral geometry but the ligand in each are different with the same metal ion.
- Ligands strength order: CN⁻ > NH₃ > F⁻
- The ligand CN will act as a strong field ligand so that the splitting is maximum when compared to NH₃ and F⁻
- If the splitting is more, the energy required for transition is more, and the wavelength is inversely proportional to energy.
- So CN complex will absorb at lower wavelength (yellow color)
The number of formula units in 2.50 mol of the compound is 15.1 * 10^23.
The question is unclear whether NaNO2 or NaNO3 is implied. However,in either case, the solution applies equally.
6.02 * 10^23 formula units of the compound are contained in 1 mole
x formula units are contained in 2.5 moles of the compound
x = 6.02 * 10^23 formula units * 2.5 moles/ 1 mole
x = 15.1 * 10^23 formula units of the compound.
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Volume of Argon V1 = 5.0 L
Pressure of Argon P1 = 2 atm
Final temperature T2 = 30 C = 30 + 273 = 303 K
Volume at final temperature V2= 6 L
Pressure at final temperature P2 = 8 atm
We know that (P1 x V1) / T1 = (P2 x V2) / T2
(2 x 5)/ T1 = (8 x 6)/ 303 => T1 = (10 x 303) / 48
Initial Temperature T1 = 3030 / 48 = 63.12
Initial Temperature = -209. 8 C
When the Pka for formic acid = 3.77
and Pka = -㏒ Ka
3.77 = -㏒ Ka
∴Ka = 1.7x10^-4
when Ka = [H+][HCOO-}/[HCOOH]
when we have Ka = 1.7x10^-4 &[HCOOH] = 0.21 m
so by substitution: by using ICE table value
1.7x10^-4 = X*X / (0.21-X)
(1.7x10^-4)*(0.21-X) = X^2 by solving this equation for X
∴X = 0.0059
∴[H+] = 0.0059
∴PH= -㏒ [H+]
= -㏒ 0.0059
= 2.23