None. It's the O horizon. Compromises of mostly decomposed plant litter.
Answer:
Five peaks
Explanation:
GaAs would show two peaks:
Ga-69 + As-75 = 144
Ga-71 + As-75 = 146
Their intensities are in the ratio 60 : 40 = 100:67.
Ga₂S₂ would show three peaks:
Ga-69 + Ga-69 + 2As-75 = (Ga-69)₂(As-75)₂ = 288
Ga-69 + Ga-71 + 2As-75 = (Ga-69)(Ga-71(As-75)₂ = 290
Ga-71 + Ga-71 + 2As-75 = (Ga-71)₂(As-75)₂ = 292
Their relative intensities are in the ratio
(60 + 40)² = 60² : 2×60×40 : 40² = 3600 : 4800 : 1600 = 75 : 100 : 33
A spectrum of a mixture of GaAs and Ga₂As₂ would show five peaks:
Two at m/e 144 and 146 with intensity ratios 100: 67 and
Three at m/e 288, 290, and 292 with intensity ratios 75:100 :33.
We cannot predict the absolute sizes of the peaks because we do not know the relative amounts of GaAs and GaAs₂.
Given :
0.00072 M solution of
at
.
To Find :
The concentration of
and pOH .
Solution :
1 mole of
gives 2 moles of
ions .
So , 0.00072 M mole of
gives :
![[OH^-]=2 \times 0.00072\ M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D2%20%5Ctimes%200.00072%5C%20M)
![[OH^-]=0.00144\ M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.00144%5C%20M)
![[OH^-]=1.44\times 10^{-3}\ M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.44%5Ctimes%2010%5E%7B-3%7D%5C%20M)
Now , pOH is given by :
![pOH=-log[OH^-]\\\\pOH=-log[1.44\times 10^{-3}]\\\\pOH=2.84](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D%5C%5C%5C%5CpOH%3D-log%5B1.44%5Ctimes%2010%5E%7B-3%7D%5D%5C%5C%5C%5CpOH%3D2.84)
Hence , this is the required solution .