Excited state , as stated in quantum mechanics
Answer:


Explanation:
Given that;
diameter (d) = 10cm/2 = 0.1m/2 = 0.05 m
length (l) = 10 cm = 0.1 m
porosity = 50%
height (h) = 30 cm = 0.3 m
time (t) = 5 s
volume (v) = 60 cm³ = 60 × 10⁻⁶ m³
Q (flow rate) = 
Q = 
Q = 
From constant head method, we use the relation K =
to determine the hydraulic conductivity ; we have:


Seepage velocity 
where; velocity = 
=
= 


Reorder 4Fe and 3O2.
3O2 + 4Fe
It’s baron I got it right on edg
Answer:
1.427x10^-3mol per L
Explanation:

I could use ⇌ in the math editor so I used ----
from the question each mole of Y(IO3)3 is dissolved and this is giving us a mole of Y3+ and a mole of IO3^3-
Ksp = [Y^3+][IO3-]^3
So that,
1.12x10^-10 = [S][3S]^3
such that
1.12x10^-10 = 27S^4
the value of s is 0.001427mol per L
= 1.427x10^-3mol per L
so in conclusion
the molar solubility is therefore 1.427x10^-3mol per L