The concentrations : 0.15 M
pH=11.21
<h3>Further explanation</h3>
The ionization of ammonia in water :
NH₃+H₂O⇒NH₄OH
NH₃+H₂O⇒NH₄⁺ + OH⁻
The concentrations of all species present in the solution = 0.15 M
Kb=1.8 x 10⁻⁵
M=0.15
![\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D%5Csqrt%7BKb.M%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.15%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B2.7%5Ctimes%2010%5E%7B-6%7D%7D%3D1.64%5Ctimes%2010%5E%7B-3%7D)
![\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21](https://tex.z-dn.net/?f=%5Ctt%20pOH%3D-log%5BOH%5E-%5D%5C%5C%5C%5CpOH%3D3-log~1.64%3D2.79%5C%5C%5C%5CpH%3D14-2.79%3D11.21)
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Answer:
the original answer is 38.9km (3sf)
Answer:- The natural abundance of
is 0.478 or 47.8% and
is 0.522 or 52.2% .
Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:
Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)
We have been given with atomic masses for
and
as 150.919860 and 152.921243 amu, respectively. Average atomic mass of Eu is 151.964 amu.
Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of
as n then the abundance of
would be 1-n .
Let's plug in the values in the formula:

151.964=150.919860n+152.921243-152.921243n
on keeping similar terms on same side:


negative sign is on both sides so it is canceled:



The abundance of
is 0.478 which is 47.8%.
The abundance of
is = 
= 0.522 which is 52.2%
Hence, the natural abundance of
is 0.478 or 47.8% and
is 0.522 or 52.2% .
In order to find the answer, use an ICE chart:
Ca(IO3)2...Ca2+......IO3-
<span>some.......0..........0 </span>
<span>less.......+x......+2x </span>
<span>less........x.........2x
</span>
<span>Ca(IO₃)₂ ⇄ Ca⁺² + 2 IO⁻³
</span>
K sp = [Ca⁺²][IO₃⁻]²
K sp = (x) (2 x)² = 4 x³
7.1 x 10⁻⁷ = 4 x³
<span>x = molar solubility = 5.6 x 10</span>⁻³ M
The answer is 5.6 x 10 ^ 3 M. (molar solubility)