1 wavelength, 2 crest, 3 trough, 4 wave height <3
The answer for the following question is explained.
<u><em>Therefore the number of electrons present with the values n = 5, l = 2, m = -2, s = +1/2 is</em></u><u> </u><u><em>one(1).</em></u>
Explanation:
Here;
n represents the principal quantum number
l represents the Azimuthal quantum number
m represents magnetic quantum number
s represents spin quantum number
n = 5,
l = 2,
m = -2,
s = +1/2
Here, it implies 5d orbital.
In the 5d orbital, 10 electrons.
As the magnetic quantum number is -2, and so it can have 1 electron.
<u><em>Therefore the number of electrons present with the values n = 5, l = 2, m = -2, s = +1/2 is</em></u><u> </u><u><em>one(1)</em></u>
The concentration of [H3O⁺]=2.86 x 10⁻⁶ M
<h3>Further explanation</h3>
In general, the weak acid ionization reaction
HA (aq) ---> H⁺ (aq) + A⁻ (aq)
Ka's value
![\large {\boxed {\bold {Ka \: = \: \frac {[H ^ +] [A ^ -]} {[HA]}}}}](https://tex.z-dn.net/?f=%5Clarge%20%7B%5Cboxed%20%7B%5Cbold%20%7BKa%20%5C%3A%20%3D%20%5C%3A%20%5Cfrac%20%7B%5BH%20%5E%20%2B%5D%20%5BA%20%5E%20-%5D%7D%20%7B%5BHA%5D%7D%7D%7D%7D)
Reaction
HC₂H₃O₂ (aq) + H₂O (l) ⇔ (aq) + H₃O⁺ (aq) Ka = 1.8 x 10⁻⁵
![\tt Ka=\dfrac{[C_2H_3O^{2-}[H_3O^+]]}{[HC_2H_3O_2]}}\\\\1.8\times 10^{-5}=\dfrac{0.22\times [H_3O^+]}{0.035}](https://tex.z-dn.net/?f=%5Ctt%20Ka%3D%5Cdfrac%7B%5BC_2H_3O%5E%7B2-%7D%5BH_3O%5E%2B%5D%5D%7D%7B%5BHC_2H_3O_2%5D%7D%7D%5C%5C%5C%5C1.8%5Ctimes%2010%5E%7B-5%7D%3D%5Cdfrac%7B0.22%5Ctimes%20%5BH_3O%5E%2B%5D%7D%7B0.035%7D)
[H₃O⁺]=2.86 x 10⁻⁶ M
<span>AX(aq)+BY(aq)→no precipitate
AX(aq)+BZ(aq)→precipitate
this two equations imply
</span>
AX(aq) is soluble and <span>BY(aq) is insoluble
the answer is
</span><span>E. BY</span>
