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Answer:
A. One unpaired electron
B. 5 unpaired electrons
Explanation:
In A ,Fe is in +3 oxidation state and Electronic configuration- [Ar]3d5
And NO2 is a strong field ligand hence it causes pairing in t2g orbitals and results one unpaired electron in dZX orbital.
In B, also Fe is in +3 oxidation state but F is weak field ligand hence causes no pairing of Electrons hence it results 5 unpaired electrons with electronic configuration t2g^3 eg^2
Answer:
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Explanation:
Answer:
examples of type of matter that <u>c</u><u>i</u><u>r</u><u>c</u><u>u</u><u>l</u><u>a</u><u>t</u><u>e</u><u> </u>through the environment are carbon, nitrogen, and water
Explanation:
Knowing the pH, you know the concentration of protons:
−log[H+]=pH=3.7
[H+]=10−3.7 M
Now, since the weak (monoprotic) acid dissociates into its conjugate base and a proton, the mols of protons are equimolar with the mols of conjugate base---the protons came FROM the weak acid, so the conjugate base that forms must be equimolar with the protons given out to the solvent.
HA⇌A−+H+
Hence, [A−]=[H+] in the same solution volume. Using the equilibrium constant expression, we get:
Ka=[H+]2eq[HA]eq
Don't forget that the HA form of HA had given away protons, so the mols of protons given away to generate A− is subtracted from the mols of (protons in) HA.
=[H+]2eq[HA]i−[H+]eq
=(10−3.7M)20.02M−10−3.7M
Ka=2.0105×10−6 M
