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Magnetic moment (spin only) of octahedral complex having CFSE=−0.8Δo and surrounded by weak field ligands can be : Q
To answer this, the Crystal Field Stabilization Energy has to be calculated for a (d3 metal in both configurations. The geometry with the greater stabilization will be the preferred geometry. So for tetrahedral d3, the Crystal Field Stabilization Energy is: CFSE = -0.8 x 4/9 Δo = -0.355 Δo.
[Co(CN)64-] is also an octahedral d7 complex but it contains CN-, a strong field ligand. Its orbital occupancy is (t2g)6(eg)1 and it therefore has one unpaired electron. In this case the CFSE is −(6)(25)ΔO+(1)(35)ΔO+P=−95ΔO+P.
The crystal field stabilization energy (CFSE) (in kJ/mol) for complex, [Ti(H2O)6]3+. According to CFT, the first absorption maximum is obtained at 20,3000cm−1 for the transition.
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The answer is (2) 10.0 mol. The equation given is balanced. So the ratio of mol number of compound is the ratio of the number before the compound. The HCl and CO2 ratio is 2:1. So the answer is 10.0 mol.
Answer is: pKa for the
monoprotic acid is 5.<span>
Chemical reaction: HA(aq) </span>⇄ A⁻(aq) + H⁺<span>(aq).
c(monoprotic acid) = 0.100 M.
pH = 3.00.
[A</span>⁻] = [H⁺] = 10∧(-3).<span>
[A</span>⁻]
= [H⁺] = 0.001 M; equilibrium concentration.<span>
[HA] = 0.1 M - 0.001 M.
[HA] = 0.099 M.
Ka = [A</span>⁻]·[H⁺] / [HA].<span>
Ka = (0.001 M)² / 0.099 M.
Ka = 0.00001 M = 1.0·10</span>⁻⁵ M.
<span>pKa = -logKa = 4.99.
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