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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Answer:
174 kPa
Explanation:
Given that,
Initial temperature, T₁ = 25° C = 25+273 = 298 K
Final temperature, T₂ = 225°C = 225 + 273 = 498 K
Initial pressure, P₁ = 104 kPa
We need to find the new pressure. The relation between the temperature and pressure is given by :
So,
or
P₂ = 174 kPa
So, the new pressure is 174 kPa.
Answer:
c) HCl and NaCl
Explanation:
Since all the solutions are on a 1:1 mole ratio the comparison is straight forward.
The lowest pH will be for solution c) which has a strong acid, HCl, which ionizes 100 % and the neutral salt NaCl (which is neutral since it is derived from the reaction of the strong acid HCl and the strong base NaOH).
Solutions a) and b) are buffers of the weak base NH₃ and its conjugate acid NH₄⁺ and weak acid H₃PO₄ and its conjugate weak base NaH₂PO₄ respectively.
Solution c) is a basic solution being a mixture of the weak base NH₃ and the strong base NaOH
Solution e) is a mixture of a weak base NH₂ and weak acid HC₂H₃O₂
I believe a Community Pharmacy
The answer should be A because the outer planets are just gas, therefore stuff would fall through it.
