Answer:
All three are present
Explanation:
Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: .
- Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
- Secondly, addition of liquid ammonia would form a precipitate with silver: ; Silver hydroxide at higher temperatures decomposes into black silver oxide: .
- Thirdly, we also know we have in the mixture, since addition of potassium chromate produces a yellow precipitate: . The latter precipitate is yellow.
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:steeper
Explanation:if it was me thats what i would pick
Wavelength is measured from crest to crest.
The answer is B. bases measure below seven
think of it like this 7 is neutral going down from seven it rises in acidity and going up from seven increases in alkalinity