Answer:
a.all four carbon atoms.
Explanation:
Acetyl-CoA labeled with 14C in both of its acetate carbon atoms is incubated with unlabeled oxaloacetate and a crude tissue preparation capable of carrying out the reactions of the citric acid cycle. After one turn of the TCA cycle, oxaloacetate would have 14C in all four carbon atoms.
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From the calculation as shpwn in the procedure below, the equilibrium constant of the substance is 6.9 * 10^-15.
<h3>What is equilibrium constant?</h3>
The equilibrium constant for the solubility of aa solid in solution is called the solubility product Ksp. The Ksp shows the extent to which a solid is dissolved in solution.
Given that;
Fe(OH)2 ⇄Fe^2+ + 2(OH)^-
Ksp = s(2s)^2
We have s as 1.2 x 10^-5 M
So
Ksp = 4s^3
Ksp = 4( 1.2 x 10^-5 )^3
Ksp = 6.9 * 10^-15
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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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Answer:
A) The solubility of NH3 increases, and the
solubility of KCl increases.
Explanation:
