Answer: 11, Na, 23, 100, −9.529 ... phosphorus, 15, P, 31, 100, −24.441 ... manganese, 25, Mn, 55, 100, −57.706.
Explanation: Make me Brainelist
Be= Beryllium= weight 9.01
Cl= Chlorine= weight 35.45
Since there are 2 Chlorine, you have to add Chlorine twice.
9.01 + 35.45 + 35.45= 79.91
Answer: 79.91
(This answer does not include sig.figs)
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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M1m1 = M2m2
where M1 is the concentration of the stock solution, m1 is the
mass of the stock solution, M2 is the concentration of the new solution and
m2 is its new mass.
M1m1 = M2m2
.925(m1) = .35(250)
m1 = 94.59 g
Mass= number of moles multiply Mass of one mole
= 24.3 multiplied by 4
= 96.2