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kotegsom [21]
1 year ago
13

Calculate the cfse for a d^3 system in an octahedral field in units of ∆_o. In other words, do not enter "∆_o" with your answer.

For example, if your answer is −0. 8∆_o, then enter "−0. 8. ".
Chemistry
1 answer:
Rudik [331]1 year ago
8 0

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.

To learn more about crystal field stabilization energy  visit:brainly.com/question/29389010

#SPJ4

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Question 7 of 15
Crazy boy [7]

Answer: 0.4 moles

Explanation:

Given that:

Volume of gas V = 11L

(since 1 liter = 1dm3

11L = 11dm3)

Temperature T = 25°C

Convert Celsius to Kelvin

(25°C + 273 = 298K)

Pressure P = 0.868 atm

Number of moles N = ?

Note that Molar gas constant R is a constant with a value of 0.00821 atm dm3 K-1 mol-1

Then, apply ideal gas equation

pV = nRT

0.868atm x 11dm3 = n x (0.00821 atm dm3 K-1 mol-1 x 298K)

9.548 atm dm3 = n x 24.47atm dm3mol-1

n = (9.548 atm dm3 / 24.47atm dm3 mol-1)

n = 0.4 moles

Thus, there are 0.4 moles of the gas.

3 0
3 years ago
A gas quickly expands in an isolated environment. During the process, the gas exchanges no heat with its surroundings. The proce
AysviL [449]

Answer:

e. adiabatic process

Explanation:

Adiabatic process  -

In the thermodynamic system , an adiabatic process is the one which involves no transfer of mass or heat of the substance , is referred to adiabatic process.

In this process , the temperature need not be constant ,

But only the heat is transferred into or out of the system .

Hence, from the given information of the question,.

The correct option is e. adiabatic process .

6 0
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When a magnesium ribbon is heated in air the product former is heavier.on the other hand when potassium manganate(VII) is heated
vesna_86 [32]

Answer: On heating, Magnesium forms its oxide; while potassium manganate(VII) decomposes

Explanation:

Magnesium Mg, on heating forms Magnesium oxide

2Mg(s) + O2(g) --> 2MgO

Potassium permanganate KMnO4, on heating decomposes to potassium manganate K2MnO4, manganese dioxide MnO2, and Oxygen gas O2.

2KMnO4 --> K2MnO4 + MnO2 + O2

The difference in observation is that, on heating, Magnesium forms its OXIDE as product; while potassium manganate(VII) decomposes, giving OFF most of its constituents and reducing its weight.

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What can we tell from the column that an element is in?
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Answer:

The number of electrons in the outer energy level

Explanation:

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