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The [Fe(H₂O)₆]³⁺ complex requires a relatively small amount of energy to promote an electron from the t2g to the eg. Based on the UV data, predict the spin of this complex.

Answers

Answer 1

Answer:

Answer:

The spin of the complex is 5.92 B.M

Explanation:

Please see the attachments below

Answer 2

Answer:

Final answer:

The [Fe(H₂O)₆]³⁺ complex is a high-spin complex due to the relatively small energy required to promote an electron from the t2g to the eg orbital. As such, it is reasonable to predict that it has a high-spin state with five unpaired electrons.

Explanation:

The [Fe(H₂O)₆]³⁺ complex is a type of coordination complex in which the central metal atom, Fe³⁺, is surrounded by six water molecules acting as ligands. The spin state of such a complex can be determined based on the energy required to promote an electron from the t2g to the eg.

In [Fe(H₂O)₆]³⁺, the field produced by the water ligands is relatively weak, resulting in a small crystal field splitting (Aoct <P). Given that it requires less energy for the electrons to occupy the eg orbitals than to pair up, there will be an electron in each of the five 3d orbitals before any pairing occurs. Hence, for the six d electrons on the Fe³⁺ ion in [Fe(H₂O)₆]³⁺, there should be one pair (two electrons) and four unpaired electrons.

High-spin complexes are those in which the electrons tend not to pair up because the crystal field splitting is not large enough to make it energetically favorable for them to do so. Given that the [Fe(H₂O)₆]³⁺ complex falls under the categories of high-spin complexes, it is reasonable to predict that it exhibits a high-spin state with five unpaired electrons.

Learn more about Coordination Complexes and Spin States here:

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What are the starting substances (molecules) in a chemical equation called?

Answers

Answer:

A chemical reaction is the process in which atoms present in the starting substances rearrange to give new chemical combinations present in the substances formed by the reaction. These starting substances of a chemical reaction are called the reactants, and the new substances that result are called the products.

4.289x10^0 as regular numbers

Answers

4.289

Explanation:

4.289 x 10^0 can be written as a regular number by moving the decimal point to the right or left based on the exponent value. Since the exponent is 0, the decimal point does not need to be moved. Therefore, the regular number form of 4.289 x 10^0 is simply 4.289.

Calculate the entropy change for the gas. (2b)Calculate the entropy change when 14 g of nitrogen expand from a volume of 10 L to a volume of 30 L at the same temperature. Assuming ideal behaviour for the nitrogen gas

Answers

Explanation:

because T is const so

deltaS=Q/T=nRLn(V2/V1)

=(14/28)x8.314xLn(30/10)=4,567 J/k

PLZZZZZZZZZZZZ HELPPPPPPPPP BRAINLIEST FOR WHO GETS IT RIGHTTTTTTWhat is the mass of reactants in the following equation?N2 + 3H2 ----> 2NH3
Question 3 options:

34.05 amu

31.03 amu

30.02 amu

15.01 amu

Answers

Answer: 34.05

Explanation:

2N and 6H = abt 34

9. The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different reaction predominates: 4NH3(g) + 3O2(g) ⇔ 2N2(g) + 6H2O(g) When 0.0120 mol gaseous NH3 and 0.0170 mol gaseous O2 are placed in a 1.00 L container at a certain temperature, the N2 concentration at equilibrium is 2.20×10-3 M. Calculate Keq for the reaction at this temperature.

Answers

Answer: The value of $K_(eq)$ is $4.66* 10^(-5)$

Explanation:

We are given:

Initial moles of ammonia = 0.0120 moles

Initial moles of oxygen gas = 0.0170 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $(0.0120)/(1.00)=0.0120M$

Concentration of oxygen gas = $(0.0170)/(1.00)=0.0170M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial: 0.0120 0.0170

At eqllm: 0.0120-4x 0.0170-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $2.20* 10^(-3)M=0.00220M$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.00220\n\n\Rightarrow x=(0.00220)/(2)=0.00110M$

Now, equilibrium concentration of ammonia = $0.0120-4x=[0.0120-(4* 0.00110)]=0.00760M$

Equilibrium concentration of oxygen gas = $0.0170-3x=[0.0170-(3* 0.00110)]=0.0137M$

Equilibrium concentration of water = $6x=[6* 0.00110]=0.00660M$

The expression of $K_(eq)$ for the above reaction follows:

$K_(eq)=([N_2]^2* [H_2O]^6)/([NH_3]^4* [O_2]^3)$

Putting values in above expression, we get:

$K_(eq)=((0.00220)^2* (0.00660)^6)/((0.00760)^4* (0.0137)^3)\n\nK_(eq)=4.66* 10^(-5)$

Hence, the value of $K_(eq)$ is $4.66* 10^(-5)$

A chemist designs a galvanic cell that uses these two half-reactions:O2 (g) + 4H+(aq) + 4e− → 2H2O (l) Eo =+1.23V
Zn+2 (aq) + 2e− → Zn(s) Eo=−0.763V

Answer the following questions about this cell.

Write a balanced equation for the half-reaction that happens at the cathode.
Write a balanced equation for the half-reaction that happens at the anode.
Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written. Do you have enough information to calculate the cell voltage under standard conditions

Answers

Answer: The reaction is spontaneous and there is not enough information to calculate the cell voltage.

Explanation:

The substance having highest positive $E^o$ reduction potential will always get reduced and will undergo reduction reaction.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

For a:

The half reactions for the cell occurring at cathode follows:

$O_2(g)+4H^+(aq)+4e^-\rightarrow H_2O(l);E^o_(cathode)=+1.23V$

For b:

The half reactions for the cell occurring at anode follows:

$Zn(s)\rightarrow Zn^(2+)+2e^-;E^o_(anode)=-0.763V$ ( × 2)

For c:

The balanced equation for the overall reaction of the cell follows:

$O_2(g)+4H^+(aq)+2Zn(s)\rightarrow H_2O(l)+2Zn^(2+)(aq)$

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_(cell)$

For a reaction to be spontaneous, the standard electrode potential must be positive.

To calculate the $E^o_(cell)$ of the reaction, we use the equation:

$E^o_(cell)=E^o_(cathode)-E^o_(anode)$

Putting values in above equation, we get:

$E^o_(cell)=1.23-(-0.763)=1.993V$

As, the standard electrode potential of the cell is coming out to be positive, the reaction is spontaneous in nature.

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_(cell)=E^o_(cell)-(0.059)/(n)\log ([Zn^(2+)]^2)/([H^(+)]^4* p_(O_2))$

As, the concentrations and partial pressures are not given. So, there is not enough information to calculate the cell voltage.

Hence, the reaction is spontaneous and there is not enough information to calculate the cell voltage.

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