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3 years ago

The spectrochemical series is I < Br< < Cl^- < F

Chemistry

1 answer:

vfiekz [6]3 years ago

6 0

Answer:

b. The splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

Explanation:

The spectrochemical series is an arrangement of ligands in increasing order of their magnitude of crystal field splitting.

Ligands that occurs towards the right in the series are called strong field ligands and they tend to cause a greater magnitude of crystal field splitting. Ligands that occur towards the left hand side in the series are called weak field ligands and they tend to cause a lesser magnitude of crystal field splitting.

Since Cl^- is a weak field ligand, it causes a lesser magnitude of d orbital splitting compared to ethylenediammine (en) which causes a greater magnitude of d orbital splitting.

Hence; the splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

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True or False: The atomic number is the number of neutrons in an element?

klio [65]

Answer:

true

Explanation:

5 0

3 years ago

When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

Answer: The value of $K_c$ for the given reaction is 1.435

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial: 0.20

At eqllm: 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

$[N_2O_4]_{eq}=0.057M$

Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

6 0

3 years ago

Diagram of paper chromatography from a chemistry textbook

frozen [14]

Save Time & Ensure Consistent Reproducible Results With Waters LC Columns & Supplies. A Range Of Chromatography Column To Meet The Needs Of Virtually Every Application

8 0

3 years ago

(image attached)

Naily [24]

Answer:

D) He did not multiply the chlorine and oxygen atoms by the coefficient 4.

Explanation:

The coefficient 4 at the beginning of the chemical formula indicates that there are four Ca(ClO3)2 molecules. Think of this as Ca(ClO3)2 × 4. This means that he had to multiply the number of atoms for each element by 4 as well, so he should've ended up with 4 total calcium atoms (which is correct), 8 total chlorine atoms, and and 24 total oxygen atoms. He did not get all these answers because he didn't multiply the chlorine and oxygen atoms by the coefficient 4.

8 0

3 years ago

Read 2 more answers

Addition of an excess of lead (II) nitrate to a 50.0mL solution of magnesium chloride caused a formation of 7.35g of lead (II) c

KiRa [710]

Answer:

[Cl⁻] = 0.016M

Explanation:

First of all, we determine the reaction:

Pb(NO₃)₂ (aq) + MgCl₂ (aq) → PbCl₂ (s) ↓ + Mg(NO₃)₂(aq)

This is a solubility equilibrium, where you have a precipitate formed, lead(II) chloride. This salt can be dissociated as:

PbCl₂(s) ⇄ Pb²⁺ (aq) + 2Cl⁻ (aq) Kps

Initial x

React s

Eq x - s s 2s

As this is an equilibrium, the Kps works as the constant (Solubility product):

Kps = s . (2s)²

Kps = 4s³ = 1.7ₓ10⁻⁵

4s³ = 1.7ₓ10⁻⁵

s = ∛(1.7ₓ10⁻⁵ . 1/4)

s = 0.016 M

3 0

3 years ago