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Alja [10]
2 years ago
12

For each of the following pairs of complexes, identify which one you would predict to have the larger Δo value, and explain why.

(Assume an octahedral environment)
a. [Mn(H2O)6]2+ or [Fe(H2O)6]3+
b. [Fe(H2O)6]3+ or [Fe(CN)6]3−
c. [Fe(CN)6]3− or [Ru(CN)6]3-
Chemistry
1 answer:
mash [69]2 years ago
4 0

Answer:

a) [Fe(H2O)6]3+

b) [Fe(CN)6]3−

c) [Ru(CN)6]3-

Explanation:

. [Mn(H2O)6]2+ or [Fe(H2O)6]3+

The both complexes are d5 complexes with the same ligand , water. Water is a weak ligand and note that Mn^2+ often have a crystal field stabilization energy of zero hence

[Fe(H2O)6]3+ will possess a greater ∆o value.

The splitting of d orbitals according to the crystal field theory depends on the;

i)geometry of the complex

ii) nature of the metal ion,

iii)charge on the metal ion,

iv) ligands that surround the metal ion.

When the geometry and the ligands are held constant, the order of crystal field splitting is as follows;

Pt4+ > Ir3+ > Rh3+ > Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+

[Fe(H2O)6]3+ or [Fe(CN)6]3−

[Fe(CN)6]3− will have a greater ∆o because the cyanide ion is a strong field ligand compared to water. A strong field ligand causes a greater splitting of the octahedral crystal field compared to a weak field ligand.

. [Fe(CN)6]3− or [Ru(CN)6]3-

[Ru(CN)6]3- will exhibit a greater crystal field splitting. Crystal field splitting increases with the second and third row transition elements when compared to the crystal field splitting of the first row transition elements. Note that, there is an increase of approximately 30%–50% in Δo on going from a first-row transition metal to a second-row metal and another 30%–50% increase on going from a second-row to a third-row metal when they have the same geometry and oxidation state.

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The density of methanol is 0.7918g/mL. Calculate the mass of 89.9mL of this liquid.
Novosadov [1.4K]
The mass is 71.2 g.

Mass = 89.9 mL × \frac{0.7918 g}{1 mL} = 71.2 g
4 0
2 years ago
How many electrons are involved in one equivalent of oxidation-reduction?
kotegsom [21]

Answer:

1 electron is involved.

Explanation:

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In redox reactions, when therer's the necessity to know the involved equivalents, they equal the number of transferred electrons, in this case, since one equivalent is stated, one electron is transferred (involved).

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2 years ago
Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the sam
mafiozo [28]

Question:

Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the same as pure water at 20.0 degrees C (18 torr). The volume of the mixture is 1.7 L and its total pressure is 0.987 atm. Determine the number of moles of hydrogen gas present in the sample.

A. 0.272 mol

B. 0.04 mol

C. 0.997 mol

D. 0.139 mol

E. 0.0681 mol

Answer:

The correct option is;

E. 0.0681 mol

Explanation:

The equation for the reaction is

Zn + HCl = H₂ + ZnCl₂

Vapor pressure of the liquid = 18 torr = 2399.803 Pa

Total pressure of gas mixture H₂ + liquid vapor = 0.987 atm  

= 100007.775 Pa

Therefore, by Avogadro's law, pressure of the hydrogen gas is given by the following equation

Pressure of H₂ = 100007.775 Pa - 2399.803 Pa = 97607.972 Pa

Volume of H₂ = 1.7 L = 0.0017 m³

Temperature = 20 °C = 293.15 K

Therefore,

n = \frac{PV}{RT} =  \frac{100007.775 \times 0.0017 }{8.3145 \times 293.15} = 0.068078 \ moles

Therefore, the number of moles of hydrogen gas present in the sample is n ≈ 0.0681 moles.

7 0
2 years ago
What is the net ionic equation of the reaction of MgCl2 with NaOH? Express your answer as a chemical equation. View Available Hi
kari74 [83]

Answer:

Net ionic equation for the reaction between MgCl₂ and NaOH in water:

\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s).

Net ionic equation for the reaction between MgSO₄ and BaCl₂ in water:

\rm {Ba}^{2+}\; (aq) + {SO_4}^{2-}\;(aq) \to BaSO_4\; (s).

Explanation:

Start by finding the chemical equations for each reaction:

MgCl₂ reacts with NaOH to form Mg(OH)₂ and NaCl. This reaction is a double decomposition reaction (a.k.a. double replacement reaction, salt metathesis reaction.) This reaction is feasible because one of the products, Mg(OH)₂, is weakly soluble in water and exists as a solid precipitate.

\rm MgCl_2\; (aq) + 2\; NaOH\; (aq)\to Mg(OH)_2 \; (s) + 2\; NaCl\; (aq).

MgSO₄ reacts with BaCl₂ in a double decomposition reaction to produce BaSO₄ and MgCl₂. Similarly, the solid product BaSO₄ makes this reaction is feasible.

\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq).

How to rewrite a chemical equation to produce a net ionic equation?

  1. Rewrite all reactants and products that ionizes completely in the solution as ions.
  2. Eliminate ions that exist on both sides of the equation to produce a net ionic equation.

Typical classes of chemicals that ionize completely in water:

  • Soluble salts,
  • Strong acids, and
  • Strong bases.

Keep the formula of salts that are not soluble in water, weak acids, weak bases, and water unchanged.

Take the first reaction as an example, note the coefficients:

  • MgCl₂ is a salt and is soluble in water. Each unit of MgCl₂ can be written as \rm Mg^{2+} and \rm 2\; Cl^{-}.
  • NaOH is a strong base. Each unit of NaOH can be written as \rm Na^{+} and \rm OH^{-}.
  • Mg(OH)₂ is a weak base and should not be written.
  • NaCl is a salt and is soluble in water. Each unit of NaCl can be written as \rm Na^{+} and \rm Cl^{-}.

\rm Mg^{2+} + 2\; Cl^{-} + 2\; Na^{+} + 2\; OH^{-} \to Mg(OH)_2\;(s) + 2\; Na^{+} + 2\; Cl^{-}.

Ions on both sides of the equation:

  • \rm 2\; Cl^{-}, and
  • \rm 2\; Na^{+}.

Add the state symbols:

\rm Mg^{2+}\; (aq) + 2\; OH^{-}\; (aq) \to Mg(OH)_2\;(s).

For the second reaction:

\rm MgSO_4\; (aq) + BaCl_2\; (aq) \to BaSO_4\; (s) + MgCl_2\; (aq).

\rm Mg^{2+} + 2\; {SO_4}^{2-} + Ba^{2+} + 2\; Cl^{-} \to BaSO_4\; (s) + Mg^{2+} + 2\; Cl^{-}.

\rm Ba^{2+}\; (aq) + {SO_4}^{2-}\; (aq) \to BaSO_4\; (s).

5 0
2 years ago
Read 2 more answers
How much volume (in cm3) is gained by a person who gains 11.1 lbs of pure fat?
kirill [66]

The density of human fat is 0.918 g/cm^{3}. The mass of the pure fat is 11.1 lbs.

First convert mass from lb to g as follows:

1 lb=453.592 g

Thus,

11.1 lb\rightarrow 11.1\times 453.592 g=5034.875 g

Density of a substance is defined as mass per unit volume, thus volume can be calculated as:

V=\frac{m}{d}

Putting the values,

V=\frac{5034.875 g}{0.918 g/cm^{3}}=5484.61 cm^{3}

Therefore, volume gained by person will be 5484.61 cm^{3}.

6 0
2 years ago
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