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3 years ago

What is the molecular geometry of a molecule with 4 outer atoms and 2 lone pairs on the central atom?

Chemistry

1 answer:

Lera25 [3.4K]3 years ago

8 0

Answer:

See explanation

Explanation:

The valence shell electron pair repulsion theory holds that the number of electron pairs on the outermost shell of the central atom in a molecule determines the shape of the molecule.

In accordance with this theory, the electron domain geometry of a molecule having 4 outer atoms and 2 lone pairs on the central atom is based on an octahedron. The molecular geometry of this molecule is square planar with a bond angle of 90 degrees.

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Calculate the percent ionization of a 0.15 M benzoic acid solution in pure water and in a solution containing 0.10 M sodium benz

Hoochie [10]

Answer:

% ionization for benzoic acid = 0.08%

% ionization for sodium benzoate = 2.5%

The percentage ionization differ significantly because benzoic acid is a weak acid while sodium benzoate is a salt of benzoic acid. Their extent of dissociation also differ because they were compared in different solutions

Explanation:

Ka for pure water = 1.0 * 10-⁷

Ka for sodium benzoate = 6.5*10-⁵

1. For benzoic acid (C6H5COOH)

C6H5COOH ==== C6H5COO‐ + H+

0.15M 0 0

0.15-x x x

Ka = [C6H5COO-] [H+] / [C6H5COOH]

Ka = [X] [X] / 0.15 - X

1.0*10-⁷ = [X]² / 0.15 - x

But x is negligible compared to 0.15,

(1.0*10-⁷)*0.15 = x²

Take square root of both sides,

X = 1.22 * 10-⁴

% ionization = ( [H+] / [C6H5COOH] ) * 100

% ionization = (1.22*10-⁷ / 0.15) * 100

% ionization = 0.08%

2. For C6H5COONa

Note: I will not repeat the same procedure of dissociation again since they're basically the same just the difference in ions

Ka for C6H5COONa = 6.5*10-⁵

6.5*10-⁵ = [X]² / (0.10 - X)

Cross-multiply both sides;

(6.5*10-⁵ * 0.10) = X²

Take square root of both side,

X= 2.5*10-³

% ionization = (2.5*10-³ / 0.10) *100

% ionization = 2.5%

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3 years ago

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Setler79 [48]

Answer:

1 true

Explanation:

bcz it is true

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evablogger [386]

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3 years ago

How much does the Earth weigh?

ELEN [110]

5.972 × 10^24 kg

hope it helps

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3 years ago

Read 2 more answers

Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi

Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

Calculate ΔG at 25°C given the following sets of partial pressures.

Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

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3 years ago