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2 years ago

What is the bond dissociation energy for breaking all the bonds in a mole of o2 molecules?

Chemistry

1 answer:

djverab [1.8K]2 years ago

8 0

<h3>Answer:</h3>

498 kj/mol

<h3>Explanation:</h3>

Chemical reactions occur as a result of bond breaking and bond formation.
The bonds in reactants are broken and atoms are rearranged to form new bonds.
During bond breaking energy is absorbed to break the bonds of reactants while bond formation involves the release of energy during the formation of new bonds.

In our case;

In 1 mole of the Oxygen molecule, there is one O=O bond

Energy absorbed to break O=O is 498 kJ/mol

Therefore, the ΔH required to break all the bonds in one mole of Oxygen(O₂) molecules is 498kJ/mol.

Note that, bond breaking is endothermic since energy is absorbed from the surroundings.

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One of the emission spectral lines for Be3+ has a wavelength of 253.4 nm for an electronic transition that begins in the state w

4vir4ik [10]

Answer:

Explanation:

Relationship between wavenumber and Rydberg constant (R) is as follows:

$wave\ number=R\times Z^2\frac{1}{n_1^2} -\frac{1}{n_1^2}$

Here, Z is atomic number.

R=109677 cm^-1

Wavenumber is related with wavelength as follows:

wavenumber = 1/wavelength

wavelength = 253.4 nm

$wavenumber=\frac{1}{253.4\times 10^{-9}} \\=39463.3\ cm^{-1}$

Z fro Be = 4

$39463.3=109677\times 4^2(\frac{1}{n_1^2} -\frac{1}{5^2})\\39463.3=109677\times 16(\frac{1}{n_1^2} -\frac{1}{5^2})\\n_1=4$

Therefore, the principal quantum number corresponding to the given emission is 4.

6 0

2 years ago

A chemist dissolves of pure hydroiodic acid in enough water to make up of solution. Calculate the pH of the solution. Be sure yo

Gre4nikov [31]

Answer:

1.76

Explanation:

There is some info missing. I think this is the original question.

<em>A chemist dissolves 660.mg of pure hydroiodic acid in enough water to make up 300.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.</em>

<em />

Step 1: Calculate the molarity of HI(aq)

M = mass of solute / molar mass of solute × liters of solution

M = 0.660 g / 127.91 g/mol × 0.300 L

M = 0.0172 M

Step 2: Write the acid dissociation reaction

HI(aq) ⇄ H⁺(aq) + I⁻(aq)

HI is a strong acid, so [H⁺] = 0.0172 M

Step 3: Calculate the pH

pH = -log [H⁺]

pH = -log 0.0172

pH = 1.76

5 0

2 years ago

Name two compounds in unpolluted air?

goblinko [34]

Nitrogen and oxygen are in unpolluted air

6 0

2 years ago

What is the volume of 12.0 moles of Cl2 gas at STP?

Ymorist [56]

The answer has to be 22.4L

3 0

1 year ago

Given the following reaction: 2D(g) + 3E(g) + F(g) \longrightarrow⟶ 2G(g) + H(g) When the concentration of D is decreasing by 0.

Zepler [3.9K]

Answer:

Rate of reaction = -d[D] / 2dt = -d[E]/ 3dt = -d[F]/dt = d[G]/2dt = d[H]/dt

The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1

E decreseas 3/2 as fast as G increases = 0.30 M/s

Explanation:

Rate of reaction = -d[D] / 2dt = -d[E]/ 3dt = -d[F]/dt = d[G]/2dt = d[H]/dt

When the concentration of D is decreasing by 0.10 M/s, how fast is the concentration of H increasing:

Given data = d[D]/dt = 0.10 M/s

-d[D] / 2dt = d[H]/dt

d[H]/dt = 0.05 M/s

The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1

When the concentration of G is increasing by 0.20 M/s, how fast is the concentration of E decreasing:

d[G] / 2dt = -d[H]/3dt

E decreseas 3/2 as fast as G increases = 0.30 M/s

5 0

2 years ago