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liq [111]
3 years ago
8

Consider a hydrogen atom in the ground state. What is the energy of its electron?

Chemistry
2 answers:
Pani-rosa [81]3 years ago
5 0

The energy of electron in ground state of hydrogen atom is \boxed{-2.1799\times{{10}^{-18}}{\text{ J}}} , and the energy electron in excited state when n is equal to 2 is \boxed{-5.4498\times{{10}^{-19}}{\text{ J}}} .

Further explanation:

The ground state of an electron in an atom refers to its lowest energy state. The energy of the ground state is also known as zero-point energy. An excited state in an atom is a state that has higher energy than the ground state.

In case of a hydrogen atom,

The formula to calculate energy of an electron in a hydrogen atom is,

{E_n}=\frac{{\left({-2.1799\times{{10}^{-18}}{\text{ J}}}\right)}}{{{n^2}}}

Here, n is a principal quantum number or energy level of an electron.

For an electron in the ground state, the value of n is 1. Thus the energy of an electron in ground state is,

\begin{gathered}{E_n}=\frac{{\left({-2.1799\times{{10}^{-18}}{\text{ J}}}\right)}}{{{n^2}}}\hfill\\{E_1}=\frac{{\left({-2.1799\times{{10}^{-18}}{\text{ J}}}\right)}}{{{{\left(1\right)}^2}}}\hfill\\{{\mathbf{E}}_{\mathbf{1}}}{\mathbf{=-2}}{\mathbf{.1799\times1}}{{\mathbf{0}}^{{\mathbf{-18}}}}{\mathbf{ J}}\hfill\\\end{gathered}

Therefore, the energy of an electron in energy level of n is equal to 2 can be calculated as follows:

First, substitute 2 for the value of n.

\begin{gathered}{E_n}=\frac{{\left({-2.1799\times{{10}^{-18}}{\text{ J}}}\right)}}{{{n^2}}}\\{E_2}=\frac{{\left({-2.1799\times{{10}^{-18}}{\text{ J}}}\right)}}{{{{\left(2\right)}^2}}}\\\end{gathered}

Then, calculate the energy as follows:

\begin{aligned}{E_2}=-0.54498\times{10^{-18}}{\text{ J}}\\\simeq{\mathbf{-5}}{\mathbf{.4498\times1}}{{\mathbf{0}}^{{\mathbf{-19}}}}{\mathbf{ J}}\\\end{aligned}

Learn more:

1. Ranking of elements according to their first ionization energy.:brainly.com/question/1550767

2. Chemical equation representing the first ionization energy for lithium.:brainly.com/question/5880605

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Hydrogen atom

Keywords: Hydrogen atom, ground state, excited state, energy of an electron, energy in ground state, energy of electron in excited state, n=2, n=1, energy level, -5.4498x10^-19.

mr Goodwill [35]3 years ago
4 0
The energy of an electron in a hydrogen atom is:
<span>E= (-2.179 x 10 ^-18 J)(1/n^2)

</span>where n is the principle energy level of the electron.
<span>
E= (-2.179 x 10 ^-18 J)(1/4)
</span><span>E= -2.179 x 10 ^-18 J
</span>E = -5.4 <span>x 10 ^-19 J</span>
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When adding or subtracting values, you must round your answer to the same "place" as the measurement with its last significant figure <em>furthest to the left</em>.  

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3 years ago
Which best describes a hydrate?
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What kind of plant growth hormone to regular plant growth. We dissolve 5.0 mg of the solution into 2.0 kg of water. What is the
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The dosage of the solution is 2.5 parts per million.

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The solvent proportion in parts per million is equal to the ratio of solution in miligrams to ratio of water in miligrams multiplied by a million. (A kilogram is equivalent to a million miligrams) That is:

p.p.m. = \frac{5\,mg}{2000000\,mg}\times 1000000

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3 0
3 years ago
100.00 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution.
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Answer:

a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

<h3>pH = 2.04</h3><h3 />

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

<em>Where pH is the pH of the buffer,</em>

<em>pKa is -log Ka = 3.25</em>

<em>And [NaNO₂] [HNO₂] could be taken as the moles of each compound.</em>

<em />

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

<h3>pH = 3.85</h3><h3 />

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

<em>Where [OH⁻] = [HNO₂] = x</em>

<em>[NaNO₂] = 0.075M</em>

<em />

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

<h3>pH = 8.06</h3><h3 />

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

<h3>pH = 11.56</h3>
5 0
2 years ago
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