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tankabanditka [31]
3 years ago
7

Calculate the energy(J) change associated with an electron transition from n=2 to n=5 in a Bohr hydrogen atom. Put answer in sci

entific notation.
Chemistry
2 answers:
frez [133]3 years ago
8 0

Answer:

4.58*10^(-19) J

Explanation:

Using the Rydberg's equation:

\frac{1}{\lambda} = R \times (\frac{1}{n^2_{final}} - \frac{1}{n^2_{initial}})

where

\lambda is  the wavelength of the photon;

R is the Rydberg's constant = 1.0974*10^7 m^(-1)

final level is 5 and initial level is 2.

\frac{1}{\lambda} = 1.0974 \times 10^7 \times (\frac{1}{5^2} - \frac{1}{2^2})

\frac{1}{\lambda} = -2304540 m^{-1}

\lambda = -4.339 \times 10^{-7} \;m

Energy change is calculated with the next formula:

E = h*c/λ

where h is the Planck's constant = 6.626*10^(-34) J*s, and c is the speed of light = 299,792,458 m/s

E = 6.626*10^(-34)*299,792,458/-4.339*10^(-7)

E = 4.58*10^(-19) J

Oliga [24]3 years ago
3 0
The energy at n level of hydrogen atom energy level =13.6/n^2

substiture the respective n values in the equation above and find the difference in the energy levels

instagram : imrajsingh

gimme a follow^ :)
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Svet_ta [14]

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The number of neutrons present in one atom of isotope of Silicon of mass 28 amu is<u> 14 neutrons</u>

Explanation:

Symbol of Si isotope

_{14}^{28}\textrm{Si}

<u>Number of Neutron = Mass number - Atomic Number</u>

Mass number = Total number of protons and neutrons present in the nucleus of the atom.For Si = 28 amu

Atomic Number = Total number of Protons present in the nucleus.

Si = 14

Number of neutron = 24 - 14

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3 years ago
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An analytical chemist is titrating of a solution of nitrous acid with a solution of . The of nitrous acid is . Calculate the pH
Burka [1]

Answer:

pH = 2.69

Explanation:

The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>

<em />

The reaction of HNO₂ with KOH is:

HNO₂ + KOH → NO₂⁻ + H₂O + K⁺

Moles of HNO₂ and KOH that react are:

HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>

KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>

That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:

NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻

HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂

It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:

pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]

pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]

<h3>pH = 2.69</h3>
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