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1 year ago

The electron configuration that belongs to the atom with the lowest first ionization energy is?

Chemistry

1 answer:

sasho [114]1 year ago

4 0

The electron configuration that belongs to the atom with the lowest first ionization energy is francium.

<h3>What is ionization energy? </h3>

Ionization energy is defined as the minimum amount of energy required to remove the most loosely electron present in outermost shell.

<h3>Ionization energy across period</h3>

Ionization energy increase as we move from left to right in the period. This can be explained as when we move from left to right along period new electron is added to the same shell which increase the nuclear charge. Hence results int he decrease in size. Due to this decrease in size more energy is required to remove electron from outermost shell.

<h3>Ionization energy along group</h3>

Ionization energy decrease as we move from top to bottom along group. This can be explained as we move from top to bottom new electron is added to new shell. Due to addition of new shell the size of atom increases which results in the decrease in the nuclear charge. Due to this less amount of energy is needed to remove an electron.

Thus, we concluded that the electron configuration that belongs to the atom with the lowest first ionization energy is francium.

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The student ran out of time and did not do the second heating. Explain how this error would affect the calculation for the numbe

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The error caused will lower the number of moles when he did not do the second heating.This question is related to the empirical formula.

<h3>What is Empirical formula ?</h3>

A formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.

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Hence, The error caused will lower the number of moles when he did not do the second heating.

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3 years ago

Determine the energy of 1.70 mol of photons for each of the following kinds of light. (Assume three significant figures.)PART A

BabaBlast [244]

<u>Answer:</u>

<u>For A:</u> The energy of the given amount of photons for infrared radiation is $1.271\times 10^5J$

<u>For B:</u> The energy of the given amount of photons for infrared radiation is $4.026\times 10^5J$

<u>For C:</u> The energy of the given amount of photons for infrared radiation is $1.355\times 10^6J$

<u>Explanation:</u>

The relationship between energy and frequency is given by Planck's equation, which is:

$E=n\rimes N_A\times \frac{hc}{\lambda}$ ......(1)

where,

h = Planck's constant = $6.62\times 10^{-34}Js$

E = energy of the light

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of light

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

n = number of moles of photons = 1.70 moles

Conversion factor used: $1m=10^9nm$

<u>For A:</u>

Wavelength of infrared radiation = $1600nm=1.6\times 10^6m$

Putting values in equation 1, we get:

$E=1.7\times 6.022\times 10^{23}\times \frac{6.62\times 10^{-34}\times 3\times 10^8}{1.6\times 10^{-6}}\\\\E=1.271\times 10^5J$

Hence, the energy of the given amount of photons for infrared radiation is $1.271\times 10^5J$

<u>For B:</u>

Wavelength of visible light = $505nm=5.05\times 10^7m$

Putting values in equation 1, we get:

$E=1.7\times 6.022\times 10^{23}\times \frac{6.62\times 10^{-34}\times 3\times 10^8}{5.05\times 10^{-7}}\\\\E=4.026\times 10^5J$

Hence, the energy of the given amount of photons for infrared radiation is $4.026\times 10^5J$

<u>For C:</u>

Wavelength of ultraviolet radiation = $150nm=1.5\times 10^7m$

Putting values in equation 1, we get:

$E=1.7\times 6.022\times 10^{23}\times \frac{6.62\times 10^{-34}\times 3\times 10^8}{1.5\times 10^{-7}}\\\\E=1.355\times 10^6J$

Hence, the energy of the given amount of photons for infrared radiation is $1.355\times 10^6J$

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3 years ago