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Yuliya22 [10]
3 years ago
7

Given the following formula for calculating the ionization energy of one-electron species such as Li2+, He+, and H, calculate th

e ionization energy (in J/mol) for B4+. Use scientific notation in answers (ex: 1E10, 3.20E-6)
Chemistry
1 answer:
Yuki888 [10]3 years ago
4 0

Answer:

The answer is "32819.9 \ \frac{J}{mol}\\\\"

Explanation:

Boron: 5^{B}\to 1s^2 2s^2 2p^1

\Delta E=-2.18\times 10^{-18}\ \frac{J}{atom}\ (\frac{1}{\infity^2}-\frac{1}{n^2_{initial}})(z^2) (6.022\times 10^{23}\ \frac{atom}{mol})\\\\

      =-2.18\times 10^{-18}\ \frac{J}{atom}\ (0-\frac{1}{1})(5^2) (6.022\times 10^{23}\ \frac{atom}{mol})\\\\ =2.18\times 10^{-18}\times 25 \times 6.022\times 10^{23}\ (\frac{J}{mol})\\\\ =328.199 \times 10^{5}\ (\frac{J}{mol})\\\\ =32819 \times 10^{3}\ (\frac{J}{mol})\\\\ =32819.9 \ (\frac{J}{mol})\\\\

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Element Y has two natural isotopes Y-63 (62.940 amu) and Y-65 (64.928 amu). Calculate the atomic mass of element Y, given the ab
djverab [1.8K]

Answer:

Average atomic mass = 63.553 amu.

Explanation:

Given data:

Abundance of Y-63 = 69.17%

Abundance of Y-65 = 100 - 69.17 = 30.83%

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Atomic mass of Y = ?

Solution:

Average atomic mass= (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

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5 0
3 years ago
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Chloride is a considered an extended structure because in sodium chloride it forms an unending repeated pattern of ions which makes it a perfect example of an extended structure.

Hence we can conclude that Chloride can be described as an extended structure because its atoms are arranged following a repeating pattern and are of distinct ratio.

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