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3 years ago

Erbium (Er) is element 68 on the periodic table. A sample contains 5.04×1023 atoms of Er. Calculate the amount of Er.

Chemistry

1 answer:

Korvikt [17]3 years ago

3 0

Answer:

139.98 g to nearest hundredth.

Explanation:

Using Avogadro's Number:

One mole (167.26 g) of Erbium equates to 6.022141 * 10^23 atoms.

So 5.04 * 10^23 = 167.26 * 5.04/6.022141

= 139,98 g.

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Extend the aufbau sequence through an element that has not yet been identified, but whose atoms would completely fill 7p orbital

Thepotemich [5.8K]

Answer:

Number of electrons = 118

Electronic configuration: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

Explanation:

Given: A chemical element that has completely filled 7p orbital.

According to this, the principal occupied electron shell or the valence shell of such an element is 7p.

⇒ the principal quantum number (n) for the valence shell is 7.

∴ this element belongs to the period 7 of the periodic table.

Also, an element that has completely filled p-orbital belongs to the group 18 of the p-block.

Therefore, an element that belongs to the group 7 and period 18 of the periodic table, should have a completely filled 5f, 6d, 7s and 7p orbitals.

Therefore, the electronic configuration should be: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

And, the number of electrons = atomic number of radon (Rn) + 14 + 10 + 2 + 6 = 86 + 32 = 118.

Therefore, the given element has atomic number 118 and has the electronic configuration: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶. Thus the given element can be Oganesson.

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3 years ago