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jok3333 [9.3K]
2 years ago
13

In Universe , recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quant

um mechanics works just as it does in our universe, except that there are eight orbitals instead of the usual number we observe here. Use these facts to write the ground-state electron configurations of the fifteenth and sixteenth elements in the first transition series in Universe . Note: you may use to stand for the electron configuration of the noble gas at the end of the row before the first transition series.
Chemistry
1 answer:
tiny-mole [99]2 years ago
7 0

Answer:

Check the explanation

Explanation:

When talking about our universe there are 5 d orbitals. The element of first transition series moves away from the universal principles of Hund's rule and Aufbav's principle. So in order to attain stability these elements tend to form half or full filled orbitals.

In our universe the ground state electronic configuration of sixth transition metal, Iron (Fe) :  [Ar] 3d^{6} 4s^{2}

and the electronic configuration of seventh transition metal, Cobalt (Co) :  [Ar] 3d^{7} 4s^{2}

=================================

=================================

In universe L there are seven orbitals.

Ground state electronic configuration of sixth and seven transition element.

Sixth transition metal: [Ar] 3d^{7} 4s^1 or [X] 3d^{7} 4s^1

Seventh transition metal: [Ar] 3d^{7} 4s^{2}or [X] 3d^{7} 4s^{2}

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How many are molecules ( or formula) in each sample?
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Answer:

  • 4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}
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<u>Explanation</u>:

<u>Number of molecules for 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}</u>

\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:

Atomic mass of Na + H + C + 3(O)  = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }

55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)

=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}

<u>Number of molecules for for \left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}</u>

\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.

=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

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3 years ago
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