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3 years ago

Which element has a partially filled f orbital? sm os ba bi

Chemistry

2 answers:

Sladkaya [172]3 years ago

8 0

Sm is the answer

i hope i helped :)

asambeis [7]3 years ago

6 0

Answer:

Sm (Samarium) = [Xe] 4f6 6s2

Explanation:

Noble gas electron configuration of Sm (Samarium) = [Xe] 4f⁶ 6s²

Noble gas electron configuration of Os (Osmium) = [Xe] 4f¹⁴5d⁶6s²

Noble gas electron configuration of Ba (Barium) = [Xe] 6s²

Noble gas electron configuration of Bi (Bismuth) = [Xe] 4f¹⁴5d¹⁰6s²6p³

A completely filled s, p, d, and f orbitals have 2, 6, 10 and 14 electrons respectively. Half filled s, p, d, and f orbitals have 1, 3, 5 and 7 electrons respectively. Bi and Os have completely filled f orbitals. Sm has partially filled f orbital.

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<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

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We are given:

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We know that:

$\Delta U=0$ (For isothermal process)

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$\Delta U=\Delta q-\Delta W$

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Now, calculating the work done for isothermal process, we use the equation:

$\Delta W=nRT\ln (\frac{P_1}{P_2})$

where,

$\Delta W$ = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

$P_1$ = initial pressure = 100 kPa

$P_2$ = final pressure = 50 kPa

Putting values in above equation, we get:

$\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW$

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

$\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW$

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

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