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Pavel [41]
3 years ago
7

The filling of which orbital is represented by the transition metals in period 4?

Chemistry
2 answers:
Y_Kistochka [10]3 years ago
8 0

<span>The 4s orbital has a lower energy than the 3d, and so fills next. That fits with the chemistry of potassium and calcium and the entire period. So the answer is 3d</span>
Sindrei [870]3 years ago
3 0

Answer: 3d

Explantion:

1) Period 4 contains the elements with atomic numbers 19 through 36.

2) The elements with atomic numbers 19 (K) and 20 (Ca) fill the orbital 4s.

3) After that, as Aufbau's rule may help you to remember, the energy of the orbitals 3d is lower than the energy of the orbtitals 4p. So, the element 21 (Sc) start fillind the orbital 3d.

There are ten 3d orbitals, so the elements 21 through 30 fill the 3d orbitals.

Those elements, called transition metals are: Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, and Zn.

When the 3d orbitals are full, the next elements in the same period 4, fill the six 4p orbitals.

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First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

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\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

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\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

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p^o = partial pressure of gas

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X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

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p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

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