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skad [1K]
3 years ago
14

In which orbitals would the valence electrons for selenium (Se) be placed?

Chemistry
2 answers:
RSB [31]3 years ago
8 0
Selenium falls in the same column with oxygen therefore it has the same number of valence electrons with oxygen which is 6. It fills the 3d orbital but only 4s and 4p are counted. The electronic configuration is [Ar]3d^10 4s^2.
elixir [45]3 years ago
8 0

Selenium falls in the same column with oxygen therefore it has the same number of valence electrons with oxygen which is 6. It fills the 3d orbital but only 4s and 4p are counted. The electronic configuration is [Ar]3d^10 4s^2

Explanation:

Selenium has 6 valence electrons, two of which are placed in the S orbital and 4 of which are placed in the P orbital. In its electron structure, it is quite comparable to oxygen.

At the first energy level, the only orbital accessible to electrons is the 1s orbital, but at the next level, as well as a 2s orbital, there are 2p orbitals. A p orbital is formed like 2 identical balloons tied commonly at the nucleus. The orbital shows where there is a 95% chance of getting a critical electron.

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1. Compare and contrast the characteristics of metals and nonmetals.
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Answer:

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4 0
3 years ago
Read 2 more answers
Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

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