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3 years ago

8

What is the orbital diagram for silicon

2 answers:

Novay_Z [31]3 years ago

7 0

Answer: Silicon the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Silicon go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons.

Hope this helps! :)

erik [133]3 years ago

3 0

Answer:

In writing the electron configuration for Silicon the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Silicon go in the 2s orbital. The nex six electrons will go in the 2p orbital. The p orbital can hold up to six electrons

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Explanation:

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Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )

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Density of gas = $\frac{PM}{ZRT}$

$\frac{dP}{dZ} = \rho \times g$

$\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

$\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

$ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

= 0.4548

$P_{1} = P_{o} \times e^{0.4548}$

= $1.4 \times 10^{7} Pa \times 1.5797$

= $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$ .

(b) At the bottom of the tank,

$P_{2} = P_{1} + \rho \times g \times h$

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

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= 309.8 bar

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