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1 year ago

how many electrons does it take to fill the mos formed from the combination of the 3s orbitals of two atoms?

1 answer:

7 0

The number of electrons it take to fill the mos formed from the combination of the 3s orbitals of two atoms simply is 14 electrons.

<h3>How electrons are distributed in the 3s orbitals.</h3>

The 3s orbital possess two different spherical nodes which is highly connected with the principal quantum number. In order words, it has 2 radial nodes. Also the shape of the 3s orbital is spherical in shape.

That being said, from the context of the above given task, the number of electrons which fill the mos formed from the combination of the 3s orbitals of two atoms is fourteen electrons.

However, the electron configuration of an atom simply is the arrangement of electrons in the electron shell or orbit of the atom of that element.

In conclusion, it can be deduced from above s orbital has a maximum of two electrons and this energy increases as the orbitals increases.

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Starting with 2.50 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0C, a chemist first heats the gas at con

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Answer:

a) $T_b=590.775k$

b) $W_t=1.08*10^4J$

d) $Q=3.778*10^4J$

d) $\triangle V=4.058*10^4J$

Explanation:

From the question we are told that:

Moles of N2 $n=2.50$

Atmospheric pressure $P=100atm$

Temperature $t=20 \textdegree C$

$t = 20+273$

$t = 293k$

Initial heat $Q=1.36 * 10^4 J$

Generally the equation for change in temperature is mathematically given by

$\triangle T=\frac{Q}{N*C_v}$

Where

$C_v=Heat\ Capacity \approx 20.76 J/mol/K$

$T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }$

$T_b-293k=297.775$

$T_b=590.775k$

Generally the equation for ideal gas is mathematically given by

$PV=nRT$

For v double

$T_c=2*590.775k$

$T_c=1181.55k$

Therefore

$PV=Wbc$

$Wbc=(2.20)(8.314)(1181_590.778)$

$Wbc=10805.7J$

Total Work-done $W_t$

$W_t=Wab+Wbc$

$W_t=0+1.08*10^4$

$W_t=1.08*10^4J$

Generally the equation for amount of heat added is mathematically given by

$Q=nC_p\triangle T$

$Q=2.20*2907*(1181.55-590.775)\\$

$Q=3.778*10^4J$

Generally the equation for change in internal energy of the gas is mathematically given by

$\triangle V=nC_v \triangle T$

$\triangle V=2.20*20.76*(1181.55-293)k$

$\triangle V=4.058*10^4J$

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2 years ago

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Answer:

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Explanation:

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Explanation:

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