Helium is a very important element for both industrial and research applications. In its gas form it can be used for welding, an

d since it has a very low melting point (only 0.95 K under 2.5 MPa) it can be used in liquid form to cool superconducting magnets, such as those found in particle physics experiments. Say we have a cylinder of n = 125 moles of Helium gas at room temperature (T = 20° C). The cylinder has a radius of r = 17 cm and a height h = 1.64 m.What pressure is Helium gas under?

Chemistry

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Draw the Lewis structure of N₂O₄ and then choose the appropriate pair of hybridization states for the two central atoms. Your an

castortr0y [3046]

Answer:

Refer to the explanation

Explanation:

In this scenario, one must consider the valence electrons for each atom:

N => 5 electrons

O => 6 electrons

<pFor the structure $N_2O_4$ , the total valence would amount to:

$(5*2)+(6*4)=34~electrons$

Moreover, each atom needs to have 8 electrons. Therefore, for the oxygens labeled 5 and 3, there will be 3 lone pairs and 1 bond (totaling 8 electrons). Regarding the oxygens marked 6 and 4, there will be 2 lone pairs and 2 bonds, again totaling 8 electrons. For the nitrogens labeled 1 and 2, there will be 4 bonds, summing up to 8 electrons.

To determine the hybridization, we count both the atoms and lone pairs surrounding the nitrogen. There are 3 atoms and no lone pairs. Considering the applicable rules:

$Sp^3~=~4$

$Sp^2~=~3$

$Sp~=~2$

Taking this into account, nitrogen's hybridization is $Sp^2$ .

Refer to figure 1

I trust this is helpful!

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