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Sonja [21]
2 years ago
9

In solid I₂, is the distance between the two I nuclei of one I₂ molecule longer or shorter than the distance between two I nucle

i of adjacent I₂ molecules? Explain.
Chemistry
1 answer:
Crazy boy [7]2 years ago
4 0

Shorter because the I-I bond in the I2 molecule is covalent whereas the bond between adjacent molecules is due to dispersion.

<h3>What is bond length?</h3>

Bond length typically falls between 0.1 and 0.2 nm. Covalent radius is the term used to describe the length of a connection formed between two comparable atoms. The order of the bond or the quantity of linked electrons between two atoms determines the bond length.

<h3>What connection exists between the length and the strength of a bond?</h3>

The C-C bond in alkanes must be the longest since it is the weakest, and the C-C bond in alkynes must be the shortest because it appears to be the strongest, according to the data. We determined in the preceding section that the bond strength is inversely correlated to the bond length.

<h3>How can bond length be calculated in chemistry?</h3>
  • The bond gets shorter as the atoms get smaller.
  • The length of the bond decreases as bond multiplicity increases.
  • The shorter the bond, the higher the effective nuclear charges of the bound atoms.

Learn more about bond length:

brainly.com/question/13054857

#SPJ4

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Calculate the number of atoms per cubic meter in aluminum. The density and atomic weight of aluminum are 2.70 g/cm3 and 26.98 g/
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Answer:

The number of Aluminium atoms present in per cubic meter is N=6.02\times 10^{28}\ atoms/m^3

Explanation:

Given the density of Aluminium (\rho ) is 2.70\ g/cm^3

And the atomic weight of Aluminium (A_{Al}) is 26.98\ g/mol

Also, the Avogadro's number (N_A) is 6.023\times 10^{23}\ atoms/mol

We need to find the number of atoms per cubic meter (N)

N=\frac{\rho N_{A}}{A_{Al}}

Plug the given values in the formula we get,

N=\frac{2.70\times 6.022\times 10^{23}}{26.98}\\ \\N=0.602\times 10^{23}\ atoms/cm^3\\N=6.02\times 10^{22}\ atoms/cm^3\\N=6.02\times 10^{22}\times10^6\ atoms/m^3

N=6.02\times 10^{28}\ atoms/m^3

So, the number of Aluminium atoms present in per cubic meter is N=6.02\times 10^{28}

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