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lesya [120]
2 years ago
7

In addition to those in Table 14.3, other less stable nitrogen oxides exist. Draw a Lewis structure for each of the following:(c

) N₂O₃ with no N¬N bond
Chemistry
1 answer:
Natalka [10]2 years ago
5 0

Lewis structure for each of the following N₂O₃ with no N¬N bond is attached below.

Even though pi symmetry occupies the antibonding orbitals of NO, this is unimportant after the dimer forms. A sigma connection exists. The enthalpy of the newly formed sigma bond in the dimer is low because the loss of a particularly distinctive set of single-electron resonance forms that were available for no monomer offset the net gain in bond. When the whole free energy is taken into account, there is no gain because the entropic effects are on the order of 1030kJ/mol, and dimerization is entropically disfavored at G=17kJ/mol. Therefore, any little increase in enthalpy is cancelled out by the loss of entropy.

Learn more about dimer here-

brainly.com/question/17152685

#SPJ4

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A rectangular block of zinc was found to be 11.014 cm wide, 0.2481 m high, and 523.735 mm deep.
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From the stoichiometry of the problem, 2.01  × 10^5 g of zinc phosphate is produced.

<h3>What is volume?</h3>

Volume is the product of width, height and depth. The significant figures of the width, height and depth are 5, 4 and 6 respectively.The volume of the box is; 11.014  × 10^-2 m  × 0.2481 m × 523.735 × 10^-3 = 0.01431 m^3 or 14310 cm^3 or 0.5053ft^3.

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There are 195 g of zinc in  386 g zinc phosphate

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Learn more about stiochiometry: brainly.com/question/9743981

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