It is either the 3rd answer or the 4th answer. Both are correct ways to write hydrates, but I was taught the 4th way. Just make sure this answer matches the format of the examples your teacher gave you.
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.
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The reaction is
<span>Zn (s) + 2 HCl (aq) ----> ZnCl2 (aq) + H2 (g)
which is already balanced
5.4 L of 2.8 M HCl contains
5.4 L (2.8 M) = 15.12 moles HCl
The amount of Zinc that will react completely with the acid is
15.12 mol HCl (1 mol Zn / 2 mol HCl) (65 g Zn/1 mol Zn) = 491.4 g Zn
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