The internal energy of a system refers only to the energies of the nuclei of the atoms of the component molecules is the sum of the kinetic energy of all of its components is the sum of the potential and kinetic energies of the components is the sum of the rotational, vibrational, and translational energies of all of its components none of the above

The most appropriate answer here would be :

Internal energy of a system is the sum of the rotational, vibrational, and translational energies of all of its components

Explanation:

Internal energy of a system is the total energy the system possess. It is represented by U (I'll be referring to internal energy as U now). This option is particularly true for ideal gases. In ideal  monoatomic gases, U is the sum of translational kinetic energies only. In di and polyatomic gases, U is the sum of translational and rotaional kinetic energies. Also, vibrational kinetic energies come into play as we increase the temperature and this also adds to U. But, in real substances such as real gases, solids, liquids, there is also interatomic forces and these accounts for intermolecular potential energies. Intermolecular potential energies also add to U in these type of systems. But even for real gases, under many circumstances, the intermolecular potential energy can be neglected.

So, the most appropriate answer here is: Internal energy of a system is the sum of the rotational, vibrational, and translational energies of all of its components

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

the role of the microscope world in creating new materials .