The first four ionization energies for an imaginary element, xz, are e1 = 102 kcal, e2 = 186 kcal, e3 = 4021 kcal, and e4 = 4862 kcal. the number of valence electrons most probable as indicated by this ionization data is:

Two valence electrons.

Explanation: Ionization energy is the energy required to remove electrons from valence shell of an atom in its gaseous state.

More closer are the electrons to the nucleus, more energy is required to remove them. Ionization energy increases as the effective nuclear charge increases.

First two ionization energy values are low and there isn't a much difference between them but the third ionization energy is very much high as compared to first and second ionization energy values. The fourth ionization energy value is also closer to third ionization energy value.

It indicates that first and second electrons are removed from same shell. Similarly, third and fourth electrons are also removed from same shell which is different than the shell from which first and second electrons are removed.

So, outer most shell that is valence shell has only two electrons. Third ionization energy has increased a lot as the third electron is removed from the inner shell and first two electrons are removed from valence shell.

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

5.6⋅10−^19  j

do you need the explaination?