After doing some analysis, the probe determines that the gas pocket contains an element that has an ionization energy of I = 1350 kJ/mol and an electron affinity of

EA = 32.0 kJ/mol.

What is the identity of this element?

you need to give more information

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

answer: they produce more slow than unicellular organisms.