Answer:
98.9062 amu
Explanation:
The element between molybdenum and ruthenium is technetium.
Its atomic is 98.9062 amu.
Its atomic number is 44.
Its melting point is 2200 °C.
Its density is 11.5 g/cm³.
It is silver gray metal.
It is present in VII-B group.
It is d-block element.
It has seven valance electrons.
Its electronic configuration is:
Tc₄₄ = [Kr] 4d⁵ 5s²
