Answer: Be= 2, C =4, Li = 1 and B=3
Explanation:
The valence shell can be define as the outermost shell of an atom that contains the valence electrons.
Beryllium (Be), electronic configuration; 1s2 2s2, = 2 electrons in its valence shell.
Carbon (C), electronic configuration; 1s2 2s2 2p2, = 4 electrons in its valence shell.
Lithium (Li), electronic configuration; 1s2 2s1 = 1 electron in its valence shell.
Boron (B) , electronic configuration; 1s2 2s2 2p1 = 3 electron in its valence shell.
I think the answer is mercury
Assuming that the reactants are:
(NH4)2SO4 (aq) + Ba(NO3)2 (aq)
and the products are:
BaSO4 (s) + 2NH4NO3 (aq),
then you will have to determine which product is insoluble. You should have access to solubility rules to help you determine this.
According to the solubility rules, the following elements are considered insoluble when paired with SO4:
Sr^2+, Ba^2+, Pb^2+, Ag^2+, and Ca^2+
Therefore, the precipitate will be BaSO4 (s).
