Yes the answer is 1s22s22p63s23p64s23d5
Answer:
½O 2 + 2e - + H 2O → 2OH.
Explanation:
Redox reactions - Higher
In terms of electrons:
oxidation is loss of electrons
reduction is gain of electrons
Rusting is a complex process. The example below show why both water and oxygen are needed for rusting to occur. They are interesting examples of oxidation, reduction and the use of half equations:
iron loses electrons and is oxidised to iron(II) ions: Fe → Fe2+ + 2e-
oxygen gains electrons in the presence of water and is reduced: ½O2 + 2e- + H2O → 2OH-
iron(II) ions lose electrons and are oxidised to iron(III) ions by oxygen: 2Fe2+ + ½O2 → 2Fe3+ + O2-
Answer:
Scandium(III) fluoride, ScF3, is an ionic compound. It is slightly soluble in water but dissolves in the presence of excess fluoride to form the ScF63− anion.
hope it will help you......
D, carbon dioxide and water. Methane and oxygen PRODUCE carbon dioxide and water in their reaction.
