Order of metals from least reactive to most reactive: B <C <A <D
<h3>Further explanation</h3>
Reducing agents are substances that experience oxidation
Oxidizing agents are substances that experience reduction
The metal activity series is expressed in voltaic series
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>
The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent
The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent
So that the metal located on the left can push the metal on the right in the redox reaction
Let's analyze the statement in the problem
I. Only A, C and D react with 1 mol/L HCl to give H₂(e)
M + HCl ⇒ MCl + H₂(MCl : alkali, MCl₂ : alkaline earth)
A, C and D can react with 1 mol / L HCl, meaning metals A, C and D are located to the left of element H (more reactive), and B in the right of element H
II. When A is added to solutions of the other metal ions, metallic B and C are formed but not D.
This means that metal A is more reactive than metals B and C, while D is more reactive than A, so metal D is the most reactive
where N0 is the initial quantity, Nt is the remaining quantity after time t, t1/2 is the half-time. So work out the equation, t1/2 = t (-ln2)/ln(Nt/N0) = 11.5*(-ln2)/ln(12.5/100) = 3.83 days