AP Chemistry 1.5 Atomic Structure and Electron Configuration - Exam Style questions - FRQs- New Syllabus

(a)(i)
Accept one of the following:
\( 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^5 \)
or \( [\mathrm{Ar}]\,4s^2\,3d^5 \)

(a)(ii)
\( 4s \)

(b)
\( 62.673\ \mathrm{g} – 61.262\ \mathrm{g} = 1.411\ \mathrm{g\ Cl} \)

(c)
\( 1.411\ \mathrm{g\ Cl} \times \dfrac{1\ \mathrm{mol\ Cl}}{35.45\ \mathrm{g\ Cl}} = 0.03980\ \mathrm{mol\ Cl} \)

(d)
\( \dfrac{0.03980\ \mathrm{mol\ Cl}}{0.0199\ \mathrm{mol\ Mn}} = 2.00 \)
Therefore, the empirical formula is \( \mathrm{MnCl}_{2} \).

(e)
Less than. If some of the \( \mathrm{Mn}_{x}\mathrm{Cl}_{y} \) splatters out, the final measured mass is smaller, so the calculated mass and moles of chlorine are smaller.

(f)(i)
\( \mathrm{2\,MnO}_{2}(s) + \mathrm{H}_{2}\mathrm{O}(l) + 2\,e^- \rightarrow \mathrm{Mn}_{2}\mathrm{O}_{3}(s) + 2\,\mathrm{OH}^-(aq) \)
\( \mathrm{Zn}(s) + 2\,\mathrm{OH}^-(aq) \rightarrow \mathrm{ZnO}(s) + \mathrm{H}_{2}\mathrm{O}(l) + 2\,e^- \)
Overall:
\( \mathrm{2\,MnO}_{2}(s) + \mathrm{Zn}(s) \rightarrow \mathrm{Mn}_{2}\mathrm{O}_{3}(s) + \mathrm{ZnO}(s) \)

(f)(ii)
\( E^\circ_{\mathrm{cell}} = 0.15\ \mathrm{V} – (-1.28\ \mathrm{V}) = 1.43\ \mathrm{V} \)

(f)(iii)
\( \Delta G^\circ = -nFE^\circ \)
\( \Delta G^\circ = -(2)(96{,}485)(1.43)\left(\dfrac{1\ \mathrm{kJ}}{1000\ \mathrm{J}}\right) = -276\ \mathrm{kJ/mol}_{\mathrm{rxn}} \)

(f)(iv)
Disagree. The battery is sealed, so the total mass remains the same.