There is 11 valence electrons in sodium (Na)
Answer:
6 Cl0 + 6 e- → 6 Cl-I (reduction)
2 Fe0 - 6 e- → 2 FeIII (oxidation)
To determine the k for the second condition, we use the Arrhenius equation which relates the rates of reaction at different temperatures. We do as follows:
ln k1/k2 = E / R (1/T2 - 1/T1) where E is the activation energy and R universal gas constant.
ln 1.80x10^-2 / k2 = 80000 / 8.314 ( 1/723.15 - 1/593.15)
k2 = 0.3325 L / mol-s
4.1 h = 14760 s
<span>t 1/2 = ln 2 / k </span>
<span>k = rate reaction = 4.97 x 10^-5 </span>
<span>ln 0.045 / 0.36 = - 4.97 x 10^-5 t </span>
<span>2.08 = 4.97 x 10^-5 t </span>
<span>t = 41839.9 s = 11 h 37 min 19 s</span>
