Answer:
magnesium metal
Explanation:
According to the redox reaction equation, six electrons were transferred hence n=6 and F= Faraday's constant 96500C. ∆G° is given hence E°cell can easily be calculated as follows:
From ∆G°= -nFE°cell
E°cell= -∆G°/nF= -(-411×10^3/96500×6)
E°cell= 0.7098V
But for Al3+(aq)/Al(s) half cell, E°= -1.66V from standard table of reduction potentials.
E°cell= E°cathode- E°anode but Al3+(aq)/Al(s) half cell is the cathode
Hence
E°anode=E°cathode - E°cell
E°anode= -1.66-0.7098= -2.37V
This is the reduction potential of Mg hence the anode material was magnesium metal
Answer:
Mass of Rb-87 is 86.913 amu.
Explanation:
Given data:
Average mass of rubidium = 85.4678 amu
Mass of Rb-85 = 84.9117
Ratio of 85Rb/87Rb in natural rubidium = 2.591
Mass of Rb = ?
Solution:
The ration of both isotope is 2.591 to 1. Which means that for 2.591 atoms of Rb-85 there is one Rb-87.
For 100% naturally occurring Rb = 2.591 + 1 = 3.591
% abundance of Rb-85 = 2.591/ 3.591 = 0.722
% abundance of Rb-87 = 1 - 0.722= 0.278
84.9117 × 0.722 + X × 0.278 = 85.4678
61.306 + X × 0.278 = 85.4678
X × 0.278 = 85.4678 - 61.306
X × 0.278 = 24.1618
X = 24.1618 / 0.278
X = 86.913 amu
The hallogens chloride with br
