Q1. TI (210/81Thallium)
Q2.
The answers are opposite from each other
Steps 1 and 3 are correct rate
I think it's Sodium Sulfide, because Na2S contains one Sodium and two Sulfur. Plus Sulfide<span> is used to describe any of three types of chemical compounds that contain sulfur.</span>
We will assume that the solvent is water. So, if we have 100 grams of the solution, 19 grams will be sodium hydroxide, while the remaining 81 grams will be water.
The molar weight of sodium hydroxide, NaOH, is 40. The molar weight of water is 18. Finding the moles of each:
NaOH:
19 / 40 = 0.475
Water:
81 / 18 = 4.5
Total moles present:
4.5 + 0.475 = 4.975 moles
The mole fraction of NaOH is:
0.475 / 4.975 = 0.0955
The mole fraction of NaOH is 0.0955
The free energy change of the reaction; Fe (s) + Au3+ (aq) -> Fe3+ (aq) + Au (s) is calculated to be -443.83KJ/mol.
For the reaction shown in question 7, we can divide it into half equations as follows;
Oxidation half equation;
6 Al (s) -------> 6Al^3+(aq) + 18e
Reduction half equation;
3Cr2O7^2-(aq) + 42H^+(aq) + 18e -----> 6Cr^3+(aq) + 21H2O(l)
The balanced reaction equation is;
6Al(s) + 3Cr2O7^2-(aq) + 42H^+(aq) -----> 6Al^3+(aq) + 6Cr^3+(aq) + 21H2O(l)
The E° of this reaction is obtained from;
E° anode = -0.04 V
E°cathode = +1.50 V
E° cell = +1.50 V - (-0.04 V) = 1.54 V
Given that;
ΔG° = -nFE°cell
n = 3, F = 96500, E°cell = 1.54 V
ΔG° = -(3 × 96500 × 1.54)
ΔG° = -443.83KJ/mol
Learn more: brainly.com/question/967776
