Answer : Option 3) Electrolysis.
Explanation : In the process of electrolysis, decomposition of the chemical is done by passing the electric current from the liquid or the solution containing ions in it. This is the process where compound is broken into ions namely, cations and anions.
Rest all options does not involves chemical decomposition of the compound.
Chromatography is a technique where rates of solute and solvent are compared in a medium where the solute components move at different rates and the separation is obtained.
Distillation simply purifies the liquid or solution by heating or cooling process.
Filtration just involves removal of impurities by using a more porous material device to get purified compound.
A beta emission radioactive decay can pass through the body.
Ignition wires make it more accurate because it will cook it faster
stirrer would have the less results of fast
a sealed bomb may cook it fast but you would have to be careful and don't mess up
Answer:
Elements that fall between those on the left and right sides of the periodic table
Explanation:
Transition metals:
These are present at the center of periodic table.
These are d-block elements.
They include the elements of group 3 to 12 in periodic table.
They have large charge to radius ratio.
They mostly form paramagnetic compounds.
They shoes more than one oxidation state.
They form colored compounds.
They all have high melting and boiling point.
They have high densities.
They form stable complexes.
The elements of f-block are also transition but they are called inner transition.These are consist of two series lanthanide and actinides.
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
