This problem is providing us with two reactions between gold (III) chloride and cobalt in order to obtain the two possible cobalt chlorides and gold. Thus, it asks for the most feasible reaction, which is found to be the second one.
<h3>Uses of stoichiometry</h3>
In chemistry, relationships between moles and grams in chemical reactions can be studied via stoichiometry, which is based on proportional factors relating molar masses and mole ratios in chemical equations.
In this case, since we have two reactions and the same initial amount of cobalt, one can calculate the grams of solid gold, with the 1:1 mole ratio of these two in the first reaction and the 3:2 mole ratio in the second one and their atomic masses of 58.933 g/mol and 196.966 g/mol respectively:
Thus, since the second reaction would produce 13.56 grams of gold, and it is pretty much the same to the recovered amount of 13.572 grams, one concludes the second reaction took place in the experiment.
Learn more about stoichiometry: brainly.com/question/9743981
OILRIG:
Oxidation is loss (of electrons)
Reduction is gain (of electrons)
so...
The first one is an oxidation half-equation as the Sn loses electrons;
The second one is a reduction half-equation as the Cl₂ gains electrons
Answer:
0.4 * 10⁻⁶mol
Explanation:
The unit conversion required are ,
1 μ M = 10⁻⁶ M
1 L = 1000 mL
1 mol = 1000 mmol
molarity of a solution is given as -
Molarity (M) = ( w / m ) / V ( in L)
where ,
m = molecular mass ,
w = given mass ,
V = volume of solution ,
Hence , the unit is ,
M = mol / L
from the question ,
0.4 μ M = 0.4 * 10⁻⁶ M
Now,
0.4 * 10⁻⁶ M = 0.4 * 10⁻⁶ mol / L
Since ,
1 L = 1000 mL
and ,
1 mol = 1000 mmol
0.4 * 10⁻⁶ mol / L = 0.4 * 10⁻⁶ * 1000 mol / 1000 mL = 0.4 * 10⁻⁶mol