Answer:
Al is oxidized while Ag is reduced.
Explanation:
The complete molecular equation is;
3Ag2S + 2Al --> 6Ag + Al2S3
Oxidation half equation;
2Al ------> 2Al^3+ + 6e
Reduction half equation;
6Ag^+ + 6e -------> 6Ag
Overall redox reaction equation;
2Al + 6Ag^+ ----->2Al^3+ + 6Ag
Hence; Al is oxidized while Ag is reduced.
Explanation:
The given data is as follows.
= 10 mM = 
M
= 750 ml, 
= 5 ml
= ?
Therefore, calculate the molarity of given NaCl stock as follows.
= 1.5 M
Thus, we can conclude that molarity of given NaCl stock is 1.5 M.
Volume of the gas is 525 L.
<u>Explanation:</u>
It is given that the volume of the gas divided by the temperature is 1.75.
V/T = 1.75
As per the Charles law, volume is proportional to the temperature.
V ∝ T
V/T = constant
Now we have to find V, and T is given as 300 K.
So plugin the values as,
V/300 = 1.75
Rearranging the equation to get V as,
V = 1.75×300
= 525 L
Think of it this way,
Mix Iron and sulphur in a bowl. How do you separate them? Use a magnet right. Yes.
Now, mix the iron and sulphur together but know, heat them up. Let them cool for a while. After that, use a magnet to separate. You cant. This is because the compound (FeS) now has a different property from its original components.
Apply this theory onto salts.
<span>Answer is: Van't Hoff factor
(i) for this solution is 1.051 .
Change in boiling point from pure solvent to solution: ΔT
=i · Kb · b.
Kb - </span><span>molal boiling point elevation constant</span><span> is 0.512°C/m.
b - molality, moles of solute per kilogram of solvent.
b = 1.26 m.
ΔT = 101.63°C - 100</span>°C = 1.63°C.
i = 1.63°C ÷ (0.512°C/m · 1.26 m).
i = 1.051.
