This problem is providing information about the initial mass of mercury (II) oxide (10.00 g) which is able to produce liquid mercury (8.00 g) and gaseous oxygen and asks for the resulting mass of the latter, which turns out to be 0.65 g after doing the corresponding calculations.
Initially, it is given a mass of 10.00 g of the oxide and 1.35 g are left which means that the following mass is consumed:
Now, since 8.00 grams of liquid mercury are collected, it is possible to calculate the grams of oxygen that were produced, by considering the law of conservation of mass, which states that the mass of the products equal that of the reactants as it is nor destroyed nor created. In such a way, the mass of oxygen turns out to be:
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Because copper sulphate is not an ionic substance, no bonds are changed when it is dissolved. Thus it has to be a physical change
Total of 127.013 C of charge is passed
Given
weight of Ag solution before current has passed = 1.7854 g
weight of Ag solution after current has passed = 1.8016 g
Molecular mass of Ag = 107.86 g
Faraday's Constant = 96485
First of all we have to apply Faraday's First Law of Electrolysis i.e
m = ZQ
where
Z is propotionality constant (g/C)
Q is charge (C)
Hence,
Z = Atomic mass of substance/ Faraday's Constant
= 
= 0.0011178 g/C
Now ,
change in mass before and after the passing of current (Δm)
Δm = 1.8016g-1.7854g
= 0.0162g
Now amount of coulombs passed = 
amount of coulombs passed = 127.03524 C
Thus from the above conclusion we can say that amount of coulombs have passed is 127.03524 C
Learn more about Electrolysis here: brainly.com/question/16929894
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... Really wanna go sleep
Particles in a solid are fixed in place.!!
