How long would it take a current of 10A to deposit 6.36g of copper during the electrolysis of copper (ll) tetraoxosulphate VI so
1 answer:
It take 0.54 hours to deposit 6.36g of copper
<h3>Further explanation</h3>Faraday's Law I
"The mass of the substance formed at each electrode is proportional to the electric current flowing in the electrolysis
W = e.i.t / 96500
e = equivalent = Ar / valence
i = current, A
t = time, s
W=6.36 g
e = 63.5 : 2 =31.75
i = 10 A
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Explanation:
Relation between pressure, latent heat of fusion, and change in volume is as follows.
Also,
where, is the difference in specific volumes.
Hence,
As, = 22.0 J/mol K
And, ...... (1)
where, = density of water
= density of ice
M = molar mass of water =
Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.
=
=
Therefore, calculate the required pressure as follows.
=
or, = 145 bar/K
Hence, for change of 1 degree pressure the decrease is 145 bar and for 4.7 degree change dP =
= 681.5 bar
Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.