How many grams of mercury can be produced if 18.0 g of mercury (11) oxide decomposes?
1 answer:
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C. Aluminum (Al) oxidized, zinc (Zn) reduced
<h3>Further explanation</h3>Given
Metals that undergo oxidation and reduction
Required
A galvanic cell
Solution
The condition for voltaic cells is that they can react spontaneously, indicated by a positive cell potential.
or:
E ° cell = E ° reduction-E ° oxidation
For the reaction to occur spontaneously (so that it E cell is positive), the E° anode must be less than the E°cathode
If we look at the voltaic series:
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>
The standard potential value(E°) from left to right in the voltaic series will be greater, so that the metal undergoing an oxidation reaction (acting as an anode) must be located to the left of the reduced metal (as a cathode)
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From the available answer choices, oxidized Al (anode) and reduced Zn (cathode) are voltaic/galvanic cells.
Answer:
Explanation:
The breakdown reaction of ozone is as follows
It can be seen that 2 moles of ozone is required in the complete cycle
So for 10 cycles, 20 moles of ozone is required
m = Mass of = 15.5 g
M = Molar mass of = 104.46 g/mol
P = Pressure = 24.5 mmHg
T = Temperature = 232 K
R = Gas constant =
Number of moles is given by
From ideal gas law we have
For 20 cycles of the reaction the volume of the ozone is .