What are negative charged particles orbiting the nucleus called?
1 answer:
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Answer:
Explanation:
Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.
The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:
Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:
The net equation is then:
However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:
Actual anode half-equation:
Actual cathode half-equation:
Actual net reaction:
Answer:
The volume of nitrogen oxide formed is 35.6L
Explanation:
The reaction of nitric acid with copper is:
Cu(s) + 4HNO₃ → Cu(NO₃)₂ + 2NO₂(g) + 2H₂O(l)
Moles of copper are:
Moles of nitric acid are:
As 1 mol of Cu reacts with 4 moles of HNO₃:
0.697 mol Cu × (4mol HNO₃ / 1mol Cu) = 2.79 moles of HNO₃ will react. That means Cu is limiting reactant.
Moles of NO₂ produced are:
0.697 mol Cu × (2mol NO₂ / 1mol Cu) = <em>1.394 moles of NO₂</em>
Using PV = nRT
<em>Where P is pressure (735torr / 760 = 0.967atm); n are moles (1.394mol); R is gas constant (0.082atmL/molK); T is temperature (28.2°C + 273.15 = 301.35K). </em>
Thus, volume is:
V = nRT / P
V = 1.394mol×0.082atmL/molK×301.35K / 0.967atm
V = 35.6L
<em>The volume of nitrogen oxide formed is 35.6L</em>