Answer:
Oxygen
Explanation:
· Air decolorization makes use of chromophores’ instability on oxygen to decolorize the oil by air-oxidizing pigments. For example, the carotenoid and chlorophyll in the oil are very unstable because of their structure, which is easy to be discolored under the action of oxygen. However, air decolorization leads to thermal oxidation of the oil, too.
B. The partial pressure of N2 is 101 kPa
<h3>Further explanation</h3>
Given
volume = 22.4 L
1.0 mol of nitrogen and 2.0 mol of hydrogen at 0°C
Required
Total pressure and partial pressure
Solution
Ideal gas law :
PV = nRT
n total = 3 mol
T = O °C + 273 = 273 K
P = nRT/V
P = 3 x 0.08205 x 273 / 22.4
P total = 3 atm = 303,975 kPa
P Nitrogen = 1/3 x 303.975 = 101.325 kPa
P Hydrogen = 2/3 x 303.975 = 202.65 kPa
Answer:
0.9 moles of water
Explanation:
Use mole ratios:
5 : 6
divide by 5 on both sides
1 : 1.2
multiply by 0.75 on both sides
0.75 : 0.9
So the result is 0.9 moles of water
(Please correct me if I'm wrong)
Answer:
![Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)](https://tex.z-dn.net/?f=Fe%20%28s%29%20%2B%20Sn%5E%7B2%2B%7D%20%28aq%29%5Crightarrow%20Fe%5E%7B2%2B%7D%20%28aq%29%20%2B%20Sn%20%28s%29)
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:
![Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-](https://tex.z-dn.net/?f=Sn%20%28s%29%5Crightarrow%20Sn%5E%7B2%2B%7D%20%28aq%29%20%2B%202e%5E-)
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:
![Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)](https://tex.z-dn.net/?f=Fe%5E%7B2%2B%7D%20%28aq%29%20%2B%202e%5E-%5Crightarrow%20Fe%20%28s%29)
The net equation is then:
![Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)](https://tex.z-dn.net/?f=Sn%20%28s%29%20%2B%20Fe%5E%7B2%2B%7D%20%28aq%29%5Crightarrow%20Fe%20%28s%29%20%2B%20Sn%5E%7B2%2B%7D%20%28aq%29)
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:
![Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-](https://tex.z-dn.net/?f=Fe%20%28s%29%5Crightarrow%20Fe%5E%7B2%2B%7D%20%28aq%29%20%2B%202e%5E-)
Actual cathode half-equation:
![Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)](https://tex.z-dn.net/?f=Sn%5E%7B2%2B%7D%20%28aq%29%20%2B%202e%5E-%5Crightarrow%20Sn%20%28s%29)
Actual net reaction:
![Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)](https://tex.z-dn.net/?f=Fe%20%28s%29%20%2B%20Sn%5E%7B2%2B%7D%20%28aq%29%5Crightarrow%20Fe%5E%7B2%2B%7D%20%28aq%29%20%2B%20Sn%20%28s%29)