Explanation:
Bayer process is industrial method of the refining of the bauxite to produce alumina which is aluminum oxide.
As alumina is amphoteric in nature, it exhibits a higher solubility at both the extremes of pH range, it is possible to dissolve alumina in low as well as in high pH solutions.
Dissolution of the alumina at high pH is well recognized in Bayer process. Bauxite is digested in very <u>high pH solution (> 13) of alkali</u> like sodium hydroxide at temperature of about 150–250°C and pressure at 20 atm. <u>This is done so that the dissolved alumina is separated from rest of insoluble bauxite minerals. </u>
The final volume of the methane gas in the container is 6.67 L.
The given parameters;
- <em>initial volume of gas in the container, V₁ = 2.65 L</em>
- <em>initial number of moles of gas, n₁ = 0.12 mol</em>
- <em>additional concentration, n = 0.182 mol</em>
The total number of moles of gas in the container is calculated as follows;

The final volume of gas in the container is calculated as follows;

Thus, the final volume of the methane gas in the container is 6.67 L.
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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:
Add G and H to determine the atomic mass for helium
Explanation:
RADIATION. Radio waves, microwaves, IR, light, UV, x-rays, GAMMA waves etc are ALL Electromagnetic radiation. The difference between ALL the above is the frequency, I.E. The number of waves per second. The higher the frequency the more energy.