Answer:
12.5 g of Li are needed in order toproduce 0.60 moles of Li₃N
Explanation:
The reaction is:
6Li(s) + N₂(g) → 2Li₃N(s)
If nitrogen is in excess, the lithium is the limiting reactant.
Ratio is 2:6
2 moles of nitride were produced by 6 moles of Li
Then, 0.6 moles of nitride were produced by (0.6 .6)/ 2 = 1.8 moles of Li
Let's convert the moles to mass → 1.8 mol . 6.94 g/ 1mol = 12.5 g of Li
The reaction formula CH4 + 2O2 → CO2 + 2H2O shows the oxidation of 1 mole of CH4 (Methane) will yield 1 mole of CO2 (Carbon Dioxide). Since 1 mole of CH4 will weigh 12g (for the Carbon) + 4g (1g for each Hydrogen) = 16g, then 32g of CH4 will correspond to 32g / 16g/mole = 2 moles. Therefore the oxidation of 2 moles of CH4 will yield 2 moles of CO2.
Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.
Commonly, sacrificial electrodes are employed to stop another metal from corroding or oxidising. A metal that is more reactive than the metal being shielded must serve as the sacrificial electrode. Magnesium, aluminium, and zinc are the three metals most frequently used in sacrificial anodes.
Manganese-Magnesium (Mn-Mg) electrode is more suited for on-shore pipelines where the electrolyte (soil or water) resistivity is higher since it has the highest negative electropotential of the three. In order to replenish any electrons that could have been lost during the oxidation of the shielded metal, the highly active metal offers its electrons.
Therefore, Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.
Learn more about electrode here:
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- emission spectrum is the amount of electromagnetic radiation of each frequency it emits when it is heated (or more generally when it is excited). When the electrons in the element are excited, they jump to higher energy levels.
- Absorption spectrum is a technique used to find out what makes up a sample of a substance – in other words, a chemical analysis.
