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.
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The grams of water produced is c
Ductility because ductility is the ability to be stretched and be made into wires.
Answer:
c. liquid
Explanation:
because the are not very close if the said very closely packed together it would have been a solid
Answer:
The equilibrium expression is:
CoC2O4(s)⇌Co2+(aq)+C2O2−4(aq)
For this reaction:
Ksp = [Co2+][C2O2−4]=1.96×10−8
Explanation:
Batteries will not clot if cobalt ions are removed from its cells. Some blood collection tubes contain salts of the oxalate ion,
C2O2−4
, for this purpose. At sufficiently high concentrations, the calcium
and oxalate ions form solid, CoC2O4·H2O (which also contains water bound in the solid). The concentration of Co2+ in a sample of blood serum is 2.2 × 10–3M. What concentration of
C2O2−4
ion must be established before CoC2O4·H2O begins to precipitate.
CoC2O4 does not appear in this expression because it is a solid. Water does not appear because it is the solvent.
Solid CoC2O4 does not begin to form until Q equals Ksp. Because we know Ksp and [Co2+], we can solve for the concentration of
C2O2−4
that is necessary to produce the first trace of solid:
