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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8.5L H2O (1 mol H2O/22.4L)=.37946 mol H2O
.3796 mol H2O(1 mol CH4/2 mol H2O)=.18973 mol CH4
.18973 mol CH4(22.4L/1mol CH4)=4.25L CH4 gas
D will be the answer I hope that help good luck I hope. My answer help you
Answer:
Explanation:
A single replacement or single displacement reaction is a reaction in which one substance replaces another.
A + BC → AC + B
The replacement of an ion in solution by a metal higher in the activity series is a special example of this reaction type.
The relative positions of the elements in the activity series provides the driving force for single displacement reactions.
A double replacement reaction is one in which there is an actual exchange of partners between reacting species. This reaction is more common between ionic substances;
AB + CD → AC + BD
Such reactions are usually driven by;
- formation of precipitation
- formation of water and a gaseous product
Answer:
4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.
Explanation:
Half life (t1/2) = 8 days
Original mass (No) = 64 g
Elapsed time (t) = 32 days
Mass remaining (Nt) = ?
Using the half life equation we can obtain the mass remaining (Nt)
Nt = No (1/2) ^t/t1/2
Substituting the values, we have;
Nt = 64 * ( 1/2 ) ^32/8
Nt = 64 * (1/2) ^4
Nt = 64 * 0.0625
Nt = 4 g
So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.