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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Answer:
The same holds true for pH values above 7, each of which is ten times more alkaline (another way to say basic) than the next lower whole value. For example, pH 10 is ten times more alkaline than pH 9 and 100 times (10 times 10) more alkaline than pH 8.
Explanation:
Answer:
ΔH = -470.4kJ
Explanation:
It is possible to sum 2 or more reactions to obtain the ΔH of the reaction you want to study (Hess's law). Using the reactions:
1. CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s)ΔH = −414kJ
2. 6C2H2(g) + 3CO2(g) + 4H2O(g) → 5CH2CHCO2H(g)ΔH = 132kJ
6 times the reaction 1.
6CaC2(s) + 12H2O(l) → 6C2H2(g) + 6Ca(OH)2(s)ΔH = −414kJ*6 = -2484kJ
This reaction + 2:
6CaC2(s) + 3CO2(g) + 16H2O(l) → + 6Ca(OH)2(s) + 5CH2CHCO2H(g) ΔH = -2484kJ + 132kJ = -2352kJ
As we want to calculate the net change enthalpy in the formation of just 1 mole of acrylic acid we need to divide this last reaction in 5:
6/5CaC2(s) + 3/5CO2(g) + 16/5H2O(l) → + 6/5Ca(OH)2(s) + CH2CHCO2H(g) ΔH = -2352kJ / 5
<h3>ΔH = -470.4kJ</h3>
Answer:
4.17e+22 atoms of tin are present in the cube
Explanation:
We don't require the size of the cube. With the mass and the molar mass of tin = 118.7 amu we can find moles of Tin. As 1 mol = 6.022x10²³ atoms we can find the number of atoms:
<em>Moles Tin:</em>
8.21g * (1mol / 118.7g) = 0.0692 moles Tin
<em>Atoms Tin:</em>
0.0692 moles Tin * (6.022x10²³ atoms / mol) =
<h3>4.17e+22 atoms of tin are present in the cube</h3>
Bronchi that is the answer
