Answer: B. 1:2
Explanation: Beryllium and chlorine forms a binary ionic compound. Ionic compound is formed when a metal loses its electrons to a receiving non metal. Beryllium (metal) has two valence electrons while chlorine (nonmetal) has seven valence electrons, and so a beryllium atom has to give out its two valence electrons to attain a duplet stable structure while a chlorine atom will gain one electron to attain its stable octet structure. In the reaction between beryllium and chlorine, two atoms of chlorine have to accept the two electrons from one beryllium atom to attain their stable octet structure.
The formula of the compound formed is BeCl2.
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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Explanation:
Equation of the reaction:
Br2(l) + Cl2(g) --> 2BrCl(g)
The enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, BrCl.
The standard enthalpy change of formation for a compound,
ΔH°f, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of 1 atm.
This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction
1/2Br2(g) + 1/2Cl2(g) → BrCl(g)
Here, ΔH°rxn = ΔH°f
This means that the enthalpy change for this reaction will be twice the value of ΔH°f = 2 moles BrCl
Using Hess' law,
ΔH°f = total energy of reactant - total energy of product
= (1/2 * (+112) + 1/2 * (+121)) - 14.7
= 101.8 kJ/mol
ΔH°rxn = 101.8 kJ/mol.
2H2 (g) + O2 (g) -->2H2 O(g)
mole ratio of H2:O2=2:1
7.25/2=3.625
Answer: 5
Explanation: this is because the energy level of the emitted of absorbed photon increases as the number of electron shell decreases, thereby making the inner shell have higher energy than other shells
