7.47 Two atoms have the electron configurations 1s22s22p6 and 1s22s22p63s1. The first ionization energy of one is 2080 kJ/mol an
1 answer:
Answer:
2080 kJ/mol is the first ionization of 1st atom and 496 kJ/mol is the first ionization of 2nd atom
Explanation:
Given electronic configurations are :
1st:
2nd :
given 1st ionization energy are: 2080 kJ/mol and 496 kJ/mol
generally ionization energy of fulfilled orbital is more than half filled orbital and these two state are more stable.
therefore ionization energy of fulfilled is more than half filled orbital
hence
ionization energy of 1st atom will be very high because its orbital is fulfilled and less energy for 2nd atom so 2080 kJ/mol is the first ionization of 1st atom and 496 kJ/mol is the first ionization of 2nd atom.
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The limiting reagent : NH₃
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Reaction
4NH₃(g)+5O₂(g)⇒4NO(g)+6H₂O(g)
mol NH₃ :
mol O₂ :
mol ratio
NH₃ :
O₂ :
Limiting reactants : NH₃ (smaller ratio)
Explanation:
It is known that specific heat of water is 4.184 and atomic mass of tin is 118.7 g/mol. For the given situation,
Let us assume that,
= mass of Sn
= mass of
Therefore, heat energy expression for heat lost and gained is as follows.
= 0.207
For, 118.7 g the specific heat of tin will be calculated as follows.
= 24.5
Thus, we can conclude that specific heat of tin is 24.5 .