Answer:
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
Explanation:
Sodium is present in group 1.
It is alkali metal.
It has one valence electron.
The atomic number of sodium is 11.
Its atomic mass is 23 amu.
The longhand notation of electronic configuration of sodium can be written as,
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
The electronic configuration in shorthand notation( noble gas) would be written as,
Na₁₁ = [Ne] 3s¹
Sodium loses its one valence electron to complete the octet and get stable thus form +1 cation.
It react with halogen and form salt. Such as sodium chloride.
2Na + Cl₂ → 2NaCl
C, 0.746 mol Ag.
1 mol Ag = 6.022 x 1023 atoms of Ag -> 4.49 x 1023 atoms of Ag x 1mol Ag/6.022 x 1023 atoms -> 0.746 mol Ag
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Answer:</h3>
498 kj/mol
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Explanation:</h3>
- Chemical reactions occur as a result of bond breaking and bond formation.
- The bonds in reactants are broken and atoms are rearranged to form new bonds.
- During bond breaking energy is absorbed to break the bonds of reactants while bond formation involves the release of energy during the formation of new bonds.
In our case;
In 1 mole of the Oxygen molecule, there is one O=O bond
Energy absorbed to break O=O is 498 kJ/mol
Therefore, the ΔH required to break all the bonds in one mole of Oxygen(O₂) molecules is 498kJ/mol.
Note that, bond breaking is endothermic since energy is absorbed from the surroundings.
Answer:
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Explanation:
