42. How would you prepare 1.00 L of a 0.50-M solution of each of the following? A. H2SO4 from “concentrated" (18 M) sulfuric acid B. HCl from “concentrated” (12 M) reagent
C. NiCl2 from the salt NiCl2 6H2O
D. HNO3 from “concentrated” (16 M) reagent
E. Sodium carbonate from the pure solid​

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Explanation:

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c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.