A) \[{{H}_{2}}{{C}_{2}}{{O}_{4}}.2{{H}_{2}}O=\frac{126}{2}=63]\]
B) \[=0.1N\]
C) \[{{H}_{2}}{{C}_{2}}{{O}_{4}}.2{{H}_{2}}O\]
D) \[{{\beta }_{4}}=\frac{1}{dissociation\text{ }constant}\]
Correct Answer: C
Solution :
The lead of lead pencils is made up of graphite (an allotrope of carbon), melting point of which is \[N{{O}_{2}}+{{H}_{2}}S\xrightarrow{{}}{{H}_{2}}O+S+NO\]. Hence, lead of lead Dencils melts at \[MnC{{l}_{2}}\Rightarrow M{{n}^{2+}}=[Ar]3{{d}^{5}},4{{s}^{0}}\]
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